#linear-algebra

2 messages · Page 125 of 1

pastel kettle
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nw!

spiral star
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i can use /= if you prefer that

pastel kettle
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it's okay

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now i can see

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so to deduce A != I

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i need to look at the char.polynomial

spiral star
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two matrices with different characteristic polynomials cant be equal

pastel kettle
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so I has (x-1)

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but A has -x^3+1

spiral star
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identity has eigenvalue 1 but 3 times

pastel kettle
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oh yes

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and A

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has 1 and other

spiral star
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and you know the char. polynomial of A

pastel kettle
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complex number

spiral star
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yea, they cant be equal but you dont have to calculate the roots

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you just compare the polynomials

pastel kettle
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also why is it saying 3 complex number? it has 1 real and 2 compelx

spiral star
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you can compare the polynomials by their coefficients

pastel kettle
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yeah i see A!=I now

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so A^2 left

spiral star
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yea that's the only one that needs some thinking

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let's see what your note says

pastel kettle
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so 1.9.1

spiral star
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well one argument would be that you could diagonalize the matrix over C and since you have roots of unity for z³ = 1, you need A³ to get the identity

pastel kettle
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yeah A^3 i can tell that it's identity matrix

spiral star
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well if you choose the complex eigenbasis, you get a diagonal matrix

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and if you square that, it just squares the eigenvalues

pastel kettle
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oh

spiral star
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but squaring them won't give you the identity

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you need a third power

pastel kettle
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so squareing -x^3 +1

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the char polynomial

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won't give me identity

spiral star
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well squaring it wont give you the char polynomial of the identity

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but that isnt the point

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the char polynomial of the squared matrix has the squared eigenvalues

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because you can diagonalize it

pastel kettle
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oh

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so the eigenvalue will be square of 1 and the two complex numbers each

spiral star
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at least if you interpret everything over C

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just draw some matrices i guess

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if you have a diagonal matrix and you square it, it just squares the values on the diagonal

pastel kettle
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yeah i get it!!

spiral star
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you have 3x3 matrix with 3 distinct eigenvalues, hence its diagonalizable over C

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now square the diagonal matrix

pastel kettle
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yes

spiral star
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an easier way to think about it is that a diagonal matrix has always the eigenvalues on the diagonal

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so the squared matrix here has the squared eigenvalues

pastel kettle
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yes

spiral star
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well, but your eigenvalues are solutions to x³ = 1

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and the identity has the eigenvalues 1

pastel kettle
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yeah so A^3 needs to be Identity

spiral star
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yea, thats another way to get there

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but also the complex eigenvalues wont get to 1 if you only square them

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you need to take them to the third power

pastel kettle
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true

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ohhh

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thanks!!

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can i also one last question! i need some confirmation that if my work is right!

spiral star
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okay

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here is something i would like you to do so you can see even better whats going on

pastel kettle
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okay!

spiral star
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write the complex roots of that polynomial in exponential form

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that makes them really easy to multiply

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and write out the diagonal matrix over C

pastel kettle
spiral star
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yea but write them in exponential form

pastel kettle
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so like

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e^1

spiral star
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like how you would calculate them in the first place

pastel kettle
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and e^ -1+ i root3

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those?

spiral star
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$e^0$, then $e^{i \cdot \frac{2 \pi}{3}}$ and $e^{i \cdot \frac{4 \pi}{3}}$

stoic pythonBOT
spiral star
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i hope i did that right

pastel kettle
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oh

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i see what you meant

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yeah e^0 = 1

spiral star
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now if you multiply by that matrix, you see how the exponents change by exponent laws right?

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$\frac{2 \pi}{3} \mapsto \frac{4 \pi}{3} \mapsto \frac{6 \pi}{3} = 2\pi = 0$

stoic pythonBOT
pastel kettle
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yes

spiral star
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(mod 2pi)

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the same happens with the third eigenvalue

pastel kettle
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ohh yeah

spiral star
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$\frac{4 \pi}{3} \mapsto \frac{8 \pi}{3} \mapsto \frac{12 \pi}{3} = 4\pi = 0$

stoic pythonBOT
spiral star
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so after A³ you end up exactly with e^0 = 1 in each component

pastel kettle
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yes!

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omg this is brilliant work

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omg

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i got goosebump

spiral star
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lol

pastel kettle
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lol

spiral star
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that works because we have an eigenbasis for this matrix

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the power of matrix similarity

pastel kettle
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jesus like this is kinda killing

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wait how did you convert (-1+iroot3)/2

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like how to write them down in terms of e

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using the euler's foormular?

spiral star
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https://en.wikipedia.org/wiki/Root_of_unity

Root of unity

In mathematics, a root of unity, occasionally called a de Moivre number, is any complex number that yields 1 when raised to some positive integer power n. Roots of unity are used in many branches of mathematics, and are especially important in number theory, the theory of grou...

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complex numbers have a very nice property

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this is how you calculate the roots

pastel kettle
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oh

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yeah i did those when i was doing Alevel lol

spiral star
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the takeaway is, if you want to find the complex numbers z for z^n = 1

pastel kettle
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yeah?

spiral star
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then you can use the property that multiplication of complex numbers adds the angles

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and the roots are spaced out evenly on the unit circle

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you just divide the angle 2pi by n

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for example the solutions of z³ = 1 will be at 0*(2pi / 3), 1 * (2pi / 3), and 2 * (2pi / 3)

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so you can just read off the exponential representation

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you can find more info on the wiki page

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but this is how you calculate complex roots in general

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there are very nice group theoretic properties to study here

pastel kettle
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oh yes i kind remember the Z thing

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thanksss

spiral star
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but knowing the roots isnt really important for the problem anyway

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you just see -x³ = 1 so it has 3 roots in C

pastel kettle
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yeah

spiral star
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hence, your matrix is diagonalizable over C

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and when you have a diagonal matrix you can see how the char polynomial of its square has the squared eigenvalues

pastel kettle
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yes!

spiral star
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so thats one way to find the char polynomial of the square

pastel kettle
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yeah thanksss

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i couldn't tell the A^2 thing

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but now i can

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:))

spiral star
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im not fully satisfied with that answer tho since i didnt use any of your theorems

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it's correct tho

pastel kettle
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and this is other question i solved

spiral star
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uh

pastel kettle
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(a) is fine

spiral star
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which one were you doing?

pastel kettle
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but for (b) did i do right

spiral star
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yea (a) is just sum of eigenvalues and product of eigenvalues

pastel kettle
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for (a) i can tell -pi i pi i and 0

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0*

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so i can directly tell trace=0 and det = 0

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paper work is for B

spiral star
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uh okay

pastel kettle
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so i think it literally says i can write down e^(root)

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1.6.5

spiral star
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yea that should work

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because taking A^k is just taking the values on the diagonal ^k

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and the diagonal is the eigenvalues

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if you then divide by k! and take the sum k = 0 ... inf

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then its the definition of e^(eigenvalue)

pastel kettle
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but (c) i kinda have problem

spiral star
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what's the problem with (c)?

pastel kettle
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so (c) is just stating a fact?

spiral star
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well that (c) works is kinda obvious

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you are in the eigenbasis of A

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then exp(A) has the e^(eigenvalue) on the diagonal

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the eigenvectors of A in the eigenbasis are (1,0,0) (0,1,0), (0,0,1) and their scalar multiples

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exp(A) is still diagonal, so the eigenvectors stay eigenvectors

pastel kettle
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ohh

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so which means

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eigenvalues stay

spiral star
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eigenvectors for A are also eigenvectors for exp(A)

pastel kettle
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yessss

spiral star
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because exp(A) is diagonal in the eigenbasis

pastel kettle
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and for (d)

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according to thi

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rank(A) = 3

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cuz all eigen values are non-zero

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right??

spiral star
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uh yea, basically A doesnt have full rank if 0 is an eigenvalue

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because Ker(A - 0I) = Ker(A)

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if 0 is an eigenvalue then Ker(A) has dimension >= 1

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but now you look at exp(A)

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and none of its eigenvalues is 0

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so it has full rank

pastel kettle
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if rank(A) = 2, it means A has one eigenvalue=0, and other two are non-zero?

spiral star
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rank(A) = 2 means dim Ker(A) = 1

pastel kettle
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nullity

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yes

spiral star
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that means the geometric multiplicity of 0 in A is 1

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hence, 0 must have at least algebraic multiplicity 1

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and it turns up exactly once

pastel kettle
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yes

spiral star
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so A has definitely rank 2

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and exp(A) cant have rank 2, because 0 is not a root

pastel kettle
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ohhhhhhhh

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yes A has rank two because one of eigenvalue is 0

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but exp(A) has 3 of non-zero eigenvalues

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as we showed on paper

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oh my god Flow

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you're absolutely geneius

spiral star
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lol

pastel kettle
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love you flow

spiral star
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diagonal matrices are your friends

pastel kettle
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you're my love

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loool jk x

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those two questions kept me awake until 4:39am in korea

spiral star
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all this jazz with eigenvalues is usually about choosing an appropriate basis and relying on similarity

pastel kettle
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i seee

spiral star
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once you have similarity to a matrix with a nice layout, you can just throw determinants at it

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the matrices you want always have triangular form

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because for all triangular matrices the determinant is just the product of the elements on the main diagonal

pastel kettle
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can similarity shown by having same determinant

spiral star
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so if you want to prove anything about characteristic polynomials you show that the matrix is similar to some matrix in triangular form

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and then they have the same char polynomial because of similarity

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and for the triangular matrix its very easy to reason about because the determinant is found via the diagonal

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that works really well for a lot of proofs about the char. polynomials and eigenvalues

prisma pier
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can similarity shown by having same determinant
@pastel kettle similar matrices need to have the same determinant but I don't think it's enough to prove similarity

spiral star
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yea you need more

pastel kettle
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ohh

prisma pier
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showing that they have the same eigenvalues isn't enough too I think

spiral star
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the usual way is to find a similar matrix that is easier to work with and then use the fact that they have the same determinant and char polynomial etc.

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not the other way around

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not the other way around

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not the other way around

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(because it doesnt work)

pastel kettle
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ohh

spiral star
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there are some stronger arguments

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where knowing that same eigenvalues + additional properties implies that the matrices are similar

pastel kettle
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ohh

spiral star
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but it's easy to construct matrices with the same determinant that are not similar, like smh said

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so that direction doesnt work in general

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one example for a stronger requirement is orthogonal similarity or symmetric matrices

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two symmetric matrices (or hermitian if you prefer C) are similar iff they have the same eigenvalues and for all eigenvalues they also have the same geometric multiplicities

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so in this particular case you can compare eigenvalues and their geometric multiplicities to get similarity

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but that's specific to symmetric / hermitian matrices

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i guess you know that already but the whole point of studying eigenvalues is to characterize similarity

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find certain classes of endomorphisms (and their transformation matrices) where you can characterize similarity quite well

pastel kettle
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kkkkkkkkkkkkkkkk

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fuck

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typo

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my keyboard wasn't working

spiral star
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lol

pastel kettle
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so i was keep typing that

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sorry

spiral star
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i should probably stop bothering you anyway, it's late for you

pastel kettle
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where are you?

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im reading over and over again !

spiral star
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,ti

stoic pythonBOT
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The current time for Flow is 09:52 PM (CEST) on Sat, 29/08/2020.

spiral star
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i think you can check anyone's time with ,ti @username if they have it set in the bot

pastel kettle
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what country though

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lol

spiral star
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germany

pastel kettle
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ohh

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i've been to berlin only

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because i studied in London

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then went tropical island for summer

spiral star
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lol

pastel kettle
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danke flow

spiral star
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❤️

pastel kettle
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love you

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now im off to bed

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lol i screenshotted all maths thing we've talked cheers mate

spiral star
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lol

latent ledge
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Suppose $T \in\mathcal{L}(V)$ has a diagonal matrix A with respect to some basis of V and that $\lambda\in F$. Prove that $\lambda$ appears on the diagonal of A precisely $\dim E(T -\lambda I)$ times.

stoic pythonBOT
latent ledge
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is $T-\lambda I$ similar to $A-\lambda I$

stoic pythonBOT
west spade
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Yes

real mulch
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I need help with some algebra and number systems I’m getting stuck

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Hello

half ice
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@real mulch
JustAsk

half storm
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I'm pretty sure I get it.

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But I'm just asking for validation

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We know that given the product of two matrices, we know that the rank of their product is bounded above by the ranks of the indvidial matrices

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So we know that $rank(AB) \leq rank(B) \leq n$

stoic pythonBOT
half ice
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Weird question. Any matrix can be written as the sum of rank 1 matricies.

Just take the matrix with all zeroes except for one entry. It's rank 1 but can clearly be used to create anything

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Sum of n matricies I missed that bit

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Dorp

stoic pythonBOT
cunning tinsel
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Can anyone confirm if this is correct index/suffix multiplication?

📎 image0.jpg
native rampart
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What is $x_i$ or $x_j$?

stoic pythonBOT
cunning tinsel
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Ah, should x_i*x_j = x_ij?

native rampart
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No,how do you define them?

cunning tinsel
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The components of a position vector x = (x1, x2, x3)

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i,j =1,2,3

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Pls

cunning tinsel
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<@&286206848099549185>

median forum
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where did the kronecker delta come from

limber sierra
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kronecker

median forum
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Im talking about the picture

muted holly
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lmfao

pastel kettle
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hey guys

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and using rank nullity theorem, i can tell that nullity is also 1

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but when i read " nullity of A equals "number of columbs without leading entries"

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i just get confused all of sudden

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if it's like this

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because of K, the nullity is 1?

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i mean J

cunning tinsel
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where did the kronecker delta come from
@median forum isn't it *(d_ik)(d_kj) = d_ij?

marble lance
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@pastel kettle the first column doesn't have a leading one, so it contributes one to the nullity. The second column does have a leading one, so it contributes one to the rank. So both the rank and the nullity are 1.

pastel kettle
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what if

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this situation?

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does this have leading entry

marble lance
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The second column has a leading one (the leading one of the second row), but technically that is not in rre form since the zero rows should be at the bottom

pastel kettle
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ohh

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if it's

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1,1, on top

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1,1
0,0

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like this

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it has two reading entries?

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so nullity is 0

marble lance
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No

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The second column does not have a leading one

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Only the identity matrix would have a nullity of 0 in rre form

pastel kettle
marble lance
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A leading one is the first 1 in a row. So that second one is NOT leading.

pastel kettle
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what about this

marble lance
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There are literally only 4 possible 2x2 matrices in rre form: (0 0, 0 0), (1 0, 0 1), (1 0, 0 0), (0 1, 0 0). The first one has nullity 2 rank 0, second one has nullity 0 rank 2, last two have nullity 1 rank 1

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Where before comma is first row, after is second row

pastel kettle
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(1 0, 0 1)

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isn't it rank 2

marble lance
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That's what I meant, sorry

pastel kettle
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(1 1, 0 0)

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in case of this

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this cannot be considered?

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this has rank 1 right

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but without using rank-nullity theorem

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how do we tell it has nullity 1?

dusky epoch
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the vector [1; -1] is in its nullspace

marble lance
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There are literally only 4 possible 2x2 matrices in rre form: (0 0, 0 0), (1 0, 0 1), (1 0, 0 0), (0 1, 0 0). The first one has nullity 2 rank 0, second one has nullity 0 rank 2, last two have nullity 1 rank 1
Yeah, I left out (1 1, 0 0), dont know why. Sorry.

dusky epoch
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you left out $\bmqty{1 & a \ 0 & 0}$ for all real $a$ except 0

stoic pythonBOT
dusky epoch
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these are all inequivalent rrefs

marble lance
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Indeed...

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Thanks for the correction

pastel kettle
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so what does not rref tell us?

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(sorry for dumb questions)

native rampart
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Not much?

pastel kettle
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ugh why im so confused lol

dusky epoch
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\\

stoic pythonBOT
native rampart
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You convert a thing to rref because it usually makes your life easier

dusky epoch
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$\bmqty{1 & 1 \ 0 & 0}, \bmqty{1 & 2 \ 0 & 0}, \bmqty{1 & 3 \ 0 & 0}$ and $\bmqty{1 & -\sqrt{\pi} \ 0 & 0}$ are four matrices each in RREF and not row-equivalent to one another

stoic pythonBOT
marble lance
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What is bmqty short for? Can't figure it out

dusky epoch
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bracket matrix quantity

marble lance
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Ah thanks

pastel kettle
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ohhh

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can't imagine life without you guys

marble lance
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I can't imagine life without me either

viscid kernel
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I have got 3 questions

  1. What is a euclidian space ( intuition, geometric view,... )

  2. What is a prehilbert space

  3. What is an inner product, what is the difference between dot products and inner products

native rampart
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The inner product is a function,which takes in 2 vectors and spits out a scalar,satisfying some conditions

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An inner product is a dot product in some basis.

viscid kernel
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@native rampart whats the geometric interpretation of the innerproduct ?

native rampart
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Same as dot product? Just change your basis to an orthonormal one(basis, where the inner product behaves as a dot product is called that)

dusky epoch
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the concept of "inner product" generalizes that of the dot product

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inner products definitionally have all the same properties that make the dot product important: bilinearity, symmetry and positive-definiteness

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an inner product space also naturally comes with a norm: |x| = sqrt(<x,x>) - this is in some sense a generalization of the distance formula in R^2 or R^3

viscid kernel
#

aaaaah

native rampart
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Think of it as a dot product,but designed for a different basis,rather than standard one

dusky epoch
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you should abstract away from trying to view it purely geometrically

viscid kernel
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So its just an approximation of the norm, you say ?

dusky epoch
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no?

native rampart
#

How?

dusky epoch
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where did i say approximation

viscid kernel
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nvm, you didnt, I just made that unconsiously up.

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Ann

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Could you also explain those three things ( bilinearity, symmetry and positive-definiteness )

dusky epoch
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i take it you'll reject my attempts to write out their definitions?

native rampart
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Are we talking about inner products in real vector spaces?

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Or complex ones?

dusky epoch
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i'm doing the real case

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<au + bv, w> = a<u,w> + b<v,w> [linearity in 1st argument]
<u, av+bw> = a<u,v> + b<u,w> [linearity in 2nd argument]
<v,w> = <w,v> [symmetry]
<v,v> >= 0 for all v, and <v,v> = 0 implies v = 0 [positive-definiteness]

native rampart
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For the complex case,the difference is $\langle{\alpha,\beta}\rangle=\overline{\langle{\beta,\alpha}\rangle}$

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And change linearity of second argument condition accordingly

dusky epoch
#

bruh why did you put the equals sign outside of mathmode

stoic pythonBOT
queen gate
#

I have a little question about this type of writing :
What is the différence between
$ this : [0 ; 1) and this :[0 ; 1] $

stoic pythonBOT
native rampart
#

In 1st 1 is not included,later 1 is included

queen gate
#

ok but for "not included", isn't it : $[$

stoic pythonBOT
native rampart
#

Both symbols are used equivalently

dusky epoch
#

are you french

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in france they use [0,1[ but in anglophone countries they use [0,1)

queen gate
#

yep, I m french

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because in France, " ) " says "can't catch" so we use the ) for the infinite like : [0 ; +inf)

native rampart
#

Why not just use [ for everything,if you are using that?

queen gate
#

because in france, if you write [ , you imply you can catch the number... but you can't catch infinite, so you write ) 😅 France is strange

west spade
#

Is there an area form for n vectors in a larger than n dimensional space? I see no reason determinants need to be for square matrices only

half ice
#

Put those vectors on their own orthonormal basis, and now the volume they enclose is the determinant of the matrix they form. Basically, cut the dimensions you don't need

spiral star
#

@half storm i saw that question about writing a matrix product as a sum of rank 1 matrices. i approached it in a more general setting since there is something to be learned about linear maps. thought you might be interested in this as well. the statement about matrices follows directly by taking an isomorphism between the matrix spaces and the spaces of linear maps

📎 unknown.png
half storm
#

Thanks

spiral star
#

i guess the key takeaway here is that if some matrix has rank r, then you can write it as a sum of r rank 1 matrices. writing it as a sum of n >= r matrices is not a problem, since any more matrices you add can just be in the span of the first r matrices. so for the remaining matrices you just choose a matching linear combination. i chose mine such that they just add up to what i had originally, but you can get really creative here...

half storm
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I've actually proven that first Lemma

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It's in my text book lol

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It's stated a little differently in the form of linear transformations but same drill.

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I didn't prove that there exists a basis that such that you express the function as a linear combination of the others it gave it to me and showed me to prove it was possible

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I've done problems in the book that are basically drawing on these ideas but they were just different.

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Like every vector space is isomorphic to a space of functions.

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But now I'm seeing it being used for something

spiral star
#

yea

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well spaces where the vectors are linear maps are quite common

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and if those maps go between finite dimensional spaces, then those spaces are isomorphic to spaces where matrices are vectors

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once you choose a basis, you can define an isomorphism between them

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and the isomorphism is just the obvious one

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but yea, L(V,W) type vector spaces are quite common, and you will do more with those when you study dual maps, bilinear forms etc.

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or when you go towards tensor products

queen cave
#

1th, have you guys ever heard of wedge product? I'm here to actually say that I finally can appreciate the existence of the cross product because of this, and that we should all study this union of dot and wedge product. (inner/outter product).
2nd, @dusky epoch nice hs pic.

dusky epoch
#

thank you

wintry steppe
#

the wedge product is pretty cool

simple depot
#

does [A]11 mean the element of A that is in position 1,1?

half storm
#

Yea

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Your notation seems non-standard; never seen it before.

simple depot
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oh i dont know

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its just the way they write it at my university

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im in south america if that means anything idk

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but thanks!

half storm
#

np

wintry steppe
#

i usually just see A_{11} or A_{1,1}, or even in weird diffgeo cases A^1_1, never with square brackets around it

half storm
#

Yea and his set builder notation is something that I've never seen before too

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the / instead of :

gray dust
#

| is used too, not far from /

wintry steppe
#

inverse of a matrix is the same as switching the rows and columns correct?

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ok guess not from an easy google search

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that sounds like transpose

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yeah for some reason I thought it made sense, but it was a bad thought process

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transpose isn't always the same as taking the inverse. for one, you can take the transpose of a nonsquare matrix, but not the inverse

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I see that M*M^-1= I

prisma pier
#

wait to clarify, the * is for times, right? Because some people denote the transpose with star

wintry steppe
#

should be

prisma pier
#

^

wintry steppe
#

so we can find the inverse of a matrix by ... deviding the unit matrix I by M?

#

uh

#

and yes * is times

#

short answer: you don't divide by matrices

#

ah you don't? pretty sure I did often

prisma pier
#

you could say that multiplying by the inverse is like dividing by a matrix

wintry steppe
#

well maybe that was computer software, could be different I guess

#

you can multiply something by the inverse of a matrix and call it division, if you want

#

oh i got sniped

#

so my definition is wrong

prisma pier
#

I see that M*M^-1= I
^this is right

#

but it also needs to be true the other way to be a true inverse (M^-1)M=I

wintry steppe
#

what definition is wrong? that MM^(-1) = I? this is correct; some work shows that the other way works

#

other way?

#

that M^(-1)M = I

#

uh

#

as long as rows and columns are the same should work I guess

#

?

prisma pier
#

^

wintry steppe
#

do you mean number of them?

#

that didnt make sense, we're talking about the same matrix lol

#

only square matrices (number of rows and columns are equal) have inverses

#

oh

#

otherwise the products MM^(-1) and M^(-1)M don't even make sense

#

well look I proved it 😂

#

i do have some bad background about matrices

#

but how would you calculate inverse of M , if you only know M?

#

there are many ways

#

the most common is by doing a row/column reduction on the augmented matrix (M | I) until you get something of the form (I | A); the matrix A will be the inverse of M

#

will be honest, I never saw that notation

#

if A and B are n by n square matrices, then (A | B) means the n by 2n matrix formed by placing A first and then B second

#

ah i see

#

i.e. first n columns are those of A, rest are those of B

#

i would prove this but im on my phone so

#

don't worry, just promise me you're telling the thruth

#

necessary and sufficient condition for invertibility is nonzero determinant, so you might want to check that before you try inverting any matrices

#

i think im not bullshitting lol

#

i used to use it a lot so i hope i remember it correctly

#

you can probably find the augmented matrix thing on wikipedia or in any LA book

prisma pier
#

I see no bullshit here

wintry steppe
#

yeah I did search it up, was just a joke haha

#

no need to prove it

#

alright, I'll head back to work

#

there are probably other ways to calculate the inverse of a matrix, but i think that's the most straightforward

#

👍

jolly turtle
#

Excuse me ;-;

#

MAY I GET HELP ON MATRICES PLS

wintry steppe
#

@jolly turtle The way things work is you ask a question and someone might answer you.

jolly turtle
#

I can't get help

wintry steppe
#

Asking generically for "help" is a major turn-off. Please bear in mind that people on these forums are not working to provide homework help. So it's up to you to present your problems in a way which would interest people in helping you.

#

If you ask an interesting question and show an intent to learn, people will be more likely to offer help.

jolly turtle
#

I'm just panicking

#

Idk what to do

wintry steppe
#

It's much harder to learn things when your panicking.

#

how about you take a break from your test, don't do anything that'll get you banned, and come back to your test when you've calmed down

jolly turtle
#

Ok 😞

vague sundial
#

Hello, I'm struggling understanding something here

#

It says prove that set P (...) is not a vetor subspace of R^3

#

I don't understand the transition to the last line

median forum
#

br?

#

@vague sundial last line they test if the sum is in the space

vague sundial
#

Why is (u1+v2) squared in the first place?

#

@median forum are you asking me if I'm Brazilian?

median forum
#

because for every element of that set, the second coordinate is equal to the first coordinate squared

#

$x_1^2 - x_2=0 \iff x_1^2 = x_2$

stoic pythonBOT
tidal bronze
#

you found out that u + v is that vector

#

then you test if the elements in there is equal to 0 or not

vague sundial
#

Oh, so the last line is what it is supposed to be in case it belonged to the set?

tidal bronze
#

yea

#

if it is, then that axiom is satisfied

#

then you move on to the next one

#

but because it's not, you can say that P is not a subspace

vague sundial
#

That is very helpful. You guys are very helpful here, I admire the time you spend teaching us 🙂

tidal bronze
#

happy to help :D

median forum
cerulean quest
#

can anyone explain how to read these sets? its too confusing for me... i believe option 1 is a plane but i don't know how to read the other options

📎 unknown.png
half storm
#

You're right the first one is a plane. It is the vector equaiton for a plane. The 2nd one is also a plane ; Look up the scalar equation of a plane, and the 4th one is also a plane ; the xy plane.

cerulean quest
#

(x, y) is a plane where x and y are arbitrary parameters? they only used x and y to confuse us?

half storm
#

I guess you could say that. If you think about what it's saying, it just means the set of all ordered pairs such that x and y are real numbers; that's the whole xy plane. Or, if prefer it to be in a form that you recognize i.e. the vector equation form that you have for number 1, you rearrange the equation as $$\begin{pmatrix} x \ y \ \end{pmatrix} = x \begin{pmatrix} 1 \ 0 \ \end{pmatrix} + y \begin{pmatrix} 0 \ 1 \ \end{pmatrix} $$

stoic pythonBOT
half storm
#

Hopefully now it's unambiguous that it's a plane.

cerulean quest
#

oh i see!!!

#

yeah it's very clear thank you! why is the second one a plane though if there are 4 variables and 1 equation? that gives us 3 parameters correct..?

half storm
#

The second one is a plane because it is the scalar equation of a plane. There are two ways of viewing planes via their scalar equations and their vector equations.

#

The first answer is a vector equation of plane, but there is another formulation and is expressing the plane and it's general form is given by $A_1 x_1 + A_2 x_2 + \cdots + A_n x_n = D$ where $\langle A_1, A_2, \dots , A_n \rangle$ is the normal vector of the plane.

stoic pythonBOT
median forum
#

second is a plane?

half storm
#

I believe so.

#

The scalar equation of a plane

median forum
#

but

#

thats an equation in R4

half storm
#

Oh I see.

median forum
#

it lowers one degree of freedom

#

should be a 3-hyperplane

#

the third should tho

#

me thinks

half storm
#

I think the question asks though which of the following correspond to a plane in their respective space.

#

For the third one, it seems a bit strange because you have a plane in $R^3$ given by the first equation and then a line given by the second equation yea? So then, we would want to find the solution space of the set of equations given by
$w - 5y - z = 3$ and $x + 2y = 5$ yea?

stoic pythonBOT
half storm
#

Yea the third one is

#

i actually just did it out and you can parametrize it so that it is the vector equation of a plane in R^4

#

I did not understand really the difference between planes and hyperplanes but now Ido.

#

I thought I did

median forum
#

a hyperplane is just the generalization of the plane

#

an n-hyperplane is like a linear bisection of R^(n+1)

#

you can probably generalize that to convex vector spaces with some properties

#

but cba

old flame
stoic pythonBOT
cerulean quest
#

i actually just did it out and you can parametrize it so that it is the vector equation of a plane in R^4
@half storm Sorry, so is the 2nd one not a plane and the third is?

half storm
#

Yea, lol it's really my fault.

#

third is and 2nd isn't.

cerulean quest
#

Yea, lol it's really my fault.
@half storm No no not your fault D: thank you so much for the help

half storm
#

No problem

old flame
#

<@&286206848099549185>

gaunt field
#

For b), I thought it could be exactly the same plane as A and that would work (but im not sure), but I was wondering if there was any other solution

native rampart
#

So in axler he has a chapter on polynomials. I don't really understand the second half of the proof, when he chooses c such that the coefficients are equal, why is that polynomial smaller than q-sr
@old flame what is degree of (2x^2+1)-2x(x) ? Is it less than 2x^2+1?

#

He is doing exactly that but for general cases

old flame
#

@native rampart So for this case deg r is 2 and deg p is 1, z^j is x and c is 2 ?

#

Yeah, that is deg 0 while the original polynomial is 2

native rampart
#

Yea

old flame
#

So he is consider the case where we add an extra term cz^j and compare this to the original r and he found that this new polynomial is even smaller than r ?

native rampart
#

Yes

#

But r is the minimal deg polynomial according to the original condition

#

So, Contradiction

old flame
#

Oh alright, so basically z^j is the required polynomial to match degree of p to r and c be the coefficient to make the term of same degree cancel ?

native rampart
#

Yes

old flame
#

Ohhhhhhhhhh okok thank you so much

native rampart
#

He kinda makes it explicit

#

First line of p67

old flame
#

Well the notation confused me lol, cause I couldn't visualise how it worked

#

Well I know what he wants to do with the explicit statement but I just couldn't see it

native rampart
#

Play with examples,then

#

If you don't understand the proposition even a bit, that's the best thing to do

old flame
#

But I guess u do have to understand the statement in the first place in order to make a correct example

native rampart
#

You know p,q,r are polynomials so take some 3 random polynomials

#

The 2nd part says deg(r) is not less than deg(p)

#

So take accordingly and check

gaunt field
gaunt field
#

😦

gaunt field
#

I'm little confused by this cuz it says 3 equations in 4 variables so 3 x 5? And I thought elementary matricies are n x n so I thought they would be size 5 x 5 instead of 3 x 3

wintry steppe
#

the augmented matrices are 3x5 matrices

#

but the elementary matrices are 3x3

#

so when you left-multiply the augmented matrix by the elementary matrix, you get a 3x5 matrix, same as before

gaunt field
#

ah okay

old flame
#

Okay thanks @native rampart

wintry steppe
#

what is the worst mixture problem you've ever seen

native rampart
#

What do you mean by that?

prisma pier
#

as in the kind of differential equation question?

wintry steppe
#

linear algebra

#

only requires linear algebra

limber sierra
#

If a polynomial function from an n-dimensional space to itself has Jacobian determinant which is a non-zero constant, show that it has a polynomial inverse

wintry steppe
viscid kernel
#

Is there a way to recognise immediatly that this, is a rotation matrix without computing its determinant ?

📎 image0.jpg
native rampart
#

The three column vectors are orthonormal

#

And AAt is I

viscid kernel
#

And how did you know its orthonormal, I see that the norms of the columnvectors are one but not that it is orthoganl.

native rampart
#

Take 2 column vectors and apply dot product

viscid kernel
#

Aight gotcha

native rampart
#

It's a shorthand way of doing AA(t)

viscid kernel
#

Thanks

#

Ah wait

#

So if you do AA(t) do you get the zero matrix ?

native rampart
#

You get I

viscid kernel
#

Ah true

#

I understand thanks

little frigate
#

If I wanted to rotate a figure around a point

#

how could I know if I have to apply the translation first or the rotation

viscid kernel
#

@little frigate good question, tbh, I need an answer on that too

little frigate
#

I was thinking about just dealing with inverses but idk

native rampart
#

Make that point your zero point first(translation) and use rotations

little frigate
#

haven't really seen them in class but I know one or two stuff

#

alright I'll try that, thanks

#

I mean I already tried did but maybe I did something wrong

elfin locust
#

anyone know how how to get a normal in triangle (cross product)?

#

in game is with relative or absolute positions?

#

My normals are weird

#

The ideia is to have a vector that faces a triangle

#

Like this

#

i see in this site

#

but i cant recreate idk why

stoic pythonBOT
barren cove
#

I have tried to solve this question 6 times already. I'm trying to inverse A using matrix transformation, what am I doing wrong? what rules am I breaking? I'm pretty sure it's not the calculations because I've triple checked every calculation, although there's a chance that its the calculation. Please help

stoic pythonBOT
autumn basin
#

Hi just a quickie, does this notation mean y is a column vector with those entries vertically? I'm needing to use it in matrix multiplication so I assumed so, just wanna be sure

📎 unknown.png
prisma pier
#

👍

#

yes

autumn basin
#

cool, ty 🙂

gray dust
#

this requires knowing how to write a linear system of eqns as a matrix eqn & vice versa

hollow finch
#

do this

#

$\begin{bmatrix}1&3&-2\0&-1&4\0&9&-11\end{bmatrix}=\begin{bmatrix}1&0&0\-2&0&1\3&1&0\end{bmatrix}A$

stoic pythonBOT
hollow finch
#

then multiply the 2nd row by -1 and you have your leading 1

#

no fractions necessary

#

should make the remaining reduction easier

elfin locust
#

The normals seems pretty nrmal

#

but idk why is not working

#

Wrong Output

#

the idea is to remove the triangles behind and that you can't see

pallid rampart
#

I mean nobody knows the problem unless you post your code

#

The best we can do right now is to guess the problem but I don't see why it isn't working

elfin locust
#

my cross function seems normal

#

im normalizing too

#

@pallid rampart Do u need something more?

#

relative position

pallid rampart
#

I don't know what you're doing so

#

What language is this first of all

elfin locust
#

java

#

Im trying to remove the triangles that is behind the view of the player

pallid rampart
#

Ok so your cross function is not correct

elfin locust
#

O.o

#

i triple check ;_;

#

whats wrong?

pallid rampart
#

On the second line, the x has already been changed

elfin locust
#

so what i need to put?

pallid rampart
#

It's not the original x anymore

elfin locust
#

omg -.-

#

oh god

#

yes sure

#

im so dumb

#

thks man

pallid rampart
#

sure

#

Also it's probably better to use a static function that takes in two vectors instead of having a function that takes in one vector and modify the vector

#

It is just a bit weird to do it that way

elfin locust
#

yes...

pallid rampart
#

Because for your current implementation, after
Vec3 normal = line1.cross(line2);
the variable line1 will be the same as normal

elfin locust
#

yes

#

that is not an issue

#

because i will not use before

pallid rampart
#

Yeah that is not an issue

#

So I said "it's just a bit weird to do it that way"

elfin locust
#

yup

#

ok the output is still not correct

#

but its better

#

probably the triangle order is not correct

pallid rampart
#

The normal might be facing the wrong way

elfin locust
#

yes

pallid rampart
#

Yeah it seems like it renders that triangle when looking at it from behind

elfin locust
#

yap

#

@pallid rampart thks so mutch

#

im blind with that xyz in cross

#

-.-

pallid rampart
#

lmao it's a very very common mistake

elfin locust
#

the problem is that i have been in this problem for over 3 hours

pallid rampart
#

and for this reason you shouldn't write your code that way

#

oh

#

rip

elfin locust
#

I'm writing it to work temporarily

#

I'm just applying some new theories

pallid rampart
#

still it is this kind of small details that build up

#

but eh

#

it's ok if you know what you're doing

elfin locust
#

I appreciate your concern but I will design this in a much simpler way 🙂

#

❤️

pallid rampart
#

Nice nice

#

One of the things I've always wanted to do is also to build a rendering engine

elfin locust
#

Im doing that 😛

#

but i will add shaders texture and that stuffs too

#

not a simples rendering engine

#

but with time xDD

pallid rampart
#

Nice

lucid cedar
#

I am struggling to understand how the basis was found in this question. I can clearly see the pattern, but that's not really that useful. I can also see why the basis is only 3 vectors; because, there are only 3 variables after making substitutions and combining. Thats about as far as my reasoning goes here though.

📎 Capture.PNG
gray dust
#

recall defn of basis. B is a basis of U if span(B)=U & B is linearly independent

#

after doing algebra you should find any vector in U is expressible as a linear combo of 3 vectors v1,v2,v3, hence span{v1,v2,v3}=U. check that {v1,v2,v3} is lin indep and that's a basis of U

lucid cedar
#

I am not smart

#

I get exactly what ur saying and i cant believe i had to ask

#

thanks for helping

gray dust
#

you're welcome, and no shame in asking

jolly turtle
#

excuse me, does anyone have a good video for matrices? :c

wintry steppe
#

anything specific regarding them that you need to learn?

half osprey
#

3blue1brown

jolly turtle
#

multiplication please, so i can take notes ;w;

half osprey
#

is there only one operator within a certain eigenvalue and eigenvector or there can be multiple ones?

wintry steppe
#

consider T(x,y) = (x, 2y) and S(x, y) = (x, -2y) on R^2

#

they share the eigenvector (1, 0) corresponding to the shared eigenvalue 1

#

but they are not the same operator

half osprey
#

oh, of course

#

Thanks mate

wintry steppe
gray dust
#

@half osprey or take any T with eigenvalue L & L*identity

honest egret
#

I've been scratching my head at this for 20 minutes, can someone explain this, preferably in normal english?

📎 unknown.png
wintry steppe
#

that is in normal english

honest egret
#

I absolutely hate it when they take they explain this stuff in the most convuluted way possible

#

It's been happening more and more often and it's getting really frustrating

wintry steppe
#

so, in this case, the parallelogram rule for addition says that c is given by adding the two vectors that represent the other vertices (not the opposite one, at the origin)

#

since these vertices lie on the lines spanned by u and v, you should be able to guess from the picture what those vertices are

#

then you add them. the parallelogram thingy tells you that when you add these vectors, you get c

#

the idea for the other vectors you're asked to find is similar

#

to push even closer to the answer: you should see immediately that one of them is 2u

#

i have not explained why the parallelogram rule works, only how to apply it in this scenario

honest egret
#

The top right one, is 2u I presume?

wintry steppe
#

yes

#

the picture's a little cut off for the other one but i think you can give a good answer

honest egret
#

How are you supposed to figure that out just from the sentence "The parallelogram with one vertex at the origin and the opposite vertex at the point b has a vertex in the u/v/whatever direction"?

#

After staring at it for quite a while I think I got it. It's asking "where on line u/v/whatever does it line up with b"

#

LIke, why can't it just say that, why does it have to take such a convuluted way in asking it

wintry steppe
#

it probably wanted you to figure that out

honest egret
#

I'm just venting now, maybe I need a break

wintry steppe
#

reading information-dense mathematical writing and figuring out what it means is part of learning

#

it is a skill that you get better at

honest egret
#

Thanks for the help... though I got at least 6 more of these so I might be back here in like less than 10 minutes and it's going to be nothing short of awkward

wintry steppe
#

nothing awkward about asking questions

median forum
wintry steppe
#

So apparently I+BA^-1 = (A+B)A^-1

#

Does anyone know how?

#

Right, so it appears some kind of factorization can lead from I+BA^-1 to (A+B)A^-1

dusky epoch
#

I = AA^-1

wintry steppe
#

Right, so can I = BB^-1?

#

I+BA^-1 = B(B^-1 + A^1)?

errant wyvern
#

okk so i have a task to show a matrix of a linear opperator in a certain basis, i found the matrix in the standard basis, and i found the change of matrix basis, whats the next step?

native rampart
#

Do the PAP^-1

gray dust
#

@errant wyvern if A is the map's matrix wrt standard basis & P is the change of basis matrix from the given basis to the standard basis, then the matrix of the map wrt the given basis is P^-1 AP

errant wyvern
#

allright anonther question, so in my exercise that i am doing i need to use the differential operator. The way our proffesor gets the differential operator is that he uses the standard base for the space of polynomials {1,x,x^2,x^3}, applies the differential opperator to each one of those and writes the result in the shape of c1*v1+c2 *v2+ c3 * v3+ c4 * v4, where v1,2,3,4 are the vectors of the basis {1,x,x^2,x^3}. now, if i want to use another basis for the differential operator, could i do the same thing? If so what are the vectors that i would apply the differential operator to, 1,x,x^2,x^3 or the vectors of the new basis? lets say {1-x,1+3x^2, 2x^2, x+x^2-x^3}

gray dust
#

by "use this basis for the operator" you mean construct the matrix of the operator with respect to that basis, you already asked this in #help-9 and i answered but you never replied.

errant wyvern
#

i feel asleep didnt mean to be rude, but let me clarify the question

gray dust
#

you can either do what i suggested back then or, seeing as you already got the matrix wrt the standard basis, compute a change of basis matrix then this formula gives the matrix wrt the other basis

if A is the map's matrix wrt standard basis & P is the change of basis matrix from the given basis to the standard basis, then the matrix of the map wrt the given basis is P^-1 AP

errant wyvern
#

ok so this is how our prof, gets the matrix of the differential opperator, to simplify the question, if i was to use a different basis, would i, while doing this proces, apply D to the standard basis, or the new one?

gray dust
#

i answered in #help-9 earlier

let B=(b_1,b_2,b_3,b_4) be the given basis. the i'th col of the matrix of G wrt B is the coords of G(b_i) wrt B

errant wyvern
#

yeah i see now, i was having trubble understanding what you wrote. My bad, didnt mean to be annyoing once again

gray dust
#

not understanding is generally ok, not asking me to clarify what i said earlier is bad

errant wyvern
#

im telling you i fell asleep, i find it amazing that you are still here, i had like a nice 6 hours of sleep lul, i thought that i wouldnt get a reply if i asked under questions l channel

gray dust
#

it's fine, just make sure to check replies after waking up. back then i said something else on computing a matrix representation of a linear map but it isn't totally relevant to the hw

errant wyvern
#

So this is how differential matrix would look like under the other basis right?

gray dust
#

yes

#

though i called the basis vectors b1,2,3,4 and you called them v1,2,3,4 out of nowhere

errant wyvern
#

So i need to use kronecker capelli theorem to solve this syatem of linear equations, meaning i need to check if rankA=rank(A|b)

#

where A is the matrix we get by representing the system of linear equations as A*x=b, b is the solution of the sistem, x is the vector of x,y,z like on the picture

#

but im lost at what i need to do now

#

i dont think i can solve it by doing row reduction, so maybe i should try and get the detA

#

im lost i got no clue, any ideas?

#

"discuss and solve the system based on the variable m"

errant wyvern
#

If we have linnearly dependant vectors {x1,x2,...,xk} how can we prove that dimSpan{x1,..,xk}<k?

#

Is it enough to point out that if a system is linnearly dependant we can write one of the vectors from it as a linbear combination of others, ans then the dimSpan{x1,...,xk} would be less then k?

subtle walrus
#

i mean sure

#

you can write one of the vectors, say wlog x_k, as a linear combination of the others

#

so Span{x1,..,xk} = Span{x_1, ..., x_{k-1}}

#

so a basis of that span has at cardinality k-1

queen gate
#

Hi, can anyone explain to me this symbol ?
$\bigodot$

stoic pythonBOT
limber sierra
#

as opposed to $\odot$?

stoic pythonBOT
limber sierra
#

since i havent seen specifically big odot be used

#

often times $\odot$ is used to represent arbitrary multiplication in a vector space (or other mathematical structure), especially where the $\cdot$ symbol may be ``overloaded"

stoic pythonBOT
limber sierra
#

for example, we may define multiplication in a ring by $a \odot b = a \cdot (2b + ab)$ or whatever

stoic pythonBOT
limber sierra
#

(ignore that that doesn't follow the ring axioms, its just an example)

#

here we use the $\odot$ symbol just as a reminder" that this isnt conventional" multiplication

stoic pythonBOT
limber sierra
#

that's one possible use

#

there may be others though

#

in what context did this symbol arise?

#

in a physics context M_odot is often used to represent solar mass

queen gate
#

Hey, thanks for your answer, I wrote \bigodot but I think \odot is the same thing. I use this symbole on this context : (2 sec, I take a screenshot)

#

So, we have theta - n/sqrt gt + e \odot g_t

#

for example we can simplify such as : $\theta_t - \eta \odot g_t$

stoic pythonBOT
limber sierra
#

ah, hm, i'm not familiar with that i'm afraid

#

hopefully someone else knows

queen gate
#

ok, but thanks for your answer

#

I find a not too bad answer on wikipedia : Hadamart product

#

And i think it s better to use only one cricle and not circle + point because tropical maths used the same symbol

viscid kernel
dusky epoch
#

a

viscid kernel
#

Why ?

dusky epoch
#

write out the equations

#

none of them will involve a

#

therefore the value of a will have no effect on the system

viscid kernel
#

Thanks, now it makes sense

#

Also, how do you know when a matrix is not diagonalisable ?

gray dust
#

if it's n by n and doesn't have n linearly independent eigenvectors

wintry steppe
#

I was given this question:

|a b c|            |d e f|
|d e f| = 6, find  |g h i|
|g h i|            |a b c|
#

I interpreted incorrectly, and thought I have to make a matrix of det 6

#

I pulled another big brain move, and decided to use the volume of a parallelogram

#

I ended up with a pre wacky looking formula:

#

$$det = \frac{\left|a\right|\left|b\right|\cdot \sqrt{1-\frac{a\cdot b}{\left|a\right|\left|b\right|}}}{\left|c\right|}$$

stoic pythonBOT
wintry steppe
#

The formula doesnt look too terrible for me to assume its outright wrong

#

I was pre careful making it so I think it should be okay (maybe the |c| part might be wrong)

dusky epoch
#

bruh what?

#

this is a mega bruh moment

wintry steppe
#

yea ikr, is there a easy way make a bunch of matrices of like 4x4 with some arbitrary det?

#

hmm yea nvm i guess i messed something up, thanks ann 👍

subtle walrus
#

consider that calculating det of diagonal matrices is pretty easy

#

(or triangular for that matter)

pulsar turret
#

Guys how would make an n by n matrix, that has 1 when n/i and n/j are divisible, and -1 where it’s not divisible. I and j are the rows and columns. So how would such a matrix look like for n= 5? I had a problem knowing when certain numbers are divisible and other not. I tried making a boolean, but no luck.. any help? It’s more about the reason and logic behind this question, more so than the code. But if it helps, we needed to use matlab. To set up this matrix

native rampart
#

Can't you just enumerate and count?

#

For n=5,except for (1,1),(1,5),(5,1) and (5,5) all will be -1.

pulsar turret
#

Well it’s for all indices, so i goes from 1 to n. And j goes from 1 to n

dusky epoch
# 
A = -ones(n);
for i=1:n
  for j=1:n
    if (mod(n,i) == 0 && mod(n,j) == 0)
      A(i,j) = 1;
    end
  end
end
#

this code constructs your matrix

pulsar turret
#

Wow that so stupid, I didn’t know mod existed

dusky epoch
#

Google is your friend lol

#

i just looked up "matlab modulo operation"

pulsar turret
#

You’re pretty much spot on, shame I couldn’t think of it during the test

#

Welp, thanks ann

wintry steppe
#

ok lol, im pre confident that this is an unnecessarily long version of the formula for 3x3 det:
$$det = \left|a\right|\left|b\right|\left|c\right|\cdot :\sqrt{1-\left(\frac{a\cdot :b}{\left|a\right|\left|b\right|}\right)^2}$$

stoic pythonBOT
spiral star
#

lol

#

you dont need any computation for this problem

#

that's why it is a mega bruh moment

wintry steppe
#

i know, but i stopped looking at the problem a while ago

#

im just enjoying how wacky it got

spiral star
#

lol

wintry steppe
#

while im not confident, just adding |d|, to the formula might work for 4x4 det?

median forum
#

unlikely

#

is that absolute norm?

wintry steppe
#

i have no idea, i just got it from messing with parallelograms

median forum
#

yes, but is the absolute value symbol representing a norm there?

wintry steppe
#

oh yea, magnitude

median forum
#

ok

wintry steppe
#

it'd be cool if it worked for 4x4, but i wouldnt be surprised if it didnt

median forum
#

unlikely

wintry steppe
#

i imagine it would just mean: multiply a parallepiped made of a,b,c by the magnitude of d

median forum
#

are you trying to use the fact the determinant will mean oriented volume with the canonical basis?

wintry steppe
#

yea, i guess

median forum
#

not sure if thats correct either

#

the expression I mean

wintry steppe
#

neither

#

anyway gonna test it, not that it'll be proof

#

yea it blows up for 4x4 haha

#

owell was pre ineresing

viscid kernel
#

Can someone check if I made a mistake somewhere I had to find the inverse of matrix M.

📎 image0.jpg
dusky epoch
viscid kernel
#

Ah thanks, can you now explain me, why you were so fast ??!??!!!

half storm
#

She has familiarity with matrices lol.

dusky epoch
#

your work was laid out nicely

#

and all your row ops were carefully annotated

viscid kernel
#

But still tho

dusky epoch
#

so it wasn't hard for me to check your work

wintry steppe
#

ann large brain

#

that's why

viscid kernel
#

Even if Ive been doing linear algebra for 10 years, I think I wouldnt be so fast lmao.

half storm
#

lol you would, everyone does after time

viscid kernel
#

@half storm I hope so 😦

viscid kernel
#

Finally !!! Been trying to find my mistake for 30 minutes. The negative sign I forgot, ruined my mood.

📎 image0.jpg
lucid cedar
#

This might be a stupid question, but I dont see how it follows that if each vector can be generated indivdually then it implies that linear combinations of vi and vj can also be created.

📎 Capture.PNG
dusky epoch
#

f followed by fi ligature
not even using the ffi ligature despite its existence

#

[judges you silently]

old flame
#

On the chapter of polynomials. The question is "Prove that every polynomial with odd degree and real coefficients has a real root." Heres my proof. Since $p \in P(\mathbb{R})$, by theorem, we could write p as $p(x)=c(x-\lambda_1)...(x-\lambda_m)(x^{2}+\alpha_1x+\beta_1)...(x^{2}+\alpha_M x+\beta_M)$, for $c,\lambda_1,...,\lambda_m \in \mathbb{R}$. Consider the case where $m=0$,
then $p=c(x^{2}+\alpha_1x+\beta_1)...(x^{2}+\alpha_M x+\beta_M)$, therefore $deg p=2M$, which is even, hence this cannot be the case. Now consider the case for $m=1$. $p=c(x-\lambda_1)(x^{2}+\alpha_1x+\beta_1)...(x^{2}+\alpha_M x+\beta_M)$, $deg p=2M+1$ for $M=0,1,...$.
Hence p is an odd power polynomial. Thus, odd power must have at least one real root.

stoic pythonBOT
old flame
#

o.o this is cool, rather than text

low sentinel
#

Hey guys, I just started Linear Algebra and have a noob question: What determines the dimensionality of a matrix, i.e. R^n. Is it rows or columns?

native rampart
#

Dimensions of both come out to be same

#

I think you meant the rank of a matrix

low sentinel
#

Oh right, the rank of the matrix.

limber sierra
#

doesnt matter

#

it's a theorem of linear algebra that row rank = column rank

low sentinel
#

What if I have a 2x3 matrix?

limber sierra
#

same thing applies

#

the rank of said matrix will be 2, 1, or 0 (only if its the 0 matrix)

#

you can check that both the rows and the columns have the same amount of linearly independent vectors

low sentinel
#

Mhm okay

#

Because there is this one task that goes like "Suppose T: R^5 -> R^2 and T (x )= Ax for some matrix A and for each x in R^5. How
many rows and columns does A have?"

#

And the solution is 2 rows and 5 columns. Therefore I thought that rows would determine R^n

limber sierra
#

wait

#

i missed the start of this convo

#

why is rank being discussed at all

#

since rank doesnt relate to that question

low sentinel
#

I think I dont understand the basics yet. I dont understand how R^n relates to a matrix is basically my question I think.

#

And what it says about a matrix if R^n is 3 for example. That there are three rows?

limber sierra
#

the question is saying that T(x) = Ax

#

in other words, that the matrix A represents the linear transformation T

#

note that:

  • it must be possible to multiply a 5-entry vector "x" by A on the left
  • the result must be a vector with 2 entries
#

note that x will be a column vector, and when you multiply A * x, we take each element of a_row_ of A and multiply it by an element of the corresponding column of x; this only makes sense if A has the same number of columns as x has rows

#

so A must have 5 columns (since x has 5 rows)

low sentinel
#

Right. I got that part (how to calculate). But then I started wondering about why R^5 would go down to R^2 if we did that. Since I noticed that the rows equal 2 I wondered why the rows would relate to R^n.

limber sierra
#

i'm not sure exactly what you mean

#

what are the dimensions of A?

#

we already know one dimension; it must have 5 columns

#

so how many rows should it have, if we want our resulting product to have 2 rows?

low sentinel
#

Oh wait, you are right. I read the question wrongly, confused vectors with matrices.

#

Thank you for your explanation.

south orchid
#

Hi there, Im struggling with a problem, can someone help me out

#

Solving the linear equation set:

#

A + 5C - D + E = 7

#

2A - B + 6C + 3E = 8

#

So far the steps Ive taken have been to use gaußian elimination to rewrite it to:

#

A + 5C - D + E = 7

#

-B - 4C +2D + E = -6

#

and Identified that A and B are non-flexible variables

#

and C, D, E are flexible variables leading there to be infinite solutions

#

but Im stuck

gritty frigate
#

What happens if a homogeneous system of ecuations has det(A) = 0 ?

south orchid
#

Sorry english isnt my first language

#

is a homogenous set a set like:

#

where the leading variable in each equation proceeds the next by one?

limber sierra
#

Homogenous means equal to 0.

#

In the case of a system, all equations are equal to 0

south orchid
#

Ok I see

#

infinite solutions?

gray dust
#

speaking of the dimension of a spanning set S doesn't make sense unless S itself is a vector space

#

to find the dimension of a vector space, count the cardinality of any basis for it

#

then you don't have a sufficient defn of dimension

#

if you learned linear independence & span, then that's enough for a defn of basis. a basis of a vector space V is a subset S of V where span(S)=V and S is linearly independent, equivalently stated as any vector in V is expressible as a unique linear combo of the vectors in S

#

if S is a basis of V then you can further define the dimension of V as the cardinality of S, ie how many vectors are in S

gaunt field
gaunt field
wintry steppe
#

would the minimum be -2 or 2?

zealous schooner
#

2 because it's in absolute value
also we're in the same class lol
check my profile on mutual server

#

@wintry steppe

#

How do we do #2 with the triangle inequality I don't know how I think I have everything else done though

rugged arch
#

can someone tell me what it means for a matrix to commute?

#

thanks in advance

slow scroll
#

A commutes with B if AB=BA

rugged arch
#

much appreciated

#

thx

slow scroll
#

np

knotty raven
#

<@&286206848099549185> how do I go from a solution parameterization to an echelon form matrix?

chrome pond
#

what do you mean by solution parametrization?

knotty raven
#

Say you have 4 variables and 2 equations, but you substitute some parameters for the 2 extra variables and continue producing the solution

quick acorn
#

gaussian elimination?

knotty raven
#

yeah