linear-algebra

[wintry steppe]

up to a constant factor?

[native rampart]

Yes

[wintry steppe]

or with tr(I) = n, i guess it really is unique

that's cute

https://math.stackexchange.com/questions/1359118/show-that-trace-is-a-unique-linear-functional

very nice, i didn't know this

[round coral]

@wintry steppe traceless matrices form a Lie algebra , right? It looks like it when I was studying a bit

[wintry steppe]

yes, i think that the traceless matrices are the lie algebra of the lie group SL(n) (overkill answer but why not lmao)

[hollow finch]

in the 2x2 case, the subspace of skew symmetric matrices is one dimensional, meaning QQ' is some scalar multiple of I for any skew symmetric matrices Q and Q'. Then... multiplying some skew symmetric matirx Q' on both sides of A=QS gives us Q'A=S' which implies that the product of a matrix with a zero trace with a skew symmetric matrix always gives a symmetric matrix. is this just a wonderful result of 2 dimensions making things beautiful or does this generalize in any meaningful way?

[wintry steppe]

non overkill answer: the commutator AB - BA always has trace zero, and is bilinear, alternating, and satisfies the jacobi identity (long computation), so yes

[round coral]

thanks I get the second answer a bit

lack a lot of knowledge still

[wintry steppe]

well id write down the jacobi identity but im on my phone so it would be crappy

the lie group answer should be taken as more of a fun fact if you don't know that stuff

[round coral]

yeah, I just saw the top of ice cream, I still have to eat it

thank you

[wintry steppe]

no problem

[wintry steppe]

sorry for burying, you can repost if you'd like

[hollow finch]

ye np

in the 2x2 case, the subspace of skew symmetric matrices is one dimensional, meaning QQ' is some scalar multiple of I for any skew symmetric matrices Q and Q'. Then... multiplying some skew symmetric matirx Q' on both sides of A=QS gives us Q'A=S' which implies that the product of a matrix with a zero trace with a skew symmetric matrix always gives a symmetric matrix. is this just a wonderful result of 2 dimensions making things beautiful or does this generalize in any meaningful way?

tl;dr: if A is a 2x2 with a trace of zero and Q is skew symmetric then QA is symmetric. does that generalize?

[tame mural]

Is it possible to discover a solution for roots of polynomials for degrees higher than 5 for a specific degree?

[subtle walrus]

what do you mean by 'discover'?

[tame mural]

I mean, could there be a formula to find the roots of polynomials for a <specific> degree that's higher than 5?

[subtle walrus]

not in radicals

and not in the general case

[round coral]

@hollow finch I did think about it , but I have not come at a solution yet , I thought that we have A as an nXn matrix of trace 0, and Q as a skew symmetric matrix of order also nXn , so we know the trace 0 matrices can be written like A = BC - CB, for some nX n matrices B and C. I proved this a month ago, it is a bit long, (QA)* = (Q(BC-CB))* = (BC-CB)* Q* = ( (BC)* - (CB)* ) Q* = ( (CB)* - (BC)* ) Q now if QA is symmetric QA = (QA)* => Q(BC-CB) = ( (CB)* - (BC)* ) Q

for this to happen we need to see first if ( (CB)* - (BC)* ) Q is commutative ? also, CB and BC must be skew symmetric for this to hold, but is this always true ?

another thing I did, was I took two matrices X and Y say , of same order , and Y is skew symmetric, so for XY to be symmetric , (XY) = (XY)* = -YX* => Y(X+ X*) =0 which means either Y=0 i.e. a 0 matrix , or X+ X * = 0 which is X = -X* i.e. X is skew symmetric

similar to what I did above, Y is not 0 for sure, so X is skew symmetric for it to work which meaning (BC- CB) must be skew symmetric, is it always the case?

this is all what I worked and thought, if I see anything more, I will add it up

[opaque plover]

Hello I got a question: I got a 4x4 martix A, which rank is 4, so each column is linear independent. Now I need to find a invertable sub matrix of A of max. size. I know that the rank of a matrix is the max. size of an invertable submatrix with max. size. So is now matrix A already a submatrix or does A do not have a invertable submatrix of max. size? I would say it's the first statement, but I am not sure about it

[wintry steppe]

A is the largest invertible submatrix of itself.

Because A is already invertible.

[opaque plover]

thanks for the confirmation

[thorny hemlock]

i dont get the part

where

it says

"at most dimRangeT - 1 non zero vectors in Tv1 .... Tvn"

[round coral]

follows from the definition of rank of matrix / dim (Range(T))

[thorny hemlock]

err

this ?

[remote gorge]

Hi I have a question about linear transformations. Suppose T: V--->W is a linear transformation between two vector spaces with basis B and C respectively. Now the matrix of the linear transformation is the matrix A that satisfies [Av_B]= [Tv]_C. My question is suppose instead we just calculated Tv (without changing the basis to C) what does that represent?

[wintry steppe]

...it represents Tv in W? i'm not sure what you're asking

or maybe you could say it's the unique element of W corresponding to the vector Av_B under the isomorphism determined by the basis C (but that's what you wrote, hmm) (i'll elaborate on this if you want)

[remote gorge]

Yes, right but what is Tv in W?

[wintry steppe]

without knowing what W is (and what V and T are) it's hard to say more than just "image of v under T"

[remote gorge]

yeah I think conceptually we could say that [Tv]_C is the image of v under T wrt C but the interpretation I have this is the "correctly transformed" vector in W by T

[wintry steppe]

could you elaborate on what you mean by "correctly transformed?"

[remote gorge]

Suppose T is a dilation then in my head the "right" (natural) way to properly dilate v in W is given by [Tv]_C

You could also say that I'm asking why is the definition of the matrix given as above instead of the one representing [Tv]

Like what's the intuitive reason

[limber ember]

hi, can someone help me with a math question

[nocturne jewel]

the answer is 7, provide more details next time

[gray dust]

@remote gorge [Av_B]= [Tv]_C is wrong, A actually satisfies [Tx]_C=A[x]_B forall x in V

hard to know wym intuition but you can read the eqn as: once i fix bases B of V & C of W, i can use A to blindly plug some x into T. i first get the coords of x wrt B, [x]_B, then left multiply by A, and i get the coords of Tx wrt C, [Tx]_C. then to get Tx i take a linear combo of the vectors in C with coefficients given by [Tx]_C. compactly the process is convert x to B-coords, transform by T (represented by A), then convert Tx away from C-coords

[wintry steppe]

rokabe typing an essay rn

rokabe teach me LA

[remote gorge]

Yeah actually that's what I intended by my notation A multiplied by the representation of x in terms of B.

hard to know wym intuition
I'm not sure how to communicate this but what I meant why is using the definition of A defined that way [Tx]_C=A[x]_B more useful than defining A satisfying [Tx]=A[x]_B.

[gray dust]

@remote gorge that's the same eqn, also i feel you ignored the entirety of my msg explaining how we can read the eqn which also gives a sense of why it's natural to define A that way

[remote gorge]

@gray dust I think I understand what you mean. You're basically saying what this diagram in my book is saying:

that's the same eqn,
How is this the same equation? The second one is defined in terms of Tx not written in terms of the basis C

[gray dust]

not exactly the same as the book

once i fix bases B of V & C of W, i can use A to blindly plug some x into T. i first get the coords of x wrt B, [x]_B, then left multiply by A, and i get the coords of Tx wrt C, [Tx]_C. then to get Tx i take a linear combo of the vectors in C with coefficients given by [Tx]_C. compactly the process is convert x to B-coords, transform by T (represented by A), then convert Tx away from C-coords
in the book this process matches the path that goes right, down, then the last step is converting Tu away from C-coords, the opposite of 'writing Tu in C'

so this path uses A to compute Tu rather than jumping straight from u to Tu

[remote gorge]

Ok so the bottom arrow is flipped in your explanation. But could you explain why the equation [Tx]_C = A[x]_B is the same as Tx = A[x]_B because in my head those two equation would lead to a different matrix A (unless B=C).

[gray dust]

they're not, eqn 2 doesn't make sense to say

[remote gorge]

ok is that because it's not possible to find such a matrix A or because finding such matrix has no real usefulness?

[gray dust]

it's not about A. it's comparing different object types which doesn't make sense

Tx is a vector in W, A[x]_B is a tall matrix

[remote gorge]

Sorry is it because of the notation I used? Would [Tx] = A[x]_B make more sense? Surely I can see a vector represented as a "tall matrix"?

[gray dust]

doesn't help. vectors & tall matrices generally aren't the same object type (really the object type of a vector depends on the context of the vector space we look at), so equality between them makes no sense

[remote gorge]

Thanks for the explanation!

[quasi frigate]

Can somebody check this simple calculation of a matrix determinant?

[acoustic path]

inb4 its a 10x10 matrix

[quasi frigate]

det(A) is equal to 2j

and j is imaginary unit i for engineers

[muted hazel]

what's the original matrix?

[quasi frigate]

Sorry, the equation is lower

in one line

det (3j * A) is the start

I know that when you calculating determinant of a: matrix times a number, you need to also multiply by a unit matrix, idk if I did that correct

[gray dust]

@quasi frigate it's right

wait you say det(A)=2j but wrote -2j

[quasi frigate]

Yea, you're right, my bad

Thanks, more calculations coming in few minutes

[thorny hemlock]

Tv1 = (1,....0) Tv2 = (1, ..., 0) ... Tvn = (1,...,0)

Does this kinda transformation work?

im kinda confusion :/

[quasi frigate]

What is that symbol at the beginning of the 2nd line? T is in ...

[thorny hemlock]

what symbol

[quasi frigate]

before the (V, W) bracket

[thorny hemlock]

oh

all linear maps from V to W

[quasi frigate]

okay, I just started linear algebra

[thorny hemlock]

you start with determinants :o ?

[quasi frigate]

I mean, I recently started studying matrices and this stuff, I'm on 1st semester uni

[thorny hemlock]

very nice

good luck

[quasi frigate]

Thanks, there's a lot of material to process since I have two mathematical subjects. But forums/servers like this keeps me going

Sorry, I've kinda spammed your question. You might want to repost it

[thorny hemlock]

nw lol

[wintry steppe]

well you don't know that you can write the elements of W like this so you may want to be more careful

[thorny hemlock]

why not

W is finite dimensional

[wintry steppe]

even then you are given the linear transformation T and basis v_1, ..., vn, so you can't just say that Tv_1, ..., Tv_n are all the same thing

[thorny hemlock]

ah yes

yes i understand

[wintry steppe]

my suggestion is the following

if T is not the zero operator then some Tv_i is nonzero, now ||extend that to a basis of W|| (you might need to be careful with some basis ordering stuff here, i think)

and if T is the zero operator well

[thorny hemlock]

whats zero operator

[wintry steppe]

the thing that sends everything in V to the zero element of W

[thorny hemlock]

ok

[wintry steppe]

if T is the zero operator then there isn't much to prove since every matrix of T will just be the zero matrix

[thorny hemlock]

ye

[wintry steppe]

but if not then i think you can go off of what i recommended

ah, ah, i might have been a bit misleading

instead of splitting into two cases with T being the zero operator or not, you should split into the cases of Tv_1 being zero or nonzero

@thorny hemlock

that's to deal with the basis ordering nonsense, because M(T) is taken wrt the ordering (v1, ..., vn)

[thorny hemlock]

ok

slightly confused

why is the basis of the codomain space relevant here?

[wintry steppe]

well you need a basis of the domain and a basis of the codomain to calculate a matrix rep

[thorny hemlock]

i dont see where they used standard basis of F^3

[wintry steppe]

$$T(1,0) = 1 e_1 + 2 e_2 + 7 e_3$$ so the first column of $\mathcal{M}(T)$ is $$\begin{pmatrix}1 \ 2 \ 7 \end{pmatrix},$$ where here $e_1 = (1,0,0),e_2=(0,1,0),e_3=(0,0,1)$.

[stoic python]

TTerra

[wintry steppe]

that is just off of the definition of M(T)

[thorny hemlock]

ooooh yes yes

silly me lmao

[wintry steppe]

same reasoning for the second column

[thorny hemlock]

ye

[tame mural]

In the case of $F^1$, if we have the matrix product $[1][2]$, is that the same as the dot product $[1] ⋅ [2]$?

[stoic python]

meow

[native rampart]

The first one would be a matrix,The second would be a scalar

Take matrices A and B,if Rows of A are labelled as r_1,r_2...r_n and columns of B are labelled as c_1,c_2...c_n, The element in ij position of AB is dot product of r_i and c_j

[tame mural]

I see, thx

I should have phrased as [ [1] ⋅ [2] ]

[vocal isle]

Hey guys, I'm trying to solve a multi DOF damped system. I've always been a bit confused why damped systems can't be solved using the following approach? Is there something that I've done wrong here? Is it true that because [C] is not symmetrical then it can't have distinct eigenvalues / eigenvectors, making it not possible to diagonalize? But even if that's true, e^[A]t can still be found without diagonalizing, right?

[thorny hemlock]

what does M is an isomorphism mean?

we can represent a linear map as the matrix and an inverse of it exists?

[native rampart]

Well,you can represent any linear map as a matrix

[shy atlas]

every linear map has a matrix form and vice versa

[native rampart]

It means your linear map has an inverse

[round coral]

the definition of isomorphism is given before in the book

did you skip it

[thorny hemlock]

lol

no

[shy atlas]

[thorny hemlock]

So basically just representing the linear map as a matrix and proving that its invertable and linear

says Tvk = 0 for k= 1, ....,n

am aware the theorem of injective if and only if kernel = {0}

but

its mapping multiple different vectors to same vector in codomainspace??

<@&286206848099549185>

[shy atlas]

its mapping multiple different vectors to same vector in codomainspace??

why do u think that ?

[thorny hemlock]

idk

Tvk = 0 for k = 1... n

sooo

Tv1 = 0

Tv2 = 0

Tvn = 0

@shy atlas

so v1 .. vn \in kernel ???

[shy atlas]

but ur vectors in this case are linear transformations from V to W

[thorny hemlock]

errr

v1 ... vn are just vectors in V

[shy atlas]

and they're saying no two such linear maps will have the same matrix

yeah but ur not proving M is an injection from V to something, are you ?

[thorny hemlock]

are we not?

[shy atlas]

no

M is an injection from L(V, W) to F^{m*n}

[thorny hemlock]

3 vector spaces?

[shy atlas]

[thorny hemlock]

lmao

so

T: V -> W

[shy atlas]

ye

[thorny hemlock]

the matrix of this transformation

has all its entries equal

0

[shy atlas]

mhm

now u gotta prove T is the 0 transformation

and that will prove Ker M = {0}

[thorny hemlock]

yeah and how is that injective

its not one to one that way

[shy atlas]

the 0 transformation isnt one to one, yes

but M is tho

ur mixing up V and L(V, W)

[thorny hemlock]

uhh oh

so

er

theres only 1 transformation V to W that M(T) = 0

[shy atlas]

mhmm

[thorny hemlock]

oh

ok

so

M is m by n

that makes sense i guess

[shy atlas]

M is the mapping

M(T) is m by n

[thorny hemlock]

ok i think i got it

[shy atlas]

nice

[thorny hemlock]

[quasi frigate]

Can smb help me calculating rank of matrix A? I'm stuck

[round coral]

just do gaussian elimination until to obtain RE form or RRE form , to see the number of linearly independent rows

it is the most convenient way when you can't see clearly

@quasi frigate

[thorny hemlock]

@shy atlas the dimrange of M equals the dimension of L(V,W) right

[shy atlas]

dim( range M) = dim F^{m * n}

cuz yknow F^{m * n} is the codomain of M

but becuz M is a bijection F^{m * n} is the range

oof nvm dim(L(V, W)) = dim(F^{m * n}) so its ok anyways

[quasi frigate]

Okay, and what If I have something like that?

row filled with zeros

[round coral]

@quasi frigate do you know how to do gaussian elimination ?

[quasi frigate]

I'm reading about it on wiki

I also used an online calculator with steps, and I don't understand the last step in it

I don't get it, why the rank is 2

[round coral]

there are only 2 linearly independent rows

the 0 ones are redundant

[quasi frigate]

Yea sorry, I get it now

[round coral]

online calculators use different methods to find the solution quickly

[quasi frigate]

So if there are rows filled with zeros, I must don't care about them?

[fallen maple]

is hom the same thing as a dual space?

[acoustic path]

@quasi frigate yea

ohm? @fallen maple

[fallen maple]

ohm?

[acoustic path]

nvm

[round coral]

for a vector space V over field F, Hom(V, F) = V* where V* is the dual space of V. Is this what you want to know? @fallen maple

[quasi frigate]

not gonna lie, Gaussian elimination takes a lot of time, and it's not that intuitive for me, but It worked

is there any other method that you use?

[fallen maple]

yes, thank you very much

[thorny hemlock]

how did they use 3.64 owo

[severe magnet]

Tv_i are vectors for which u can use the 3.64 equality again

unless I miss context

[dreamy iron]

more context.

[wintry steppe]

this should probably say M(Tv_k) in
-last sentence before the blue box
-last part of the blue box
since the columns of M(T) are supposed to be the m by 1 matrices of the Tv_k's wrt (w_1, ..., w_m). even worse, the proof of 3.69 that @Yes#0553 posted supports this

axler....

some part of me imagines this typo wouldn't be present in the second (read: good) version of 🪜

[round coral]

@quasi frigate there maybe. I usually use gaussian elimination only, it's mostly the best and efficient way. Other ways involve determinants but determinants can only be of square matrices, there is also one that can be used total number of non-zero eigenvalues of a matrix is equal to the rank of that matrix, you will have to find eigenvalues, what if the eigenvalues are complex , then everything goes down the drain, so direct gaussian elimination is still easier.

[quasi frigate]

Thanks, I've been calculating several examples and it seems a lot easier now

[spice storm]

Use an online calculator @quasi frigate

[quasi frigate]

Yea, I've been using it as well, but I cannot use one on exam day

[spice storm]

Is exam in class?

Practice 3 by 3 matrix questions like the hard ones. Maybe do one 4 by 4. Then you should be good @quasi frigate

[quasi frigate]

online, but still I don't want to rely on calculators all the time

[quasi frigate]

Yes, practise is the key to success

[spice storm]

Gotcha but if it’s online why not take advantage of it. But I get what you are saying

[nocturne jewel]

What is typically taught in 2nd term of 1st year LinAl?

[wintry steppe]

here's what mine did (replace primary decomposition theorem with diagonalization, though)
A theoretical approach to real and complex inner product spaces, isometries, orthogonal and unitary matrices and transformations. The adjoint. Hermitian and symmetric transformations. Spectral theorem for symmetric and normal transformations. Polar representation theorem. Primary decomposition theorem. Rational and Jordan canonical forms. Additional topics including dual spaces, quotient spaces, bilinear forms, quadratic surfaces, multilinear algebra.

[acoustic path]

good to know im only term 1

[thorny hemlock]

in the example

why is he writing (17,20) + U

(17,20) + (x,2x) ?

[wintry steppe]

because (17, 20) + U is how you write a coset

which instance of "(17, 20) + U" in the example are you referring to

[thorny hemlock]

ok

[thorny hemlock]

what about 1st term?

[wintry steppe]

ill post it when I get home

will @ you

[thorny hemlock]

ok thanks

[acoustic path]

why did they need to take the transpose

couldnt they have just rref A and B and seen that they were equal

[wintry steppe]

@thorny hemlock

A theoretical approach to: vector spaces over arbitrary fields, including C and Z_p. Subspaces, bases and dimension. Linear transformations, matrices, change of basis, similarity, determinants. Polynomials over a field (including unique factorization, resultants). Eigenvalues, eigenvectors, characteristic polynomial, diagonalization. Minimal polynomial, Cayley-Hamilton theorem.

the course was slow as fuck though so we only got up to determinants LOL

unique factorization of polynomials, diagonalization, CH, and minimal polynomials all came in LA2

[thorny hemlock]

nice

how long is 1 term?

my uni is 2.5 months a term

[wintry steppe]

roughly 4 months, one week break about halfway through, where the last 2 and a half weeks or so are examinations (no classes)

[thorny hemlock]

ow

thats long

[wintry steppe]

it's really more like 3 months

also @acoustic path this buried your post a bit, sorry, so you can repost if you'd like

[acoustic path]

na its fine

its kinda basic, would you be able to help?

like it just seems weird to take the transpose

[thorny hemlock]

what book is that ?

[acoustic path]

lol 3000 solved problems in linear algebra

[thorny hemlock]

[wintry steppe]

same rref implies that the row spaces are equal (because you perform row operations to get there, which preserve the row space), not necessarily the column spaces, so they had to take the transposes

[acoustic path]

ohhhhhh

yea u got it

thx

[wintry steppe]

no problem

[dreamy iron]

oh.....OH!!! hawt damn.

i just looked in the errata, and yeah, you're right: https://linear.axler.net/LADRErrataThird.html

@thorny hemlock might be interested to know there's a typo in our book (3rd edition)

TYVM TTerra

[wintry steppe]

👍

[round coral]

@wintry steppe that's really slow we covered diagonalization over complex vector spaces, jordan form , a bit of determinants and little about tensors too

I am talking of the first semester

But that's good, I am also eager to study functional analysis, have always liked dabbling with functional equations

[wintry steppe]

take note that's only the first semester

the second semester, which i posted waaaay above, was a bit more reasonable i feel

[round coral]

yeah, it is

[wintry steppe]

honestly most of the linear algebra i actually use i picked up from other things

i don't think i appreciated most of my first year linear algebra class at the time

tfw low mathematical maturity

[round coral]

maybe, but it was enjoyable for me, we had such good homework questions way more to think and do than axler questions

[wintry steppe]

http://www.math.toronto.edu/payman/mat247/main.html

axler...

well i also did have a bit of a problem with attending lectures and studying during that semester

[round coral]

we will be doing axler this year, no freidberg. But I will go over that too, Friedberg is good

[wintry steppe]

wait, kanishk, you're in first year? or are you gonna be in uni next year

[round coral]

yes, I am in first year

[wintry steppe]

i have lost track of time

the last time we talked you were going into first year i think

[round coral]

yeah

[wintry steppe]

there are like a million practice problems on the course website i linked, definitely check em out

[round coral]

yeah, sure . Thanks

what Tterra gave same

@acoustic path they used elementary row operations on the matrix which affect the column space , thatswhy they took the transpose. If you don't want to take transpose, you can do elementary column operations

[wintry steppe]

something something row operations preserving "row space"

[round coral]

thanks for telling me TTerra , I wrote that incorrectly before

[wintry steppe]

i can't recall the last time i unironically row reduced a matrix

probably in my ODEs class

or maybe once during a manifolds problem about finding a tangent basis

[native rampart]

Row reduction is useful for solving linear equations

[wintry steppe]

is it now

[native rampart]

So, probably you won't see them, unless you do engineering or applied math

Gaussian elimination is computationally more efficient than using cramer's rule

[round coral]

cramer's rule has a limitation too, it can be only used to solve when you have the same number of unknowns as the number of equations

more than 3 equations, I have rarely seen anyone using cramer

In engineering, row reduction is really useful , especially in circuit analysis

[wintry steppe]

i used cramer's rule a few times in my smooth manifolds class to show smoothness

ah yes

inversion in GL is smooth

because the entries of the inverse matrix are rational functions of the original entries or something

[modern glade]

i dont understand why there are two ways to achieve rotation.

1. multiply by a rotation matrix
2. transpose and reverse the matrix

[native rampart]

Can you explain the second one?

How is that rotation?

[modern glade]

1 2 3     7 8 9     7 4 1
4 5 6  => 4 5 6  => 8 5 2
7 8 9     1 2 3     9 6 3

first reverse up to down, then swap the symmetry

isn't that a rotation?

[native rampart]

Ok, You mean rotation in that sense

[modern glade]

there are two rotations?

[native rampart]

Well, There's the geometric one. Fix a point and rotate the plane

[modern glade]

is the geometric one where you multiply?

[native rampart]

Yes

[modern glade]

which one is the above I referred to then

i'm so confused

geometric rotation and algebriac rotation?

[native rampart]

Most of time,people refer to the geometric rotation

[round coral]

that is just another way to do rotation, you are using two steps. While in geometric rotation, you use only 1 step

so which is better?

[modern glade]

i think the second rotation is more of a computational solution for 2d matrix. it doesn't geometrically rotate at all.

so you can't really compare the two

[modern glade]

so to achieve rotation on a 3d matrix is it that we multiply by a 2d rotation matrix also?

or does that rotation matrix change it's dimension to respect the matrix we're multiplying?

[native rampart]

Well, You rotate in 3d with a 3d matrix

[modern glade]

makes sense

this is a 2d rotation matrix

so you can only rotate with a 2d matrix?

[wintry steppe]

there are higher dimensional rotation matrices

for example, $$ \begin{pmatrix} \cos \theta & -\sin \theta & 0 \ \sin \theta & \cos \theta & 0 \ 0 & 0 & 1 \end{pmatrix}$$ acts on $\bR^3$ by counter-clockwise rotation by $\theta$ about the $z$-axis

[stoic python]

TTerra

[modern glade]

i've noticed any linear transformation's origin is (0,0)

[native rampart]

All Linear transformations fix a point

[modern glade]

and is it always 0,0?

what if I want something from the center?

[round coral]

that will be an affine transformation then which isn't linear

[quartz compass]

if it doesn't fix the origin, it's not a linear transformation

you can show this creates a contradiction by the definition of being a linear operator

[modern glade]

Linear transformations have the special property that the origin is not moved by the transformation.

yep

[quartz compass]

this is something you should prove to yourself, it should take a few seconds

and if it doesn't take a few seconds, then you should definitely prove it

[modern glade]

i haven't done any proofs yet

[quartz compass]

I don't mean anything scarier than doing a little algebra

that counts as proving stuff you know

[round coral]

in fact you don't need to prove it, it follows from the contrapositive, if L is a linear map, then L(0) =0 , what will be its contrapositive?

[quartz compass]

I think it's fun enough to demonstrate it

suppose L(0)=v and v is not 0

[modern glade]

i don't fully understand the proofs tbh

my motivation is to just see the actual solutions to problems using methods

so I can convert it programmatically

[quartz compass]

that's all I'm asking you to do really

I shouldn't have said 'proof' it caused you to run away like a little baby lol

[modern glade]

lol

[quartz compass]

just convince yourself or a regular person with a little shuffling of algebra symbols

[modern glade]

are you referring to this?

suppose L(0)=v and v is not 0

[quartz compass]

yeah

here I'll just show, take a matrix A and then multiply by the 0 vector

A0=v let's say it doesn't map 0 to 0, but some other v

well since 0=k*0 by a scalar

A(k0) = kA0 = kv

we can pull the k out on the other side of the matrix

but now we have v=kv

oops

only vector this can be true for is v=0

just playing around with the properties a bit until you find something that works out

[modern glade]

i lost you here: "well since 0=k*0 by a scalar"

where did you come up with that

[quartz compass]

I was looking at A0=v and thinking

I could scale the 0 vector by any number k

and linearity lets you pull it out and so it would really be scaling v

if the stuff I'm saying is kind of confusing, just spend some time going back to the properties of linearity, I think what I said will help build a bit more familiarity with what's going on here

you can think of linearity as just being: how basis vectors transform is enough to know how all vectors transform

[modern glade]

properties of linearity, is this part of linear maps?

we're still referring to linear function here, right

[quartz compass]

yeah

when I say linearity I mean it follows two rules

L(a+b)=L(a)+L(b)

and

L(ka)=kL(a)

if you have never seen these then I guess I will have to say more but

probably you'll just come across it naturally in your book before too long so just keep on with what you're doing hah

[modern glade]

the former, you distributed, the latter you asscoiated?

i have seen those in a book before, yes

[quartz compass]

first one distributed, second one is more like I commuted

here k is a scalar and a is a vector

[modern glade]

never bother fully going into it

[quartz compass]

you can scale a vector before or after multiplying a linear operator, that's all it is

[modern glade]

L(a+b)=L(a)+L(b)

I'm going to try to decode this: adding two vectors first and then transforming is the same thing as transform each vector individually and the adding both transformed values?

[quartz compass]

exactly

[modern glade]

what are these rules useful for though?

[quartz compass]

in general you can think of yourself having some set of basis vectors a,b,c,... and then you represent an arbitrary vector as some combination of them

$\vec v = r * \vec a+s * \vec b+ t * \vec c$

[stoic python]

Merosity

[modern glade]

looks like the dot product?

[quartz compass]

then to know what happens to v, you trickle through to see how it acts on each of the basis vectors a,b,c

[modern glade]

maybe i should do some excercises

and apply those properties

[quartz compass]

like geometrically speaking you can represent reflections, rotations, scaling, shearing, projection and more with linear transformations

[modern glade]

right

[quartz compass]

and more, like derivative is a linear operator even

[modern glade]

so if any problem given that says "prove if it's linear"

we just need to check with those props above?

[native rampart]

Well, You could also state it as
L(cx+y)=cL(x)+L(y)

[quartz compass]

yep

[tame mural]

$Φ(σ_1 \vec{v}_1 + ⋯ + σ_n \vec{v}_n) = σ_1 Φ(\vec{v}_1) + ⋯ + σ_n Φ(\vec{v}_n)$

[stoic python]

meow

[tame mural]

I like it this way because it shows a mapping between coordinates

[modern glade]

Determine whether the following functions are linear transformations

how would i interpret the above problem?

[tame mural]

if a matrix can do it then it's linear

so ask whether a matrix T exists

[quartz compass]

you could break it up like how we've been describing it

$$\begin{bmatrix}x\y\end{bmatrix} = x \begin{bmatrix}1\0\end{bmatrix} + y \begin{bmatrix}0\1\end{bmatrix}$$

[stoic python]

Merosity

[modern glade]

where did you get 1,0?

[quartz compass]

but maybe write this as $x \vec v_1 + y \vec v_2$

[modern glade]

and 0,1

[stoic python]

Merosity

[quartz compass]

@tame mural wanna explain it? I'm going to bed

[tame mural]

two approaches jump to mind

[dire thunder]

[tame mural]

first is find the matrix

second is prove linearity from the 2 rules you just learned

[quartz compass]

we're trying to build off what he just learned here

no matrices

[modern glade]

yeah so i have the two props in mind right now

[tame mural]

you're learning linearity without matrices?

is this Axler?

[modern glade]

i have that book, yeah.

but im not following any acadmeics or books here.

[tame mural]

ah

[modern glade]

i'm come from a comp sci bg so just trying to learn things

[tame mural]

it's not wrong to just use the 2 rules of linearity

first test for preservation of addition

[dire thunder]

ok, so you have T(x,y) = (x+y, y)

[tame mural]

next test for homogeneity of scaling

[dire thunder]

have you learnt defn of a linear map?

[modern glade]

i have, above

you mean the two properties?

[dire thunder]

yes

[modern glade]

L(a+b) = La + Lb
L(ka) = kL(a)

yep

[dire thunder]

T(ax)= aTx, and T(x+y)=Tx+Ty

show that these hold on T(x,y) = (x+y, y)

[modern glade]

hmm let me thinnk here

[dire thunder]

i mean it is pretty straightforward

[modern glade]

how?

i'd like to learn to proving it

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

are you assuming the vectors equal to arbitarry values?

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

is that how you start?

[dire thunder]

yes

well in ur particular case at least

[modern glade]

T(a(v1,v2))

that's how i see it

[dire thunder]

yes you are correct

but you can also move a inside

[modern glade]

yep

[dire thunder]

because scalar multiplication

[modern glade]

T(av1 + av2)

[dire thunder]

no

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

oh!

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

T(av1,av2)

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

i'm lost there

huh?

[dire thunder]

https://media.discordapp.net/attachments/540211747613704221/793053409867071498/694042020100046990.png?width=959&height=113

[modern glade]

is this a new problem?

[dire thunder]

https://media.discordapp.net/attachments/540211747613704221/793053502103617556/694042020100046990.png?width=959&height=113

compare these two

[modern glade]

ok so do we assume

x, y are vectors?

[dire thunder]

in these case x,y are points

they are coordinates*

[modern glade]

right ok

[tame mural]

x,y together is the vector

(x, y)

T([x, y])

[modern glade]

i dont understand what you're asking for here

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[dire thunder]

this is defn of your linear map

https://media.discordapp.net/attachments/540211747613704221/793050937642975242/unknown.png

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[modern glade]

i mean i understood the substitution

and those props

but i still can't see the intuition behind it

maybe i need to do more of these to understand

[dire thunder]

what do you mean

no intuition whatsoever

just applying definitions here

https://images-ext-1.discordapp.net/external/SCaxngnsTxD4aQfpuuIRmXjN-EG9FVkgvCtWHH6FTzA/%3Fwidth%3D959%26height%3D113/https/media.discordapp.net/attachments/540211747613704221/793053409867071498/694042020100046990.png

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[dire thunder]

if you want so

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[tame mural]

Perform some concrete tests

[modern glade]

ok it's repeating the property i learned above

but ughhh

hmm

[dire thunder]

i mean yes they just use definitions directly

[modern glade]

T(x,y) = (x+y, y)

Tv = (x+y,y)

does that even make sense

[tame mural]

3 T(x, y) = T(3x, 3y) ??

Try this concrete test

as a sniff test for linearity

[dire thunder]

yes if you denote v = (x,y)

[modern glade]

meow, is that aribtirary?

you just substitued 3?

[tame mural]

No, this is just a concrete sniff test

If this function is linear

[modern glade]

i need to sniff it?

[native rampart]

Yes,3 is arbitrary

[modern glade]

Oh ok

lol

[tame mural]

then multiplying the output by 3

is the same as multiplying the arguments by 3

that's the rule for homogenuous multiplication

Next one to concretely test is addition

Then ask how you can make the proof general

because that's a specific case

[modern glade]

so can I solve what you've given above?

[tame mural]

Based on the information you have, is it even true that multiplying your output by 3 is the same as multiplying your inputs by 3?

[modern glade]

3T(x,y) = T(3x,3y)
3Tx, 3Ty = T3x,T3y

?

does that make sennse

[tame mural]

well, compute it

With some concrete values

Even though these are specific cases, it should become clearer how to generalize this

[modern glade]

should I take x,y aribtarary?

[tame mural]

3 T(pi, e) = ?

is that the same as T(3pi, 3e)?

[modern glade]

yep

[tame mural]

yup nice

time to confirm addition

"preservation of addition"

[modern glade]

would you want me to write it down?

[tame mural]

Nah I trust u... 😄

your big wall of latex shows you have diligence

To test the preservation of addition

[modern glade]

we just proved the second property?

which is scalar multiplication?

[tame mural]

We proved the "homogeneity of vector scaling"

Another way people say it is "Multiply now or multiply later, you choose"

[modern glade]

L(ka) = kL(a)

is this what we did?

[tame mural]

Yes

[modern glade]

Yes, finally some english

ok so should I prove the additive property? with the concrety problem you suggested above?

[tame mural]

Yes

I'll also tell you one last definition for linearity

Linearity is a function between two vector spaces which preserves the properties of the vector spaces you care about

also known as linear homomorphism

[modern glade]

3T(Pi+e) = 3T(Pi) + 3T(e)

is this what i'm proving now?

[tame mural]

Forget the 3

It's boring now

[modern glade]

Ok, forgotten

T(Pi+e) = T(Pi) + T(e)

[tame mural]

The problem is this

[Pi, e] is actually your input

So you need another vector just like that

[Pi^2, e^2]

don't compute these values of course, that's too painful

So my question now is

let's say [Pi, e] is your input

Is it true that T([Pi, e] + [Pi^2, e^2]) = T([Pi, e]) + T([Pi^2, e^2])

?

[modern glade]

yep

[tame mural]

Then u win hehe

[modern glade]

lol that's it?

[tame mural]

yup

[modern glade]

what do these show though?

yes, they obey the linearity property

and we know it's a linear function

[tame mural]

Well you're learning a bunch of stuff in linear algebra right now

you need to know when it's safe to use your knowledge

this test will say so

[modern glade]

are these proofs used anywhere other than academic tests asking to prove linearity

[tame mural]

This is something expected to be baked into intuition

it's very important

[modern glade]

i see

[tame mural]

more important than tests or proving

I recommend 3B1B on youtube for intuition

he gives a very visual metaphor for what a linear map does

[modern glade]

oh? i didnt see that part then

watching nonw

[tame mural]

https://www.youtube.com/watch?v=kYB8IZa5AuE&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=3

[modern glade]

watching now

T(x,y) = (x^2, y^2)
(Tx, Ty) = (x^2, y^2)

this isn't linear, right?

[tame mural]

Yup, you can tell right away

very good

[modern glade]

but isn't it obvious though?

[tame mural]

yes, because it fails preservation of addition right away

but it doesn't fail homogeneity of scaling

[modern glade]

doesn't fail homogenity of scaling? i believe it does.

unless im missing something

[tame mural]

maybe I'm just wrong

[modern glade]

T(B) = BA
TB = BA

this isn't linear either, right?

[tame mural]

all matrix operations are always linear

btw u r right it fails scaling too

[modern glade]

how do the two properties define rotation, refleciton, translation, etc.. though?

where do those come to picture

[tame mural]

The question is why are these linear operations

And that guy earlier was just listing the kinds of things linear algebra can do

it can do a lot

[modern glade]

which guy? the 3b1b guy?

[tame mural]

the guy who talked about rotation, reflection, etc.

These rules don't imply rotation, reflection, etc.

Rather, you explore rotation and ask whether it's linear

For example

let's say that R(1) means to rotate once

is R(1 + 1) = R(1) + R(1)

[modern glade]

yep

[tame mural]

And then we can test for the scaling too

then suddenly we know rotation is linear

that means we can study it via linear algebra

it also means we can find a linear function to do rotation

So the rules of linear algebra won't make you think of rotation right away

I think it's the reverse

You're going to bump into linear phenomena all the time

and you'll have to ask yourself whether it smells linear

[modern glade]

and then i perform the sniff test

[tame mural]

yeah

[modern glade]

is affine transformations part of LA?

[tame mural]

Yes and no

It's technically not part of linear algebra

but they'll teach it to you anyway

and they're pretty important

[modern glade]

affine transformations look cool and there's a problem i want to solve with it

that's why i got into LA in the first place

[tame mural]

you're in bio, right?

r u a student?

[modern glade]

biology?

[tame mural]

did you say you were a bio student?

i forgot

[modern glade]

nope

[tame mural]

oh

[modern glade]

comp sci

[tame mural]

ohhh

[tame mural]

ahhh

is this advent

[modern glade]

I got these tiles which I want to rotate with affine transformation

correct!

[tame mural]

did you learn linear algebra for advent...

i mean

that's so crazy lol

[modern glade]

well not really for advent. it's mainly because I do a bit of shaders

and i thought LA was where I should start to fully understand a few things

[tame mural]

i see

r u still a student?

[modern glade]

here's a shader i computed

which takes x,y positions of pixels

and fluidly rotates

[tame mural]

ic

you know

3b1b is teaching a course on this stuff

where he integrates linear algebra with julia hehe

but that's probably a lot to jump into

[modern glade]

yea i know

but i dont do this with Julia

this is javascript, webgl stuff

[tame mural]

whoa, very impressed

how is the api for webgl

[modern glade]

i can't say pretty neat

but they're coming up with something call webgpu

which is mind blowing

[tame mural]

Is there a de-facto winner in this space for interfacing with webgl?

i've been wanting to get into web visualizations for math lessons stuff

[modern glade]

threejs is popular

d3 is popular for svg if you're into that

this is what i've built a month ago using svg

and css transforms

uses d3 as an interface to svg

[tame mural]

I see

you got a snippet of code?

I'm curious how hard/easy it is

[modern glade]

i have a notebook with notes on it

I'll DM you

[tame mural]

ah that sounds more painful lol

sure

[modern glade]

it's an interactive notebook lol

you can view the code

easily

like Jupyter notebook

[tame mural]

ic

btw, r u still a student?

[modern glade]

no

[tame mural]

ahh I see

[modern glade]

just to understand vectors I built a small library so it can do all manipulations

next goal is to visualize it using 2d canvas

[tame mural]

in JS?

[modern glade]

yeah

[tame mural]

ahhh

[modern glade]

you can do P+Q which adds it

[tame mural]

is it on github?

[modern glade]

nope, it's offline since it's not done

i dont understand some concepts to fully 100% code it yet

[tame mural]

I also did the same thing as you

but you know

[modern glade]

yeah?

[tame mural]

I gave up encoding the types

~_~

it gets reallly painful

[modern glade]

you mean the macros?

[tame mural]

No, you represent your vector as an object

[modern glade]

yeah

[tame mural]

I'd be very very curious to see how your progress as you continue

I have the anticipation you'll run into troubles with your Vector and Matrix object type

[modern glade]

well you knw it's not possible to do this right

[1,3]+[2,3]

in JS

becuae it doesn't understand

[tame mural]

yes but I assume you intend to build it all

[modern glade]

so I transformed all macros and converted it to this...

Vector.Mul(new Vector(1,3)+new Vector(2,3))

this is painful to write

[tame mural]

Yup that's what I mean

dangerous!!

I recommend using arrays

Just arrays

Why? well

there's this operation called transpose

[modern glade]

yea of course but also i'm not doing this for it to be officially used or anything

It's for fun

[tame mural]

ofc

I think it's a great learning exercise

[modern glade]

tokenizing everything and then replacing it

some of this stuff is a fork of Geometric algebra library

but yeah

[tame mural]

u r pursuing geometric algebra?

[modern glade]

yea lol

i tried

[tame mural]

@_@

[modern glade]

i dont get half of it

but this library makes it easier

[tame mural]

Hmm

[modern glade]

so I proved this theorem

[tame mural]

btw

have you learned about matrices yet?

[modern glade]

i have

trannsposinng

and stuff

[tame mural]

oh ic

[modern glade]

i haven't gotten deep into it

i hope to find perspectives/answers from doing proofs or theorems honestly

prolly makes me solve problems quicker

[tame mural]

imho, doing proofs and theorems takes a lot of work

and isn't always the fastest way to learn

[modern glade]

do you know about the Yoneda perspective?

[tame mural]

from category theory?

[modern glade]

yep

that's what i keep in mind when i come across new concepts. If I dont get it right away... I keep moving and maybe there's another way i can look at it.

[tame mural]

nope, I haven't gone much into category theory

[modern glade]

half ass it basically lol

are you student?

[tame mural]

nope, work in software hehe

[modern glade]

oh nice!

[agile flicker]

Can somedy help me with a really basic question i have

[feral minnow]

Yeah

[agile flicker]

one sec pls

[feral minnow]

Ok

[agile flicker]

i am doing a project in linear algebra

[feral minnow]

Ok

[agile flicker]

a basic predator - prey system

[feral minnow]

Yeah

What help do you want?

[agile flicker]

i am trying to find c1, and c2

in relation with the x0 initial population matrix

[feral minnow]

Ok

[agile flicker]

i am also supposed to do this in rpackage but i dont know how to organize it

i have written the code for finding eigen values and vectors

and set up a function for graphing

just cant find these c constants

[thorny hemlock]

i dont understand how got that

((v-w) + u) is stuff that they could do since U is a subspace?

[limber sierra]

((v-w) + u) is in U because:

• u is in U (by definition)
• v-w is in U (since we're assuming (a) holds)
• subspaces are closed under addition, so the sum of u and v-w is in U as well

hence w + ((v-w)+u) is in w + U

since we're just adding w

@thorny hemlock

so yeah, its because U is a subspace

[thorny hemlock]

Cool

but

Where do they prove that two affine subsets could be disjoint

v+U intersect w+U is not equal to null

meaning that they arnt equal for every choice of vectors in U but equal for possibly 1 or more ?

@limber sierra

[limber sierra]

it means they share at least one common element, yes

[thorny hemlock]

i guess the part c doesnt mean they cant be equal

[limber sierra]

right

and in fact they ARE equal

since (c) is equivalent to (b)

[thorny hemlock]

yeah

[limber sierra]

but a priori you dont know that

until youve proven it

[thorny hemlock]

ok just to confirm, when none of those equivalences hold, two subsets are disjoint @limber sierra

[limber sierra]

if none of those hold, then v + U and w + U are disjoint

since that means (c) doesnt hold

[thorny hemlock]

yes

[limber sierra]

so yeah, parallel affine subsets are either equal or disjoint

this should match your visual intuition

[thorny hemlock]

:o subspaces?

[limber sierra]

ops

fixed

but yeah, speaking of visual intuition

if you think of, say, R^2

the one-dimensional subspaces of R^2 are straight lines through the origin

so parallel affine subsets are just parallel lines

so in that context, this theorem is just saying "two parallel lines never meet unless they are the same"

obviously this theorem is far more general than that specific statement

[thorny hemlock]

yes

i understand

[limber sierra]

but my point is that it should jive with your visualization

[thorny hemlock]

yep

thank you very much for help :)

[thorny hemlock]

i dont see how nullpi = U

[brittle juniper]

do you see what is the zero vector in V/U?

[thorny hemlock]

no, -U ?

whatever U happens to be, the negative of it ?

[agile flicker]

hello

[brittle juniper]

U and -U are the same vector space

[thorny hemlock]

yeah

[agile flicker]

can somebody help me prove the relation i have writtne is true?

it is really short

[thorny hemlock]

so nullpi is just U then ?

[brittle juniper]

yes, v+U is U if and only if v is in U

[thorny hemlock]

okay, how did they make use of 3.85 then?

[brittle juniper]

they're just saying that U being the kernel of π is immediate from the definition of π

[thorny hemlock]

this is 3.85

[brittle juniper]

you can see Ker(π)=U in the equivalence between (a) and (b) with w=0

[thorny hemlock]

AHHH

tysm!

[brittle juniper]

You're welcome!

[thorny hemlock]

In proving that this is a linear map : $T\tilde(v + nullT + u + nullT) = T(v + u) = Tv + Tu$ where v,u in V

[stoic python]

Yes

[rose umbra]

what is the T ?

[thorny hemlock]

is this correct for proving additivity?

[thorny hemlock]

Transpose

[rose umbra]

what does it means in the definition?

[thorny hemlock]

A = A transposed

[rose umbra]

ahh transposed matrix

thanks

[willow palm]

actually, it also makes sense in a Hilbert space (a space equipped with a inner product), and the transpose of a operator is just in the normal sense when we consider about the transpose of a operator in normed vector space

[native rampart]

Pretty sure,Hilbert spaces have more structure than inner product spaces

[hidden zodiac]

anyone know T1 a) and T2 a)?

[native rampart]

Let A be that given matrix,note $A(a_1,a_2,....a_{10})^{T}$=0 is a system of linear equations which have a non trivial solution

[stoic python]

MoonBears-D-

[native rampart]

Which is possible iff det(A)=0

@hidden zodiac

Same process applies for T2 a)

[quasi frigate]

Good evening, can someone guide me on how to solve this?

Express the variables x, y, z, t depending on the variables p, q, r by performing the multiplication of appropriate matrices.

[hidden zodiac]

@native rampart okay thanks!

[acoustic path]

just substitute @quasi frigate

like x=2(p+q+2r)+3(3p-q-r)

[soft burrow]

seems like the exercise asks to write it as a matrix product though

[stoic python]

derivada.schwarziana

derivada.schwarziana

[quasi frigate]

I think I do, I’m gonna solve this after my dinner. Thank you very much

[acoustic path]

oh, my b

i didnt read the performing the multiplication of appropriate matrices part

[quasi frigate]

Hey, just wanted to let you know, that I solved that

[hollow finch]

Is it possible to simplify
$$|v|^2|w|^2-(v\cdot w)^2$$?

[stoic python]

nix

[stoic python]

Kanishk

[wintry steppe]

[nocturne jewel]

@round coral magnitude of v x w*

[round coral]

I didn't understand. My solution is wrong, right @nocturne jewel

oh I get it , I see, typo

it must be the written as magnitude of (v X w)^2

[nocturne jewel]

yeah

since the original thing is a scalar, and squaring vectors isnt defined

[round coral]

yeah, thanks for pointing that out

[quasi frigate]

Can someone help?
For matrices, derive the general formula for A^n, n is a Natural number

[wintry steppe]

try to diagonalize or put it in jordan form

[quasi frigate]

They told me to use mathematical induction as well

[wintry steppe]

well, you can use diagonalization/jordan form to guess a formula, then prove it works with induction

this matrix is diagonalizable so it shouldn't be too bad

[stoic python]

TTerra

[quasi frigate]

Thanks, I will read about this things since it's new to me. I will let you know when I solve this

[nocturne jewel]

Or just do a couple examples and form a guess from that

A^2, A^3 (I personally had to do A^4)

[quasi frigate]

Ohh, yea good thinking

[wintry steppe]

that works too

just, diagonalizing or using jordan form is usually how these ones are done

but for simpler looking matrices like this one, probably not too bad

[quasi frigate]

Yea, I will def read about these as well. Thanks

[nocturne jewel]

Yeah took me like 2 minutes to get the pattern for the 1st row

but that's cause it was curve fitting

@wintry steppe I see that smh

[wintry steppe]

[nocturne jewel]

shhhh it's not my fault i couldnt recognize all the numbers were double the bottom row's

[wintry steppe]

allow me to trivialize the problem

,w {{2,2},{1,1}}^n

[stoic python]

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Requested by TTerra#5291

[wintry steppe]

lmao the 0^n's

[nocturne jewel]

imagine factoring out the 1/3 and making the matrix entries look nice, #3^(n-1) gang

[blissful pagoda]

Let's say that {u,v,w} is a set of linearly dependent vectors in R^n and that A is an nxn invertible matrix. How do I know whether the set {A^Tu, A^Tv, A^tw} is linearly dependent or independent?

I think they are linearly independent because you're just taking the transpose of a column (or row) vector giving you a row (or column) vector

and it's still linearly independent

[wintry steppe]

A^T is also invertible so it should be linearly dependent

eh you don't even need invertibility for that

the image of a linearly dependent set under any linear transformation will be linearly dependent

@blissful pagoda

[blissful pagoda]

Wait sorry

I meant linearly dependent yea

Because u,v,w are linearly dependent

[thorny hemlock]

if i can get dim(Vj) for all j's thats enough to prove its finite dimensional right?

[wintry steppe]

if you can get what about dim(V_j)?

[thorny hemlock]

dim(V1 + ... + Vm ) = dimV1 + ... + dimVm

basis exists for all vector spaces so finite dimensional right?

[wintry steppe]

you can use that formula if you already know that the V_1, ..., V_m are finite-dimensional that doesn't make sense

but you are being asked to prove that

[thorny hemlock]

really

what if we define

[wintry steppe]

i believe so

[thorny hemlock]

function