linear-algebra

[old flame]

cause your third definition looks very similar with the definition of evaluating polynomials

[spiral star]

if you equip the endomorphisms on some vector space with the addition of linear maps and the function composition as multiplication

then they form a unital ring

so all the definitions above apply

we can construct a polynomial ring End(V)[X]

[old flame]

oh okay, I see

[spiral star]

so now the polynomials are certain sequences where the elements are endomorphisms

we can do this for any ring

endomorphisms are the ring elements

basically, view the definitions from the beginning but replace $R$ with $(End(V), +, \circ)$

[stoic python]

Flow:

[spiral star]

nothing changes

[old flame]

In summary, a polynomial is a sequence of finite support, where these elements could form a polynomial ring, with the operations of addition and multiplication. We could evaluate polynomials from elements in the unital ring, namely by substitution. Since the set of endomorphisms equipped with the correct operations results in a ring as well, the above apply and we obtain polynomials of operators right ?

[dusky epoch]

essentially yes

[spiral star]

yep

and then you can evaluate this polynomial by substituting in a linear operator

and then the evaluated polynomial will be a sum of compositions of linear operators

which is again a linear operator

so the only question that remains is "why can we plug an operator into a polynomial over the field F"

and i tried to give that answer in my last remark

[dusky epoch]

the answer, imo, is that the most obvious way to do so (with the slight caveat of interpreting the constant term as a multiple of the identity operator) behaves almost as you would expect

[spiral star]

yea

that's basically what i wrote as well xd

we can map elements of the field to an operator by $\lambda \mapsto \lambda \id_V$

[stoic python]

Flow:

[spiral star]

that's how we get from polynomials over the field to polynomials over operators

and then we can just evaluate as usual there

[old flame]

so the remark is basically saying that if we create a mapping between the field F and the set of endomorphisms, the map is actually a ring homomorphism as it satisfies the conditions. Thus, this map implies that the polynomial over F is sufficient to take in elements from End(V) ?

[spiral star]

yea, we can always plug an operator into a polynomial over F if we identify each field element $\lambda$ with the operator $\lambda \id$

[stoic python]

Flow:

[old flame]

is that some sort of trick involved ?

[spiral star]

no, that's straight forward

[old flame]

so this suggests that every field has an identity operator ?

[dusky epoch]

no

we have a field F and a vector space V over F and the ring of endomorphisms End(V) whose unity is the identity operator on V

[copper pulsar]

Is $\vec{r}-\cos(\varphi)\vec{r}=\vec{r}(1-\cos(\varphi))$?

[stoic python]

Anton S.:

[native rampart]

Yes

[old flame]

ignore my stupid message just know. Instead it works because if we map each element to its identity operator we essentially have a ring homomorphism, therefore we can evaluate the polynomial of an operator. Rephrasing the remark, take the inclusion map to bring each term of the polynomial to an operator, then take the evaluation map to show that the sum itself is a operator as well right ?

[spiral star]

you dont have to show anything

but that's how you evaluate

[old flame]

evaluate as in "calculating the substitution" ?

[spiral star]

uh i guess?

evaluate as in plug an operator into the polynomial to get an operator as a result

map each field element to a multiple of the identity operator, then use the evaluation map for polynomials over rings

as usual

[old flame]

oh so you map first, then plug

ahhhhhhh okay, now I know what the evaluate do LOL

thank you so much for your time 😹

[spiral star]

i can give you an example i guess, maybe that helps

so if you start if a polynomial in F[X] like 1 + 2X + 3X²

then you first map it to the polynomial in End(V)[X]: (1 id) + (2 id)X + (3 id)X²

and then you evaluate it by plugging an operator T in

$(1 \id)\circ T^0 + (2 \id) \circ T^1 + (3 \id) \circ T^2$

[stoic python]

Flow:

[spiral star]

note that $T^0 = \id$ and $T^2 = T \circ T$

[stoic python]

Flow:

[old flame]

so the inclusion maps the polynomial $a_i$ while the evaluation maps $X$

[stoic python]

Otoro:

[spiral star]

well, strictly speaking, no

but you get how to use it, right?

[old flame]

with your example then kinda yeah ?

[spiral star]

good, i think that should be enough for now, since you didnt have ring theory and the book you read doesnt need it either

the inclusion map i defined does not map between the polynomials

it maps between the field and the endomorphisms

but i can map between the polynomials by applying the inclusion map element-wise

i think the main problem is that all my definitions need justifications and proofs and i left them out

i just claimed that a bunch of functions are homomorphisms

[old flame]

So from the book, the $a_i$'s in p is different from $a_i$'s in p(T) right ?

[stoic python]

Otoro:

[spiral star]

idk

i didnt read enough of the books to know how he defines polynomials

if its a polynomial function, then the a_i are the same

if its a formal polynomial then they are technically different, but it doesnt matter for how you use them

[dusky epoch]

@old flame no, they're the same a_i

[old flame]

@dusky epoch is it because the $a_i$ is scalars in this case ?

[stoic python]

Otoro:

[dusky epoch]

yes

they are scalars

[old flame]

Thank you very much

i think the main problem is that all my definitions need justifications and proofs and i left them out
@spiral star maybe once I start learning ring theory, the definitions would start to make more sense and I may try to prove them with enough tools

I guess this is quite an interesting introduction sort of to rings lol

[thick kite]

I have a question can someone please help me

I’ve been trying for soo long

Q17

[thick kite]

I have the solutions but I can’t make sense of it

Can someone help me out.... how did they determine which was the hard and soft earth

[old flame]

@thick kite soft earth is the blue area

[thick kite]

How did you know that though??

[old flame]

it says as B is hard

[thick kite]

You know how in the solutions when calculating the total cost they times 100 by CE

When did they considered just CE as soft

Why not the whole thing

[old flame]

I think thats a typo, it should be 100 times x

cause it should be AC + CB

and btw @thick kite this question should be in #calculus

[thick kite]

Ahhhh okayy makes sense now

My bad I avoided posting there since they were in the middle of a discussion... didn’t wanna interrupt

[old flame]

ah I see, no worries

[wintry steppe]

I need help in proving that the given set of basis is a spanning set for the vector space of polynimials(degree <= 3).

I need help in proving that the given set of basis is a spanning set for the vector space of polynimials(degree <= 3).
@wintry steppe V is the vector space of polynimials (deg <=3)

[native rampart]

You mean you have to prove the given set is a basis?

[wintry steppe]

You mean you have to prove the given set is a basis?
@native rampart yes

I don't kmow how to show that the set spans the space.

I've proved that it is linearly independent

[native rampart]

Show 1,x,x^2,x^3 are in the span

[wintry steppe]

oh

So you mean to say that represent 1, x, x^2, x^3 as a linear combination of the elements of the give basis

I hope I didn't ask a stupid question

@native rampart

[native rampart]

Actually,You get 4 independent vectors,right

They span a subspace of the required space

So,They should be a basis

@wintry steppe

[wintry steppe]

Actually,You get 4 independent vectors,right
@native rampart yes

So,They should be a basis
@native rampart oh

Clever

[native rampart]

Look up Lagrange polynomials

[wintry steppe]

I didn't think in this way

Thanks man @native rampart

[native rampart]

This is clever because it tells you that to define a polynomial,you just need outputs at n+1 different points

[wintry steppe]

This is clever because it tells you that to define a polynomial,you just need outputs at n+1 different points
@native rampart 👍 got it. Thanks again.

[solid bough]

Kinda stuck on showing this. Any tips or help?

[terse vortex]

hello people

I need help with some math and idk if im wrong or not

Determine the degree of the polynomial and the coefficients

I know the degrees

[native rampart]

Wrong group

Ask in math help

[terse vortex]

thx

[stoic python]

alef0:

$\begin{equation}
\begin{aligned}
&\operatorname{If} M \in M_{n \times n}(K) \text { and } A \text { y } B \text { are square } M=\left[\begin{array}{cc}
A & 0 \\
0 & B
\end{array}\right] \text { then }\\
&\det (M)=\det(A)\det(B)
\end{aligned}
\end{equation} 

I try this   \det(M)=\det(\begin{bmatrix}
A & 0 \\
0 & B
\end{bmatrix})=a_1\det(B)=\det(A)*\det(B) $
```Compile error! Output:

! LaTeX Error: Bad math environment delimiter.

See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...

l.54 $\begin{equation}

Your command was ignored.
Type I <command> <return> to replace it with another command,
or <return> to continue without it.

[dry pulsar]

I try this

[stoic python]

alef0:

[dry pulsar]

For 2x2 matrix $\det(M)=\det(\begin{bmatrix}
A & 0 \
0 & B
\end{bmatrix})=a_1\det(\begin{bmatrix}
A_{11} & 0 \
0 & B
\end{bmatrix})+(-1)^{3}a_2\det(\begin{bmatrix}
A_{12} & 0 \
0 & B
\end{bmatrix})=a_1a_4\det(B)-a_2a_3\det(B)=\det(B)(a_1a_4-a_2a_3)=\det(B)\det(A)$

[stoic python]

alef0:

[dry pulsar]

I have to do induction on the size of matrix A and im stuck to prove it is valid for a matrix $A_{s+1\times s+1}$

[stoic python]

alef0:

[dry pulsar]

<@&286206848099549185>

[old flame]

@spiral star quick question, whats the difference between the left and the right side ?

[native rampart]

You are defining what (pq)(z) means

As that expression on the right side

[torn ermine]

Apparently it's a good idea to put my question from #help-5 here

I want to find the transpose of this:

The two 'outer' terms for C are orthogonal matrices

I tried this and:

• switched Sigma_X and Sigma_Y
• switched ^-1/2 for ^1/2

Looking at a known calculation, it appears that my second change was incorrect

However, I can't see any issues with what I've done

Oh and we know have Sigma_XY^T of course

[old flame]

@native rampart well the book asked me to verify that $(pq)(T)=p(T)q(T)$ and Im not sure how

[stoic python]

Otoro:

[merry tree]

does it make sense to talk about normed and inner product spaces over some arbitrary field rather than R or C?

i guess the field needs to have some analogue of the absolute value/modulus

[empty copper]

I think the only other thing that your arbitrary field needs is a partial order

As you need to somehow define positive definiteness in your field

[ocean sequoia]

So i am just suppose to show that it doesnt satify one of the 3 conditions for being an inner product space correct?

which are 1. linearity 2. Symmetric 3. Positive Definite

[empty copper]

that's right

[dry pulsar]

I forget to put I have to prove it by induction on size of Matrix A

[ocean sequoia]

Im assuming it doesnt satisfy linearity because it works for the other two

[empty copper]

Provide a counter-example

[ocean sequoia]

any negative scalar should work right?

[empty copper]

Seems like it would

Good job!

[ocean sequoia]

thanks lol

[ocean sequoia]

@wintry steppe this should go in algebra/prealgebra i can help you there

[wintry steppe]

ok ty

[hallow kite]

quick question about determinants

i have a 3x3 matrix and i have a 0 in the bottom rightmost element,so i wanna switch columns so its on the left

i would have to 2 permutations and multiply the determinant by (-1)² right?

[wintry steppe]

yes

[hallow kite]

ty!

[wintry steppe]

swapping any two rows or columns changes the sign

so swapping twice multiplies by (-1)^2

you are right

[hallow kite]

i can only swap adjacent rows or columns right?

[wintry steppe]

they do not have to be adjacent, but it certainly makes computations easier (and maybe longer) if you do thay

there's something to be said more generally about the determinant as an alternating multilinear function of the columns (or rows) of your matrix, but i can't tex it rn

[rare swallow]

So, I was thinking about determinants the other day.
Let V be an n-dimensional vector space, let 𝛬ⁿ(V) denote the vector space of alternating n-linear forms on V and let F : V -> V be an endomorphism
Then F induces a pullback Fᵀ : 𝛬ⁿ(V) -> 𝛬ⁿ(V) given by, for 𝜔 ∈ 𝛬ⁿ(V) and v₁, ..., vₙ ∈ V, as
(Fᵀ𝜔)(v₁, ..., vₙ) = 𝜔(Fv₁, ..., Fvₙ)
Now Fᵀ is a linear map and 𝛬ⁿ(V) is 1-dimensional, so we can ask about the eigenvalue of Fᵀ. This eigenvalue is then the determinant of F (I haven't verified, was busy brushing my teeth when I came up with this, should hopefully be correct).

Over the complex numbers, we have this cool identity which says that the determinant of an endomorphism is equal to the product of its eigenvalues

But this obviously does not work over arbitrary fields

I wonder, is there a way of constructing some similar identity by considering the pullbacks on the vector spaces of multilinear alternating k-forms?

[rare swallow]

The older editions of Axler had this construction for the determinant of an endomorphism over the reals, where he factored the characteristic polynomial to a product of one and second degree polynomials and then proclaimed that the determinant is the product of the constant terms

[wintry steppe]

ugh sorry this is being a pain in the ass, when i fix my latex i'll post this

[rare swallow]

@wintry steppe Yes indeed. The nice thing about the construction I did is that it can be done in a basis-independent way.

But it might just be shifting the mess elsewhere

Likely to the part where 𝛬ⁿ(V) is 1-dimensional needs to be proven

[wintry steppe]

𝛬^k(V) for V n-dimensional has dimension n choose k

...so 𝛬ⁿ(V) is a 1-dimensional space.

shifting the mess elsewhere
not sure what this means. proving that 𝛬^k(V) has dim n choose k isn't that bad

[rare swallow]

I guess not. Still nicer than constructing the determinant of a matrix first and then showing that it is actually basis-independent and thus a property of the parent linear map.

https://math.stackexchange.com/a/21617

Hmm, apparently the Cramer's rule can be proven this way

[wintry steppe]

tbh this is the best defn of the determinant

[rare swallow]

Yes indeed. I am almost afraid to guess as to why most textbooks do not take this route.

[wintry steppe]

because developing exterior algebra is way too much work for the definition of the determinant most people use

it's my favorite definition, but for having a working definition of the determinant, you can go with easier

[rare swallow]

What do people actually use the determinant for?

I guess computing the characteristic polynomial

[wintry steppe]

that

change of variables regarding integration as well

[limber sierra]

defining the special linear group

[wintry steppe]

testing linear independence

so many things

[limber sierra]

the determinant is also a useful invariant in things like checking whether matrices are similar

[rare swallow]

computing-with-numbers things it seems

[wintry steppe]

namington is more knowledgeable than me so i will leave this to them

[limber sierra]

it is true that many of the determinant's applications are more computational, yes

however there are some nice more "pure" framings

[wintry steppe]

continuity of the determinant function is an important theoretical one iirc

[rare swallow]

@limber sierra But, computing the determinant for large-ish matrices is infeasible anyway

[limber sierra]

it's very fast for computers

and oftentimes determinants are more "heuristics"

like computing the exact value of a determinant may be annoying

but determining, say, its sign

might be easier

[rare swallow]

Computers compute the determinant by factoring the matrix involved though

So if you like, need to check for linear independence, you do not have to do the final step

[limber sierra]

if you want a less computational example, let $\mathbb{F}, M(\mathbb{F})$ be monoids under $\cdot$ in $\mathbb{F}$; then every homomorphism $M(\mathbb{F}) \to \mathbb{F}$ can be written as $\varphi \circ \det$ for some monoid homomorphism $\varphi\colon\mathbb{F} \to \mathbb{F}$

[stoic python]

Namington:

[limber sierra]

this means that every monoid homomorphism from square matrices to their scalar field "factors into determinant"

algebraically this is a very handy propery as it makes the determinant, in some sense, a "canonical" homomorphism

at least multiplication-wise

when extracting scalars from matrices

[rare swallow]

Interesting. I should probably study some more abstract algebra sometime.

[limber sierra]

note however that for n * n matrices where n > 1

there is no ring homomorphism from this space of matrices back into F

[main widget]

If you guys are done, I have a question..

[limber sierra]

ie theres no way to get a homomorphism compatible with both addition and multiplication

unless of course n=1, since then you have 1x1 matrices

and the identity map (converting matrices to scalars) does the trick

also, the leibniz definition of the determinant hints at some group theoretic connections (ie permutation groups)

which may be worth exploriing

@main widget go ahead

[main widget]

I'm stuck with this problem. I've recalculated it like 3x even using online calculator for cross product and still somehow get it wrong. Any idea?

Basically the way I did it was first to cross the two intersecting plane to get a vector parallel to the intersecting line. Then, find one point from the two plane equation laying somewhere in the intersection line. After we get that, we now have the vector going the same direction as the intersecting line (from cross product of the two planes) and we can find the other vector from connecting the point that's given to the one in the intersecting line. After that, I cross both vectors to get the normal vector perpendicular to the plane that we're trying to find. Once I get that, I just use the general A(x-x0) + B(y-y0) + C(z-z0) = 0 formula.

<@&286206848099549185>

[wintry bridge]

yo can anyone help me out with algebra 1 hw rn? <@&286206848099549185>

[wintry steppe]

just ask

and

it probably doesn't belong here it it's alg1

[wintry bridge]

where at?

[wintry steppe]

#prealg-and-algebra maybe, or one of the question channels

[devout void]

is this a good place to ask for matrices

[native rampart]

Sure

[devout void]

alright needed just to be sure rn no specific question tho aside from what do matrices really mean

i get they express how i j k and other base vectors will transform into a specific way

but need to do some more learning

[bold ivy]

@versed topaz Do you agree with me, if an determinant of an marix is 0, it has an area 0

[versed topaz]

Matrices don't have areas

[bold ivy]

but thats' what is determinant, so far what I understand

[versed topaz]

So like, if think of all the columns as vectors in space, they form a paralleliped, and the volume of that is the determinant

[bold ivy]

( 1 0
0 3
) this means you multiply x with y which will get you 3 which means this matrice has the area of 3

3bluebrown1 is really good at explaining this, so the yellow is represented to be the area.

so when a matrice has a determinant 0, i hat and jhat overlap which means they dont have any solutions apperantly.

[devout void]

3 blue 1 brown is god of linear transformation

[bold ivy]

ye

eitherway

If determinant is 0 of a system equation, then it has no solutions

[native rampart]

Or infinite solutions

[bold ivy]

?

[native rampart]

x+y=2
2x+2y=4

[bold ivy]

if ihat and jhat is pararell then I think there are no solutions tho?

Solutions means there is a change to the plane ?

no solutions with 3d space, means khat doesn't do any change on its axis

there is no change in 3d space

it is infinite

WTH

how is that possible?

but if you gonna do gausselimination with that system equation

You will just get 0 = 2 which is a contradiction?

no

you will get

0 = 0

in the last row

you will only get x + y = 2 which is infinite solution yeyeyeyye

@native rampart but are you saying that a matrice that has det 0 can be both have no solutions and have infinite solutions?

[native rampart]

Either no solutions or infinite

[bold ivy]

ye but it doesn't make any sense, DET 0 just means they overlapp

[native rampart]

Don't rely on geometric intuition,in places where there are none. You would have to abstract at some point

[bold ivy]

but

Can you figure out if it has a solution or not based on column space or (rank) ?

[native rampart]

You can

[misty sparrow]

Hey I have this problem, I'm hoping to get some ideas on how to solve it

There's a 4x14 rectangle and I need to find how many ways there are to fill it entirely with 1x2 rectangles (the solution is preferrably using matrices, but without matrices is fine too)

[bold ivy]

@native rampart how would you do it?

[native rampart]

There's this thing called Row reduced echleon form

Given a system Ax=b you mutliply both sides with P,such that C=PA is row reduced echleon
Cx=Pb is simple to solve,and you can check the number of soultions with that

And if the det is non zero A is invertible,and C will be Identity

[bold ivy]

and if cx = pb is false then it has no solutions+

[native rampart]

Yes

[bold ivy]

I've done Row reduced echleon form

but

I was thinking perhaps you could perhaps use determinant eithweay

You can use row reduced echelant form adn then multiply the diogonal and get the determinant

[native rampart]

If Invertible,Cx=Pb will always be true for a particular value of x

The process doesn't preserve determinants

[bold ivy]

wdym ?

[native rampart]

If you take 2I and row reduce it you get I

Determinant of original is 2^n and latter is 1

[bold ivy]

are you saying upper triangle method doesn't work in all cases?

[native rampart]

Ok, You could do that

[bold ivy]

oh I see, we missunderstood each other.

[native rampart]

Row reduce and find the determinant of the row reduced form
Is that what you mean?

[native rampart]

Well,If you do a few more steps you would get row reduced form

[bold ivy]

yeye

but

Can You use upper triangle method in nth matrices?

[devout void]

is this method aka gauss method ?

[native rampart]

A part of gauss

[devout void]

full gauss wants the factors to be 1 aswell if my correct

the diagonal

[native rampart]

Yes

Or 0

[devout void]

and gauss jordan is like total obliteration

[native rampart]

idk why you would use gauss over gauss Jordan

[devout void]

sometimes gauss jordan will end up using a lot of fractions

depends

gauss jordan fs is more useful

but might be unpractical

[old flame]

May I ask how is at least one j not injective ?

[native rampart]

If all were injective,The only vector that would be sent to 0 would be 0

So, Atleast one has to be non injective,since a non zero v gets sent to 0

[old flame]

so for the final line, could we treatt the factors with operators function composition ? or whats going on witht the factors with v

[native rampart]

ABv means apply B then A to result

Yea, composition

[old flame]

So it is a composition, where we apply the transformation $(T-\lambda_1) \circ (T-\lambda_2) \circ \cdots \circ (T-\lambda_m) (v)$ right ?

[native rampart]

Yes

[old flame]

so one of them lets say j would be zero, $(T-\lambda_j I)v=0$ ? so it corresponds to $Tv=\lambda_j v$ hence the existence of one eigenvalue $\lambda_j$

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

ok thank you

[native rampart]

Sorry no,not like that

You can't conclude that way

@old flame

[old flame]

oh whats wrong

[native rampart]

The composition of other operators put the vector into the null space of $(T-\lambda_i)$

[stoic python]

DrunkenDrake:

DrunkenDrake:

[old flame]

so following the usage of the jth term, as the jth composition produces zero, this puts the previous compositions in null(T-\lambda_j I) ?

[native rampart]

Yes

[old flame]

so this shows that $(T-\lambda_j I)$ is not injective right ?

[stoic python]

Otoro:

[old flame]

since $null (T-\lambda_j) \neq {0}$

[native rampart]

Yes, because it has a nullspace

[stoic python]

Otoro:

[old flame]

then how can we conclude it has eigenvalue ?

[native rampart]

(T-a)v=0 for some v and a

take a to be $lambda_j$ and v to be that resultant

[stoic python]

DrunkenDrake:

[old flame]

so its just using the fact that it has some non-zero vector that maps to 0, then just rearrange and done ?

[native rampart]

Yes

[old flame]

thanks

[bold ivy]

I dont understand why they remove row 2 so it becomes 3x3 matrice

[wintry steppe]

give more context

this makes no sense without context

[wintry steppe]

ah it's a determinant calculation

[bold ivy]

ye

[prisma pier]

looks like the minors method

[bold ivy]

why did they remove row 2

[prisma pier]

they remove column 3

and multiply by -1

look up the minors formula, also called the laplace expansion

[wintry steppe]

lmao im braindead

forgot that straight bars on a matrix denotes determinant sometimes

[gray dust]

det is invariant wrt row/col operations, and a tip is to laplace expand along a row/col with many 0s

[bold ivy]

???????

[wintry steppe]

to expand on what rokabe and smh were saying, if you expand along the third column, the result will be -1 * det(matrix with second row and third column deleted)

which is what happened

[bold ivy]

But that is because 1 is -1 in that scheeme

[gray dust]

did you google laplace expansion formula?

[wintry steppe]

ah, schemes, everyone's favorite LA topic

[bold ivy]

Yes

https://www.cliffsnotes.com/study-guides/algebra/linear-algebra/the-determinant/laplace-expansions-for-the-determinant

I I didn't know you could do it in this way,

but apperantly you can do that

I think you it is fine using laplace expansion with 4x4 but if you have more than 5x5 then I you are being better of just using gausselimination / upper triangle method.

[wintry steppe]

if you have a 5x5 matrix or larger you use wolframalpha

[bold ivy]

Im just speaking general

[cosmic crow]

Anyone have an idea on how to solve this problem. need some help!

[gaunt field]

https://gyazo.com/231150787ce2e619098b0b43ca3ace7b

Can someone help with the very last question after d?

I feel like it's not possible for a square matrix to satisfy b)

[errant wyvern]

Can someone help me understand this proof? A is a lin operator A (V->V), detA=/=0 then det A* =/=0. Proof : we assume detA*=0 , then there is such y, y=/=0, Ay=0, we multiply by x from the left side <x,Ay>= <x,0> or <Ax,y>=0. But since A is a regular operator, ImA=V, so there is such vector x that Ax= y, so from before we get <y,y>=0, y=0. Contradiction, detA*=/=0

why does y=0 cause a contradiction here?

[native rampart]

Can you use latex pls.

[half storm]

What is A*

[wintry steppe]

y = 0 causes a contradiction because you assumed y is non-zero

[gray dust]

A* is the adjoint of A

[half storm]

Never learned what that was in my LA class

[native rampart]

do you know inner products?

[wintry steppe]

unique linear operator with <Ax, y> = <x, A^*y> for all x,y

iirc

sniped

wait

you deleted it

[gray dust]

it's whatever you got it 1st

[errant wyvern]

ayy thanks got it now

[gray dust]

btw \* to avoid error

[errant wyvern]

il go through latex next time

[gray dust]

don't feel peer pressured to latex i like this

[native rampart]

You like not being able to read?

[gray dust]

i read it just fine

[errant wyvern]

im allready begging for help, might as well make it easier for people to help me

[dusky epoch]

see that's the spirit

[errant wyvern]

If L is a subspace defined as $ L = { A + A^{T} : A \in R^{2x2}}$

how would a basis of said subspace look like?

[stoic python]

Lonac:

[errant wyvern]

$ span{ \begin{pmatrix}
2 & 0\
0 & 0
\end{pmatrix} \begin{pmatrix}
0 & 1\
1 & 0
\end{pmatrix} \begin{pmatrix}
0 & 1\
1 & 0
\end{pmatrix}\begin{pmatrix}
0 & 0\
0 & 2
\end{pmatrix} }$

[charred nacelle]

can someone help me determine if (2a, b, a+1) is a subspace of R3

[wintry steppe]

@errant wyvern use \{ and \}

fuck

[rose coral]

@errant wyvern I think your second matrix isn't in the subspace

[stoic python]

Lonac:

[errant wyvern]

ayyy i wrote it

[wintry steppe]

@charred nacelle the channel is occupied, so you should ask one of the help channels. that said, this should be straightforward. just check the vector space axioms and see if they work. i think that one of them fails

[rose coral]

Yeah that looks better, but when you write the basis no need to include the second matrix twice

But otherwise that works as a basis. Do you need to prove it is a basis?

[errant wyvern]

Im supposed to check if Linear operator of orthogonal projection on $L^{\perp}$ is a unitary operator

[stoic python]

Lonac:

[rose coral]

I'm guessing you are using the standard inner product from R^4

[errant wyvern]

uuu thats got me thinking, yeah the inner product isnt defined, so is there a list of standard inner products based on the space its performed in?

[rose coral]

I mean the problem isn't well defined unless they give you an inner product

My guess is that they want the standard inner product like dot product from R^4

[errant wyvern]

$x=\begin{pmatrix}
x_1 & x_2\
x_3 & x_4
\end{pmatrix} ; y=\begin{pmatrix}
y_1 & y_2\
y_3 & y_4
\end{pmatrix} then $$\langle x,y\rangle$$ = x_1*y_1 + x_2*y_2+x_3*y_3+x_4*y_4$

[wintry steppe]

use _ for subscripts/indices

and \langle \rangle for inner product brackets
e.g. \langle x_{test}, y_3 \rangle

[stoic python]

TTerra:

[wintry steppe]

like that

nice

[errant wyvern]

why does x3*y3 show as x3 "x" y3

i typed it the same way as every other multipliction

[rose coral]

I don't know actually

[wintry steppe]

strange

try putting the inner product part in double dollar signs

[stoic python]

Lonac:

[rose coral]

Wait unitary means that it preserves the inner product right?

[errant wyvern]

$U*U^{*}= I$

[stoic python]

Lonac:

[wintry steppe]

which is equivalent to what vman said

[errant wyvern]

$\langle Ux, Uy \rangle = \langle x, y \rangle $

[rose coral]

Ok cool just making sure

So the projection can't be unitary

[stoic python]

Lonac:

[rose coral]

Right, so in particular it should preserve lengths/the norm. Try to show that the norm of some matrix changes after applying the projection

[errant wyvern]

could you elaborate?

[rose coral]

Sure

So if the projection was unitary, it should preserve the norms of all vectors.

But I'm saying the projection changes the norm of some vectors

So to show that the projection is not unitary it is enough to find some vector (matrix) that has its norm change after applying the projection

[errant wyvern]

ah i see, so i can just pick a vector, apply the projection in it and show that the norm is different after the projection

[rose coral]

Yeah

Well some vectors will have the same norm

But as long as you find one whose norm changes you're good

[errant wyvern]

Ok while still on this topic for said L how would i find $L^{\perp} $

[stoic python]

Lonac:

[rose coral]

There are two approaches I know of

The easiest might be to find an orthogonal basis for L and then extend it to an orthogonal basis for all of V = R^{2 x 2}. The basis vectors not in L should be a basis for the perpendicular subspace

[errant wyvern]

oh so when we did a simmilar task we would create a matrix from the vectors that span L and then for that matrix find ker, and it would be the orthogonal complement, but i couldnt do that here because my vectors are in matrix form themselves right?

[rose coral]

No, you can use the same thing

But you need an orthogonal set of vectors that span L

I think

[errant wyvern]

im confused, how would the matrix created from the vectors that span L as i found above look like?

[rose coral]

You treat the matrices like vectors in R^4 and then the projection map would be represented by a 4 by 4 matrix

For example if I wanted to represent the transposition map on 2x2 matrices, I can choose a basis like
$$\Bigg{\begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \ 0 & 1\end{bmatrix}\Bigg}$$

[stoic python]

The_Vman:

[rose coral]

And then transposition with would be represented by
$$\begin{bmatrix}1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1\end{bmatrix}$$

[stoic python]

The_Vman:

[rose coral]

In the chosen basis

As transposition exchanges the 2nd and 3rd basis vectors while keeping the others the same

[errant wyvern]

$\begin{pmatrix}2 & 0 & 0 & 0 \ 0 & 1 & 0 & 1 \ 0 & 0 & 0 & 2\end{pmatrix}

[stoic python]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[errant wyvern]

so the matrix would look like this ?

[rose coral]

You are trying to represent the projection to L?

[errant wyvern]

im trying to get a matrix of L so i can find kerL to get vectors that span $L^{\perp}

yeah i guess

[rose coral]

So if you want to find the space perpendicular to L, you can't just take any map the goes into L, you need the orthogonal projection to L. I believe to find the orthogonal projection you first need an orthogonal basis for L, which the one you have is, but then the map isn't simply putting that basis in the columns

I think the orthogonal projection onto the space spanned by an orthogonal basis ${v_1, \dots, v_k}$ is given by
$$P(x) = \sum_{i = 1}^k \frac{\langle x, v_i\rangle}{|v_i|^2} \cdot v_i$$

[stoic python]

The_Vman:

[rose coral]

So if you want to represent the orthogonal projection onto L you need to find where the basis vectors
$$\Bigg{\begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \ 0 & 1\end{bmatrix}\Bigg}$$
go when you apply $P$.

[stoic python]

The_Vman:

[rose coral]

It is a bit of a process

But the first and last basis vectors are conveniently already in L so they should be kept the same. That means you only need to compute P of the second and third basis vectors

And there two of the inner products you have to take will be zero so that should make the computation easier

[errant wyvern]

ok

[rose coral]

You can do it if you want the practice, but otherwise the Gram-Schmidt should be easier

[errant wyvern]

Ayy while reading and writting this on the side i did what u said first

[rose coral]

Oh nice

[errant wyvern]

Found that for x thats all ones

like $x=\begin{pmatrix}1 & 1 \ 1 & 1 \end{pmatrix}

[stoic python]

Lonac:
Compile Error! Click the reaction for details. (You may edit your message)

[errant wyvern]

that its norm is different when i apply the orthogonal projection operator on it then its own

so eyy that works

[rose coral]

Awesome

[errant wyvern]

I will still try to do but if you dont mind entartaining me let me attempt to clear this up in my own head

[rose coral]

You can also argue using just theory

Sure

[rose coral]

I was just gonna say that since the projection is onto a lower dimensional space, the projection has a non-zero kernel and so can't preserve the norm

[errant wyvern]

so we had a simmilar example to this in our notebooks, so im wandering if we have vectors that are written as 2x2 matrices, how would their form look like when written as a matrix, like in the picture above they littelarly put the vectors that span W into a matrix to form it

[rose coral]

Yeah you should be able to put the matrices as a row vector with 4 entries

Like you did above

I think I got a bit mixed up thinking about the projection map

It does make sense now that I think about it

So like
$$\begin{pmatrix}2 & 0 & 0 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2\end{pmatrix}$$

[stoic python]

The_Vman:

[errant wyvern]

yeah thats what i was wandering

thx for helping out really appriciated it

[rose coral]

No problem

[errant wyvern]

So if i have vectors x= 1-3t^2 and y=0, how do i find d(x,y)?

[gray dust]

how's d defined

[errant wyvern]

im not sure how to extrapolate that from the exercise im trying to figure out so let me get the exercise

In space of polynomials with degree equal or lesser then 3, we are given some scalar product, matrix B=f', subspace L=span {1,t^2} and we need to figure out based on the scalar product matrix A thats orthogonal projection on L. Task is given to us to find the distance between B(A(f)) and A(B(f)) for f=1+t-t^3

so i calculated B(A(f)) and A(B(f)) i just need the way to get the distance between them and i forgot how you do that

oh yeah scalar product is inner product or dot product in english litterature

[gray dust]

the problem or your book should define a metric on such spaces

[rough olive]

$\begin{bmatrix}1//2//3\end{bmatrix}$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

why does this happen

$$$\begin{bmatrix}1/2/3\end{bmatrix}$

[errant wyvern]

using the wrong slash

[gray dust]

\\ not //

[errant wyvern]

\ /

[rough olive]

oh

[stoic python]

Resident Tracksuit Advisor:
Compile Error! Click the reaction for details. (You may edit your message)

[gray dust]

$\begin{bmatrix}1\2\3\end{bmatrix}$

[stoic python]

RokettoJanpu:

$\begin{bmatrix}1\\2\\3\end{bmatrix}$

[rough olive]

thanks

:)

[gray dust]

no prob

[errant wyvern]

in the past we used $d(x,y) = \sqrt{ (x_1-y_1)^2+...+(x_n-y_n)^2}

[gray dust]

that's a usual metric on R^n but this is a different space

[errant wyvern]

on the side note why am i not managing to call the texit ?

what did i mistype

[gray dust]

if you got an inner product then that defines a usual norm by ||x||=sqrt(<x,x>) which in turn defines a usual metric by d(x,y)=||x-y||

[errant wyvern]

ohh so id need to find ||x-y||

ight gonna go do that quick, we were given an inner product in the exercise it self

thx

[gray dust]

you're welcome

[dark harbor]

Why am I wrong? I have tried multiple ways of putting it in. Please help.

3, 6, 15
0, 0, 0

[gray dust]

you messed up signs

[dark harbor]

It wanted it reduced.

Thank you.

[gray dust]

uh sure np

[wintry steppe]

why are there two points labelled C

also this might belong in #geometry-and-trigonometry instead of here

[twilit bronze]

ah sorry, ill move it there than

[dry pulsar]

$\delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{s-1}\
U+kV \
\vdots \
V_{s+1}\
\vdots \
V_n
\end{array}\right)= \delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{(s-1)} \
U\
\vdots \
V_{s+1}\
\vdots \
V_{n}
\end{array}\right)+k\delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{s-1}\
V \
\vdots \
V_{s+1}\
\vdots \
V_{(n)}
\end{array}\right)$

[stoic python]

alef0:

[dry pulsar]

This is the def of a n-linear function

What are all 1-linear functions of $\delta(M_{1x1}(K) to\ K$

[stoic python]

alef0:

[dry pulsar]

I think are the functions of the form $\delta(A_{11}) = cA_{11}$

For some scalar c

[stoic python]

alef0:

[wintry steppe]

what is delta(M_{1x1}(K))?

are you just asking what the linear maps from M_{1 x 1}(K) to K are?

if that is the case then you are right

[dry pulsar]

$\delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{s-1}\
U+kV \
\vdots \
V_{s+1}\
\vdots \
V_n
\end{array}\right)= \delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{(s-1)} \
U\
\vdots \
V_{s+1}\
\vdots \
V_{n}
\end{array}\right)+k\delta\left(\begin{array}{c}
V_{1} \
\vdots \
V_{s-1}\
V \
\vdots \
V_{s+1}\
\vdots \
V_{(n)}
\end{array}\right)$ the domain of this all Matrix nxn and the functions that they comply with that are called n-linear functions

[stoic python]

alef0:

[dry pulsar]

Oh thank you!

[wintry steppe]

if T : M_{1 x 1}(K) -> K is linear, then it's equal to scalar multiplication by T(1)

where, if you want to be pedantic, that 1 denotes the 1x1 matrix containing a 1

[dry pulsar]

Oh thank you!

[fallow patio]

is the fact that a vector space being closed under addition an axiom in itself, or is it derived from sth else?

[limber sierra]

it's part of how the operation of vector addition is defined

its usually not considered "its own axiom" so to speak

its just

when you say $+$ is a function from $V \times V \to V$

[stoic python]

Namington:

[limber sierra]

this requires closure

otherwise it wouldnt be a function into V

the key takeaway: you do need to check that your vector space is closed under vector addition (and scalar multiplication) unless it's obvious

[fallow patio]

thanks, that's a really good explanation

[limber sierra]

as an aside, it will usually be "obvious" that a vector space constructed "directly" from a field is closed under +

like we know R the real numbers are closed under addition

so of course R^n will be

since vector addition in R^n is just adding up real numbers repeatedly

the most common case where closure "isn't obvious" is when we're trying to verify whether something is a subspace

[dusky epoch]

let A, B be two nxn matrices both with positive determinant. is it true that det(tA + (1-t)B) > 0 for any t in [0,1]?

wait no

obviously not

[crystal oracle]

Hi. I've been thinking about how invariant subspaces of diagonalizable linear operators (over finite dimensional real and complex vector spaces) look like. Does an invariant subspace of a diagonalizable operator necessarily have a basis of eigenvectors?

[wintry steppe]

@crystal oracle The restriction of the linear operator to the subspace is itself diagonalisable (prove this) and therefore there exists a basis of eigenvectors

[crystal oracle]

@fallen sigil I know that being diagonalizable is the same as having a basis of eigenvectors.

Anyway, I've resolved my question for arbitrary field thanks to https://math.stackexchange.com/a/78106

[split heart]

Are all symmetrical matrixes invertibles?

[umbral smelt]

I don't think so?

Take zero matrix for example

[split heart]

Yes, so not all symmetrical matrixes are invertibles

I got it, thank you

[umbral smelt]

You're welcome!

[split heart]

One more thing I've a doubt with: Is the transposed of the inverted equal to the inverted of the transposed? Sorry by mistakes or conventions, I speak spanish.

[gray dust]

@split heart you can try proving this

[split heart]

I haven't go through proves yet 😟

[gray dust]

you can do it if you know properties of transpose & definition of inverse. take an invertible matrix A, prove (A^T)^-1=(A^-1)^T

[split heart]

Alright, I'll try

[half storm]

Does anyone have an nice proof of showing that the nonzero rows of a row reduced echelon form a matrix are linearly independent?

It says it's clearly true in my book and if you look at any examples of a RREf it's clear that the nonzero rows are linearly independent but I was wondering if there was a way to prove it really nicely.

I may just take it's word for it.

[wintry steppe]

in every nonzero row there's exactly one 1 and all other 0s, and that's the only 1 in its column

so if you took a linear combination equal to 0, all the coefficients would have to be zero

kind of handwavy, maybe you can use induction or something to do it properly

nvm that deleted message is super wrong, ill have to take a second to phrase it properly

[vital beacon]

Anyone available to help me get started on this?

[wintry steppe]

just check the definition of a vector space

there should be a list of things to check to see if that set is a vector space

[vital beacon]

Okay, thanks

[wintry steppe]

To show subspaces you just gotta show it's closed under addition and scaling not all the axioms of a vector space

[wintry steppe]

for a, are z and w free variables?

[spiral star]

doesnt matter. you can pick any 2 variables as free and express the other two in terms of those

i think choosing x and z as free makes it very easy to generate a basis

[gritty sapphire]

I have a question about triple cross products. Suppose u and v are vectors in 3–space where u = (u1, u2, u3) and v = (v1, v2, v3). Evaluate u × v × u
and v × u × u.

It's not said what part of the cross product is to be done first. So no brackets.

Do i then assume v x u x u = (v x u) x u and u x v x u = (u x v) x u ?

Because what i think this question wants me do is expand both u × v × u
and v × u × u to show their not equal?

Just for clarity's sake

[gray dust]

i don’t think you can do anything beside mark this as a badly written question

[gritty sapphire]

Do i then assume v x u x u = (v x u) x u and u x v x u = (u x v) x u ?
@gritty sapphire

I think i'll just to this

Because it's 10 marks of a 50 mark assignment

[gray dust]

or ask the prof to clarify

[old flame]

Could I ask how is $T-\lambda I$ not surjective ?

[stoic python]

Otoro:

[rose coral]

Since $\lambda$ is an eigenvalue of $T$, $T$ sends some nonzero vector $v$ to $\lambda v$. But then $T - \lambda I$ sends $v$ to $0$.

[stoic python]

The_Vman:

[rose coral]

Do you see why that makes T not surjective?

[spare hornet]

If this is the standard matrix of a linear transformation, how do you describe what the transformation does? Is it just nothing?

[old flame]

@rose coral so therefore, there is at least one v that maps to 0, hence the mapping cannot cover the entire space V itself right ?

[rose coral]

But why does sending some nonzero vector to zero mean it's not surjective?

[old flame]

@rose coral Is it because T and $(T-\lambda I)$ is technically the same ? cause its just an rearrangment

[stoic python]

Otoro:
Compile Error! Click the reaction for details. (You may edit your message)

[rose coral]

No, we're not thinking about $T$ really just the map $T - \lambda I$, which we could rename $S$. The question now is just about a generic linear map $S$ on a finite dimensional space which sends a nonzero vector to 0

[stoic python]

The_Vman:

[rose coral]

And we want to say S can't be surjective

[old flame]

is this related to the null ?

[rose coral]

Yes

If you want I can give you a hint

[old flame]

Is it because $null (T-\lambda I) \neq {0}$ since T is an operator, it is not injective and hence not surjective ?

[stoic python]

Otoro:

[rose coral]

Yeah exactly

Essentially rank-nullity which works on finite dimensional spaces

[old flame]

ah alright, cause I recall theres a theorem about an operator being injective is the same as not being injective and not inveritble thank you so much

[rose coral]

No problem, just wanted to make sure you knew which theorem the text was referencing

[old flame]

I didn't thought of it at first, but then suddenly rmb that $(T-\lambda I)$ is an operator as well lol, thanks

[stoic python]

Otoro:

[rose coral]

Glad I could help

@spare hornet Being very technical I wouldn't say the transformation "does nothing." But that's being kinda technical. I would say that it's adding a dimension to the space you are talking about.

[spare hornet]

@rose coral thanks for the answer! My homework asks me to "describe the transformation". Would it be accurate to say that it adds a dimension to the vector?

[native rampart]

It maps the vector to a vector space, whose dimension is greater than the original one by 1

[spare hornet]

thanks!

[stoic python]

Gas:

Gas:

[half storm]

Prove it however you would prove the kernel of any other linear transformation.

Let $ f \in ker(T)$. $T(f(x)) = \int_{0}^{x} f(t) dt = 0$ ; this is what it means for $f \in ker(T)$.

[stoic python]

TheDon:

[half storm]

Think about the fundaemntal theorem of calculus, and that should help you.

[stoic python]

Gas:

[half storm]

Seems right to me.

That's it.

Just the the set of continuous functions such that F(x) = F(0)

where F is the antiderivative of f(x)

[stoic python]

Gas:

Gas:

[half storm]

Yea

np

[old flame]

Is the underlined part a typo ? is it suppose to be $span(v_1,...,v_{k-1})$ if its not a typo, could you explain why ?

[stoic python]

Otoro:

[dusky epoch]

the representation of Tv_k in the basis {v_1, v_2, ..., v_n} is just the k'th col of that matrix, is it not?

[old flame]

yes

[stoic python]

Otoro:

[old flame]

so is it because $\lambda_k=0$, then the transformation of the kth basis vector is just the combination of vectors $v_1,...,v_{k-1}$ ?

[dusky epoch]

lambda_k=0

[stoic python]

Otoro:

[old flame]

sorry I typed wrong lol

[dusky epoch]

yes...

[prisma cairn]

I don't get how to deduce that last iff (T* being the adjoint here)

@old flame reading Linear Algebra Done Right?

[old flame]

@prisma cairn yeah you ?

[prisma cairn]

same haha

(also, just realised not the best channel, oops)

[old flame]

seems like you're way ahead of me though lol

[dusky epoch]

use the defn of adjoint

twice

[prisma cairn]

oh lol, thanks Ann

[vital verge]

If you have a parabola with a focal point in (1+sqrt(3), -1+1/sqrt(3), -2/sqrt(3)) and line (translation said scroll?) (1, -1, 0) (that is the direction of the line ) and throug point q (-1-7/sqrt(3), 1-3*sqrt(3), 8/sqrt(3))

How do i create the coordinatematrix

Cause i remember using 2 focal points to create its direction vector to have it "appointed" as z axis

For an ellips

Then finding a vector thats orthogonal to it

Abd then last orthogonal to previous two

But how do i start here?

[lucid cedar]

I am trying to show that a particular transformation is linear if and only if b=c=0. The transformation is as follows:
T(x,y,z) = (2x-4y+3z+b,6x+cxyz)

My attempt was to write out:
T(x1+y1,x2+y2,x3+y3) = T(x1,x2,x3) + T(y1,y2,y3)

In this form it is easy to show that b = 0 because one side has b and the other has b + b; however, I am unsure how to show that c = 0. Using the second term in the transformation gives the statement:

c((x1+y1)(x2+y2)(x3+y3)) = c(x1)(x2)(x3) + c(y1)(y2)(y3)
This is ignoring the "6x" in the second term. I do not know how to show from here that c must be 0.

I solved it. Homogeneity property makes c = 0 easy to show.

[gaunt field]

Anyone know where I can get the strang book for free

[native rampart]

Libgen

though,You get this particular book if you literally just google

[hollow finch]

@lucid cedar Easiest way would be to make sure that T(0)=0

That tells you immediately that b=0

By inspection we can see that cxyz makes it nonlinear as well

[lucid cedar]

That is the path that the solution manual took

i did not really follow it

the c part i mean

[hollow finch]

then we can probably just do something easy like show T(1,1,1)+T(1,1,1) is not equal to T(2,2,2) to show c=0

how did the solution manual do it?

or better yet, show that kT(x,y,z) is not T(kz,ky,kz)

[lucid cedar]

that was how i did it

kT(x) =/= T(kx)

The solution manual I have did T(1,1,0) + T(0,0,1) and showed c = 0

[hollow finch]

maybe im stupid but wouldnt both of those inputs make the c term zero?

[lucid cedar]

looking back at it, I do get what they did, but i just dont see how that method is a reliable skill. What if the example that reveals c = 0 is rather difficult to find with a more complicated Transformation

This is the solution for reference:

[hollow finch]

oh now i get it

T(1,1,1)=T(1,1,0)+T(0,0,1)

i thought you meant just looking at the right side

[lucid cedar]

yea sorry i left part of it out

This sort of solution just feels cheap because it seems like its difficult to generalize this process to any transformation. I tend to feel this way when specific vectors are used rather than general ones. What if for thousands of examples the property holds but for only a few it breaks? It could very well be possible that I am missing something about the nature of linear algebra that should lead me to, intuitively, find solutions like this or that this method is more powerful than I give it credit for.

I guess a good question I should ask myself is "Can a non linear transformation appear linear for some number of specific examples?"

[hollow finch]

@lucid cedar Well, unless it's obnoxiously obvious, I think it's possible to find an infinite number of examples that make a nonlinear transformation look linear. It's probably not easy but it's probably possible.

I mean unless you're a computer, an algorithm to determine if a transformation is linear is a waste of time. With some practice you can look at a nonlinear transformation and show it's not linear in one or two steps.

The way in which it is nonlinear makes certain properties easier to prove than others.

If a constant is being added, just show that T(0) is not 0

if terms are multiplied then homogeneity may be the fastest way

showing that T(a+b) is not T(a)+T(b) may be faster in some cases though

with a lot of these upper level math things, you have to assess the situation and pick the best tool. You're not in elementary algebra where every single problem has the same easy 5 step solution anymore. some tools make some problems nearly trivial and others obnoxiously computational. it really depends.

[lucid cedar]

That's true. Thanks for talking me through it.

[hollow finch]

np

good luck

[austere cedar]

@wintry steppe If it helps put your mind at ease, here's a plot.

I'm self banned from anything linear algebra, due to none competence or I'd try and help

[half forge]

can someone help me with this

im not udnerstanding the notation stuff

[flat sedge]

You could probably take an arbitrary element in W1, an arbitrary element in W2 and apply the definition of their sum. Given this new element, take another element of that set and apply vector addition. Then take a scalar and apply vector multiplication. And then check the 0 vector exists. And probably containment.

[true talon]

Is this right... or am I missing something

[half forge]

i tried this but idk if its right

[wintry steppe]

it's not really clear what you're doing. did you mean "suppose W_1, W_2 ⊂ V" ? what does WNTS mean? why are you considering x, y like that? what are a,b,c,d? (presumably elements of W_1, W_2, but you should spell this out)

[half forge]

hmm not sure, im confused on this stuff

wnts means we need to show

[wintry steppe]

alright

your end goal is to prove that W_1 + W_2 is a subspace of V

so it should be "WNTS W_1 + W_2 is a subspace of V"

your next step should be to go right to the definition of a subspace

there are a few things, then, that you have to check about W_1 + W_2

1. it's a subset of V (no one will ever do this, practically, but for the sake of following definitions i include it)
2. it's closed under vector addition (presumably this is what you were trying to do in the picture)
3. it's closed under scalar multiplication. that is, if v is in W_1 + W_2 and c is a scalar, then cv is in W_1 + W_2

you can check all of these by using the definition of W_1 + W_2 given in the picture. some might say that you should verify that the zero vector of V is also in here, but that follows from (2)

it might make it easier to split your proof into three steps, each verifying the three things i listed above

[robust pond]

so my book is telling me to consider the geometric interpretation of this parametric system but im not really clear on what they mean

how do you look at something like this and interpret it geometrically?

is that too general and bad of a question to ask i wasnt sure what to ask

[gray dust]

x is in the form r+x3*v where r,v are vectors. this matches the form of position vector tracing a line

[robust pond]

is the intuition that this is tracing a line built through doing a bunch of these examples or is there a standard table or something of forms people consult

[half storm]

It just comes out of the geometric interpreation of vector addition.

You usually learn about it either in this class or multivariable calculus.

[robust pond]

hmm

[median forum]

its like

you fix a vector (-2,1,0)^T

and change the parameter x3

in the direction of the other one

[robust pond]

maybe ill ask the teacher

oh wait i think i get what you mean

[median forum]

youre scaling the vector being summed by the parameter

[harsh parrot]

I have solved part a) i)

But I am completely unsure as how to approach part a) ii) and the intuitive reasoning behind it. I would very much appreciate an explanation.

[dusky epoch]

you've found the time $t_0$ at which the altitude is 60 meters in part (a)(i)

[stoic python]

Ann:

[dusky epoch]

you now need to find the angle between $\mathbf{r}(t_0)$ and the $xy$ plane

[stoic python]

Ann:

[harsh parrot]

Most definitely 🙂 I'm struggling with the application of finding the angle itself @dusky epoch I've been thinking to find the dot product between r(t0), and r(t0) without the k component, but I'm not too sure if that's the case, and why it is.

[dusky epoch]

there are two possible routes

either you find the angle between r(t0) and k, then subtract it from pi/2

or you project r(t0) onto the xy plane, and find the angle between that and r(t0)

[harsh parrot]

What's the logic behind subtracting the angle with the k unit vector from 90o?

Ohhhhh

I think i get it, that's very smart 😄

Is it because the entire angle from the k axis to the xy plane constitutes pi/2? @dusky epoch And so, we subtract to find the remaining angle

[dusky epoch]

yes

[harsh parrot]

I'm not too sure about how to project it onto the x-y plane, though

I know vector projections involve multiplying twice by the unit vector in a specific direction

Wait a second, would we find the dot product between r(t0) and r(t0) in the x-y plane (so just remove the k component). If so, that now makes a lot of sense, as that would give the line of the vector without it's altitude. @dusky epoch

Thank you for provoking my thought-process btw, instead of simply giving me the answer.

[harsh parrot]

How would we do part b)

I made the i component equal to 0, and i got an invalid solution

[dusky epoch]

the i component of what

r(t)?

[harsh parrot]

The i component of r(t) @dusky epoch

[dusky epoch]

why

[harsh parrot]

Should it have been velocity?

[dusky epoch]

yes of course it should've been velocity

what you found is when the particle touches the y axis

or goes through

[harsh parrot]

Completely right. Thank you.

[harsh parrot]

I'm quite confused about parts a) and b). I used constant acceleration formulas, but I don't think I was close to showing it.

[wintry steppe]

@harsh parrot dude you there?

Anyway, you got it wrong, the speed at which the car took off the top of the cliff doesn't matter at all in the first part. While you're considering vertical motion you gotta consider initial velocity in vertical direction i.e. y-axis which is 0 in this case. So all you gotta do is :

Ignore the top couple lines

[harsh parrot]

@wintry steppe thank you :)

I'm assuming for part b, we can just multiply the initial speed by the time (4 seconds)

Given it is in the horizontal direction

[wintry steppe]

No need to thank mate. We're here to help each other out right?

[harsh parrot]

Of course, I appreciate that man.

[wintry steppe]

Yeah you can take t = 4 in the second part too since it won't be travelling horizontally after bumping into the ground ;)

[harsh parrot]

Thanks again mate, I hope you get the same help I just did when you need it.

[wintry steppe]

Anytime mate :)

[ember wedge]

hey guys

can someone help me with this question?

[wintry steppe]

wrong channel

#precalculus maybe, or one of the question channels

[ember wedge]

but this is for my uni course

isnt this for early uni?

[wintry steppe]

well, it's not linear algebra

just because it's a university course doesn't mean it can't go in the "pre-university" channels

[ember wedge]

aii

[wintry steppe]

what am i doing wrong

[limber sierra]

where did you get your numbers from?

[wintry steppe]

the number of columns that are not pivot columns

[limber sierra]

no, thats what it means to be a free variable

basic variables are variables that are not free

so columns that are pivot columns

[wintry steppe]

oh

i thought it was asking for free variables

[agile tangle]

Hi! I have quite the trouble understanding generalized vector spaces. Could anyone tell me why the objects inside a generalized vector space are called vectors but don't actually requires to be vectors?

Why not call them Sets or something that doesn't have anything to do with vectors?

[wintry steppe]

ignoring what a vector space is

what is a vector, to you?

i ask because you say "objects inside a generalized vector space are called vectors but don't actually requires to be vectors"

[agile tangle]

For me vectors are a quantity with a direction, or a segment with direction

[wintry steppe]

okay

[agile tangle]

But in my textbook (Advanced engineering mathematics, by Greenberg)

They say that a vector can be anything

like a matrix or a function

[wintry steppe]

well you need to understand a vector space before that

so

among other things, vectors can be added and multiplied by scalars in ways that obey certain laws (associativity, commutativity, etc.)

the idea of a vector space takes this and generalizes it

[gray dust]

the definition of vector space lets one see objects beyond boxes of pointy arrows as vector spaces all the same with the similar structure we like from the original boxes

[agile tangle]

I understand pretty well the concept of vectors in higher dimensions which can't be represented as arrows, so could I say that the set of vectors of 4 dimensions is a vector space?

[gray dust]

nope we're not restricting our talk to R^n

[wintry steppe]

yes, but if you want to understand the idea of an abstract vector space, you should completely forget the idea of vectors as a thing with direction and magnitude

anyways i am going to leave this to rokabe because i know he doesn't like having multiple people in one channel

[agile tangle]

So as long as any mathematical objects follows the properties of vector addition and scalars multiplication can be called a vector?

it can be called*

[gray dust]

we have the so called vector space axioms which is basically a list of criteria for a set, with given definitions of addition & scaling of elements in that set, to qualify as a vector space

after all that, elements of that vector space can be called vectors

[agile tangle]

So, if I understand,the set of all 2D vectors, is 1 particular Vector Space

and same for the set of all 3d vectors

etc...

but Anything else which satisfies these axioms is also a vector space

[gray dust]

since we're talking of the definition of vector space, i'm going to ask you to specify how you're defining addition & scaling on those spaces, and what you mean by 2d

[agile tangle]

Ahhh we call the 2d arrows vectors because they are part of a vector space

I think my mind was going the other way around

[gray dust]

So, if I understand,the set of all 2D vectors, is 1 particular Vector Space
there are gaps in what you said, namely
i'm going to ask you to specify how you're defining addition & scaling on those spaces, and what you mean by 2d

[agile tangle]

Additon is when you take a vector, add it to another vector

and the result is a vector

it is commutative and associative

[gray dust]

1st clarify wym 2d

[agile tangle]

Well I mean the 2d arrows, like U = (u1, u2)

[gray dust]

you don't have to talk of them visually as arrows

by 2d space you mean the set of all ordered pairs with real entries?

[agile tangle]

Yeah I guess

[gray dust]

the set of all pairs (x,y) where x,y are real numbers

[agile tangle]

Yeah

[gray dust]

we call that R^2 which is shorthand for RxR, a cartesian product of R with R

now how do you like to define adding pairs ie (a,b)+(c,d)=?

[agile tangle]

(a+c, b+d)

But I think I understand better now

[gray dust]

how about scaling a pair ie c*(a,b)=?

[agile tangle]

(ca, cb)

[gray dust]

now keep your definitions of addition & scaling on the side

have you seen the vector space axioms?

[agile tangle]

Yeah

[gray dust]

have you proved smth is a vector space using em?

[agile tangle]

Yeah

I couldn't get into my head that matrices could be called vectors but I think I am starting to get it now

[gray dust]

so this shouldn't be new. make sure R^2, along with how you define addition & scaling, satisfies all the axioms, then you can call R^2, with the same definitions of addition & scaling, and with scalars presumably coming from R, a vector space, and any element in R^2 can be called a vector

yes you can do the same thing with matrices, say the set of all n by m matrices with real entries, with addition & scaling defined the usual entrywise way, with real scalars. go through the axioms with that set, that's a vector space too

[winged belfry]

I think I'm stupid but how in the world do you graph this

[wintry steppe]

draw lines

[winged belfry]

like just random lines on my paper?

[wintry steppe]

no

each equation corresponds to a line in the plane

$x_1 + x_2 = 4$ to the line $x_2 = -x_1 + 4$, and $x_1 - x_2 = 2$ to the line...

[stoic python]

TTerra:

[wintry steppe]

once you find that second one you should be able to graph both lines

[winged belfry]

why subtract x1 instead of x2

[wintry steppe]

because i am rearranging the equation $x_1 + x_2 = 4$ for $x_2$ in terms of $x_1$

[stoic python]

TTerra:

[lament lily]

when you guys are done would you mind if I asked a question

not sure if discussion is still ahppening

[wintry steppe]

it is up to apollo

[lament lily]

👍

[wintry steppe]

if they don't respond in like, 10 minutes, then just ask

[winged belfry]

ok so lets say I want to solve for x2 for the 2nd equation, so it would be

$x_1 - x_2 = 2$ then $-x_2 = 2 - x_1$ then $x_2 = -2 + x_1$

[stoic python]

Apollo:

[wintry steppe]

yes

[winged belfry]

oh lol

and it would get me the same result regardless if i solved for x1 or x2

right?

[wintry steppe]

the same solution, yeah

[lament lily]

I am wondering if the solution set lies in a plane because we are confined to (b1 -1/2b2 + b3) = 0 or because we only have 2 pivots?

I feel the former makes more sense because if we have 3 pivots then b1, b2, b3 could be anything which allows for coverage of all of R^n

am I thinking of this right

[tall thunder]

Can anyone explain to me how I go about solving a question like this ? This linear algebra class literally makes no sense to me ☠️

I don’t even know what R3 means .. all real numbers cubed??

[limber sierra]

$\bR^3$ is the space of vectors with three real number elements

[stoic python]

Namington:

[limber sierra]

so for example

$\begin{pmatrix}5.2\0\-\pi\end{pmatrix}$ is in $\bR^3$

[stoic python]

Namington:

[tall thunder]

Just because it has 3 things on it ?

[limber sierra]

whereas $\begin{pmatrix}7\-3\-14.349034\0\end{pmatrix}$ is not, since it has 4 entries

[stoic python]

Namington:

[tall thunder]

Ah ok

[limber sierra]

and $\begin{pmatrix}4+i\31.2\-i\end{pmatrix}$ is not, since $i$ is not a real number

[stoic python]

Namington:

[limber sierra]

intuitively

you can think of $\bR^3$ as ``three-dimensional space"

[stoic python]

Namington:

[limber sierra]

each entry in the vector corresponding to one of the coordinates in (x, y, z)

[tall thunder]

So when the question asks to find two points that are not in the set ... does it mean like two separate r^3 sets or 2 individual things in the set (like x and y)

[limber sierra]

two things that are not in the set l

your image does not include what l is

so its pretty hard to answer that

but presumably l is a separate set, defined in the question or a previous part somewhere

[tall thunder]

So I already tried to solve for x and y and have z free , right ? At least I think that’s right , so from how I understand I can pick any point on z

So if I can just pick any point on z, why can’t I just pick any numbers for x and y that don’t match up to Z for the same number and that not be in the set

[limber sierra]

im not quite sure what you mean

you have free variables, which means that what you choose for z will affect the values of x and y, but it will do so in a predictable way

[tall thunder]

Yeah. So if I say z = 1 and that gives me a certain value for x and y , then I can say that perticular x and y does belong to L

So why can’t I say something along the lines of x and y being something else while z still being 0 then that set shouldn’t belong to L because it’s not true

[limber sierra]

for example, lets say i had the system:

[ x + y = 1 ]
[ y + z = 0]

then the solution vectors will look like $\begin{pmatrix}x\y\z\end{pmatrix} = \begin{pmatrix}z+1\-z\z\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

where the value of z is chosen arbitrarily

for example, if i choose $z = 5$, then that particular solution is $\begin{pmatrix}6\-5\5\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

i.e. x = 6, y = -5, z = 5

here the set l would represent the set of all possible vectors which we get by choosing any value of z

(naturally your set will look a bit different since your system is different, but the same ideas apply)

[tall thunder]

Ok so , since the question is asking to find something that does not belong to that

Can I say x = 7 , y = -4 , z = 5

[limber sierra]

for the example i gave, yeah

i havent solved your system

[tall thunder]

I mean it doesn’t belong to the set , because x and y are wrong ?

[limber sierra]

so idk if that works

[tall thunder]

Yeah I mean for your example tho

[limber sierra]

right, then yeah

that would be an answer to #4

(well, one of the two vectors you need for an answer)

[tall thunder]

So all I have to do is pick numbers that are wrong ?

[limber sierra]

yep

basically, numbers that don't solve the system

[tall thunder]

Ah okay awesome

[limber sierra]

at least that's what i'm assuming, you cut off some of the text in your image

[tall thunder]

And then for parellel lines , xD how do I do that lol ?

[limber sierra]

but i think i can guess what it says

do you know how to get a line from a vector?

[tall thunder]

Uhhhhhhhhhh

Lemme think for a second lol

Isn’t it like the solution of 2 of them or something ?

Idk ☠️

Or can I just arbitrarily put in values for z and then maybe make a line out of the points ?

[limber sierra]

sorry, have to leave right away - hopefully someone else is available to help - but the key term to look up here is "vector equation of a line"

your textbook might have something on it

[tall thunder]

Ah okay, thanks for the help !

[livid ravine]

it’s about sub vector space, can somebody explain to me which of the following sets are subspaces of real numbers ^n, the rule: n element of natural numbers with n >= 2, i‘m a bit lost

[dusky epoch]

do you know the definition of a subspace

this exercise is only a matter of applying that

[livid ravine]

only a bit

[dusky epoch]

do you need me to remind you of it & take you through checking U and V against it?

[bold ivy]

I'm wondering what a 2D plane represent? Does it represent all combinations of vectors in a 2d plane? Or potential vectors?

[half storm]

A plane in the traditional sense is just a flat two-dimensional surface existing in $\mathbb{R}^3$ and has scalar equation $Ax + By + Cz = D$ and vector equation $ \langle a_1,a_2,a_3 \rangle + t\langle b_1, b_2, b_3 \rangle + s \langle c_1, c_2, c_3 \rangle$ $t, s \in \mathbb{R}$. A plane, in general, is a vector space having dimension 2. However, a hyperplane generalizes the concept of a plane. A hyperplane is a subspace of a vector space of dimension $n$ that has dimension $n - 1$. What we think of as a plane in the traditional sense, is a hyperplane of $\mathbb{R}^3$. All of $\mathbb{R}^2$ is a plane in the traditional sense, but if you want to think of what a plane is with respect to $\mathbb{R}^2$ then that's a hyperplane and it's any subspace that has dimension 1. Meaning that lines are "planes" i.e. hyperplanes in 2D spaces.

[stoic python]

TheDon:

[gritty frigate]

Hey guys

Once I was shown here a picture that talked about determinants and what its value mean

I was wondering if any of you know what picture I m talking about

[native rampart]

The volume interpretation?

[gritty frigate]

Nono, it was a list of consequences

Det = 0 then:

Something like that

[gray dust]

https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

[errant wyvern]

So if $$ \overrightarrow{c}= 3*\overrightarrow{m} + 3*\overrightarrow{n} and \overrightarrow{d} = \overrightarrow{m} - \overrightarrow{n} $$ then what is $$ |-1/2 \overrightarrow{c} \times \overrightarrow{d}| $$ equal to? $$\overrightarrow{n} & \overrightarrow{m}$$are unit vectors

[stoic python]

Lonac: