linear-algebra

[fleet sun]

Sure, glad it helped

[open locust]

You did 😀

@fleet sun I have a related question

Question: We have theorems about linear transformations $A: U^{(m)} \to V^{(n)}$, and also transformations from $A: U^{(m)} \to U^{(m)}$ (i.e. from a vector space to itself). What happens with the case $A: U^{(m)} \to V^{(m)}$? (I.e., what if we have a transformation from a vector space to another vector space with the same dimension?) Do we need new theorems for that?

Or, maybe I should say: which theorems from the first two cases apply to the third case, and what are the minimum necessary modifications if any?

[stoic python]

joesmith1042

[fleet sun]

well, if two vector spaces have the same finite dimension, they are isomorphic

[open locust]

Indeed.

[fleet sun]

so it's not essentially different from going from U to itself

[open locust]

I guess the one we just discussed wouldn't hold at all since we're not talking about the identity transform, nor change of basis.

In that way, it is essentially different

I suppose the implication is that if we go from matrices to transformations, we have to stipulate that even if we have a square matrix, that it represents a transformation from one space to that same space.

[fleet sun]

yeah

[open locust]

I.e. if we took the situation we were just discussing, even if a matrix were similar to the identity matrix, unless we stipulate this interpretation we couldn't say it represents the identity transform (perhaps this is obvious to most people but I wanted to get it down "on paper" so to speak).

[fleet sun]

By the way, I had a funny feeling about what I was saying before

and the fact is the only matrix similar to the identity matrix is the identity matrix

[open locust]

I don't think that's correct.

Er, I should clarify

Your statement about "the only matrix" is true for similar matrices, but not for equivalent matrices

But a lot of people don't make that distinction so I was glossing over that terminology difference

$A_2 = P A_1 P^{(-1)}$ iff $A_2$ and $A_1$ are similar. $A_2 = Q A_1 P^{(-1)}$ iff $A_2$ and $A_1$ are equivalent.

[stoic python]

joesmith1042

[open locust]

(where $P$ and $Q$ are invertible matrices)

[stoic python]

joesmith1042

[fleet sun]

right

[open locust]

If two linear transformations are isomorphically equivalent, then their matrices are equivalent. But I think we can only say the converse of that statement if we make the stipulation that I was mentioning before.

(not sure about my last sentence there actually, reconsidering it now)

[fleet sun]

I'm not sure what you mean by transformations being isomorphically equivalent

I could guess

T1: U1 -> V1

and T2: U2 -> V2

now if there are isomorphisms U1 -> U2 and V1 -> V2

such that the square of maps commutes

you could say T1, T2 are isomorphic maybe

[open locust]

@fleet sun I don't know what the "square of maps" is but I meant that $T_2 = P T_1 Q^{-1}$ in your example.

[stoic python]

joesmith1042

[open locust]

(where $P$ and $Q$ are the isomorphisms)

[stoic python]

joesmith1042

[fleet sun]

This square of maps

[open locust]

Nice picture, thank you!

[fleet sun]

Yeah I think it is the same as equivalence like you said

[open locust]

It is

So if we take the matrix equation $A_2 = Q A_1 P^{(-1)}$ where $Q$ and $P$ have the same dimensions and are invertible, then if we used $A_1$ as the identity matrix, there is certainly an $A_2$ that isn't the identity that satisfies that equation.

[stoic python]

joesmith1042

[open locust]

So I think we just proved: If a matrix is equivalent to the identity matrix, then it is a representation of the identity transform. (and vice versa)

Do you think that is correct?

[fleet sun]

Let me think about that

It's been a while since I was in linear algebra, need to brush up on this stuff

[open locust]

That's okay! Honestly my brain is a little tired from thinking about this right now

[fleet sun]

But I think yes you're right

[open locust]

Okay cool, thanks. I'll revisit this tomorrow and if I find I was wrong, I'll sign in here and let you know 😀

Thank you again for all the help, thinking, etc.

[fleet sun]

Wikipedia says

"Equivalent matrices represent the same linear transformation V → W under two different choices of a pair of bases of V and W, with P and Q being the change of basis matrices in V and W respectively."

[open locust]

Well... this takes us back to the beginning unfortunately.

To my original question.

I'm actually going to stop thinking about this now so I don't get more confused, I'm low on mental energy, I will revisit this again tomorrow including the Wikipedia article.

I'll let you know my thoughts then 😀 Have a good day or evening, whichever it is for you.

[fleet sun]

Sure, have a good one

[elfin grove]

my prof says that if you are finding a basis for a plane, any two non parallel vectors will automatically span the vector space (plane). Why is this? And does this always apply?

[dusky epoch]

a plane is a space of dimension 2, by defn

[hardy inlet]

im not sure how to get this positive

[quasi vale]

@hardy inlet Recall expansion of (a+b+c)^2

[hardy inlet]

is it that i assume?

[quasi vale]

Yes

[hardy inlet]

ah cool

ty

[hardy inlet]

this calculation of -0.01 the same as <Av,v> right

Av.v is what thats an attempt at

and i did v.Av which should be the same

Im strugging with: Find an example of a positive matrix some of whose entries are negative.

[tame pond]

You can take any upper triangular matrix with positive entries on the main diagonal, 0 on the lower, and a negative number on the upper

[hardy inlet]

i tried that earlier and got this, idk how this is positive, can it be factored or something

oh wait now this looks promising

isn't that (x-y)²?

[hard drum]

ye

[hardy inlet]

cool thanks guys

[hardy inlet]

🤔 it doesn't say prove true, so would there be a counterexample?

[spare widget]

If T is diagonalizable then this is the case

Since 1 / lambda is positive

idk about the non-diagonlizable case

And you can have invertible non-diagonalizable matrices

min_x |[A 1] x - b|^2

once you find the optimal x by solving

[A 1]^T[A 1] x = [A 1]^T b

then the residual is r = b - [A 1] x

Where x is the solution

[fleet sun]

I think I found the answer to this. But you can keep thinking about it if you prefer

[spare widget]

Am I tripping or does the question not make sense?

It's always I, no?

He argues about other bases, but a matrix transformation is of the form P^{-1}AP, so it's definitely always I even taking this into account.

If we have T(v) = v, the matrix for T is the identity regardless of the basis

P^{-1} I P = I doesn't really change that

So this is correct

[fleet sun]

As I understand, the condition he's talking about means the matrix can be row reduced into the identity matrix

[spare widget]

This is not

[spare widget]

what I glanced is the user believing that [I]_B1 and [I]_B2 are different

They are not

It's the same identity matrix

[fleet sun]

Well, consider the identity transformation from R^2 to itself, but where the domain uses the basis {(1,1),(2,1)} and the codomain uses the basis {(-1,1),(3,2)} or something

I don't think it will look like an identity matrix

[spare widget]

because that's not the identity transformation

What you are describing is a matrix that maps from B1 to B2

e.g. the P that I have

change of basis matrix

seems like some confusion regarding bases and what the identity map is

[fleet sun]

I think we have to conceive of the identity transformation as ignoring any choice of basis

it just sends v to v

but if the first v is written in one basis, and the second in another

that matrix doesn't look like the identity matrix

[spare widget]

That's not how this works

It's clear that [v]_B1 doesn't have to equal [v]_B2

This doesn't say anything about the identity

It just says that the coefficients in different bases of the same vector do not have to agree, but that's trivial

[fleet sun]

Don't matrix representations care about that though

[spare widget]

you can think of those as defined wrt some basis

But when you want to change the basis of a matrix it goes like this:

P^{-1}[A]_B1 P = [A]_B2

[fleet sun]

right

[spare widget]

And it's clear that with A = I

This is I regardless of the basis

Because P^{-1}P = I

And it should be clear that this needs to be the case

even without the above

Since you want T(v) = v

Well say [T(v)]_B1 = [A]_B1 [v]_B1 = [v]_B1

For all v

Then it's clear that A= I, and that's regardless of the chosen basis

[fleet sun]

I think the change of basis thing with P^-1 A P only has the effect of changing basis in the codomain

if you want to change in the domain as well it's like P^-1 A Q

matrix equivalence vs similarity

But I'm realizing I never really learned this stuff as well as I should have

[hardy inlet]

spectral theorem

[fringe fjord]

No, P^-1 A P changes the basis for both the input and the output of the linear transformation represented by A.

For P^-1 A P, the columns of P should be the new basis vectors expressed in the old coordinate system.

[fleet sun]

Yeah I said it wrong

I was thinking of this:

"The notion of equivalence should not be confused with that of similarity, which is only defined for square matrices, and is much more restrictive (similar matrices are certainly equivalent, but equivalent square matrices need not be similar). That notion corresponds to matrices representing the same endomorphism V → V under two different choices of a single basis of V, used both for initial vectors and their images."

[fringe fjord]

Oh right, if the doman and codomain are two different vector spaces (so it doesn't even make sense to use the same basis for both in the first place) then P^-1 A Q is what you want.

[hardy inlet]

If T*T = TT is T self-adjoint/hermatian?

[fringe fjord]

Yes, by definition.

[hardy inlet]

by definition of what, i know T=T* is the def of Selfadjoint but does that 2nd T have any issues

[fringe fjord]

No, you can replace equals for equals anywhere in an expression.

If T* and T are the same thing, they can't yield different things when multiplied by T.

[hardy inlet]

so this is sufficient in red?

[fringe fjord]

Hmm, that seems to assume that T is self-adjoint instead of proving it.

No wait, it doesn't assume it.

[fringe fjord]

Oh shit, I completely misread your question. Everything I said was nonsense, please ignore.

[hardy inlet]

also i think this was supposed to be an I

[fringe fjord]

Once you have T*T = T², multiply by T from the right on both sides and use T²=I.

[hardy inlet]

T*TT = TTT? What does that give us

T*I = TI

oh i see

[hardy inlet]

what does an orthonormal basis of P_2 look like? i feel like ive done it before with gram schmidt but dont wanna do it again

[fringe fjord]

What is P_2? I suppose polynomials of degree <= 2, but with which inner product?

[fringe fjord]

You could use a and bx as your first two basis elements, for appropriate normalization constants a and b. These are clearly orthogonal.

For the third, something of the form c(x²-d) would automatically be orthogonal to bx (because they are even and odd, respectively), and you can then solve for d to make it orthogonal to a. Finally normalize it by selecting c appropriately.

[hardy inlet]

im not sure where we need this orthonormal basis; i skimmed over the book and dont see anything about it

[hardy inlet]

so we have the list from another example in the gram shmit section; but what use it is

[lavish jewel]

you can note the relationship between the squared singular values and the eigenvalues

[hardy inlet]

yeah i thought it was just the sqrt of the evs of A^T A

i dont see why the basis for A matters

[lavish jewel]

presumably to compute A^T A

[hardy inlet]

but A^T A is easy with the standard basis

i thought this would 'work'

[pastel mesa]

This is my first time here so I am not completely sure how to do this, but could someone please help me get started on this problem?

[twilit anvil]

what do you know about the determinants of similar matrices?

here, we can see A and B are similar matrices, so it might help to look in your notes or book for something related to similar matrices and determinants

[pastel mesa]

They are the same

Oh, I think I see, so I just need to find values for x that would make the determinant of A = 0?

I got it, thanks @twilit anvil !

[timid sage]

So I've been looking at this question.

Am I right in thinking that (v-vp)^T is orthogonal to the nullspace of A?

[twilit anvil]

did you mean v-v_p? and what makes you think its orthogonal to A?

[timid sage]

I don't have a good mathematical justification for it, but drawing the vectors involved shows a right angle

I've just projected v onto an arbitrary vector A instead of the basis for A's nullspace for simplicity

At this point I haven't worked out how to justify it, but it looks as if v-v_p is orthogonal to the arbitrary A

Which makes me think that for the actual question, v-v_p would be orthogonal to the nullspace of A

[wintry steppe]

Yes

algebraically we know every vector has an orthogonal decomposition, v = v_A + v_Aperp

so v-v_perp=v_A

[timid sage]

thanks

[wintry steppe]

My intuition is telling me that the rank of the new matrix is n-2

then again idk

[spare widget]

Since vp is the projection of v onto the nullspace then (v-vp) is in R(A^T), so I think it's linearly dependent with the row-vectors of A and the rank doesn't change. What bothers me is why they mention that the nullspace is 2-dimensional, as this is irrelevant as far as I can tell. The rank of the new matrix is equal to the rank of A.

[timid sage]

Yeah thats it

For justification of my answer,

I can see that a space and its orthogonal complement make up a complete basis for R^n

So I can say that because (v-vp) is orthogonal to Null(A), it is therefore within the rowspace of A?

As the orthogonal complement of the rowspace of A is Null(A)

[spare widget]

Yes N(A) \perp R(A^T)

And N(A) + R(A^T) = R^n

$v_p = (I-A^{+}A)v \implies v-v_p = A^{+}A v \implies (v-v_p) \in R(A^T)$

[stoic python]

criver

[spare widget]

(v-vp) is the projection of v onto R(A^T)

[timid sage]

What's A^+? I haven't seen that notation before

[spare widget]

but since rank A = dim R(A^T) and (v-vp) in R(A^T) then then R(M^T) = R(A^T) -> rank M = dim R(M^T) = dim R(A^T) = rank A

[spare widget]

Moore-penrose pseudo-inverse

https://en.wikipedia.org/wiki/Moore–Penrose_inverse#Projectors

You define it like so:

[timid sage]

ah yeah

And I'd assume R(A^T) is the rowspace of A right?

I haven't seen that either

More commonly just R(A) or C(A^T)

[spare widget]

$A = U\Sigma V^T \implies A^{+} = V \Sigma^{+} U^T \ (\Sigma^{+}){ii} = \begin{cases} 0 & (\Sigma){ii} = 0 \ \frac{1}{(\Sigma){ii}} & (\Sigma){ii} \ne 0 \end{cases}$

[stoic python]

criver

[spare widget]

yes

U\SigmaV^T being the singular value decomposition of A

[spare widget]

This should be enough though

The point is that the projector on N(A) is I - A^{+}A

So

$v-v_p = v - (I-A^{+}A)v= (I-I+A^{+}A)v = A^{+}Av$

[stoic python]

criver

[spare widget]

But A^{+}A is the projector on R(A^T)

So (v-vp) is lin dep with the rows of A

[timid sage]

Alright yeah

Thanks for all that

I'll need a sec to get fully across everything here lol

But yeah it looks good

[spare widget]

The main idea is that v-v_p is not linearly independent from the row vectors of A

So rank M = dim R(M^T) = dim R(A^T) = rank A

To show that it is not lin indep from them you can use that it is the projection of v onto R(A^T)

[plush dust]

Hi guys, can you help me with this question?

[spare widget]

use the fact that swapping two rows results in a -1 factor for the determinant

Then count the number of swaps necessary

[plush dust]

Ah yeah, right, thanks, so

[spare widget]

As an example swap v1 and v2

Then v1 and v3

Then v1 and v4

Etc

Until it ends up in the last row

[plush dust]

Its (n-1) swaps

[spare widget]

Yes

[plush dust]

So answer should be (-1)^(n-1)

Thanks

[plush dust]

Can you check this one also? Thanks in advance

[lavish jewel]

is this an exam?

[plush dust]

A previous years once, for practice, I read the rules 🙂

[lavish jewel]

ok. yeah that seems correct

[plush dust]

Thanks, just I did a quick check witn B1 = {(1,0),(0,1)} and B2 = {(0,1),(1,0)}

Wanted to be sure that its correct answer

Little question, here p1(x) means space {1, x}, p2(x) means {1,x,x^2} and so on yes?

[lavish jewel]

what i understand from this is that p_i(x) is some arbitrary polynomial

[pseudo maple]

How did they create the matrix

i know we do the derivative for each element

[plush dust]

Arbitrary polynomial in {1, x, x^2, x^3, x^4}?

[lavish jewel]

i don't see why that would be necessary

[plush dust]

Just its subspace of P(4), thats why I thought that way

[lavish jewel]

all they tell you is that the set {p_i(x)} forms a basis, so the p_i(x) are lin indep

you don't know anything else about them

[plush dust]

Ok

[plush dust]

So, as its subspace and they are linearly independent, its basis should have dim = 3, which is the case C as I see

[lavish jewel]

yeah. you can check ann's image for more details

[plush dust]

Thanks

[plush dust]

Ok 🙂

[lavish jewel]

and then you indeed notice that the solution includes a polynomial that was not part of the canonical basis

[plush dust]

You are about that x^3 have coefficient 2?

[lavish jewel]

mhm

[plush dust]

Ok, yeah, that thing I remember good, that coefficient do not play major role here hehe 😁

[lavish jewel]

i'm not sure what you mean by that

the key takeaway should really be that you only knew there were 3 lin indep polys, and that is all

[plush dust]

Nevermind, I got what you say)

[lavish jewel]

the solution could've been {e, pi + 20x, x^3 + 300x^2}

[plush dust]

Btw, its okay that there is no x^2 in the answer?

[lavish jewel]

that is exactly my point, so it seems you have not yet gotten it

P_4 has dim = 5, and from the definition of W given in the problem, W can be any dim 3 subspace of P_4

[plush dust]

Hmm, so the exponent of x is not the key part here

[lavish jewel]

it plays a role but only in the dimension of W

read ann's comments

[plush dust]

Okay, so if understand right, as its subspace of W, its dimension should be at least 3 yes?

[lavish jewel]

what is a subspace of W?

[plush dust]

Wait, subspace is always only 1 dimension lower?

[lavish jewel]

nothing we are discussing in this problem should be a subspace of W, and there is nothing with dim >= 3

no

all of that is wrong

[plush dust]

Oh wait, as basis of W is 3 polynomials, its dimension is 3 right?

[lavish jewel]

mhm

[plush dust]

Ok, now I got it, only one thing, why B is not a basis?

Cuz of exponent of x?

Like in both places, its 1

[lavish jewel]

because the polys are not linearly independent

recall that the definition of a basis necessitates that the elements of the spanning set be linearly independent

so if the tell you "the basis has 3 elements", you immediately know the dimension is 3

but if you are given an arbitrary (spanning) set of 3 vectors, they don't necessarily form a basis

[plush dust]

Okay, I understood, thanks 🤗

[lavish jewel]

in this case, you can pick any 2 of the polys and take linear combinations of them to form the 3rd

[plush dust]

Ah, yeah, now I see

And is this one correct?

[lavish jewel]

doesn't seem so

[plush dust]

Hmm

[lavish jewel]

wait it's the augmented matrix, i misread

then yeah

[plush dust]

Great

[neon lake]

Is there a more intuitive way to determine the position between three planes rather than memorizing the positions based on ranks of the system formed by those planes?

[native rampart]

Context?

[neon lake]

For instance, the last drawing represents the position of the 3 planes when the rank of the coefficient matrix is 3 and the rank of the augmented matrix is 3 too

[spare widget]

think about the dimension of the null-space

It is the dimension of the intersection (if any)

If the null-space is trivial (i.e. full rank) then all hyperplanes intersect into a single point

Specifically

Ax = b -> x = A^{-1}b

If A is not full rank then you have two options: non-trivial intersection, or no intersection

the dimension of the nullspace gives you the dimension of the intersection if any

So if rank is n-1 -> leaves out 1 dimension for the intersection -> the planes intersect in a line or they do not have a common intersection

n-2 -> the planes intersect in a plane or they do not have a common intersection

Whether there is a common intersection or not depends on the rhs - if it is in R(A^T) then there is a common intersection having the dimension of the nullspace,otherwise the intersection of all of the hyperplanes is empty

[plush dust]

Any ideas which one is correct? Im lost

[spare widget]

Assume v in N(B) then Bv = 0

then ABv = A(Bv) = A0 = 0

Then it follows that v in N(AB)

This shows that N(B) <= N(AB)

Now let v in N(AB) -> ABv = 0

but A is invertible, so A^{-1}ABv = Bv = 0 -> v in N(B)

So N(AB) <= N(B)

From those two it follows N(B) = N(AB)

2. Now let A be singular

It is clear that if v in N(B) then v in N(AB) since Bv= 0 -> ABv = 0

So N(B) <= N(AB) and 2 is false

[plush dust]

Wait, N is null space right?

[spare widget]

yes

[plush dust]

And v is?

[spare widget]

a vector

[plush dust]

arbitrary vector?

[spare widget]

v in S means that v is an arbitrary element of S

[plush dust]

Ok

[pseudo maple]

how do we do matrix representation of polynomial linear transformation

[plush dust]

Lost in this part, can you explain once more?

[pseudo maple]

<@&286206848099549185>

[dusky epoch]

you have your basis, yes?

you're writing the matrix of the derivative operator wrt the monomial basis for P3

[pseudo maple]

im not sure if my basis is correct

[dusky epoch]

wym

[pseudo maple]

(0,b,2a)

[dusky epoch]

that's not a basis

Standard basis: 1, x, x^2, x^3

D(1) = 0 = 0*1 + 0*x + 0*x^2 + 0*x^3 => first column [0; 0; 0; 0]
D(x) = 1 = 1*1 + 0*x + 0*x^2 + 0*x^3 => second column [1; 0; 0; 0]
D(x^2) = 2x = 0*1 + 2*x + 0*x^2 + 0*x^3 => third column [0; 2; 0; 0]
D(x^3) = 3x^2 = 0*1 + 0*x + 3*x^2 + 0*x^3 => fourth column [0; 0; 3; 0]

do you understand this?

[pseudo maple]

thank you im having a read now

[dusky epoch]

https://www.youtube.com/watch?v=kYB8IZa5AuE

may be worthwhile to watch this

[pseudo maple]

awesome thank you will do!

[spare widget]

A set S is a subset of a set P if for every v in S it follows that v in P. That is, if every element of S is an element of P then it's clear that P is a superset of S (it contains all of its elements).

N(D) = {v : Dv = 0}

It's clear that from Bv = 0 it follows ABv = 0

Thus if v in N(B) it follows v in N(AB) and thus N(B) <= N(AB)

[plush dust]

Got it, thanks

And did i got this correct answer?

[spare widget]

You should be able to check this yourself

how would you go about this?

Generally if you claim something is false then you should be able to construct counterexamples

[plush dust]

Just, as I remember, there is no any connection between determinant and matrix and its rank

Thats why I chose 2 false

[spare widget]

Try constructing counterexamples

For 1) you can try with diagonal matrices to make it easy

For 2 you can try having singular matrices with the same number of linearly independent rows

[plush dust]

ok

[spare widget]

1 counterexample is enough

[restive raft]

There is a small yet important connection with rank and the determinant, if the determinant is non-zero then the matrix has full rank. But if it is zero the rank could be anything less then full

[plush dust]

Is this right guys?

[stuck tendon]

It is

[plush dust]

Thanks

[spare widget]

It's better if you prove that it is right

[wintry steppe]

characteristic polynomial of A is p_A(t) = det(tI - A). how to prove it really is a polynomial?

[spare widget]

using the definition of the determinant

It involves only multiplication, addition and subtraction operations

[wintry steppe]

thanks, @spare widget :)

[plush dust]

Not sure about this one, is it correct?

[spare widget]

What have you tried?

Have you tried writing the general solution?

[plush dust]

I just removed all other possible variants

[spare widget]

Try writing the general solution

If you're not sure what I mean refer to a textbook in the section on solutions of augmented systems in rref form

e.g. hoffman and kunze

[plush dust]

You mean this?

[spare widget]

yes

[plush dust]

Yeah, did this

[spare widget]

So what did you get?

[plush dust]

e1, e2 and v0

[spare widget]

Yes what values

Either way, you get some e1, e2 vectors

Those vectors span some 2d subspace of R^4

[spare widget]

You can pick any other f1, f2 from said subspace and the only difference would be the coefficients

However if you were to take vectors outside of the span of the e1, e2 that you have, then that will not be a solution

And they emphasize e1, e2 being vectors from an arbitrary basis of R^4

It cannot be arbitrary

[plush dust]

Hmm, so its not correct answer

[spare widget]

It must be a basis for the subspace spanned by the e1 and e2 you computed here

It is not a correct answer

[plush dust]

So its either B or E

Im not sure about B cuz can't really understand what it means

[spare widget]

you really ought to try and understand things rather than try to guess it by eliminating answers

I guess that's what happens with multiple options exams

[plush dust]

I would like to, can you explain the statement B, cuz I cant understand the statement

[spare widget]

Do you know how the nullspace for a matrix is defined

[plush dust]

Just to compute null space dimension?

[spare widget]

N(A) = {v : Av = 0}

[plush dust]

Ok, yeah, know this

[spare widget]

and we know that N(A) = N(rref(A))

[plush dust]

So

[spare widget]

So what is the dimensionality of the solution space of Av = 0?

How many basis vectors do you need to define the solution?

[plush dust]

Its equal to n - rank no?

[spare widget]

It is equal to n - rank yes

[plush dust]

So, rank is number of non zero rows

In our case its two

And n is equal to 4

[spare widget]

Wait no

[plush dust]

So nullity is 4 - 2 = 2 != 3

[spare widget]

Yeah

All is good nvm

[plush dust]

So, its E

Non of the above

[lavish jewel]

in the original mat, not the augmented one

[spare widget]

Yes in A not tildeA

You already derived the vectors though

So you could have argied just from those

That since you had e1, e2 it's 2d

[spare widget]

It even says null(A) if you look here

[plush dust]

Ah yeah, missed that

Thanks guys

[wintry glen]

Hi! The question is to show that if $A \subseteq B$ and $B$ is affine, then aff $A \subseteq B$. Is my proof below correct?\

Suppose $x \in$ aff $A$, then $x$ is an affine combination of vectors in $A$, and is thus an affine combination of vectors in $B$, since $A \subseteq B$. Then, as $B$ is affine, any affine combination of vectors in $B$ is also in $B$, so $x \in B$. Hence, aff $A \subseteq B$.

[stoic python]

PhenomPlasma

[spare widget]

Yes

[pseudo maple]

can anyone help me understand how they achieved this answer

Idk what im suppose to do with the T=x (d/dx) - x(d^2/dx^2)

but ik how to create matrix represntation for standard basis

[fringe fjord]

Apply T to each of the elements of the standard basis in turn; write down the coefficients of the result.

[pseudo maple]

cool, ill have a go thanks

[pseudo maple]

how would i do that?

[spare widget]

Use the definition of T in b)

Then feed it 1, x, x^2

The point is

T(a x^2 + bx + c) = linearity = a T(x^2) + b T(x) + c T(1)

So once you know what happens to the basis you're done

[twilit minnow]

how can you prove that if a matrix has a left inverse, it is also injective

[pseudo maple]

but how did that come about

[spare widget]

This is b)

They are proving linearity there

The linearity that I used for c

[pseudo maple]

so i differentiate teh basis and then just plug it into this?

and then ill know the matrix

[spare widget]

??m

[pseudo maple]

sorry im rlly bad at this specific topic

[spare widget]

You need to compute T(x^2), T(x), T(1)

Using the definition of T

Once you do that you will get some polynomials of the form T(1) = a11 + a12 x + a13 x^2, T(x) = a21 + a22 x + a23 x^2, T(x^2) = a31 + a32 x + a33 x^2

The coefficients are then you matrix coef

But you have to compute T(1), T(x), T(x^2) to figure those out

The definition of T is in b)

[pseudo maple]

aah ok i think i get it, ill give it a go thanks

[spare widget]

btw as I wrote them the matrix is the transposed

[broken notch]

Can someone check my jupyter solution where I answer this problem

"(𝑒) Calculate the numbers 𝜅(𝐴),cos𝜃 and 𝜂 , which control the condition numbers for the problem (see note 17). Indicate the corresponding upper limit of the condition number for how change in 𝐴 affects change in the calculated 𝑥 in part (c).

Use this to explain how accurate you can expect the calculation of 𝑥 to be."

Bcs I am getting values that look wrong but I follow the procedure given to me to the teeth

[plush dust]

Can I use arbitrary numbers to prove or I need to prove by variables?

[lavish jewel]

using arbitrary numbers is not a proof

[hard drum]

What is the definition of a linear transformation

[plush dust]

So need to prove by a and b?

[lavish jewel]

as potato says, you have to use the definition of linear transformation

[plush dust]

We mostly just tried to prove that sum of transformation is equal to transformation of sum

[lavish jewel]

in fairness, for b you can construct a counterexample. but for a, you need to do it in general

[plush dust]

And also that c * T(v) = T(cv)

[hard drum]

Yes, you have to prove that for all a,b in R^2 and all c in R we have that T(a+b) = T(a) + T(b) and T(ca) = cT(a)

i.e. exactly that T is linear

[plush dust]

So, basically, just prove the same with a and b

[hard drum]

Note the for all

Yes

[plush dust]

Ok

[lavish jewel]

note that it won't suffice to use just a and b, since you have to show properties involving addition of vectors

so you'll need at least 4 scalars

(5 considering scalar multiplication)

[plush dust]

Something like this

?

[lavish jewel]

this shows additivity, yes

[plush dust]

Sorry, this way

[plush dust]

Are these statements enough or I'm missing something? 😅

[lavish jewel]

you didn't show part a though, presumably they want you to justify it

[plush dust]

Need to prove this prepositions?

Like as there is no 0 vector, there are not equal, none of them is scalar multiple of another, (skipping 4,5) => They are independent?

[lavish jewel]

it would be enough to evoke how polynomial addition works

and then the usual definition of "scalars not all 0..."

[plush dust]

Ok, and about task b)

What I should also mention?

[hardy inlet]

You probably need to use the fact that a spanning set of length dimV is a basis

And to show S spans V u probably need to show a the linear combination that gets it.

$ax^2 + bx+ c = \alpha f_1 + \beta f_2 + \gamma f_3$

[stoic python]

MattDog_222

[hardy inlet]

for a concrete example if I wanted 21x^2 it would be 7f_1

[plush dust]

Ok, thanks

And maybe you can give a small hint, from what to start this exercise?

I'm just good at computations, not at proving 😅

[wintry steppe]

Am I doing this right??? How do you get the basis from this??

[plush dust]

As I understand, I need to prove these properties for task a)

[wintry steppe]

Yeah you gotta prove thise 3

[silk valley]

there's a slick way to prove sub-anything but i forget

[spare widget]

You have to prove that these 3 imply the axioms for something being a vector space

Because a subset of a vector space that is also a vector space is a subspace.

[plush dust]

Ok, and for task b)

?

[spare widget]

Wait

[plush dust]

Can I choose the standard bases?

Or it will not work 😄

[spare widget]

This is not the exercise?

[fringe fjord]

It sure looks to me like it is.

[spare widget]

Then forget what I said, you can directly use this if it is not the exercise

[spare widget]

I think he's referring to this

[fringe fjord]

It explicitly says: "Prove that U is a subspace ..."

[spare widget]

yeah I misunderstood that he has to prove the above definition

[spare widget]

Disregard this

[plush dust]

Erm, so should I prove that 3 properties?

[spare widget]

You should prove the 3 properties for a)

[plush dust]

Good

[spare widget]

i.e. that they hold for a)

[wintry steppe]

Isn't it like fu + fg = f(u+g) kinda

Using the matrix they gave you

[plush dust]

I think you are talking about transformation no?

[wintry steppe]

It's similar to that

I believe

[plush dust]

Hmm, maybe, not sure

[wintry steppe]

Because you're trying to show if you add them you get the same form

[spare widget]

You just have to prove that the 0 matrix isin U

That if you pick u,v in U

Then u+v in U

And alpha * u in U

[plush dust]

So, the first one is obvious, if a = 0, b = 0, we will have 0 matrix

[spare widget]

for example use a,b for u and c,d for v

Compute the sum

Then do two variable substitutions to get it in the desired form

For alpha * u do 2 variable substitutions too

[plush dust]

This way?

[spare widget]

Yes

[plush dust]

And like we substitute ⍺ = [a + c] β = [b + d]

[spare widget]

yes

[plush dust]

We will have similar to the initial one, got it

[wintry steppe]

Yep same form

[spare widget]

You need to check scalar-matrix multiplication also

[wintry steppe]

Am I approaching this right?

How do I get the basis from this??

[spare widget]

I am confused

Why did you not compute S^TAS

[plush dust]

I guess the same way yes?

Oh, I think its better to use variable instead of 5

[wintry steppe]

I did compute it three times I thought

[spare widget]

What they want you to do is S^TAS = alpha1 * B1 + alpha2 * B2 + alpha3 * B3, where B1, B2, B3 are the given basis matrices

[spare widget]

yes

[plush dust]

And for task b)?

[spare widget]

Compute the product S^TAS

Note that A is symmetric 2 x 2

[wintry steppe]

Yes that's gonna be a b, b c

Since it's similar

Err symmetric

[spare widget]

I don't see the abc in your write up

[wintry steppe]

I haven't gotten there yet

[vital drum]

i have a rather open-ended question: what's interesting about annihilators?

[wintry steppe]

I'm not sure about the alphas

Is it 1, 9, 6?

[spare widget]

choose a basis

There's an obvious one

Split the matrix as a sum of matrices multiplied by a and b

[spare widget]

Write out the product

Shouldn't be too hard

multiply 3 matrices

[fringe fjord]

Please don't have these equals signs. They are lies: What is to the left of the = is not equal to what you have written to the right of it!

[spare widget]

Compute S^TAS

Where for A you'd use [a b; b c]

Once you compute it you can decompose it

[hardy inlet]

can someone please check if these absolute values make sense (and the proof in general), I hate complex stuff.

[fringe fjord]

But that's what he did!

[wintry steppe]

I'm trying to find the basis in part a

[spare widget]

He did?

[wintry steppe]

The kernel is just st (a b, b c) s

[plush dust]

Sorry for handwriting

[fringe fjord]

That looks to me like it's exactly what the handwritten part of his image is.

[spare widget]

I see no abc anywhere here

[spare widget]

you need a sum of two matrices

Take the initial matrix and write it as asum of 2, one containing only a, the other only b

Then finally take out a and b

[fringe fjord]

Basis underlined in red, S^T and S in green.

The general element is a linear combination of the basis vectors.

[spare widget]

ah, ok I get what he's doing

triple work

[plush dust]

Like this?

[spare widget]

Yes and now take a out and b out

That will leave you with the 2 basis matrices

[wintry steppe]

That's the basis right?

[fringe fjord]

You keep posting that -- is there a difference between the copies you post that you want us to notice?

[wintry steppe]

Yes... at the bottom....

[wintry steppe]

Sorry ill stop asking

[plush dust]

These two are the basis matrixes?

[spare widget]

Yes

[plush dust]

Great

[fringe fjord]

Oh right. What you have at the bottom now is not a "basis" but the matrix representation of the transformation L in the given basis for V. That is good as far as it goes.

[plush dust]

Two questions here:
In a) is the cofactor = 2
In b) in A and B I can use Laplace Expansion, but what to use for C?

[wintry steppe]

Is the answer i got I'm trying to find :p

I'm assuming b is a typo

[fringe fjord]

Ah I see.

Comparing here, we do see a difference.

You seem to have swapped rows and columns when you wrote down the matrix.

[hardy inlet]

idk what u mean but expand on row 1. u get 1 times the determinant of the bottom right 3x3, and then all zeros

and for the bottom right u get 1 * det(2,0;2;2) + a bunch of zeros

[fringe fjord]

On the other hand the typeset solution has the basis vectors in the wrong order, compared to the image with your handwriting on it ...

[wintry steppe]

So it's prob an exam solution error

[plush dust]

Its for B, but what to do for C?

[spare widget]

Notice tha C has two equal rows

[plush dust]

Oh, true

But what it gives?

[fringe fjord]

Right, but your solution ought to have been $\begin{bmatrix}0&0&0\1&9&6\0&0&0\end{bmatrix}$.

[stoic python]

Troposphere

[spare widget]

What happens when yoy have linearly dependent rows/cols?

What is the determinant equal to

[plush dust]

Determinant is 0!

[wintry steppe]

So I need to get the transpose.

[fringe fjord]

Yes. When you represent a linear transformation with a matrix, each column of the matrix represents the output of applying the transformation to one of the basis vectors.

[hardy inlet]

sry to repost but is the part below the red correct in its |absolute values| and square rooting arithmetic?

[fringe fjord]

(It's a bit tedious that the notation in linear algebra has ended up being such that matrices operate on columns of coordinates instead of rows, which would have been easier to type -- but there's no helping that now).

[wintry steppe]

Bam!!! Sorry man if I'm dumb. I hate my school, we get exam problems Notting like our homework. So I'm trying to learn previous years problems. Tyty

[lavish jewel]

multiply vectors from the left 😌

[spare widget]

v^T M^T

[fringe fjord]

Yes, but matrix multiplication might have been defined such that it takes rows from the right operand and columns from the left. Far too late for that now, though.

[lavish jewel]

certainly

[spare widget]

they actually have the two conventions for differentiating in matrix calculus

It's very confusing

the conventions are transposes of each other

[spare widget]

To be sure, not mult from the right by row vector, just transposes

[plush dust]

What matrix operations change determinant?
Multiplying by k -> det = k * det
Swapping rows or columns -> det = -det
Row operations -> no effect
What else can change it?

Inverse -> det = det * 1/det

[native rampart]

If A -> det
A^k -> det^k

[plush dust]

Oh yes, forgot about it, thats all right?

[hardy inlet]

is this algebra correct 🥺

[native rampart]

This should be multiplying one row or column by k

[plush dust]

And what if whole matrix is multiplied by k?

[hardy inlet]

is bad

[native rampart]

k^(dim matrix)

[plush dust]

Huh?

det = ?

[native rampart]

If matrix is 2x2 det -> k^2 det

[plush dust]

Ah, ok

[native rampart]

If matrix is nxn det-> k^n det

Yea that should be it

[spare widget]

Are you worried about the +-? The singular values are non-negative.

[hardy inlet]

oh right i kinda forgot about the \pm, but i was more concerned with the modulus/abs

[spare widget]

What's wrong with the modulus?

[hardy inlet]

is it maintained after the sqrt of the square

[spare widget]

Substitute r^2 = |x|^2, -> sqrt(r^2) = r

why would there be anything wrong with the modulus

[vital drum]

should i ask the question i asked earlier again or did ppl ignore it for good reason?

[hardy inlet]

i feel like it should be okay but absolute values are kinda weird sometimes so I wasn't sure

[pure crypt]

Can someone give me a hint on how to do this? People are telling me its really easy but i don't see it.

[vital drum]

let x be an element in both null(S) and null(T). what happens if you apply S+T or S-T?

[spare widget]

v in null(S) cap null(T) -> S(v) =0, T(v) =0 -> (S+T)(v) =0, (S-T)(v) =0 -> v in null(S+T) cap null(S-T) -> null(S) cap null(T) <= null(S+T) cap null(S-T)

Now do the opposite

To show the opposite inclusion

Then the two together imply equality

The opposite is a little more complicated since you get a system, but it's trivial to get S(v) =0 and T(v) =0 from it

[pure crypt]

oooh i see, ty ty

[twilit minnow]

what would be the best way to prove that

if A has a left inverse => rank(A) = n

[spare widget]

what is n?

Is A square?

I am guessing columns

Let x be in the null-space of A

Then Ax = 0

let B be its left inverse

Then x = BAx = 0 -> x= 0 -> kernel is trivial -> full column rank

[twilit minnow]

A is m>=n

[spare widget]

did you get the idea for the proof

[twilit minnow]

yeah

[twilit minnow]

how about going the other way

rank(A) = n => left inverse

[spare widget]

(A^TA)^{-1}A^T

[twilit minnow]

can u show it with row operations instead

and row reduction

[spare widget]

Isn't the above enough? (A^TA)^{-1}A^T * A = I

[stoic python]

pecfex

[spare widget]

And you can argue that A^TA is invertible because A has full column-rank

[vital drum]

@wintry steppe here's a hint

[spare widget]

I am assuming you can do it with row ops, but I haven't thought about it

[vital drum]

also you're gonna have to use the commutative diagram

[wintry steppe]

Yeah

I understand the commutative diagram

[twilit minnow]

i havent actually learned it that way in class

[wintry steppe]

how did you know that it was equivalent to show that? @vital drum

[twilit minnow]

the whole A^TA way

couldnt i just show that it can be reduced to In

[vital drum]

by exercise 17

[spare widget]

Well it is clear that (A^TA)^{-1}A^TA = I

[twilit minnow]

i dont see it

[spare widget]

B^{-1}B = I

[wintry steppe]

I dont understand

this is q17

[spare widget]

The only thing you have to prove there is that A full column rank means A^TA is non-singular

[vital drum]

part b is what you need

[wintry steppe]

I mean, how did you go from rank(T)=rank(L_A) to equivalently showing that phi_gamma(im(T))=im(L_A)?

[vital drum]

do you see it, jswatj?

[spare widget]

And I think you can prove tgis using the svd

[wintry steppe]

how did you know to do that?

[twilit minnow]

also, the kernel of A is trivial in this case

right

[spare widget]

Knowing that the columns are lin-indep

[wintry steppe]

I see where it can be used down the line i think

[stoic python]

pecfex

[wintry steppe]

Oh I see

wait im still a little confused

[vital drum]

yes?

[wintry steppe]

Oh wait

is phi_gamma(im(T)) a subspace of im(T)?

[vital drum]

no, it's the image of im(T) under phi

[wintry steppe]

so how did you use part b) to show equality?

oh wait nvm

oh

[vital drum]

i can type it out in case you still don't get it

[wintry steppe]

yeah im still sort of lost

cause this question came up as a theorem later in the book

so i need to understand it

[vital drum]

you know that im(T) is a subspace of W, right?

[wintry steppe]

Yup

[vital drum]

yea ok so by exercise 17

dim(im(T)) = dim(phi(im(T)))

[wintry steppe]

Oh

[vital drum]

phi is an isomorphism, that's why exercise 17 applies

[wintry steppe]

OHHH

you replaced T with phi in exercise 17

[vital drum]

yup

for the case of nullities, it's the same story

[wintry steppe]

and im(T) is a subspace of W

[vital drum]

yea

[wintry steppe]

ohhh that makes sense

what about the im(L_A) part?

[vital drum]

and now we prove that part by using the commutative diagram

[wintry steppe]

oh because

[wintry steppe]

phi_gamma leads to F^m

and the im(L_A) also leads there?

[vital drum]

more specifically this

the composed maps don't just have the same codomain, in fact they're the same map

that's what the diagram is saying

[wintry steppe]

wait but in our case aren't things different?

We're working with A=[T]_beta^gamma

and its only im(L_A)

how does that connect with their conclusion

[vital drum]

what do you mean more precisely?

ooh

wait what?

[wintry steppe]

LMaooooo

I understand their conclusion cause the standard rep of beta outputs vectors in F^n and L_A sends them to F^m

same with the other side

[vital drum]

yea

[wintry steppe]

how does that relate to our case with phi_gamma(im(T))=im(L_A)?

[vital drum]

well what's phi(im(T))?

[wintry steppe]

Sorry

[vital drum]

i could have said it better

[wintry steppe]

how did you use this conclusion to conclude they are the same

[vital drum]

just didn't want to give away too much

it's easy to see that im(T) = T(V), right?

[wintry steppe]

Uh

yes

[vital drum]

you can start from there

[wintry steppe]

Okay

[vital drum]

also the person who reacted, isn't this just definitional? or am i wrong?

[wintry steppe]

Oh i think i see

rearrange for T

then phi_gamma^-1 L_A phi_beta = T

so plugging in V for both sides gives

wait no

u don't have to rearrange

u can just plug it in

cause phi_beta gives F^n

L_A(F^n) = im(L_A)?

and that's equal to phi_gamma(im(T))?

thats wild

[wintry steppe]

I get this now

[vital drum]

alright

[wintry steppe]

thats so crazy

[brazen sinew]

Hey guys

Any strategies on how to start on this problem? Specifically, assuming S is infinite, how do i show that the given set is not a basis?

[hard drum]

Have you shown that if S is finite then it's a basis? Try to see where you used finiteness and that should lead to it

Or you can consider a more familiar example, like F = S = R and some standard functions to get intuition

[spare widget]

What definition of basis do they use in your book?

I am assuming they have something about ghe sum being finite, in which case you can just give a function that wpuld require an uncountably infinite sum of indicator functions

[brazen sinew]

A basis is a linearly independent set which spans the vector space

[brazen sinew]

No, im not sure how to unpack/use the set chi_s for s in S

[sinful bane]

is my book giving this a different name? this is apparently "reduced row echelon"

[spare widget]

I haven't formally studied functional analysis so I can't say much, but your book may be using a definition of hamel basis, or schauder basis, or something else.

I am assuming your book has the words finite linear combination somewhere in there

Which you could use for a counterexample

[wintry steppe]

linear combinations are always finite

[spare widget]

Schauder bases have a countably infinite sum in there

[wintry steppe]

that's not a linear combination

[spare widget]

I have seen people refer to it as an infinite linear combination, but as I mentioned I have not studied functional analysis formally

[wintry steppe]

"linear combination" alone always means finite

[pure crypt]

Anyone know how to find the eigenvalues of a non-finite matrix?

[wintry steppe]

this isn't a matrix

if you have a scalar c, you need to solve D(f) = cf for f

if you can find such a non-zero f, then c is an eigenvalue and f is an eigenvector. if not, then c is not an eigenvalue

[pure crypt]

ok so in over words, this doesn't have any eigen values then? since no such "c" exists?

i did write down Dx = \lambda x down on my paper but i thought i was doing something wrong

so i was trying to do the whole det(D - \lambda I) = 0 thing

[wintry steppe]

i didn't say anything about whether such a c exists

[wintry steppe]

this doesn't work because the determinant of an operator on an infinite dimensional space is not well defined

[pure crypt]

ooh i see

[pure crypt]

wait ok but if D(p) is the derivative of polynomial p, there shouldn't exist a constant such that D(p) = cp right? since u can't change the exponents of a polynomial by only scalar multiplication?

[wintry steppe]

yup, but you're forgetting one small exception

your reasoning works if c is non-zero, but what if c is zero?

[pure crypt]

oooh so the only eigenvalue for it would be zero

oh shit acutally that makes sense, cuz then in the form det(D - \lambda I), if lambda = 0, than the det = zero

i get it! Thanks a lot @wintry steppe

[wintry steppe]

no determinants!

[pure crypt]

okok fine

i just think about it like that cuz thats how i intuitively understand eigenvalues

[subtle gust]

Hello ...

I need to know if this sentence is accurate

"There exist only one unique matrix L and one unique matrix U for every n such that A=LU and diagonal(L or U)={n,n,n..........n} for n=/0"

[tacit pelican]

to do this you literally just add the rows $\begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 1 \ 0 \end{pmatrix}$ right

[stoic python]

sean

[tacit pelican]

or am i not getting it

my notes say to just do this

[fringe fjord]

Yes, that will work.

It's easiest to see that you get a basis if you follow the advice in the notes to insert your new rows between those you already have, such that the result is an upper triangular matrix with non-zero diagonal elements.

[tacit pelican]

alright thank you

[woeful saffron]

Helloo, can someone help me pls? I need to prove that given A a point in space and v a vector in space, so there is only one representation of v starting at A. How can I do this?

[plush dust]

Guys hello, I'm a bit confused here, can you say from what I should start?

[lusty swallow]

after a few days, I think he can solve that
because 2x2 matrix space and 4x1 vector space are isomorphic

[subtle gust]

hey everyone

i need to know if this sentence is correct or not ..

"There exist only one unique matrix L and one unique matrix U for every n such that A=LU and diagonal(L or U)={n,n,n..........n} for n=/0"

i'm writing smthng and i'm trying to be as accurate as possible

ummm anyone?

[eager grail]

Does it feel correct to you?

[open locust]

I'm confused about the relationship between invertible matrices, change of basis matrices, and the identity transformation. I understand that a change of basis matrix is a matrix representation of the identity transformation, with particular "to" and "from" bases. I also understand that any invertible matrix can be seen as a change of basis matrix for one particular set of to and from bases. But it seems pretty clear that not every invertible matrix represents the identity transform. What am I missing here?

[spare widget]

wdym to and from bases

The identity map is T(v) = v

Whatever basis you use, it's the identity matrix

[open locust]

@spare widget Like, a matrix is a representation of a linear transformation given some bases

[spare widget]

e.g. say I have a vasis B

Then

[open locust]

@spare widget I don't think it is

[spare widget]

[T(v)]_B = [I]_B [v]_B = [v]_B

But only the identity matrix maps any vector v to itself regardless of the basis B

[open locust]

@spare widget But if the matrix representation of the identity transformation is the identity matrix no matter what bases you choose, then every change of basis matrix is the identity matrix.

[spare widget]

I think you agree T(v) = v for any v represents the identity map?

[open locust]

Because a change of basis matrix is a matrix representation of the identity matrix with respect to particular to and from bases.

[spare widget]

no

[open locust]

I agree with this

[spare widget]

A change of basis matrix is such that

[v]_B2 = P [v]_B1

It says nothing about v irrespective of basis

It only gives the relationship for the coefficients wrt sone bases

[spare widget]

Then do you also agree with this

[open locust]

@spare widget So a change of basis "transformation" is not the identity transformation?

[spare widget]

it is not

A change of basis matrix changes your basis

But [v]_B2 and [v]_B1 are the same vector written wrt different bases

e.g.

$\vec{v} = \sum_{i}\alpha_i \vec{e}i = \sum{i} \beta_i \vec{f}_i$

[stoic python]

criver

[open locust]

I think the thing that's confusing me is that if we write Tu = x, I always try to think of that as "basis-free", but it seems like it's not possible to do that.

[spare widget]

why not possible?

It is possible

T(v) = v is coordinate free

[spare widget]

But this isn't

[open locust]

@spare widget So why isn't the change of basis transformation the identity transformation? It doesn't move or change the vector it takes in.

[spare widget]

I think you're mistaking the coefficients alpha for the vector v

The vector v is not the coefficients alpha

[open locust]

@spare widget What I'm saying is, I have a point in space. I do a "change of basis" transformation on it. Does it move the point, or no?

[spare widget]

the alpha coefficients are coordinates of v wrt E

The beta are coordinates of v wrt F

[open locust]

I understand that.

Does the change of basis transformation move the physical point in space, or no?

[spare widget]

A change of basis matrix operates on the coord representations

[open locust]

I didn't say matrix. I said transformation.

[spare widget]

It does not

[open locust]

Okay. So the transformation is the identity transformation.

[spare widget]

No

It doesn't move the point, it changes wrt which basis it is expressed

[open locust]

How can it be that a transformation that takes a vector and doesn't change it, be something other than the identity transformation?

[spare widget]

Imagine

[open locust]

But we aren't using bases when we talk about transformations, right?

[spare widget]

I have ruler

in meter and inches

I measure something

[open locust]

We're just saying stuff like, Tu = x, which you said can be expressed in a basis-free way.

[spare widget]

It x inches and y meters

[open locust]

So, our change of basis transformation T is Tu = u, right?

[spare widget]

its length didn't change

The way I measured it did

[open locust]

I know.

I'm not talking about specific bases here.

Tell me why the change of basis transformation does not take the form Tu = u in a basis-free setting.

[spare widget]

Your change of basis transformations are:

[open locust]

Please don't write anything about matrices in your next sentence.

[spare widget]

f1 = T1(e1,....,en), ..., fn = Tn(e1,....,en)

[open locust]

Okay, that makes sense.

[spare widget]

Makes sense?

ok

[open locust]

Yes I'm with you.

[spare widget]

Now this T1 has coefficients

To be precise it looks like this

$\vec{e}i = \sum_j a{ji} \vec{f}_j$

[stoic python]

criver

[open locust]

Yes I'm with you.

[spare widget]

The matrix A is the change of basis matrix

[open locust]

We have defined this transformation with respect to particular to and from bases.

[spare widget]

as you saw it's n maps, and they consume multiple vectors

It's not what you're used to

T(v) = w

Instead it's ei = Ti(f1,...,fn)

[open locust]

Well I'm waiting to find out where this is taking us.

😋

[spare widget]

Well now write out v in the E basis

[open locust]

I'm not really sure if it's $n$ maps, this is the same thing as how any linear transformation has to be defined by its actions on particular to and from bases.

[stoic python]

joesmith1042

[open locust]

E.g. $\forall j \quad A(e_j) = \sum_{i=1}^n a_{ij} f_j$

[spare widget]

$\vec{v} = \sum_i \alpha_i \vec{e}i = \sum_i \alpha_i \sum_j a{ji} \vec{f}_j\ = \sum_j (A\vec{\alpha})_j \vec{f}_j = \sum_j \beta_j \vec{f}_j$

[stoic python]

criver

joesmith1042

[open locust]

@spare widget I follow you.

[spare widget]

Well that's it

That's everything

[open locust]

I just wanted to point out that this isn't different from what I'm used to, this happens with every single linear transformation.

[spare widget]

beta = A * alpha

But alpha and beta are the coefficients of the vectors wrt the bases E and F

They are not v itself

There's the notion of active transformation

Then you can do

[open locust]

@spare widget Well, what you're saying is just that the change of basis matrix acts on the coefficients of $v$ in the old basis, and produces coefficients in the new basis.

[spare widget]

[w]_E = [Av]_E which actually gets you another vector

[stoic python]

joesmith1042

[open locust]

Which is the same for any linear transformation, right?

[spare widget]

But typically change of basis is understood as a passive transformation

Where

[open locust]

Oh, except that the new coefficients aren't usually coefficients of $v$. They're usually coefficients for some new, different vector.

[spare widget]

[v]_F = A [v]_E

[stoic python]

joesmith1042

[open locust]

Right, I see how that's the difference.

Like, for any transformation Av = w , w isn't equal to v in general.

But, I'm still not sure why this says that the change of basis transformation is not the identity transformation. I understand everything you said and I agree with it.

But I could make exactly the same argument about the identity transformation, and say it all in the same order, with the words "change of basis transformation" replaced with "identity transformation."

Like, I take the identity transformation, and do the things with to and from bases, and in the end write Av = v, etc.

Do you see what I'm saying?

[spare widget]

A change of basis matrix acts on the coordinate representations

It doesn't modify the vector itself

[open locust]

@spare widget Every matrix only acts on the coordinate representations.

[spare widget]

The maps from which this matrix is derived are ei = Ti(f1,...,fn)

[open locust]

Given a pair of to and from bases, there is an isomorphism between the set of linear transformations from U^m to V^n, and the set of linear transformations from F^m to F^n

[spare widget]

yes, but some you interpret as modifying the vector, and those you interpret as changing its basis

active vs passive transformation

[open locust]

So if you take a vector in U^m, you have its coordinates with respect to the to basis. You multiply the matrix representation of A by that, and you get the coordinates with respect to the from basis.

Well wait a minute, the transformations from U^m to V^m never change the scalar multiples themselves directly, the transformation always acts on the basis vectors

[spare widget]

[v]_F = A [v]_E is nit the same as [w]_E = A [v]_E

[open locust]

@spare widget I agree with that statement

[spare widget]

To be sure the coef of [v]_F and [w]_E are equal

But they are to be interpret wrt different bases

[open locust]

Is it normal I feel confused about this? I don't really get how to nail this down

[spare widget]

$\vec{v} = \sum_j [v]_{F, j} \vec{f}_j = \sum_j (A[v]_E)_j \vec{f}_j \ \vec{w} = \sum_j (A[v]_E)_j \vec{e}_j$

[stoic python]

criver

[spare widget]

The first is change of basis

The second is an active transformation

Note the bases

[open locust]

Why is w still in terms of the original basis?

[spare widget]

Because we didn't use A as a change of basis matrix

We instead used it as a matrix modifying v

To get w

[open locust]

But we don't have any choice of how A is "used." It just has to follow the rules.

[spare widget]

[w]_E = [v]_F

[open locust]

Like, (\forall j \in [m] \quad A(e_j) = \sum_{i=1}^n a_{ij} f_i)

[spare widget]

But those are two different vectors

[stoic python]

joesmith1042

[open locust]

That's how every linear transformation has to be defined, right?

[spare widget]

A(e_j) = \sum_i a_ij e_i

[open locust]

That's if you take the to and from bases to be the same.

[spare widget]

if you don't then you're implicitly putting in a change of basis

Then you mix the map wuth a change of basis

So you essentially compute PA instead of A

If you define your input and output to be wrt different bases then there is an implicit change of basis

If you define it wet the same basis, then this change of basis matrix is not there

[open locust]

This is the definition I'm working from.

[spare widget]

well it's because you have potentially different vector spaces

So you have a change of basis matrix there too

[open locust]

@spare widget Yes, I'm trying to bring my definitions in line with the ones you are stating systematically

[spare widget]

It's the identity if you take the same basis