linear-algebra

[north hedge]

i didnt scroll up :p

[wintry steppe]

Let $U$ be a subspace of $V$. If $U \neq V$, then $\dim U < \dim V$. Suppose that $U \neq V$. We know that $\dim U \le \dim V$. Suppose that $\dim U = \dim V$. Then let the basis of $V$ be $u_1,\dots,u_n$ and the basis of $V$ be $v_1,\dots,v_n$. Append vectors $v_j,v_{j+1},\dots,v_{j+k}$ to the basis of $U$ such that $U = V$.

We then know that $\mathrm{span}(u_1,\dots,u_n,v_j,\dots,v_{j+k}) = V$ and that all vectors in this list are linearly independent, as none of the $v_i$ were in $U$ for them to not be equal. But observe that the length of this list is $n + j + k$, but we know that length of $\dim V = n$, so we have that $n = n + j + k$ which implies that $j + k = 0$. Clearly, $j,k > 0$ which implies that $j = k = 0$ which is a contradiction to the assumption that $U \neq V$ as this would imply that the two subspaces are equal without appending any vectors to $U$ from $V$. Thus, $\dim U < \dim V$.

Does this proof work?

[stoic python]

n/c

[wintry steppe]

<@&286206848099549185>

[wintry steppe]

Can anyone check my proof?

[hard drum]

Would it not be more straightforward to do the contrapositive i.e. if dim U = dim V then U = V?

[dusky epoch]

@wintry steppe you should be explicit about the fact that "if U != V then dim(U) < dim(V)" is a goal, not an assertion. it reads as an assertion rn.

and also you need V findim as a hypothesis. if V is infinite dimensional it can have infinite dimensional proper subspaces.

also, "the" basis? there's (almost) no such thing as a vector space with one and only one basis.

also what's that with appending vectors to the list of u's "such that U = V"?

and why is this new list LI?

also the length of your new list isn't n+j+k, it's n+k+1.

[empty copper]

What's the question?

[waxen flume]

https://i.imgur.com/PD3ZXYm.png

can anyone help me with this

I'm getting back A, but matlab is marking it wrong

[shadow gyro]

Can the characteristic polynomial of an Endomorphism be defined without any choice of Basis or usage of matrices?

[lavish jewel]

vici, that is not an SVD

you will have to read the definition to see why, but first of all, you have a 0 vector on the leftmost matrix

by definition, the SVD has an orthonormal basis for the image of the original matrix as well as its orthogonal complement (i.e. in an M x N matrix, the leftmost matrix is M x M with M orthonormal columns) as the leftmost matrix, and yours doesn't

since the 0 vector is linearly dependent

[waxen flume]

oh nvm

i see

[flint jackal]

@waxen flume Also the reason you're getting A is because you're multiplying by the identity matrix. Any matrix times the identity matrix, you get the same matrix, kinda like any number multiplied by 1, you get itself

[nocturne oracle]

0

[waxen flume]

I redid my work

but now im stuck on finding u_2

because if i do 1/0 its division by 0 obviously

https://i.imgur.com/tI9ZvKE.png

what is wrong with my Us

I dont get it

[flint jackal]

Do AA^T, plug in the eigenvalues and that's the U matrix

[waxen flume]

?

[flint jackal]

You did A^(T)A, and found the eigenvalues for the V matrix, do the same process except AA^(T)

[waxen flume]

https://i.imgur.com/ADlagc7.png

https://i.imgur.com/oZPLYWI.png

lol.

[flint jackal]

Or you can do the method I said too, it works

[waxen flume]

but bro

https://i.imgur.com/vUF9xjS.png

{0; 0} is not in the U

from wolfram

[flint jackal]

First do AA^(T)

Plug in the eigenvalues, and then find the eigenvectors

[waxen flume]

nvm

i used symbolab

to calculate the u1

[flint jackal]

Those vectors are in the span of U

[waxen flume]

i guess they did it wrong

because when i did it in my calculator

i got the same as wolfram

BUT

https://i.imgur.com/ADlagc7.png

for the second U column

this makes no sense

because

1/0 * [A] * [v2]

[flint jackal]

Which is exactly why I said do my method, because you don't get a divide by 0 error

If you do it my way

[waxen flume]

whats your method

[flint jackal]

AA^(T), plug in each eigenvalue, find the eigenvector corresponding to that eigenvalue, and that's the U matrix

[waxen flume]

are you talking about A^T * A?

[flint jackal]

Nope, that's to find V

A*A^T will find U

[waxen flume]

A^T * A != A * A^T?

I didnt know this lol

[flint jackal]

Work it out yourself and you'll see that it's not equal

It's the same concept as A * B != B * A

[waxen flume]

thats big

didnt know that

https://i.imgur.com/UpyQ9YO.png

now what?

[flint jackal]

Plug in the eigenvalues and find the eigenvectors

didnt know that
@waxen flume How long have you been doing matrix multiplication/linear algebra to not know that?

[waxen flume]

i got

[waxen flume]

we dont talk about that

[flint jackal]

You messed up when finding the eigenvector for 4

You forgot a negative sign

You do need that negative sign, because if you don't, when you do UΣV, you get a positive 2 in the A matrix instead of negative 2

[waxen flume]

there is no negative sign

i just did the whole process again

[flint jackal]

Multiply UΣV, and see what you get

[waxen flume]

https://i.imgur.com/B0aBiHI.png

even calculator doesnt have any negatives

[flint jackal]

I'm telling you, do it by hand

[waxen flume]

bruh

i just did xd

[flint jackal]

Multiply UΣV

Do you get the same A matrix you started?

[waxen flume]

1 sec im doing it

https://i.imgur.com/DV3yit0.png

no

https://i.imgur.com/mtO323T.png

but i dont get it

where is that -1 from??

[flint jackal]

Because it needs to be a -1 to produce a -2 in the A matrix

[waxen flume]

okay

but i did the whole eigen system process

and i didnt get any -1 in my eigen vectors

[flint jackal]

Technically, you used the negative. You choose the 0 to be negative

If you choose the 1 to be negative, that's how you find it or you trial and error to find out which term should be negative

[waxen flume]

there has to be a reason why 1/sigma * A * V_i didnt work in my case

there's gotta be easier way to compute this lol

[flint jackal]

Because the σ was a 0

That's why it didn't work

Or if you know, the eigenvectors should be orthogonal so you can guess the missing vector

And you're normally not going to do SVD like that, unless you're doing image compression. Other decompositions are faster. Those are just problems to help you understand the concept of SVD

[waxen flume]

whats the other decomposition

A = QR

something like that

[flint jackal]

LU decomposition, V * Λ * V^-1 (which is eigen decomposition)

[waxen flume]

thanks @flint jackal

[flint jackal]

LDU is another one

[waxen flume]

im gonna watch more examples of the svd decomposition

[flint jackal]

It's simple if you do the process I told you, it's tedious but you can get all the vectors easily

[waxen flume]

the thing is with your process for the U part

i got the eigenvalues/vectors

but that negative im clueless on still

[flint jackal]

Or to find the U matrix, you can do (A * Vi) /(Norm(A * Vi))

but that negative im clueless on still
@waxen flume Like I said, if you multiplied out UΣV, and did not get A, that means one of the vectors has the wrong sign

[waxen flume]

ohhhh

i see

guess and retry?

is what ur saying

for the sign

[flint jackal]

Yeah

Normally, a SVD problem shouldn't have that many 1s but because it was a simple question that produced matrices of 1s, just trial and error

[waxen flume]

thanks again

i have been doing linear algebra hw since 3pm, it is now 2am

[flint jackal]

Linear algebra is easy

[waxen flume]

Normally yes, but the computations can be tedious.

its definitely easier than calculus 2 and 3

[flint jackal]

Normally yes, but the computations can be tedious.
@waxen flume Which is why you should use a calculator for matrices bigger than a 3 x 3

[frosty vapor]

svd decomposition

rip in peace

smh my head :3

[novel hamlet]

How do you make jordan decomposition

[waxen flume]

why

[waxen flume]

https://i.imgur.com/2UqzCA8.png

my eigenvalues are -3, -4, -5, 0
does that make it negative definite?

[novel hamlet]

No, its negative semi definite

[waxen flume]

thanks.

[novel hamlet]

how do i calculate ker(A)

for 2x2 matrix?

is ker = nullspace?

[zealous junco]

ye

[novel hamlet]

kinda forgot how to calculate nullspace

been 9 months since i needed that last time

[zealous junco]

you get rref form right

and like when you see any number to the right of pivot 1

then those numbers could be set to anything

not really helpful i guess unless it reminds you of the method

[novel hamlet]

yeah found out how to do it, im trying to do some jordan decompositions

and im tilted af on these

wtf happens in these jordan youtube tutorials

[zealous junco]

i wish i knew jordan decomp

🥲

[novel hamlet]

oh ffs, this is for when we know A

[zealous junco]

but this just looks like

eigendecomp

which is a special case

just solve for eigenvalues and find eigenvectors, then you find p

note P contains the 2 eigenvectors as columns

[novel hamlet]

back to youtube i go

[nocturne jewel]

Computing kernel is the same as solving for the homogeneous system Ax=0

[novel hamlet]

that youtube was not really helpfull, since i need to do B=PAP^-1

and in that case I had A

[nocturne jewel]

$\begin{bmatrix} 3&4&0\3&4&0\end{bmatrix}\to\begin{bmatrix} 3&4&0\0&0&0\end{bmatrix}$

[stoic python]

moshill1

[nocturne jewel]

so the eigenvectors are vectors [x,y] which satisfy 3x+4y=0

[novel hamlet]

yeah i have done eigenvalues for my matrix M now i just needd to figure out how to do jordan for it

wait a second, couldnt i do it this way take M -> eigen values diagonally = middle matrix in jordan -> do nullspace and get P and then inverse P to get P^-1?

oh yes its all coming together now

[flint jackal]

If you put the eigenvalues in a diagonally matrix, it'll be an n x n matrix, right? Won't it have dim(ker(A)) = 0?

[novel hamlet]

wait a second, how i determine for 3x3 matrix if one of the 2 eigenvalues is double eigen?

[surreal otter]

doesnt matter

you put them in too from what I know

similar to repeated roots of a polynomial

[novel hamlet]

i got this situation i know my Jordan should have -1,-1,-0,5 or -1,-0,5,-0,5

but im unsure on how to determine the correct choise (spoiler it is the first one)

[surreal otter]

by hand, that is?

[novel hamlet]

yeah

[surreal otter]

could try Newton-Raphson

[novel hamlet]

or any reasoning i could use to justify selecting one

[flint jackal]

@novel hamlet Can you post your original problem? I want to see what the goal/problem is

[novel hamlet]

jordan decomposition for that

jordan normal form J and then decomposition PJP^-1

[flint jackal]

And that's what it says to use? Jordan decomposition?

[novel hamlet]

no method spesified as long as i can get it to jordan normal form and get that decomposition out of it

[surreal otter]

pardon my ignorance, but isnt that just diagonalizing?

[novel hamlet]

I prefer to call it diagonalization with extra steps

[surreal otter]

so what are your steps so far?

[novel hamlet]

calculate eigenvalues for A, then proceed to calculate Ker(A-eigen x I)

then combine null spaces to matrix P

[surreal otter]

I meant as in your working

[novel hamlet]

i managed to do eigenvalues for A, but im only getting 2 instead of 3 i need

[surreal otter]

probably one that is repeated

[flint jackal]

There's a repeated eigenvalue

[novel hamlet]

yes, but i dont know how to determine wich one is repeated

[surreal otter]

solving the cubic is a good start

[flint jackal]

,w solve -x^3 - 5/2 x^2 - 2 x - .5 == 0 for x

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dldh06#2659

[novel hamlet]

ah that was fast

you can form it to this

[flint jackal]

Well, it's the -1 that's repeated

[surreal otter]

yup, was looking for that, now you know which one is repeated

[novel hamlet]

yeah this should help, next i just need to do Ker(A-(-1)I)

[surreal otter]

Did it on a graphics calc, just to confirm

[novel hamlet]

how did you get that-1.5?

[surreal otter]

If you solve the second one you showed . . .

I think

Idk

[novel hamlet]

oh i got typo

[surreal otter]

In yours or mine?

Yeah just checked, it's -1.5 or -3/2

So -1, -1 and -3/2

[flint jackal]

@novel hamlet Your Characteristic Polynomial is wrong

[novel hamlet]

yeah i accidentally put wrong value

noticed it already

hmmm, come to think of it the nullspace of that is trivial

wait this method does not work

cant go via ker(A-eigen x I) route here

i dont get why J gets that poisitive 1 on 2nd row

[frosty vapor]

oh nothing lol i was just poking fun ahah ur good. SVD has D for decomposition so i thought it was funny to see svd decomposition in the same way as rip in peace. ur good lol

[lavish jewel]

singular svd value decomposition

[novel hamlet]

edd, yuou got any idea where im getting it wrog?

[flint jackal]

Are you still doing that problem?

[novel hamlet]

yeah, cant figure out, since null space is trivial

and not 100% sure where that positive 1 comes from

since if null space is trivial this whole thing does not work out

[novel hamlet]

do you know how to get P when we know A and J for A=PJP^-1

[flint jackal]

Isn't P, just the eigenvector matrix?

[novel hamlet]

no, at least when you look the above photo, from wolfram the eigenvectros dont hit anywhere

[flint jackal]

I haven't done much with with Jordan decomposition so not sure

[novel hamlet]

usual algorithm for this would be, get eigenvalues for A, get ker(A-eigen x I)

and J comes from egenvalues of A and is diagonal P is some kombination of null spaces

but since my null space is trivial this thign does not work

[flint jackal]

What do you mean by this exactly?

[novel hamlet]

I'd get 0,0,0 as my vector and that does not appear as column anywhere on PJP^-1

for matrix P

[flint jackal]

Not 100% sure then, sorry

[novel hamlet]

please somepne ping me if you could help

[lavish jewel]

what's the q

[novel hamlet]

what Q?

[lavish jewel]

question

[novel hamlet]

jordan decomposition for

I have done eigenvalues and got the J

but when i try to do ker(A- eigen x I) i get that the result is trivial

when using eigen = -1

[lavish jewel]

i'm afraid i can't help

[novel hamlet]

rip

you were my only hope

[lavish jewel]

i don't know what a jordan decomp is 😛

[gray dust]

decomp using jordan normal form

[novel hamlet]

yeah basically get jordans normal form (done) and then find common eigenvectors for P and then invert P

using

where A is square matrix € C and lambda is eigenvalue of it € C

[faint lintel]

I am confused as to how the matrix transformations on the left of each 3 x 6 matrix work in relation to the right hand side

Like the first change

how does 1,2 and 3,2 change by the right hand side remains as the identity matrix

like the first operation is adding the left most column to second column

But then the second transformation of adding row 1 to row 2 changes the right hand side matrix

I don't get it

Yea I'm stuck for 2c

<@&286206848099549185>

[flint jackal]

Not entirely sure why those operations are legal but can't combine columns like that because the first column is x while the second is y, you just can't have (x+y) in the column

[golden kelp]

I haven’t tried the question but if you get λ = -1 then you need to consider ker(A - - I) = ker(A+I), I think you’ve done ker(A-I)

[faint lintel]

yea me either but that's what the book says >_>

[edgy kelp]

https://i.imgur.com/kXtIJis.png

Does anyone know how I can rref this?

[flint jackal]

Row reduce like normal, except you'll have an h term

[golden kelp]

r2= r2-r1 seems like a good start

[edgy kelp]

[1;2;0;0;h;1;0;0;1]

The idea is to get the h in a position where it would have an effect on the pivot in last column

[golden kelp]

That’s not rref

[edgy kelp]

No I know

[golden kelp]

One second

[edgy kelp]

https://i.imgur.com/QrSj49t.png

I'm trying to answer that ^

[flint jackal]

Separate with commas and semicolons

[novel hamlet]

Gona try dat next just finishibg my food

[edgy kelp]

I can't get rid of the h

I need the h

to figure out the values

[golden kelp]

Sorry small mistake for h=0 case

Row 2 should be 0,0,1

[edgy kelp]

how did you get 0h0

in last row

[golden kelp]

Step 2 is r3 = r3 - r2

[edgy kelp]

oh i see

[flint jackal]

You should try again, there's errors in the first step

[golden kelp]

No there isn’t

[flint jackal]

There is

[golden kelp]

Where

Apart from the h=0 case I mentioned

[flint jackal]

Element 2,3 should be -1

[edgy kelp]

Is this a homogenous system willrum

non?

[golden kelp]

Why?

[edgy kelp]

he did r2=r2-r1

its 1

[golden kelp]

I am correct my friend

[edgy kelp]

He is correct

This is homogenous right willrum

[flint jackal]

I did r2 = r1 + -r2

My bad

[golden kelp]

If h is not 0

[weak blaze]

Hello

Prove that the determinant and trace of the linear transform A are independent of the base selection in the range set of the transformation.

[golden kelp]

If h=0 there is nontrivial kernel

[weak blaze]

İs there anyone solve this?

[edgy kelp]

Ah okay @golden kelp

[golden kelp]

You can do this a couple of ways. You can show that the characteristic polynomial is independent of basis and then the fact that the constant term in the characteristic polynomial is the determinant and the coefficient of x^(n-1) is the trace

[weak blaze]

Hmmm

I don't know much about the characteristic polynomial.

Can you explain Liitle more? :)

[golden kelp]

https://math.stackexchange.com/questions/72303/proving-the-trace-of-a-transformation-is-independent-of-the-basis-chosen

This explains it well

[weak blaze]

Ohhh thank you so much 👍👍

[snow jetty]

A Matrix M has an eigenvector v if M*v = λv, where v is a vector and λ is a scalar. This means that Mv - λv = 0 and the vector belongs to Ker(M - λI).

[golden kelp]

History?

Sorry?*

[snow jetty]

That's why a characteristic polynomial has all eigenvalues as roots. As it is exactly det(M-λI) which is zero if you plug in an eigenvalue.

[golden kelp]

I dont think that is the question they were asking...

But yes you are correct

[snow jetty]

👀 ps

It's about proving that two similar matrices have the same characteristic polynomial and determinant?

[weak blaze]

Aha ı got it. Eigenvectors and eigenvalues ​​are in the last chapter in the book I'm working on. But you explained it well. I think in this way, we can see that the trace is calculated independently from the range set.

[golden kelp]

Have the same determinant and trace. I mean what you said was correct but the question wasn’t about the eigenvalues per se

[weak blaze]

I will learn eigenvalues and eigenvektor

[snow jetty]

That's because A and B are similar iff there is invertible P s.t. P^-1 * A * P = B

In fact, A and B do the same application. P "rewrites" the vector into another basis

And P^-1 gets back to the initial basis.

And det(A^-1)=1/det(A) for inversible A.

That can be proved by considering the determinant follows det(AB) = det(A)*det(B). You can prove it if you consider det as a multilinear function (just a long calculation with a lot of Σ)

That's why similar matrices have same determinant

[weak blaze]

But the problem is not about having the same determinant. Sorry maybe ı got you wrong

[snow jetty]

As to the characteristic polynomial: consider det(M - λI). Then consider det(P*(M-λI)*P^-1).

[weak blaze]

Okayyy okayyy now ı get you

Woow that's so Cool way if you are right

[snow jetty]

(let me think, I am improvising 🙃 )

[weak blaze]

Thank you so much;)

Let me think and learn hahah

[snow jetty]

Yes indeed you can do so.

[snow jetty]

This 👆 is the same expression as det(P^-1 *M * P - λΙ)

[weak blaze]

Yeah I understood how proof it is. So now I have to study eigenvalues ​​and eigenvectors. Good by and thanks a lot 🙏

[snow jetty]

👍

[edgy kelp]

https://i.imgur.com/jycF0VN.png

Does anyone know how to find this change of coordinates matrix

Is it just [4;5 , 5; 1]

[flint jackal]

Don't you need to find the inverse?

[wintry steppe]

the one you said goes from the new basis to the standard so ^

Suppose $U_1,\dots,U_m$ are finite-dimensional subspaces of $V$ such that $U_1 + \dots + U_m$ is a direct sum. Prove that $U_1 \oplus \dots \oplus U_m$ is finite-dimensional and [ \dim U_1 \oplus \dots \oplus U_m = \dim U_1 + \dots + \dim U_m. ]

[stoic python]

n/c

[wintry steppe]

First we will show that if $U_i$ are finite-dimensional subspaces of $V$ for $i = 1,2,\dots$, then so is their direct sum. We will use induction.

Clearly, $U_1$ is finite-dimensional, and so its direct sum with no other subspace is finite-dimensional. We assume that $\displaystyle\bigoplus_{i=1}^k U_i$ is finite-dimensional for some $k \ge 1$. This implies that it has some basis, say $e_1,\dots,e_n$. Then, if $U_{k+1}$ is a finite-dimensional subspace of $V$ with basis $f_1,\dots,f_k$ such that $U_{k+1} \cap \displaystyle\bigoplus_{i=1}^k U_i = {0}$, then clearly $U_{k+1} + \displaystyle\bigoplus_{i=1}^k U_i$ is direct and has basis $e_1,\dots,e_n,f_1,\dots,f_k$, which is finite-dimensional.

We will now show that [ \dim \bigoplus_{i=1}^m U_i = \sum_{i=1}^m \dim U_i. ] We example the left-hand side: Suppose the bases of $U_i$ are given as
\begin{align*}
U_1 &= (e_{11},\dots,e_{1x_1}) \
U_2 &= (e_{21},\dots,e_{1x_2}) \
& \vdots \
U_m &= (e_{m1},\dots,e_{1x_m}).
\end{align*} We are given that the sum of the $U_i$ is direct, and therefore [ \dim \bigoplus_{i=1}^m U_i = \dim (e_{11},\dots,e_{1x_1},\dots,e_{1x_m}) = x_1 + \dots x_m. ] We now examine the right-hand side. We have, by definition, [ \sum_{i=1}^m \dim U_i = x_1 + \dots x_m, ] and so we are done.

[stoic python]

n/c

[wintry steppe]

Does this proof work? I know it's kind of long sorry

[lapis osprey]

Can someone explain how to solve this question? I am not sure what to do.

[faint lintel]

You want to find a vector v such that Pv = v

well lets do some rearranging

Pv - v = 0

(P - I)v = 0

v is non-zero

so you want to find non-zero vectors v such that (P - I)v = 0

and then remember that it's a probability vector

so you have to normalize it

@lapis osprey

Ok now I have my own question

the last part of part c

answering this conceptual question of notions of diagonalization

Very lost

[faint lintel]

<@&286206848099549185> plz part c in that question above ^

Ok I think I have it

is it because symmetric matrices in R are orthogonally diagonalizable?

ie nxn matrices over R are symmetric iff they are orthogonally equivalent to a diagonal matrix?

[wintry steppe]

<@&286206848099549185> Can someone look at the proof I posted above?

Why is this a proof?

I don't see how that proves that there does exist a unique linear map that does exactly that

[gray dust]

@wintry steppe this only defines a suitable T. it's really up to us to show T satisfies the above. we must also show T is unique, ie if T & S are maps satisfying the above then T=S

[wintry steppe]

Yes, but how do we know the map exists at all?

Serious question.

[gray dust]

the map is provided in the 'proof'. it's now our job to show it actually satisfies the above properties

[wintry steppe]

I don't know if I can agree with that

We're not shown that it will always be the case that such a map will exist

[gray dust]

we ARE. it's explicitly defined right there

[wintry steppe]

No?

[gray dust]

https://i.gyazo.com/316ce0baeba5d36116109f88a353981d.png

[warped cape]

technically you do need to show that T is well defined, and that it is a linear map

but both are obvious, since v_i are a basis

[gray dust]

i already pointed out we must show T satisfies the above properties

they're caught up on explicitly defining T

[wintry steppe]

What properties?

We haven't shown a map exists

[warped cape]

T is a map

therefore it exists

[gray dust]

there are 2 parts to showing some object satisfying some property exists

[warped cape]

every x in V can be written as a linear combination of the basis vectors, so T(x) = T(c_1v_1 + … + c_nv_n) = c_1w_1 + … + c_nw_n, and that is the definition of the map

[gray dust]

explicitly define/provide the object, then show it satisfies the desired property

so to do the existence part of the proof, we must define T in such a way that we can then show T satisfies the desired properties

[wintry steppe]

But we don't know this can be done??

[gray dust]

the 'proof' just defines T. we're left to show T satisfies the properties

[gray dust]

by definition of basis, any x in V is a unique linear combo of the vj's

[wintry steppe]

I know that

[warped cape]

which part do you not know it can be done?

it literally just defines what the image of each vector in V is

and that's all that a map is

[gray dust]

@warped cape if you're being so tight about well defined, you failed to mention the linear combo is unique ensuring Tx is unique

[warped cape]

that is true

[wintry steppe]

Again I'm just saying we don't know that we can write it in terms of a linear combination in W

[gray dust]

let x in V. since the vj's are a basis of V, there exist unique scalars c1,...,cn where x=c1v1+...+cnvn. then c1w1+...+cnwn is a uniquely defined vector in W, and we choose to define Tx as that

[verbal pivot]

Keep in mind, you're not writing an element of V as a linear combination in W, you are defining a map, a correlation between elements of V and elements of W

[gray dust]

this uniquely defines Tx for each x in V, thus defining T as a map V->W

[hollow willow]

If I wanted to reduce the following into a basis

I would make a matrix with each row representing one element (matrix) as follows:

1 0 0 0 1 1
0 1 1 0 0 0 
0 1 0 1 0 0 
1 1 0 0 0 0
0 0 0 0 1 0
0 1 0 0 0 0 
0 0 0 1 0 0```

then I would reduce the matrix.

I am looking for reassurance that I have the right idea.

[gray dust]

@hollow willow redo row 4 but yes

[hollow willow]

ahh yes typo on my part but I'm glad I got the idea right. I wasn't sure if I would rewrite them as rows or columns

[gray dust]

doesn't matter, we can row reduce em as rows or col reduce em as cols

[hollow willow]

bless

thank you

[dawn fractal]

I need help with Question 4

[v]_B1 is the coordinate matrix of v w.r.t. B1

Let $v\in V.$ From the definition of coordinate matrix,
$$[v]{B_1} = [\alpha_1 \ ...\ \alpha_n ]^T$$
and
$$[v]{B_2} = [\beta_1 \ ...\ \beta_n ]^T$$
for all $\alpha_i ,\beta_i\in\mathds{F},\ i\in\mathds{Z}_{n+1}^*.$

[stoic python]

!superficialsicko

[dawn fractal]

idk where to proceed from here

[dusky epoch]

$\bZ_{n+1}^*$?

[stoic python]

Ann

[dawn fractal]

by that i mean the set {1,2,...,n}

[dusky epoch]

why not write {1,2,...,n}

or 1:n for short

[dawn fractal]

it was our convention to write {...,-1,1,...} as Z*

but anyway

[dusky epoch]

it's unnecessary and distracting

just write $i=1,\dots,n$

[stoic python]

Ann

[dusky epoch]

anyway you will also need the change of basis matrix

[dawn fractal]

the question was asked before that topic

so how could i proceed without using that

i mean, is there any relation between alpha_i and beta_i?

scratch that question
here's a new one

how do i prove that
$${B_2}P{B_1}[v]{B_2} = [v]{B_1}?$$

[stoic python]

!superficialsicko

[dusky epoch]

can you write out your definition of the CoB matrix

you'll need to use that

[dawn fractal]

i already have an intuition for what change of basis does to vectors in R^2 when i want to write them as a linear combination of the new basis vectors, but

i can't grasp the concept in general

[dusky epoch]

it's gonna take some index fuckery and keeping careful track of summations

[dawn fractal]

so to prove the theorem i need the fact that every vector v in V is a (unique) l.c. of vectors in B_1 or of vectors in B_2?

and not by actually explicitly proving the theorem

[wintry steppe]

Finding the eigenvalues of a matrix is easy, but finding those for transformations for some reason is so much more complicated for me

anyone know how to tackle this?

[dusky epoch]

for what values of lambda does f' = λf have a nonzero solution?

[wintry steppe]

Yea I got to that step

But I have no idea how to continue from there

[dusky epoch]

this is a differential equation

[wintry steppe]

Ooh

[wintry steppe]

How do I go about solving this?

[halcyon pollen]

guys i got this doubt

are gaussian elemination and row echelon form are the same?

oh they're the same lol

[dusky epoch]

no, they are related but they are not the same

[halcyon pollen]

like ?

In mathematics, Gaussian elimination, also known as row reduction, is an algorithm for solving systems of linear equations. well i googled it

confused 😅

[dusky epoch]

"gaussian elimination" refers to the process

"row echelon form" refers to a form matrices can be in

[halcyon pollen]

oh so what's gaussian elimination is generally?

[dusky epoch]

?

do you want me to describe the gaussian elimination algorithm in full or something

[halcyon pollen]

well kind of confused but it's a process right and generally what does that does?

just say it simple i'm not asking it in brief

[dusky epoch]

it's a sequence of row operations applied to the (augmented) matrix of a linear system in order to solve the system

[wintry steppe]

Process that lets you solve a system of linear equations

[dusky epoch]

if you ask me it's just a certain kind of algebraic manipulation except it's written in a more streamlined format

[halcyon pollen]

yeah now got it thanks 😄

[wintry steppe]

gaussian eleemeenation

[novel hamlet]

let A be square NxN matrix (A€C), find smallest k€N that Rank A^m = rank A^k, for all m > k, how should i approach this?

[strange delta]

for part b) how comes the span of p_1, p_2, p_3 is equal to this

i get the (x-2)^2, (x-2)^3, but idk why (x-2) and 1 is in the span

[dusky epoch]

it doesn't say "Show {p1, p2, p3} is a spanning set for V", does it?

i mean, ok, hold on

$p_1(x) = (x-2)^2, p_2(x) = (x-2)^3, p_3(x) = x(x-2)^2$

[stoic python]

Ann

[dusky epoch]

(x-2) and 1 are not in V

[strange delta]

he didn't say the span was equal to it but yeah

[dusky epoch]

{1, (x-2), (x-2)^2, (x-2)^3} is a basis for R_3[x].

not for V.

[strange delta]

ah ok

[dusky epoch]

@wintry steppe try some rotation and reflection matrices maybe?

[wintry steppe]

what is the matrix of moments associated to legendre polynomios?

[halcyon pollen]

guys what's crout's method?

[flint jackal]

Google is your friend, I suggest that you look that up and read about it. But from my research, it looks like it's a algorithm for LU decomposition

[halcyon pollen]

yeah LU decomposition method learning it

[flint jackal]

It looks like Crout's method returns a lower triangular matrix and a unit upper triangular matrix

[halcyon pollen]

yeah well i don't get this

https://www.youtube.com/watch?v=rhNKncraJMk

i'm referring to the MIT courseware video tbh

i think i'll watch another video

confused enough lol

actually how they're calculating ?

[flint jackal]

Calculating what exactly?

[halcyon pollen]

well never mind lol i were confused in the row operations

understood the LU decomposition 😄

[novel hamlet]

Can someone tell how rank(a) works with A^m

Since rank a =< min (z,y) where a €R z*x

[limber sierra]

you introduced a lot of variables without clarifying what they are

are a and A meant to be the same thing? what are x, y, z?

[novel hamlet]

A is same just phone autocorrect

Z and y are the dimensions of matrix a

X was just typo

[modern palm]

I know I have to prove the 10 axioms. But I’m not sure what an element of V would look like?

For example, how does u+v look like

[wintry steppe]

an element of V is a positive real number

[dusky epoch]

with vector addition being ordinary multiplication

[modern palm]

so if u = 2 v = 3, then u+v = 2*3=6 and uv=3^2 = 9

[dusky epoch]

uv doesnt make sense

you cant multiply two vectors in a vector space

[modern palm]

oh right.

[dusky epoch]

you will need to distinguish 3 as an element of your vectorspace and 3 as a scalar

granted, this construction is a bit unorthodox but its purpose is to illustrate that vector spaces do not always look familiar

[modern palm]

okok thank you

so elements of V are single positive real numbers so it can't be like (2,3), but are still considered vectors.

[wintry steppe]

you could write them like (6) if you want

[dusky epoch]

"vector" just means "element of a vector space"

so yeah

[modern palm]

ok that helps alot

thanks again

[strange delta]

how do we know that the only solution for the spanning set is c_1,c_2,c_3 = 0

oh actually nvm thats a dumb question

is this question just the definition of a basis, just do reduce row echelon and pick the pivots with 1 and the columns that correspond to the basis, don't really know how to approach this

[wintry steppe]

insofar as it should be done using the definition of a basis, yes

try for small matrices first

you do not need to do any row reduction for this problem.

Mat_{nxn}(R) is that same thing as R^(n²)

[strange delta]

same as a polynominal field?

[gray dust]

yes they both denote the n by n matrices over R

[wintry steppe]

hmm usually R^(n²) would denote lists of n² elements over R. which are indeed the same thing as vector spaces

[gray dust]

oh i read that as R^(nxn) while you meant tuples. but yes they're isomorphic

[strange delta]

welp i tried to do the question

but pretty sure i proved absolutely nothing

big f

damn 2nd line doesn't even make sense

meant to say it's a basis of r^n

[wintry steppe]

V consists of matrices, not polynomials...

so your basis should consist of matrices

[strange delta]

how should i denote my matrices?

[nocturne jewel]

Not as monomials

[strange delta]

like when i'm writing them out as a span

[nocturne jewel]

$E_{i,j}$ is standard notation for the elementary matrix with 1 in the (i,j)-entry 0 else

[stoic python]

moshill1

[wintry steppe]

give them the answer

[nocturne jewel]

[strange delta]

ah ok lol i forgot about that

is my apporach ok?

[wintry steppe]

no, because Mat_{nxn}(R) doesn't contain polynomials

its elements are matrices

so your basis should consist of matrices

[strange delta]

alright

there we go

[wintry steppe]

this is only a basis for the space of diagonal matrices, but you're on the right track

not sure what "where S is an identity matrix" means since it's a set of matrices. you've checked linear independence - good - but why does this span?

a basis of V is, by definition, linearly independent and spans V

[strange delta]

i was trying to say when they added up it makes a diagonal matrix

but i don't think i needed to say that

it's kinda self explanatory

[novel hamlet]

how do i get eigenspace of matrix A if i have done row reduce and i got eigenvalues?

[wintry steppe]

if c is an eigenvalue find ker(A - cI)

[novel hamlet]

yeah i know of that method, but that is probably worst idea since i have to use R to do this

I know i can deduce the matrix eigenvectors from this

but i have forgotten how

[strange delta]

i've got an idea, list every single matrix not just the diagonal and show that it's linear independent

[novel hamlet]

wrong matrices

[wintry steppe]

every single matrix? be more precise, i'm sure that's not what you mean

[strange delta]

like E_1x1, E_1x2,..E_1xn

and all the way so when they add up

they make a filled up matrix

and i can show linear independence and so it's a matrix

works?

[wintry steppe]

you tell me

[strange delta]

yes

[strange delta]

thought you was the teacher here

[wintry steppe]

[gray dust]

tera's making sure you know definitions and using them correctly, not holding your hand

[vague rampart]

I don't understand this question. It is mostly the basis that I am confused about (I have read quite a few places and watched videos). I am not completely sure about egenvectors either. (egenvektorer=egenvectors)

I think I am supposed to make vectors out of the columns

[limber sierra]

can you find the eigenvectors of A?

[vague rampart]

no i am not sure

some gibberish i made

i was trying to isolate this

[flint jackal]

I don't understand that graphic

But do eigenvalue decomposition

Det(A - λI)

[vague rampart]

ah i did that part

[flint jackal]

Find λ

[vague rampart]

2 and 0

[flint jackal]

You should have 3 roots

[vague rampart]

2,2,0

[flint jackal]

So say for instance $λ_1 = 2$. Do $A - λ_1 * I$

[stoic python]

dldh06

[flint jackal]

Find the null space

To find the eigenvector

[vague rampart]

null space?

[flint jackal]

Kernel

[vague rampart]

ahhhh

[flint jackal]

They both mean the same

[vague rampart]

ok i understand kernel somewhat. i just learned about it though.

[flint jackal]

Work out what you think you do, and post your work and I'll see if it's right or not and help you that way

[vague rampart]

so here i have a reduced echelon and the koordinates. then i would get the kernel through that, right?

[flint jackal]

Which value of λ did you do first?

[vague rampart]

oh hmmm. think i am confused.

[flint jackal]

No need for determinate

[vague rampart]

yea but what i posted before should be the lambda=0 then

[flint jackal]

Is this your answer that you found or the book/solution key?

[vague rampart]

no i wrote all of it

i am just majorly unsure about what i am doing

[flint jackal]

So this matrix is correct but the solution is wrong. Set that equal to 0 and then solve for the values

So for the first row, it would be $1x_0 + 0x_1+ 0x_2 = 0$, correct?

[stoic python]

dldh06

[flint jackal]

You understand how I got that for the first row?

[vague rampart]

hmmm i was trying this

i understand how you got the row

[flint jackal]

Ignore the λ

[vague rampart]

ok

[flint jackal]

That is what I mean

[vague rampart]

Just $A\vec x=\vec 0$?

[stoic python]

Sim

[vague rampart]

and we get the kernel from that, because that is the standard way to get the kernel?

[flint jackal]

Specifically, $(A-λ_1)\vec x=\vec 0$

[stoic python]

dldh06

[flint jackal]

Yes

[vague rampart]

but i thought i should ignore the lambda... shouldn't lambda occur somewhere in what you wrote?

[flint jackal]

Those λ are the roots, when you did $det(A - λ*I)$

[stoic python]

dldh06

[vague rampart]

ah yes it is zero now. i see

[flint jackal]

So for instance, the first one you did. You did $λ_1 = 0$

[stoic python]

dldh06

[flint jackal]

Then use this

[vague rampart]

yea, so i get to keep it for the other one

but i am slightly confused that we are gonna have two kernels then

[flint jackal]

2 kernels but the one with repeat roots (λ = 2 because it's a repeated root), you can create two different vectors from the same kernel

For when $λ_1 = 0$, get this, understand?

[stoic python]

dldh06

[flint jackal]

I did $x_2 = 1$

[stoic python]

dldh06

[vague rampart]

this is helpful

[flint jackal]

Do the same for $λ_2 = 2$

[stoic python]

dldh06

[vague rampart]

ok, so now i have two kernels, right? i am slightly unsure about that we a dealing with two different x vectors

[wintry steppe]

two kernels??

[flint jackal]

Do it one step at a time, show your work for $(A - 2 * I)x = 0$ first, because I'm confused on how you got to the second line

[vague rampart]

idk T_T

[stoic python]

dldh06

[flint jackal]

Not exactly, you can still reduce it a bit more

Also, when you find the kernel, it should be in Reduced Row-Echelon Form

[vague rampart]

ah so it is zero on two rows. so i it supposed to be reduced echelon?

[flint jackal]

Yes, exactly

When you have this, what should the solution be? The set of equations like we did for the first value of λ

[vague rampart]

❤️

[flint jackal]

You're missing one more set

[vague rampart]

i only had two lambdas

[flint jackal]

And that first line is incorrect

[vague rampart]

oh

x_0=0?

[flint jackal]

Yeah

When there are rows of zeros, those variables will be free variables

[vague rampart]

so we just can't know for this one? so we put the two solutions together to know?

[flint jackal]

That should be the set of equations from that matrix

[vague rampart]

ok

[flint jackal]

Now because we have a lot of free variables, we can't say $x_0 = 0$ $x_1 = 0$ $x_2 = 0$

[stoic python]

dldh06

[vague rampart]

ofc. gotcha

[flint jackal]

Make one of the variables equal to 0, and the other 1, and that's one eigenvector for when λ = 2

Then because λ = 2 is repeated, switch the 0 and 1 to get the other eigenvector

[vague rampart]

yes so now i have two eigenvectors

[flint jackal]

So $x_0 = 0$ $x_1 = 1$ $x_2 = 1$, is one vector and the other vector is $x_0 = 1$ $x_1 = 0$ $x_2 = 0$

[stoic python]

dldh06

[flint jackal]

Right?

And those 3 vectors you found are the eigenvectors

[vague rampart]

wasn't the other one like this?

[flint jackal]

It's not $x_1 = -x_2$, it should be $x_1 = x_2$

[stoic python]

dldh06

[vague rampart]

hmmm i feel a bit stupid atm

[flint jackal]

That's the matrix, right?

[vague rampart]

ah no this was the previous one

oki

yea then i agree with (1,0,0) or (0,1,1) if it was only for the last one

[flint jackal]

So you should have all of your eigenvectors with the eigenvalues

[vague rampart]

ok 3 vectors.

[flint jackal]

So the problem was the basis consisting of the eigenvectors, correct?

[vague rampart]

yes

[flint jackal]

So you found the eigenvectors, which is the basis

[vague rampart]

so i make a span of the 3 i found?

[flint jackal]

Yep, exactly

[vague rampart]

and then nothing more?

[flint jackal]

Nothing more, because it asked for the eigenvectors, which you found

[vague rampart]

thanks a lot! 🙂

❤️ that was super helpful.

[flint jackal]

No problem

[safe locust]

not sure how to do this, anyone willing to help?

<@&286206848099549185>

[nocturne jewel]

do you know Least Squares solutions?

[safe locust]

no?

not sure what that is

i think i need to project it but im not sure what to do

[flint jackal]

You need least squares method

[safe locust]

ight

so what do i do?

[surreal otter]

Hiya, I was wondering if anyone could point me in the right direction with this; I am trying to generalize my notion of transforms and functionals to linear maps/operators and was wondering if someone could help me make the connection between these. So I know what a linear operator is now, and that it has a matrix representation (which you get by letting it act on the basis vectors of the initial vector and then find with respect to the second vector´s basis- correct me if Im wrong). I´ve been introduced to coordinate transforms, and within that intro we were given a transformation matrix which is unitary. How might I visualize this in terms of a linear operator? Im kind of new to this generalized approach so i cant yet fully visualize it. Below is an example of a matrix ive been given which i want to connect with the operator concept:

[flint jackal]

If you don't know about least squares, I suggest you watch this https://www.youtube.com/watch?v=WABC6wmuLOk&list=PLkZjai-2Jcxlg-Z1roB0pUwFU-P58tvOx&index=28&ab_channel=JeffreyChasnov. Also, you were kinda right about projection but not exactly

[tawdry bramble]

If x^TAx=x^TUDU^Tx where A is symmetric and D is diagonal, does x^T have the same dimensions as U?

If not would x^TU=(U^Tx)^T?

[wintry steppe]

How do I show that S(T_1 + T_2) = ST_1 + ST_2

Where S, T_1, T_2 are linear maps?

I can show that (S_1 + S_2)T = S_1T + S_2T

But we only defined addition of maps (S + T)v = Sv + Tv

And so I am not sure how to do this way

[nocturne jewel]

(S+T)(v) is a linear combination of vectors, assuming both S and T act from V to W

so if S is linear, then you can apply linearity again

[flint jackal]

What is this in relation to?

[crisp cloud]

Can someone check my work? I don't know if I did this correctly

[crisp cloud]

<@&286206848099549185>

I think I need to change my signs for V4 and problem 2

[onyx zinc]

If I had a short proof from my linear algebra class is this the right spot to ask if someone can look at it and verify?

[wintry steppe]

yes

[onyx zinc]

Cool! here's a pdf; there are 4 proofs, each about a page long. Our class is using Otto Bretscher linear algebra with applications 5th edition.

thank you so much!

If you only have time to look at the first one that would be more tan enough help, I don't want to hog anyone's time

[tawdry bramble]

I need help showing that if A is positive semi-definite, then there exists a matrix B such that A=BB.

[tawdry bramble]

<@&286206848099549185> im going to cry

[lavish jewel]

you probably meant BB^T

or B^T B

[tawdry bramble]

That is true if B is symmetric, which I think it is since A is symmetric since A is positive semi-definite. Unless, I'm wrong about that

[dusky epoch]

arent all symmetric matrices diagonalizable or something

[lavish jewel]

A is symmetric, B doesnt have to be

and yes, diagonalize it as suggested

[tawdry bramble]

Yeah, I have because A is positive semi-definite, then x^TAx\geq 0 for all xin R^n

A=UDU^T where UU^T=I

If A=B^TB then D=(BU)^T(BU), but I don't know how to get to that without assuming what I'm trying to show

[lavish jewel]

backwards

D is diagonal, so you can easily say D = sqrt(D) * sqrt(D)

then just associate U sqrt D

[tawdry bramble]

You can take square roots of matrices?

[lavish jewel]

yes

and particularly for diagonal matrices, you just take the square root of each element

you can prove that yourself, it isn't difficult

[zealous junco]

edd my man

multivar king can u help in multivar? lol

w/ my question

[tawdry bramble]

Thanks

[zealous junco]

wait i figured out

im so dumb

[lavish jewel]

[tawdry bramble]

What if D=C^2 and B=UCU^T?

[lavish jewel]

you're making a mess

i'm in a meeting right now though, hopefully someone else can give you a hand

you only need B = UC tho

[sleek spruce]

wait

can someone tell me why the rhr is used

i asked a long time ago but forgot about it

when looking for orthogonal vectors

how do we determine which vector to use

since there are two orthogonal vectors

the positive and negative

i understand how to get the multiplication table for i * i and j * j and etc

but i don't understand why some are negative

[zealous junco]

Its like a physis convention, you use right hand rule as convention to choose what it means by positive value

[sleek spruce]

and some are positive

i have it in my book but

it doesn't explain why

[zealous junco]

to for example encode which direction something rotates in

[sleek spruce]

but if you have two vectors cross producted

aren't there two different vectors

that could arise

as there's two different orthogonal vectors

[zealous junco]

well depend on the order of crodd product, like AxB = -BXA

but thats consistent with RHR

[sleek spruce]

is this like linear algebra

like orientation?

wait

this is linear algebra

bruh

this is vector calc

so is this linear algebra

[zealous junco]

um i guess it is yea but just 3D linear algebra

[sleek spruce]

that makes so much more sense then

thank you

[minor bluff]

Could someone explain what the R^n and R^m notation actually means? I'm so confused about it because I'm not sure what it represents, I thought it was like the dimensions of a matrix or vector but I'm not sure 🤔

[sleek spruce]

dimensions

it's the dimensions of the vector

iirc

so like R^3 mapped to R^2 means mapping 3d to 2d

I may be incorrect but that's what I recall

[minor bluff]

So would that be a 3x3 to a 2x2 matrix

[sleek spruce]

I don't really remember if they have to be square matrices

I just remember that they're two separate dimensions

If i proceed anymore, i may confuse you so I'll stop here for the best

[minor bluff]

Yeah I'm confused about this

[spiral star]

R^n and R^m are vector spaces

a linear map does map between vector spaces and is linear

R^n is the set of n-tuples of real numbers and R^m is the set of m-tuples

the associated matrix must have m rows and n columns

so if x is in R^n then Ax is in R^m

[minor bluff]

Hmm I think I understand a bit better now - so each one is a vector space with m x n rows right

[spiral star]

no

vector spaces dont have rows or columns

a matrix does have rows and columns

[minor bluff]

Is it one matrix transformed between 2 vector spaces?

[spiral star]

the matrix is just convenient notation for the linear map T

in this context

the matrix is a way to represent a function

if you want, you can say that T transforms R^n to a subspace of R^m

that just means you apply T to all elements of R^n

[minor bluff]

Ahhh so can I think of it as a transformation that maps x to Ax ?

[spiral star]

yea, T maps x -> Ax

[minor bluff]

The first q asks how many pivot columns are in A, but I don't understand how this would be related to vector spaces if vector spaces don't have rows or columns

[spiral star]

its just related to A

the question is equivalent to asking whether T is an injective map

but this is the case exactly when the image has the same dimension as the domain

the columns of A are the images of basis vectors

so this is also equivalent to asking whether A has n linearly independent columns

and that is the number of pivot columns

[minor bluff]

So just to make it clear, n and m of a matrix have nothing to do with R^n and R^m right

[spiral star]

they have something to do with them

the dimension of the vector spaces determines the number of rows and columns of A

[minor bluff]

How?

[spiral star]

if you have T: V -> W with dim(V) = n and dim(W) = m, then any associated matrix A will have m rows and n columns

[minor bluff]

so in this case, x has n columns and Ax has m rows?

[spiral star]

what is x

[minor bluff]

x in R^n

[spiral star]

if x is in R^n then Ax is in R^m

i think you should go back to matrix multiplication

[minor bluff]

I think I sorta understand it, thank you for your time :))

I think I got confused because for some reason I'm thinking of vectors as having rows and columns which is wrong

[wintry steppe]

Question:
When finding a standard basis of a subspace in P2. 2 conditions must be satisfied.

1. The vectors have to be linearly independent
2. They have to span the entire subset.

However, I've read somewhere that the basis vectors have to span the entire space of P2 (not just the subspace).
Is this true, and is this specifically for subsets of polynomials?

@wintry steppe if you want a basis for P2 it has to span P2. If you want for a subspace it has to span the subspace.

Dosen't matter if it's a subspace of P2? In other words, the vectors dosen't necessarly have to consists of x^2, x and a constant.

[marble lance]

P1 is a subspace of P2, and a basis for P1 would be {1, x} which spans P1 not P2.

[wintry steppe]

Makes sense, thanks

[slate siren]

$x=\sqrt{1}$

[stoic python]

kanga

[wintry steppe]

I got a question, again.

W = p(x) = a + bx^2
How to form an orthogonal basis from subspace W, in P2?

[dusky epoch]

do you mean W = {a + bx^2 | a, b in R}?

[wintry steppe]

Yes

[dusky epoch]

and orthogonal in what inner product?

[dusky epoch]

okay

{1, x^2} is clearly a basis for W, so orthogonalize it

using that ^ as your inner product

[wintry steppe]

How could you see that 1, x^2 is a basis, that easly?

easily*

[dusky epoch]

you literally defined W as the span of those

[wintry steppe]

Can't seem to understand the basics

:p

nvm, I'm slow

Okay @dusky epoch How do you go from that orthogonal basis to an orthonormal basis? Do you have to normalize it?

[dusky epoch]

{1, x^2} isn't orthogonal

but yes, after you orthogonalize, if you want not just an orthogonal but an orthonormal basis then you will only need to normalize each vector.

but you set out to just get an orthogonal basis, not an orthonormal one. so technically you don't need that.

[wintry steppe]

ok, thanks

[wintry steppe]

I know, it was just a follow up question, just to map everything in my head.

[dawn fractal]

does this mean _B1P_B2 is equal to its own inverse here?
why does _B3P_B4 become equal to (_B1P_B2)^-1 and not simply _B1P_B2 ?

where _B3P_B4 is the transition matrix from B4 to B3

and m(f)_B2,B4 is the matrix representation of f relative to B2 and B4

[gray dust]

@dawn fractal looks like a typo. the leftmost matrix should be _B1P_B2. now the inverse of a transition matrix is another transition matrix causing the 'reverse' change of basis, so _B1P_B2=(_B2P_B1)^-1 which is what i think they meant to say

[dawn fractal]

any particular name for this theorem?

[dawn fractal]

same thought

[sonic aurora]

guys

can someone please provide a solution to this

actualy no i solved this haha

for question 1

[gray dust]

i'd call it a formula for the matrix representation of a linear map under a change of basis of the domain & codomain

[dusky epoch]

we don't give out answers here.

[sonic aurora]

ok thank you

[dawn fractal]

ok thx

[gray dust]

write W as the span of a set of vectors then use any procedure to pick out a basis of W and hence find dim(W)

[dawn fractal]

@sonic aurora you can use the membership rule of W to form the basis

e.g. a basis of P_3(R) is {1,x,x^2,x^3} =: B

and every vector v in P_3(R) is a (unique) linear combination of the elements of B

v = a(1) + b(x) + c(x^2) + d(x^3)

so to form the basis you just need to express each vector w in W as a linear combination of the terms in w in W

[gray dust]

they dm'd me asking for a full solution. i don't think they intend to read any of this

[dawn fractal]

and of course prove that that potential basis is a basis by proving it is linearly independent

@sonic aurora

it's why i have my privacy settings turned off xD

[gray dust]

they're banned

[dawn fractal]

sad

[gray dust]

@dawn fractal i won't say specifically, but the severity can be compounded by poor attitude etc

[wintry steppe]

Question:
What is the length of a polynomial vector?
Is it sqrt of inner product(vector, vector)?
😄

[gray dust]

since they went 'under the table' to me after just being told not to do so, i placed a longer ban

[limber sierra]

vector spaces a priori dont have a notion of "length"

you can give them one, but you have to specify how

the method youre referring to is getting an induced norm from an inner product

but of course, this requires you to have an inner product on your vector space

[strange delta]

to show an isomorphism do i need to show injectivity, surjectivity and also linear mapping?

[limber sierra]

which you dont have "automatically" either

[limber sierra]

yes.

[strange delta]

hmm

[limber sierra]

but you already did all that in (a), (b), (c)

(d) is just "putting it all together"

[strange delta]

oh

answer is no

it fails surjectivity

[limber sierra]

right.

[strange delta]

would the right answer be it's not because b,c is not mapped in the image

[limber sierra]

right, it's not surjective because its image is not all of $\mathrm{Mat}_{2\times 2}(\bR)$; for example, $\begin{pmatrix}1&1\1&1\end{pmatrix}$ is not in the image

[stoic python]

Namington

[strange delta]

yeah

[limber sierra]

because the "b" and "c" entries are always 0.

[strange delta]

i don't get the need for the other lines, isn't the 1st line already sufficient to state the kernel?

[dawn fractal]

u wouldn't be able to set a=d=0 without the 2nd line

[hoary osprey]

thats just the defining the kernel for the map

[strange delta]

hmm ok

in my solutions i just jumped right to the 3rd line

[nocturne jewel]

1st line is definition
2nd line just evaluates the linear mapping
3rd line is the kernel for this specific mapping

[strange delta]

yeah makes sense

if we have dim v> dim w

then there exist no linear map that is surject & injective right

[wintry steppe]

no injective one from V to W, and no surjective one from W to V. so certainly no invertible one either way

[stoic python]

slimevesus

[acoustic zodiac]

Is there a reason as to why a change of basis is defined as:

$$e^*_i = c^j_i e_j$$?

I suppose it's because every vector of the new basis is ultimately just a linear combination of the vectors of the previous basis; is this reasoning somewhat correct?

[stoic python]

rcatalang

[quartz compass]

that is exactly correct

there has to be an invertible linear transformation between bases

[acoustic zodiac]

Nice, thanks!

[quartz compass]

you're welcome

[wanton fiber]

sometimes, when finding the inverse of a matrix, my gaussian elimination steps are all mathematically correct, but the inverse i obtain is wrong

why is that?

how do i know the correct steps?

[hollow garnet]

example?

[wanton fiber]

this is the correct answer

and the above is mine

both follow the same steps but in a different order. but shouldnt the answer still be the same

[wooden wave]

for part c, I get that (cf)(3) = cf(3) = c+cf(-5), but I don't understand why that doesn't work. why do we expect cf(3) to also be 1 + cf(3)?

I think I'm missing something about (cf)(3) = cf(3), what does (cf)(3) really mean in comparison to cf(3)?

[hollow garnet]

because you made a mistake

[wintry steppe]

(cf)(3) is defined as cf(3)

[hollow garnet]

-2+1 = -1 which is not 1

[wintry steppe]

even better, 0 \neq 1 + 0 (re rohak's question)

[nocturne jewel]

Yeah I was about to say 0 isnt in option c

[wooden wave]

hmm I don't understand what you're trying to say here

[wintry steppe]

subspaces contain 0

this doesnt

[nocturne jewel]

For it to be a subspace it needs the 0 vector (0 function in this case)

[wooden wave]

ohhhh yea, gotcha

[wintry steppe]

it's probably the first thing you should check if you're ever given a "is this a subspace?" question

[wooden wave]

I meant it as kind of a vehicle to explain cf(3) here doesn't equal (cf)(3)

[wintry steppe]

that's the definition of (cf)(3)

[wooden wave]

why is it that we expect 1+ cf(5) though

isn't c * f(3) literally multiplying the output of f(3) by c

[wintry steppe]

we shouldn't expect that to be true. take c = 0.

it's literally not true. the set here isn't a subspace

[wooden wave]

the solution I have is this for reference:

I understand that it isn't a subspace, I just want to understand this hah sorry!

[wintry steppe]

that's kind of poorly written

but

ok, i guess it's saying that if f is in the space, and c is a scalar, then cf is in the space only (something something)