linear-algebra

[thorn robin]

If you want to deduce T(0.5*a) = 0.5*T(a) then try looking at T(0.5*a) + T(0.5*a)

In general you pretty much already showed that T(na) = T(a + ... + a) = T(a) + ... + T(a) = nT(a) for n = 0, 1, 2, ...

You can get it for scalars of the form 1/n using the idea I gave you for 0.5

Then do something about negative numbers and you're pretty much done

[mild current]

I see thanks so much

And is there any book that covers the topic of Cauchy equations?

[thorn robin]

I would just look up stackexchange posts

It's a really common problem to discuss

[mild current]

Oh I see, Thanks for the heads up

[wintry steppe]

Would this one be false because 0 can't be a scalar we want

[thorn robin]

Yes you're right

[wintry steppe]

okay thanks

[nocturne oracle]

yw

[hollow finch]

Idk my text has it as a standard property of vector spaces. Not an axiom but in a separate theorem in the same section. I suppose it does depends on if it's supposed to be only using axioms only or is meant to be more straightforward

[wintry steppe]

[wintry steppe]

Question: ƒ is an endomorphism of Rˆ3 and fˆ2=0. Show that there exists a v in Rˆ3 and a g: Rˆ3 -> R so that for each x in Rˆ3, f(x)=g(x)v

for now i only managed to prove that rkf<=1 through the rank theorem and the fact that imf is in kerf

[dusky epoch]

rank(f) ≤ 1?

great

this means im(f) = span(v) for some fixed vector v in R^3

[wintry steppe]

if my reasoning is good, yes

[dusky epoch]

v may or may not be 0

[wintry steppe]

okay, so i could say Imf=ßv?

ß being a real number

[wintry steppe]

also why? i dont see it

[tall oriole]

in a lecture the prof said any vector in R^3 can be written as (a, b, c) = (a,b,0) + (0,0,c) which is an element of U+W in a unique way. how is that true? couldn't (a,b,c) be written as (a,0,0) + (0,b,c)? or is it saying if U=(a,b,0) and W=(0,0,c) then the only way to write U+W is exactly that way?

is this the correct way to interpret it: >is it saying if U=(a,b,0) and W=(0,0,c) then the only way to write U+W is exactly that way?

[wintry steppe]

yea thats it, its just a matter of how you define U and W

[tall oriole]

ah i see

[wintry steppe]

you could also say any vector (a,b,c) can be written as U+W+B with U=(a,0,0), W=(0,b,0) and B=(0,0,c)

[tall oriole]

and the only unique way to write (a,b,c) with your U,W,B is exactly U+W+B ?

[dusky epoch]

the dimension of im(f) is either one or zero

[wintry steppe]

its unique because theyre linearly independant meaning ßU+∂W+kB=0 iff ß=∂=k=0

[dusky epoch]

if it's one then it's the span of one nonzero vector

[tall oriole]

okay, thanks for the clarification

[wintry steppe]

aha, so can i say there exists a certain ß so that f(x)=ßv?

then i can just say that there exists a vector w i Rˆ3 so that g(w)=ß since g is from Rˆ3 to R

[dusky epoch]

are you using ß to mean beta

[wintry steppe]

yea

just a constant

[dusky epoch]

the fact that it's actually an eszett is making me die inside a little bit

[wintry steppe]

whats an eszett lol

[dusky epoch]

german letter

[wintry steppe]

oh shi thats not beta?

[dusky epoch]

it is not beta

[wintry steppe]

ohhh lmao

[dusky epoch]

this is beta: β

[wintry steppe]

do u have a special keyboard for that?

[dusky epoch]

i have autohotkey

[wintry steppe]

thats on phone or pc/mac

[lavish jewel]

ßeta

[wintry steppe]

yea thats the same as i have, but its german

[dusky epoch]

[wintry steppe]

happy birthday ann

[dusky epoch]

thank

[lavish jewel]

what what, it's today?

congratulations!

i hope you have a great day

[drowsy flower]

Hey I have a question again about projections

So I have a subspace U and W, where U is subset of W-perp. Now, lets say I am taking a vector v such that it does not belong to the subspace U + W, but belongs to V. Then obviously it must be the case that v belongs to W-perp but does not belong to U.
So then proj_W(v) = 0, because v is orthogonal to subspace W.
But what is proj_U(v) going to be? Clearly v cannot belong to U, and U does not have enough vectors to represent v..

Would it not have any projection, hence it would be 0 as well?

[leaden tide]

What is V?

[drowsy flower]

n-dimensional hermitian inner product space

[leaden tide]

okay

Now, lets say I am taking a vector v such that it does not belong to the subspace U + W, but belongs to V. Then obviously it must be the case that v belongs to W-perp but does not belong to U.
I believe this isn't true

Take V=ℝ³

[drowsy flower]

But V = W perp + W no?

[leaden tide]

Yeah, but that doesn't mean an element of V is either in W perp or in W

[drowsy flower]

ohh

[leaden tide]

v can be an element of U+W plus something else

[drowsy flower]

Okay so then if I take a vector v which is not in U + W, then what is proj_U(v) + proj_W(v) then?

it could be any real numbeer then?

[leaden tide]

First of all, don't get confused between a projector and something that gives you a coordinate

A projector has values in V

If you slap a fly onto a wall, you can still tell where it is in space

[drowsy flower]

yeah

[leaden tide]

If you take only the U part and the W part and sum them, since they're orthogonal you're just going to be left with the U+W part

if v = u+w+x where x is something else and the other belongs to the spaces you think they belong to

Then $p_U(v) = u$, $p_W(v) = w$ so $p_U(v)+p_W(v)=u+w = p_{U+W}(v)$

[stoic python]

Syst3ms

[leaden tide]

Note that this only works because U and W are orthogonal

[still lodge]

silly question but setting this up as a system of equations would be $x_1 + 2x_2 + 3x_3$ and so on right

[stoic python]

nitezba

[leaden tide]

nope

[still lodge]

bleh

[leaden tide]

write out x1w1

[still lodge]

in my head i was writing it as one big W matrix times x_1, x_2 and x_3 in a col matrix

[leaden tide]

Oh that is the case

You may be doing your matrix multiplication wrong then @still lodge

[still lodge]

actually i think i just made the big W matrix wrong in my head

idk how to make it in latex

[leaden tide]

yeah, the w's are the columns

[still lodge]

but first row would be 1 4 7 right

[leaden tide]

yup

[still lodge]

kk

aha so setting up the system would be x + 4y + 7z

i dont like the subscripts

but got it, merci

[leaden tide]

de rien

[still lodge]

so what might be a solution to this

wait nvm i can just use row operations

is it just 0,0,0

im probably overthinking this

[copper pasture]

hi, so in an machine learning course we discussed linear regression.

Now we then discussed the two cases above. I think the first one is, at least from a mathematical point of view, a bit weird. I think they did that since in reality you do have over determined systems all the time while in math you don't - because either you just have linear dependent rows (but still full rank) or you just can't solve it.

But anyway. So they go ahead and derive the closed form of those cases which I think is the same for both no? Just the normal-equation/moore-pensore pseudo inverse. Right?

So while we only have one solution in the d<n case we have infinite ones in the later case.

Now in both cases we still have the closed form and in the first case, the closed form is the only solution and thus also the solution to the corresponding minimization problem.

But what about the second case? If we have more than one solution, will the closed form be a minimizer?

[lavish jewel]

in the second case, you get the minimum norm solution

as you pointed out, it has infinitely many solutions

it picks the one with the min norm

[leaden tide]

Yes, the point of that exercise is to show that (0,0,0) is the only solution, i.e those vectors are linearly independent

[drowsy flower]

Sorry there are couple things that I don't get

[leaden tide]

go ahead

[drowsy flower]

First of all, isn't V spanned by basis vectors from W and W perp?

[leaden tide]

indeed

[drowsy flower]

okay

next, since U is subset of W perp, this means that not all vectors in V can be written as linear combination of vectors from U and W right

[leaden tide]

Not necessarily at least, a set is always included in itself

[drowsy flower]

yeah but except from that case

[leaden tide]

yeah

[drowsy flower]

So then when we try to project a vector which cannot even be written as linear combination of U and W, isn't it meaningless then?

[leaden tide]

Project onto what?

[drowsy flower]

U, W and U+W

we have a theorem that says proj_W(x) = v - perp_W(x) right

[leaden tide]

If the fly isn't on the wall, is it meaningful to talk about slapping it onto it?

[drowsy flower]

oh

wait but

Say the orthogonal basis for U is {b_1, ..., b_k}

The orthogonal basis for W is {c_1, ..., c_l}

[leaden tide]

So then k+l<n, right?

[drowsy flower]

yeah

aren't we missing some vectors?

is my question

because look

[leaden tide]

Yeah, but that doesn't really matter

[drowsy flower]

oh

[leaden tide]

Fortunately this situation can be drawn in three dimensions

Take W a line

And U a line orthogonal to W

No matter where your point is in space, you can project it orthogonally onto U or W

or U+W which is a plane

[drowsy flower]

hmm

my friend did something like: if v is not in U + W, then the projection is meaningless, hence proj_W(v) + proj_U(v) = 0 = proj_(U + W)(v)

and I was confused on why they thought this was correct....

[leaden tide]

Lemme draw this rq

Oh yeah this is nonsense

Let's take the same argument with one less space and see why it's stupid

If v is not in W, the projection is meaningless, hence proj_W(v)=0

[leaden tide]

v not being in the plane U+W doesn't mean that it's in the line orthogonal to that

[drowsy flower]

yeaha

this is honestly such an annoying question...........

The only way I can think of proving this is by extending basis........

[leaden tide]

anyway what's the question

[drowsy flower]

this is the original question

I did the forward direction

I had help from coycoy about the backward direction, but wanted to see if there was another way to do this without extending the basis for U to W-perp

[leaden tide]

Wait, I don't see why this is necessary

Okay, Imma need some translations for some french terms, sec

[drowsy flower]

sure, let me know if some notation is confusing as well, I can put it in words instead

Of course since intersection of U and W is {0} since U is subset of W-perp

[leaden tide]

How do you call it when the intersection of two spaces is {0}

Which guarantees that the decomposition is unique

[drowsy flower]

they are orthogonal no?

[leaden tide]

No

(1,1) and (0,1) (their spans anyway) satisfy this but aren't orthogonal

The term we use here is "are in direct sum"

[drowsy flower]

In our textbook, we have a theorem that says,
If W is a subspace of Inner product space V, then W-perp is a subspace of V also and W \intersection W-perp = {0}

[leaden tide]

Yeah, orthogonal spaces are in direct sum

Quite a simple corollary

Anyway, I'll use this terminology

[drowsy flower]

So ya what coycoy had told me was that since their intersection is trivial, then basis of U and basis of W are the basis of U+W

[leaden tide]

That is true

But I feel like there is no need to reason in terms of bases

[drowsy flower]

but they also mentioned that I might have to extend the basis of U

[leaden tide]

If $E = F+G$ and F and G are in direct sum we write $E = F \oplus G$

[drowsy flower]

ah yeah we were asked by prof to use basis for this question because we are going to study the direct sum term later

[stoic python]

Syst3ms

[leaden tide]

I see

Seems like a backwards way to approach it, but eh, whatever

[drowsy flower]

ikr

[leaden tide]

I mean, the strategy with direct sums gives you an outline for how to do it with bases

So i'll do that

Because I can't be bothered to have to deal with all the indexing fuckery

[drowsy flower]

lol

[leaden tide]

We set X the space orthogonal to U in $W^\perp$, such that $U \oplus X = W^\perp$

[stoic python]

Syst3ms

[leaden tide]

So then $V = W \oplus (U \oplus X) = W \oplus U \oplus X$, since addition is associative ; would you believe it

[stoic python]

Syst3ms

[leaden tide]

So then let $v\in V$. We can set $u\in U, w\in W, x\in X$ such that $v=u+w+x$.

[stoic python]

Syst3ms

[drowsy flower]

yeah

[leaden tide]

(these are unique because the spaces are in direct sum)

So then $p_U(v)+p_W(v)=u+w$ because these spaces are orthogonal to each other

[stoic python]

Syst3ms

[leaden tide]

But then, $p_{U+W}(v)=p_{U+W}((u+w)+x) = u + w$ with some bracketing

[stoic python]

Syst3ms

[leaden tide]

Which works because $U + W \perp X$, which isn't too hard to check

[stoic python]

Syst3ms

[leaden tide]

Good luck using bases to do that

Doable but annoying

But in a nutshell if e_1 and e_2 are basis vectors of two orthogonal spaces F and G then p_F(e_2)=p_G(e_1)=0

[drowsy flower]

Couldn't I do something like this?
First prove that X is also a subspace.
Then V = Span(U, W, X),
So then V = W + W-perp
V = W + U + X
V = (W + U) + X

And on the other side, V = U + W + X

So proj(U) + proj(W) + proj(X) = proj(U + W) + proj(X)
so proj(U) + proj(W) = proj(U + W)

if I could do this life would be so easy :/

[leaden tide]

Yep, that's my proof in a nutshell

[drowsy flower]

okay gonna try to write this now thanks a lot

ill come back if I get stuck

[dreamy iron]

hi all.

I'm trying to find other examples of this "dot diagram" that dr. peyam is using? https://www.youtube.com/watch?v=tnVkVJpn-BE&list=PLJb1qAQIrmmCs0fJDQnXgeuyFR8iQDwLV&index=2

on the horizontal he puts the basis vectors v_i \in V
on the vertical, he plugs in the kronerker delta of a linear functional f_i (v_i)

i'm googling "dot diagram linear functional"

and my hits are all like: do you mean dot product? do you mean graphic linear functions (high school level)

7:19

(i'm getting strong inclinations that i have to actually learn tikz and not just rip images from the net......and i don't really wanna learn tikz right now)

[visual hatch]

Can someone help me on this probelm

I'm supposed to find if this coefficant matrix has 1, many or no solutions

[nocturne jewel]

are you given the constant vector?

[visual hatch]

No, this is all im given

[nocturne jewel]

what was part a by chance?

[visual hatch]

I was given a augmented matrix and had to find amount of sols

[nocturne jewel]

can you post the entire question, a and b?

as well as a preamble if it's present

[eternal jasper]

is there a way to graph slope fields, but facing the same direction, with a regular function that could be plotted on soemthing like desmos

im really struggling with terms here so i cant find help anywhere else online

[north anvil]

I never found one

[north anvil]

Part a) This matrix includes the solution vector.

Part b) This matrix excludes the solution vector

Those are the differences between the coefficient matrix + and the augmented matrix

[visual hatch]

So is part a no solutions and part b many solutions?

[north anvil]

I think that's correct

[visual hatch]

Why is part b many solutions thouugh?

[north anvil]

How did you come up with the answer?

[visual hatch]

idk I just guessed on part b

[north anvil]

You row reduce and you get a row of 0s

[visual hatch]

Yeah bit its an coefficient matrix not an augmented matrix

[north anvil]

Regardless, there is a free variable I believe

[hollow finch]

row of zeros on a coefficient matrix means that IF there is a solution then there will be INFINITELY many of them

because you will have a free variable

compare these two

$\left[\begin{array}{cc|c}1&1&1\0&0&1\end{array}\right]$

$\left[\begin{array}{cc|c}1&1&1\0&0&0\end{array}\right]$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

both have a coefficient matrix with a row of zeros but the first one is inconsistent. the second is consistent and because of the row of zeros has infinitely many solutions.

[jagged gulch]

@north anvil how is this augmented, or is the last column supposed to be the constant vector

[hollow finch]

when a matrix is said to be augmented the last column is assumed to be the RHS vector. drawing a bar is optional

[jagged gulch]

not drawing the bar is an act of heresy

[north anvil]

One is, one isn't

[hollow finch]

i dont disagree with you. i always draw the bar

[sturdy portal]

Why should the production vector for a given demand vector be unique?

Or equivalently, why is it the definition of a productive consumption matrix?

[hollow finch]

one thing i can say is that if (I-C) is invertible then it doesnt have an eigenvalue of 1, and since its invertible the solution to that system will be unique

but idk enough about economics to say more than that

[sturdy portal]

@hollow finch Do you anyone whom I can ask or where I should ask?

[hollow finch]

maybe someone else here knows more about it

[visual hatch]

Can I get help on part c?

I think part b is right but idk

[hollow garnet]

z can be written as linear combination of b1 and b2

[visual hatch]

So do I just multiply the vector [2 -2] by [8/7 -5/7]?

[hollow garnet]

no

what is the definition of coordinate vector?

[visual hatch]

Its like an ordered list of numbers

[hollow garnet]

If $B = \qty{b_1, \cdots, b_n}$ is the basis for some vector space $\mathbb{V}$, then every vector in $\mathbb{V}$ can be written as linear combination of basis vectors. So the coordinate vector $v$ relative to basis $B$ is given by $[v]_B = \bmqty{c_1\ \vdots \ c_n}$ since $v = c_1 b_1 + \cdots c_n b_n$

[stoic python]

zslya

[visual hatch]

So I just write it as 2[8/7] - 2[-5/7?]

[hollow garnet]

read what I wrote again and think about what 8/7 and -5/7 represents and how they are two different things

[visual hatch]

I still don't understand

[hollow garnet]

is 8/7 and -5/7 your basis vectors?

[visual hatch]

yea

[hollow garnet]

uh

Read the question for b) again

In your question the basis is B

[visual hatch]

ohh yea

[hollow garnet]

so then what is v?

[visual hatch]

vector space

[hollow garnet]

sorry I meant what is z=

[visual hatch]

Coordinate vector

[hollow garnet]

No! I am asking you what the answer to c) is then

Put the pieces together now....

[visual hatch]

I'm still lost

[hollow garnet]

Fine

v = 2 b_1 - 2 b_2

that's it

you already know what b_1 and b_2 are

plug that in...

[visual hatch]

so its 2*[2 1] -2[-1 3]?

[hollow garnet]

Yes

because b_1 and b_2 are basis vectors

and every vectors in that space can be represented as linear combination of those two vectors

[visual hatch]

ohh okay

also is my part b correct?

I just put them in a augmented matrix and solved for x1 and x2

[hollow garnet]

yes

If you are unsure, you can always check your answer

simply evaluate $\frac{8}{7} \cdot \bmqty{2\1} - \frac{5}{7} \cdot \bmqty{-1\3}$ and you should get back $\bmqty{3\-1}$

[stoic python]

zslya

[visual hatch]

okay thank you

[hollow garnet]

np

[sturdy portal]

https://math.stackexchange.com/questions/4193896/why-the-definition-of-productive-economy-in-leontief-open-model-is-such

[wraith patio]

i get the general idea for how id do part b, but that was also what i was going to do for part a. how would i start the proof for part a?
also they dont tell us that dim(V)=dim(W), isnt that necessary for part c?

[frosty vapor]

you can do A by contradiction

in both directions

kind of a cheap solution but i dont remember how to do it nicely

B is also easier with contradiction, use the definition of linearly independent/dependent along with T being linear

as for c since T is one to one and onto its bijective

so the spaces are isomorphic/same dimension

[twilit anvil]

for A, i would start with
c_1Tv_1 + ... + c_nTv_n = T(c_1v_1 + ... + c_nv_n), and then note that the LHS is 0 if and only if the RHS is also, to make a point about the possible values for the c_i. i think that will be a short proof, even after including the onto part... for the converse, i think you can just reword the proof by contradiction for a direct proof.

sorry for the slants, not sure how that even happened

[sturdy portal]

https://math.stackexchange.com/questions/4193896/why-the-definition-of-productive-economy-in-leontief-open-model-is-such

[visual hatch]

Could I get help on this problem pls?

[dusky epoch]

do you know what an orthogonal set is?

[visual hatch]

its when all the vectors dot product is not 0

[dusky epoch]

that's vague wording and incorrect

we say that a set of vectors is orthogonal if every pair of vectors in it is orthogonal

and two vectors are orthogonal when their dot product is 0

[visual hatch]

ohh yes, my bad

[dusky epoch]

so for {x, y, x×y} to be orthogonal, we need the following:

1. x and y are orthogonal
2. x and x×y are orthogonal
3. y and x×y are orthogonal

are all three of these always true?

[visual hatch]

no, they are only true if the dot product of x and y or x and x * x or y and x * x are 0

[dusky epoch]

wording!!!

2 and 3 are actually true always. a cross product is always orthogonal to both its inputs

it's 1 which is not at all guaranteed true.

[visual hatch]

How do we know a cross product is always orthogonal to both its inputs?

do we have a theorm or rule that tells us that?

[dusky epoch]

what's your definition of cross product?

[visual hatch]

a vector perpendicular to 2 given vecctors

[dusky epoch]

"x × y is a vector perpendicular to x and y"

this is not enough

but even this makes the answer to your question of:

How do we know a cross product is always orthogonal to both its inputs?
downright tautological

[visual hatch]

okay, thank you

And I saw it was your birthday today, Happy Birthday

[dusky epoch]

it was yesterday but thank you

[wraith patio]

whats the proper way to denote a basis? like if i find a basis for R(T) is {v,w} i dont say R(T)={v,w} right? would i do R(T)=span({v,w})?

[lavish jewel]

you can also just say {v,w} is a basis for (some subspace)

[wintry steppe]

just say it in words

[dusky epoch]

but words bad /j

[mild current]

what is the geometry behind transpose of a matrix?

Say, (1; 2) be matrix. Applying transpose gives
(1 2) .But what the hell does it mean geometrically?

And why is it used in inner product?

[lavish jewel]

to be formal, the transpose deals with the dual spaces of the vector spaces that the original matrix works with

in this sense, transposed vectors are elements of the dual space of the vector space V from which the original vectors come

i would say it's difficult to think of the transpose geometrically because in general, it works on a completely separate vector space from the one you start in

it's not like you could even plot transposed vectors with regular vectors because of that

[old swift]

axler readers in triumph rn

traces in chapter 10 😎

[mild current]

Then what is the difference between row and coloumn vectors (geometrically)

[scenic fulcrum]

Stop me if I make mistakes

Let's suppose you have F and G = orthogonal(F).
A is the matrix of a basis of F so each vector of F is written in the form x1.a1+...xp.ap where ai are columns of A.

Then A^T is the equation of G, i.e the set of vectors X which check the equation A^T X = 0 is exactly G.

The transposition transforms a view "F = Vect(something)" in G={X such as f(X)=0} (dual view). Two manners to explicit a vector subspace

[lavish jewel]

they are in different spaces

i don't know if thinking about it geometrically is useful at all tbh

but let's see if anyone else can give some other insight

but it's in my opinion more useful to think of transposed vectors as linear forms/functionals that act on vectors

not as vectors, but rather as some type of function that acts on vectors

[leaden tide]

As for why it's used in the inner product ; row-vector multiplication gives us the result we want

[lavish jewel]

right, you can take the inner product of two vectors by projecting one onto the other. one can do this by turning one of the two vectors into a function that acts on the other

something like that

[leaden tide]

That could be one way to look at it

[mild current]

Ok, lets I have non orthogonal coordinate system. There is Vector V. So, in a non orthogonal coordinate space, the vector V can be projected on the y axis in two ways and same for the x axis. Is this what dual space means?

[leaden tide]

Careful, projection != coordinates

[dire thunder]

two projections?

[leaden tide]

There are infinitely many ways to project

[dire thunder]

[leaden tide]

Sometimes the orthogonal projection is favored

But that's still an arbitrary choice, and not the coordinates in a non-orthogonal basis anyway

[scenic fulcrum]

When you take a basis and write matrix in this base , everything is done as if it was an euclidian space

[mild current]

This is what I am taling abt

[scenic fulcrum]

,rotate

[leaden tide]

i honestly do not know what this is, but it doesn't correspond to my understanding of dual vectors

[mild current]

have a look at this

Its more appropiate on what I am trying to say

on what I mean by dual basis vectors

[lavish jewel]

https://en.wikipedia.org/wiki/Dual_space

that's what i was talking about

[leaden tide]

While we're at it, quick intuition question about dual vectors

So I know that there isn't a privileged isomorphism between V and V*, and that it depends on a choice of basis

But if we take basis $B = {e_1,e_2}$, then we can have its dual basis $B^* = {e_1^,e_2^}$ which respectively give the coordinates along $e_1$ and $e_2$ (easy enough to interpret geometrically). Then we have $(e_1+e_2)^=e_1^+e_2^$. Is there a nice geometric interpretation for $(e_1+e_2)^$ ?

[stoic python]

Syst3ms

[jagged granite]

I wrote solution to this problem in detail. Can someone take a look and tell me if its ok?

If $A^k=0$ for $k\leq n$ then $A^n=0$, a contradiction. Suppose that $A^k=0$ for $k>n$ then the minimal polynomial of $A$ divides $x^k$. Thus minimal polynomial can only be $x^m$ for $m\leq n$. Which again implies $A^n=0$, a contradiction.

[stoic python]

bert

[wintry steppe]

@jagged granite looks good to me

nice proof

[visual hatch]

For part c of this problem, would the vector [x1 x2] just be swapped to [x2 x1]

Part c is the second one

[twilit anvil]

maybe you can add a bit more to describe what the transformation does graphically?

in any case that is precisely what it does

[visual hatch]

I'm confused what it does graphically

I noticed that it just swaps the x1 and x2 varibales around

[twilit anvil]

yeah i think that should be fine

[lavish jewel]

they don't ask you do describe it graphically 😛

but if you like, it swaps the x with the y axis

[visual hatch]

ohh okay thanks

[lavish jewel]

or in another sense

ask yourself which vectors don't change

looks like a reflection along y = x

[visual hatch]

Could I also get help on this problem?

[lavish jewel]

this sounds like something that should be in your course notes

at any rate

[visual hatch]

Does this just mean that the vector is not writen in standard coordinates but in b coordinates

[lavish jewel]

in the basis B, x = d b_1 + e b_2

and so the coordinate vector for x relative to B is the vector [d, e]

[visual hatch]

But its asking for the definition of the coordinate vector right?

[lavish jewel]

yeah, which is what i wrote

[visual hatch]

what is d and e though?

scalars?

[lavish jewel]

yes

x is a linear combination of the vectors in the basis

[visual hatch]

would I have "he vector is not writen in standard coordinates but in b coordinates" also be okay?

[lavish jewel]

that's incomplete

or rather, circular

they're asking you what it means to be written in b coordinates

and you're saying it means to be written in b coordinates 😛

[visual hatch]

Ohh okay

So i would just say that "x is a linear combination of the vectors in the basis" and "in the basis B, x = d b_1 + e b_2"?

[lavish jewel]

those two things are the same

[visual hatch]

oh okay, so I would only need one

[lavish jewel]

you still havent said what the vector is

[visual hatch]

then I would also need this "coordinate vector for x relative to B is the vector [d, e]"

[lavish jewel]

yes

i would suggest you review the definitions given in your course

[visual hatch]

okay thanks

[tropic rune]

would this be the right place to ask a question about relative entropy calculations?

[lavish jewel]

try the statistics channels

[tropic rune]

hmm okay, my issue just might be stemming from a lack of intuition about tensors

[lavish jewel]

well, when you say entropy, one immediately thinks about probability distributions, not tensors

i guess you can try here and we see what's up

[tropic rune]

the probability distribution is encoded in a tensor 😓 i'm super confused about it

okay let me try to explain

i guess my initial issue is in itself a statistics issue, because i'm working with two different descriptions of relative entropy. the first one is described as follows

so this would be for individual arrays

but the other description I'm working with doesn't have an explicit definition but rather says that if X = Y, then rel_entr(X,Y) = 0

here I'm assuming that X and Y would be referring to the probability distribution, i.e. the tensor in this case

and I'm not sure if this rel_entr(X,Y) = 0 is a result of the formula itself or if there's something else I need to consider

let me know if my question doesn't make sense ._.

[lavish jewel]

in what you wrote below, it's from the definition of relative entropy

you have this log term of the ratio of two distributions

if they are identical, you get log(1)

[tropic rune]

ah okay, so it doesn't even matter how complex the distribution is?
because like, if this definition is for arrays, then would i be computing the pairwise entropies for each row in the tensors i'm comparing?

then i would wind up with a 1D vector that i'm not sure what to do with -- sum it?

[lavish jewel]

that depends on what the elements of the array mean

do you have like many realizations of the same random process?

or each individual entry is a random variable unrelated to the others, with a different underlying distribution?

[tropic rune]

I guess each tensor would be a random variable, where each row represents a data point and each element of the row vector represents a probability of occurrence

[lavish jewel]

hmm that doesn't sound right

[tropic rune]

this might be easier if I describe it in terms of what I'm actually trying to do lol

[lavish jewel]

btw this is purely statistics, representing it as a multiway array is an arbitrary choice 😛 just for future reference

but yeah, give more info

[tropic rune]

oof alright, sorry about that lol, definitely will keep in mind for the future

[north anvil]

If I am given a subspace H, and that one particular basis for H has n vectors, does that mean all bases for H also have n vectors?

[wintry steppe]

yes

[north anvil]

Oh wow - really quick response! Appreciate that 👍

[marble lance]

This is true for vector spaces in general, the fact that H is a subspace of another space is irrelevant

[wintry steppe]

you can prove this using the replacement lemma, i think. it should be in any linear algebra book

[north anvil]

As per subspaces + vector spaces:

Not all vector spaces are subspaces, but all subspaces are vector spaces?

[wintry steppe]

any vector space is a subspsace of itself

[marble lance]

Well, all vector spaces are subspaces of themselves

[north anvil]

Or does it depend on the "outer" vector space

[wintry steppe]

sniped haha

[north anvil]

Oh okay

I shoulda known that

[marble lance]

But you don't talk about a subspace unless you want to directly reference the larger space

[marble lance]

Rood

[north anvil]

Oh aight then

Just got a final coming up and I'm asking some questions about the stuff 👀

"final"

Appreciate you guys 👍

[wintry steppe]

https://en.wikipedia.org/wiki/Steinitz_exchange_lemma

see the first sentence

[north anvil]

Clarification, given a matrix A, with dimensions m x n, let B represent a basis for col A. Is the maximum possible number of vectors in B equal to n?

Given that col A is the span of n vectors (n columns of A)

And the basis may or may not include all of them or only some of the columns of A?

[wintry steppe]

correct

[north anvil]

Shit nice - my reasoning works

[wintry steppe]

ah wait hold on one second

i might have misread

[marble lance]

The maximum is min(m,n). If m < n, then it's impossible for more than m of the columns to be linearly independent.

[wintry steppe]

thx luna

[marble lance]

np sniper

[north anvil]

min(m, n) = m if m < n,
min(m, n) = n if m > n
min(m, n) = m = n if m = n?

[marble lance]

Yes

[north anvil]

I can see that - "wide" matrix vs "tall" matrix

Ty

🙂

[marble lance]

👍

[wintry steppe]

i like my matrices like i like my men

[marble lance]

[north anvil]

😳

[north anvil]

Really big and difficult to work with?

[wintry steppe]

damn okay, call me out

[marble lance]

@mods

[north anvil]

😳

[wintry steppe]

@​Moderators.

[north anvil]

Shit, just remembered another thing. If a set of vectors is linearly independent, then of all possible linear combinations of any number of vectors in that set, none are present in said set.

Except for where the linear combination of a set of vectors has scalars of a single 1 and the rest 0?

Idk if this question makes sense

Um

Damn it

How do I phrase this?

I guess I'll try with an example

So the set (let's call it F) consisting of the 2 vectors

<1, 2, 3> and <2, 4, 6>

is linearly dependent because you can form one of the vectors in F from the other vectors in

Oh I think I get it

[wintry steppe]

[north anvil]

I guess

Um

[marble lance]

This makes no sense tho.

[north anvil]

That's what I was saying

[marble lance]

Oh

The except part kinda fixes it

Yeah, then it's fine ig

[north anvil]

Is the following true:

A set of vectors (S) is linearly dependent if there is at least 1 vector (v_k) that happens to be a linear combination of the other vectors in S?

[marble lance]

Yes

[north anvil]

There we go

I thought you could use the original vector in the linear combination

But it makes sense why not

Tyty

[marble lance]

👍

[north anvil]

Can a transformation be both onto and one-to-one?

Both a (pivot in every row) and (pivot in every column) can happen for a matrix

[marble lance]

Yes

[north anvil]

Thank you

[marble lance]

These are called isomorphisms

[north anvil]

A matrix is an isomorphism if there's a pivot in every row and every column?

or

A matrix is isomorphic if there's a pivot in every row and every column?

You're gonna get me through this lin alg test 😅

[marble lance]

A linear transformation is an isomorphism iff it is onto and one to one

The corresponding matrix is invertible

[north anvil]

Right

If a matrix satisfies at least one of the two criteria below, then it has an inverse:

1. There is a pivot in every row.

2. There is a pivot in every column.

I think

I'm pretty sure

[marble lance]

If your matrix is square

Because then those are equivalent statements

[north anvil]

Oh yeah, if a matrix is invertible, then it is square

[marble lance]

Yeah, but a pivot in every row is not enough

[north anvil]

That's right

Oh yeahh

good point

You need both

[marble lance]

You need square, a pivot in every row, a pivot in every column. Any two of the three.

And for a linear transformation (between finite dimensional spaces) you need two of three: the dimensions of the domain and codomain are equal, one to one, onto

[north anvil]

If a matrix has a pivot in every row and a pivot in every column, then it is square

[marble lance]

Yes and invertible

[north anvil]

Yup

[marble lance]

So those are the corresponding statements between matrices and the linear transformations they represent

[north anvil]

Yeah

[marble lance]

Ima sleep now, so hopefully someone else will answer anymore questions if you have

[north anvil]

If a transformation is linear, then there must be a matrix A such that T(x) = Ax

x is a vector

[north anvil]

gn - really appreciate the help

❤️

[marble lance]

If you are working between finite dimensional spaces

[north anvil]

Yeah - haven't worked with spaces that have infinite dimensions

Idk if anyone can even visualize 4 spatial dimensions

[marble lance]

Sure, but you should specify. Because there are transformations between infinite dimensional spaces, and you should be clear you are not talking about those

[north anvil]

👍

[wintry steppe]

hi all can someone pls verify this hw problem

also i feel like this is no bc the vector i got when i multiplied them is odd but

not sure

[restive raft]

looks like it is with eigenvalue of 4

[wintry steppe]

wym?

for which one?

[wintry steppe]

oh this one?

i dont think so

i think w 4 it gives -1 -1 1

[wintry steppe]

is that the same thing?

[restive raft]

I was looking at the bottom image there

[wintry steppe]

yeah

i think w 4 it gives u -1 -1 1

is that the same thing?

i assumed not

[restive raft]

if you matrix multiply the vector you get (4,4,-4) which is (1,1,-1) scaled by 4

[wintry steppe]

so 4 is an eigenvector?

value*

[restive raft]

yeh it would be

[wintry steppe]

hmm strange

bc

symbolab also only shows

-1 -1 1

also i matrix multiplied them

and got like

-5 6 21

omfg

i entered the wrong

value into my 3 x 1 matrix

tysm

[lethal wave]

Hi, i was struggling with the (b) part of this exercise, it says:
(b) Prove that there is a real matrix C such that C^2=A
Thanks in advance 😄

[wintry steppe]

perhaps you have to use part (a). what does that one say?

[lethal wave]

It asks for a complex matrix that satisfies A=B^2 and its minimal polynomial of degree 4

the eigenvalues of A are -1 with double multiplicity, 2 and 3, so a diagonal matrix with its diagonal entries being i, i, sqt(2), sqrt(3) satisfies the equation

[red prawn]

If you consider the 2x2 diagonal matrix diag(i,-i), this is conjugate to a (real) rotation matrix. So if you negate one of the occurences of i in your matrix...

[wintry steppe]

Does S' have the same span of S because the coefficients of each u_1 , u_2 , and u_4 can take any value (since we can just take a_3 = 0 and make a1 , a2 , a4 anything else)?

[marble lance]

Well, that is why span(S') is contained in span(S)

The reason why span(S) is contained in span(S') is what they explain there

[hollow finch]

so you clearly found that we just need to find a square root of the diagonal matrix A is similar to (with diagonal entriez -1,-1,2,3).
In part a you used the obvious square root which is just square rooting each diagonal entry.
Part b is essentially asking to show there is a real square root of it.
Treating it as a block diagonal matrix, you just need to find a real square root of the matrix
[-1 0;0 -1] for which at least one does exist.
Hint 1: ||the real matrix [a -b; b a] functions just like the complex number a+bi.||
Hint 2: ||So then [-1 0;0 -1] is like the number -1.||
Hint 3: ||So what are the matrices corresponding to the square roots of -1?||

[keen flame]

What does this mean in terms of matrices?

[quartz compass]

GL means general linear group, so it's a group of invertible matrices

the GL_3 means they're 3x3 matrices

the Z_3 means the entries of the matrices are from the field with 3 elements

the group in particular has the binary operation of matrix multiplication, not addition, just to clarify

[keen flame]

Really? that's it?

[quartz compass]

yep

[keen flame]

GL just means matrices

?

[quartz compass]

no

there are 3x3 matrices with entries in Z_3 that aren't in GL_3(Z_3)

[keen flame]

hmm. Like?

[quartz compass]

do you know what a group is?

[keen flame]

yes

It's a group which has a function that takes two elements that keep the ouptut of the function within that group, has a identity elements and to each element has an inverse element that results in the identity element

[quartz compass]

yeah good

so try to think of a matrix that doesn't satisfy some condition you've outlined here

[keen flame]

/
[2 0 0]
[0 0 0]
[0 0 0]

multiply

[2 0 0]
[0 0 0]
[0 0 0]

result

[4 0 0]
[0 0 0]
[0 0 0]

[quartz compass]

ok

so are these matrices in the group or not in the group?

[keen flame]

matrices group of z3

z_3

[quartz compass]

uh that was a yes/no question

what's the identity of the group?

[keen flame]

1 0 0
0 1 0
0 0 1

[quartz compass]

ok what's the inverse of
[2 0 0]
[0 0 0]
[0 0 0]

[keen flame]

1/2 0 0
0 0 0
0 0 0

Doesn't make sense

[quartz compass]

why doesn't it make sense

[keen flame]

z_3 suppose to be intergers of 0-2

[quartz compass]

1/2 = 2 mod 3

[keen flame]

ohh

I see

[quartz compass]

not yet you don't

[keen flame]

ok

[quartz compass]

this is still not the inverse

when you multiply them together what matrix do you get?

[keen flame]

1 0 0
0 0 0
0 0 0

Which isn't the identity

[quartz compass]

exactly

so that matrix has no inverse

an easy way to check is if the determinant is 0

just be sure you look at the determinant mod 3 since the determinant is an element of Z_3 as well too

[keen flame]

ok

thanks

[quartz compass]

yeah you're welcome

also you can always pick the 0 matrix as well, that's an easy example too

[keen flame]

What's a normal subgroup? I'm looking for a normal subgroup that isn't trivial.

[quartz compass]

normal subgroups are exactly the groups that are the kernels of homomorphisms

since this is linear algebra, the determinant is a pretty nice homomorphism to think about

[keen flame]

What do you mean as kernels of homomophism?

[quartz compass]

this is more of an #groups-rings-fields question than linear algebra

you should try to read about and prove several things about group homomorphisms first if you don't know what a kernel or homomorphism is

a subgroup of a group is normal if and only if it's the kernel of a homomorphism, pretty important to know that

[keen flame]

I know what kernel means

found what homomorphism means f(x)f(y) = f(xy)

[lethal wave]

oh ok ill try that out, thanks!

[twilit anvil]

hi, i would like some hints on this problem:

suppose M is an isometry of R^n (or C^n), mapping 0 to 0. prove that M is linear.

i am able to see how M is linear in addition, with some work with the inner product, but not scalar multiplication.
how should I approach M(sx) = sM(x)? I tried looking at || M(sx) - sM(x) || = 0, but was not able to show it

[wintry steppe]

how did you prove it's linear with respect to addition?

i feel like the same thing should work

[twilit anvil]

...........................im an idiot

(yes the same thing works)

[wintry steppe]

it happens

[twilit anvil]

here is a cute exercise that came from this: suppose ||Tz - Tx - Ty|| = ||z - x - y|| for any choice of z,x,y. prove that T is linear. maybe good in a first course?

[wintry steppe]

[hollow finch]

Is there any need for proof of the fact that for T: V->W
rank(T)≤dim(W)

[twilit anvil]

i dont think so, because rewriting it just gives dim Im(T) <= dim(W)

it is extremely self evident from the definition

[hollow finch]

what are you struggling with

[fallen scaffold]

dont even know where to start

[hollow finch]

maybe part a? :p

[fallen scaffold]

yeah

i need to show tp is a linear map

idk how

[hollow finch]

check your notes or text on the definition. its probably in a giant colored box or something bc its pretty important

there are 2 requirements

[fallen scaffold]

is it that linear maps need to preserve operations of addition and scalar multiplication

[hollow finch]

absolutely

homogeneity and additivity

[fallen scaffold]

yeah

[hollow finch]

looks like thats what you need to show

[fallen scaffold]

alright

so how should i start by showing that

I have p3 and p2 sets

[hollow finch]

remember when you were learning about subspaces and you had to show that they were closed under addition and scalar muliplication?

these kind of proofs are very similar

[fallen scaffold]

i wasnt sure how to do those proofs

[hollow finch]

Homogeneity: $T(ku)=kT(u)$

Addititivty: $T(u+v)=T(u)+T(v)$

[stoic python]

nix (@ me for the love of euler)

[fallen scaffold]

able to find this in my book

[hollow finch]

alright well come up with some general vectors in the domain u and v, and verify those properties hold

general vectors means that they look like a0+a1x+a2x^2 with variable a0,a1,a2. you cant do these proofs by verifying it works for some particular vectors like 1+x^2 or something

[hollow finch]

i would review those proofs because they are pretty important/fundamental. maybe ask your prof

[fallen scaffold]

so can i make p2 and p3 into general vectors

[hollow finch]

P2 and P3 are vector spaces but yeah they give you the general form of the vectors in those spaces in the problem

[fallen scaffold]

so like this?

[fallen scaffold]

what next

[hollow finch]

im assuming they want you to keep it in P2 and P3. so still as polynomials like they give you at the beginning of the problem

and i said two general vectors in the domain which is P2. P3 is the codomain

[hollow finch]

youre going to be doing stuff more like this in part b so i wouldnt erase it or anything

actually yeah i wouldnt definitely keep it

[fallen scaffold]

keep it as a polynomial right?

so i dont have to put it in a matrix

[hollow finch]

yeah https://puu.sh/HVnlM/e27d04dc86.png

[hollow finch]

not until part b

[fallen scaffold]

ok ill write that matrix in part b

[hollow finch]

you can come up with the vectors however you want btw theres no strict standard. some possibilities are

\begin{align}
u=a_0+a_1x+a_2x^2,&& v=b_0+b_1x+b_2x^2\
u=a+bx+cx^2,&&v=d+ex+fx^2\
u=a_{10}+a_{11}x+a_{12}x^2,&&v=a_{20}+a_{21}x+a_{22}x^2\
u=u_0+u_1x+u_2x^2,&& v=v_0+v_1x+v_2x^2
\end{align}

i like the first one personally

[fallen scaffold]

does it have to be b

oh wait nvmd

so i came up with a u

[fallen scaffold]

so should the v be the one from P3

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

again i said two general vectors in the domain which is P2. P3 is the codomain

[hollow finch]

youre going to do it pretty much just like this

[hollow finch]

somewhere on this page they should be telling you that u and v are in the domain (likely V or something)

[fallen scaffold]

oh ok so they can be any variables

[hollow finch]

yeah they can be just about anything as long as theyre all different (so you have 6 total in this case). the examples i gave are some of the most readable and common ways to do it though

[fallen scaffold]

This is fine right

[hollow finch]

love it

thats great

[fallen scaffold]

ok next step?

[hollow finch]

verify the properties

[fallen scaffold]

additivity and homogeneity

[hollow finch]

yep

so basically do the left side and show that you get the right side

like start with L(u+v) and show thats equal to L(u)+L(v)

or in the case of your problem

$T_p(u+v)=T_p(u)+T_p(v)$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

youll be plugging in u and v for q(x) since they define it as Tp(q(x))=p(x)q(x)

so Tp(u)=p(x)u

[fallen scaffold]

got this for additivity

[hollow finch]

well you wrote down what you want to show but you didnt show it

[fallen scaffold]

i have to expand it through

[hollow finch]

yeah. youre given an explicit definition of Tp so you need to use it

[fallen scaffold]

I expanded it

[fallen scaffold]

so how would I use it to prove additivity and homogeneity

[hollow finch]

again youre showing things you want to prove without proving it. Tp(q(x))=p(x)q(x) and i dont see p(x) anywhere in your proof

[fallen scaffold]

so i didnt need to expand this all the way right

[hollow finch]

no in fact you cant since you havent proven that you can

[fallen scaffold]

ok im erasing it then

[fallen scaffold]

so u+v = q(x)

[hollow finch]

yes

for Tp(u+v)

[nocturne jewel]

are you still showing T_p is linear?

[fallen scaffold]

yeah that its a linear map

[nocturne jewel]

Ok so for the sum property, you take 2 elements of the input space (P_3) and show that you can split the operation by the + sign

[nocturne jewel]

If $q,r\in P_2$, then $q+r\in P_2$ and $T_p[q+r]=p(x)[q(x)+r(x)]$ by definition of the operation

[jagged gulch]

For the love of Euler

[hollow finch]

you cant quite simplify the LHS on the second line quite yet. youre still trying to prove you can do that. you know what u and v are, so you need to plug them in

[nocturne jewel]

rest is straightfoward, similar for scaling

[hollow finch]

$T_p$ is $P_2\to P_3$

[stoic python]

nix (@ me for the love of euler)

[nocturne jewel]

read the wrong subscript

[stoic python]

Mosh

[fallen scaffold]

is this ok

[hollow finch]

thats great

keep going

[fallen scaffold]

so i have to keep expanding?

[nocturne jewel]

You dont need to actually write out the polynomials long form

[fallen scaffold]

really?

so what should I do

[nocturne jewel]

$T_p[q+r]=p(x)[q(x)+r(x)]$

[stoic python]

Mosh

[nocturne jewel]

how would you expand that out?

[fallen scaffold]

p(x) q(x) + p(x) r(x)

[nocturne jewel]

yes, which is..?

[fallen scaffold]

uhh

the additivity property?

[nocturne jewel]

$=p(x)q(x)+p(x)r(x) \ = T_p(q)+T_p(r)$

[stoic python]

Mosh

[hollow finch]

yes

[nocturne jewel]

since p(x) times a polynomial from P_2 is just T_p applied to that polynomial

[fallen scaffold]

but ill just get something really long

on the right side

[hollow finch]

well you dont have to expand everything. you know where you want to get to

Tp(u)+Tp(v)

you can get there in one step

[hollow finch]

from what you have here

[fallen scaffold]

so should i expand the left or right side

@hollow finch

[hollow finch]

ive told you everything you need to do and so i can only ascertain that what youre missing is conceptual understanding of what youre even trying to do in the problem.
at this point i would try rereading your notes and/or the section in the text, looking at some examples of similar problems (there should be at least a couple in the section) and understanding what theyre doing in each step, maybe taking a break, and then trying the problem again once you understand what is being asked of you.

[ruby basin]

I posted in the wrong channel oops

[fallen scaffold]

ok so im trying to get Tp(u) + Tp(v)

[ruby basin]

#competition-math

[fallen scaffold]

i dont know how to get it to that part

i know what im trying to get

[nocturne jewel]

why are you going backwards.. you had it right?

[fallen scaffold]

on the right hand side

[hollow finch]

yeah thats why i told you what i think you should do in my last message

[fallen scaffold]

mosh you were telling me i dont need to expand it

[hollow finch]

theres not much use proving something if you dont know what youre proving or why

[nocturne jewel]

well you do have to expand, you just dont need to write out an explicit version of the input vector

since you'd end up just needing to factor it all again to get the linearity obvious

More a matter of what I said would just save writing

[wraith patio]

is there any need for proof of the statement that for a transformation T: V->W,
rank(T)<=dim(W)

if so how would you even do that it? seems like almost a matter of definition. would it just be 1 line or something?

[wintry steppe]

do you know that if you have a vector space A of finite dimension and a subspace B then dim B <= dim A

[wraith patio]

yeah thats from the definition of a subset

since a subspace is just a subset

[wintry steppe]

it's a little more than that

a lot more, in fact

but the fact you want follows from that

[wraith patio]

hm then how would you articulate it? it just seems too self explanatory to me

[wintry steppe]

im T is a subspace of W and W is finite dimensional (i hope, else what you've written is nonsense) so rank T = dim im T <= dim W

[wraith patio]

well right of course. im getting caught up in the exact explanation/justification that that dim im T <= dim W. or that

if you have a vector space A of finite dimension and a subspace B then dim B <= dim A
as you said

[wintry steppe]

B cannot contain more than dim A linearly independent vectors

[fallen scaffold]

ok im expanding it

[wraith patio]

okay yeah that makes sense. i think im still just unsure of if that requires proof/how you would prove that. idk if the proof would be really long or really short either

[fallen scaffold]

im doing the (3x+2)(input vector)

[nocturne jewel]

you also dont need to write out p(x) explicitly...

[fallen scaffold]

yeah but im doing it to show the work

[nocturne jewel]

You can show the work by writing p(x)

[wintry steppe]

if B contains dim A + 1 linearly independent vectors, then so does A. but then dim A >= dim A + 1 since dim A is the maximum number of linearly independent vectors in A, contradiction

[wraith patio]

ahhh i see. thats a pretty sweet proof. thank you tterra ❤️ ❤️ ❤️

[nocturne jewel]

Also you can do the entire linearity test w/ a linear combination input and showing it's equivalent to a linear combination of mapped vectors w/ same scalars

[fallen scaffold]

this is how i expanded

[nocturne jewel]

yeah ugly

but do it the more writing way if you want

[fallen scaffold]

ok

now whats the next step

i expanded

[nocturne jewel]

simplify and show it equals T_p[q]+T_p[r]

[fallen scaffold]

i need to factor out this entire thing

[nocturne jewel]

yes

in the way that gets the desired result

[fallen scaffold]

i should have just used (u+v)

not its actual values

[nocturne jewel]

Yeah, almost like I was trying to show you the cleaner more efficient way

[fallen scaffold]

but nix was saying to expand this way

[nocturne jewel]

and helping you avoid algebra soup

yeah, cause you can do it that way, it's just a lot of writing

[fallen scaffold]

ok im erasing it and just using u+v

[nocturne jewel]

$T_p[q+r] \ =p(x)[q(x)+r(x)] \ =p(x)q(x)+p(x)r(x) \ =T_p[q]+T_p[r]$

4 lines

[stoic python]

Mosh

[nocturne jewel]

now you could just replace p, q and r with respective polynomials, but as long as q and r are defined in P_3, it's sufficient

[nocturne jewel]

Poor notation galore

[fallen scaffold]

yeah yeah im doing the subscripts now

and ()

[fallen scaffold]

how does it look now

[nocturne jewel]

Still bad

Cause nowhere did you explicitly show its pq+qr

[fallen scaffold]

or in my case pu + pv

[nocturne jewel]

Yes

[hollow finch]

No i did not. I specifically said don't expand it all the way

[nocturne jewel]

u and v also aren't defined so...

[hollow finch]

Right here

Doing it with polynomials is fine like mosh said but specifically defining vectors is a more general method that applies to pretty much any problem of this sort. It's also more familiar if you actually learned how to do subspace proofs.

But it's going to be hard either way if you don't know what you're proving.

[fallen scaffold]

is it possible if you can give an example by drawing out what you mean

[hollow finch]

Look in your textbook. If it doesn't have at least one example of proving something is a linear map then it's a pretty bad text.

Your prof should have also done an example in lecture as well

[fallen scaffold]

yeah i really couldnt find anything

ive been searching

[hollow finch]

I don't have a copy of your text so i can't say either way but I'd bet it's in there. I'd just read the section. Either way the process for these proofs are pretty much all the same. The only difference comes from what the particular vectors in the spaces actually look like. All the steps are the same though

And they're also just like verifying something is a vector space or a subspace. So maybe go back to that chapter and look it over. The difference is minimal

[fallen scaffold]

for this part (a) of the problem, i just need to prove its a linear map

[hollow finch]

That's all it asks yes

[fallen scaffold]

so just check its additivity and homogeneity

[hollow finch]

Yeah like i said

[nocturne jewel]

or just check a linear combination maps to a linear combination of mapped vectors

[wintry steppe]

proof: obvious from properties of polynomial

[nocturne jewel]

well yes but actually yes

[hollow finch]

I mean i would only do it that way if you're already comfortable doing it the standard way. They dont even know how to prove something is a subspace.

[fallen scaffold]

so again just to clarify: i can just u+v to expand to prove

[nocturne jewel]

Yeah hence why it was strikethrough

[hollow finch]

Fair

[fallen scaffold]

i dont need to use this

[nocturne jewel]

You dont need algebra soup, no

[fallen scaffold]

right

so let me try again with u+v

but im gonna sub in p(x) with 3x+2 and q(x) with u+v

[nocturne jewel]

yes.

[fallen scaffold]

but where would i show this

[nocturne jewel]

I mean I already posted a mostly complete proof

[wintry steppe]

"mostly"

[nocturne jewel]

didnt define q and r explicitly

[fallen scaffold]

this?

[nocturne jewel]

yes

it's just definition of T and the pointwise operations of polynomials/functions

[nocturne jewel]

missing brackets around 3x+2 but yes

and again havent defined u and v

[fallen scaffold]

so is it done?

oh

[nocturne jewel]

you need to write the last line, then yes

[fallen scaffold]

where do i do that

[nocturne jewel]

at the start

before you've used u and v in the proof

[fallen scaffold]

oh i did define it

i just didnt get it in the pic

hold on

[nocturne jewel]

ok then it's fine

[nocturne jewel]

yeah, then just clean up the proof for your final submission of course

[fallen scaffold]

ok what about homogeneity?

[nocturne jewel]

similar thing

[fallen scaffold]

yeah but what do i use for c

[nocturne jewel]

c is just a scalar from the field

so you use c

[wintry steppe]

[nocturne jewel]

need a lurk emote

wtf Terra

[fallen scaffold]

what do i need to sub in

i set it up for homogeneity

[nocturne jewel]

homogeneity just tests a scaled vector

$c\in\mathbb{R}, q\in P_2 \ T_p[cq(x)]$

[stoic python]

Mosh

[fallen scaffold]

is this fine?

[nocturne jewel]

q(x) isnt u+v

you only considered u+v to show additivity

[fallen scaffold]

so what should i do here

[nocturne jewel]

what do you mean..?

You just do a similar working, just w/ a scaled vector from P_2 instead of a sum of vectors

[fallen scaffold]

so i can just use u or v

[nocturne jewel]

yes....

[fallen scaffold]

so i can make q(x) = u

[fallen scaffold]

i feel like this is it

[nocturne jewel]

I mean you didnt prove anything

you just wrote out homogeneity

[fallen scaffold]

i made q(x) = u

[nocturne jewel]

yeah..

idk why that required a whole line of working

[fallen scaffold]

ok and then i subbed it in for q(x)

[nocturne jewel]

sure

you're just applying T_p to u(x), idk why you're saying sub

[nocturne jewel]

again

you have started with homogeneity, not proved it

idk why your scalar is already outside the transformation

and I thought we had gone over not needing to use explicit polynomials

[wintry steppe]

🙀

[nocturne jewel]

No, cause you don't know that T has homogeneity

[fallen scaffold]

which line did i mess up

[nocturne jewel]

2nd of proper working

$T_p c u \ =cT_p[u]$

[stoic python]

Mosh

[nocturne jewel]

that work is flat wrong

[fallen scaffold]

how many lines would i prove this in

[nocturne jewel]

4

4 showing the important steps of reasoning

[fallen scaffold]

so for the second line should i multiply it out or what

[nocturne jewel]

$T_p[cq(x)]$ then what would the next logical step be?

[stoic python]

Mosh

[fallen scaffold]

put in u for q(x)

[nocturne jewel]

Im using q

you can use whatever letter you want so long as it's defined

[fallen scaffold]

so second line i got Tp[c(u(x)]

[nocturne jewel]

that'd be your 1st if you use u for the vector

but anyway, next step?

[fallen scaffold]

remove parentheses and brackets?

[nocturne jewel]

how?

[fallen scaffold]

so that way i can rearrange the scalar out in front

[nocturne jewel]

didnt answer my question

how are you going to remove the brackets

[fallen scaffold]

just not use them

[nocturne jewel]

.....

NO

what was the 2nd step in showing T has additivity?

[fallen scaffold]

p(x) [u+v]

[nocturne jewel]

yes, use the definition of T

not remove brackets or whatever nonsense "just dont use them"

$T_p[cu(x)] \ =p(x)[cu(x)]$

[stoic python]

Mosh

[wintry steppe]

the first thing you should always do when faced with a problem like this is to think about what would happen if you just use the definition of the things involved

no fancy stuff

[fallen scaffold]

ok next step after this would be to use (3x+2)

[nocturne jewel]

sure, if you feel the need to do so

[fallen scaffold]

then can i move the scalar out in front?

[nocturne jewel]

yes, since this multiplication is associative

[fallen scaffold]

and then im done?

[nocturne jewel]

$T_p[cu(x)] \ =p(x)[cu(x)] \ =c[p(x)u(x)]$

[stoic python]

Mosh

[nocturne jewel]

yeah, still have that random 1st line floating around

then the same line twice..

[fallen scaffold]

better?

[nocturne jewel]

sure, though you still have a random q

[wintry steppe]

think you should finish the proof before polishing it

what's the last step?

also that shouldnt be there

the extra c, that is

[fallen scaffold]

yeah getting rid of that

so it would just be c p(x) (u(x))

[wintry steppe]

yup. what's p(x)u(x) equal to?

[fallen scaffold]

q(x)?

[wintry steppe]

recall the definition of T_p