linear-algebra

[round coral]

Hi

@errant cedar just take a null space, and extend its basis in two different ways. What's that hard about it?

or you can take an example where the null space is 0. there are so many such linear maps that you can make with the given condition. Meditate on the nature of linear maps

[steady fiber]

maybe it's a polynomial ring :^)

and q can

[wintry steppe]

not possible. if T has trivial kernel, then by rank-nullity, rank T = 5, impossible since rank T is at most 2

[round coral]

that's right. I made a mistake

[wintry steppe]

happens to us all

[steady fiber]

why did you sully me ttera

smhmh

[round coral]

how is 1/x^2 a polynomial, you got that wrong

[wintry steppe]

[errant cedar]

Its not looks hard but i just started to learn and dont know exactly how to do it

[round coral]

that I said wrong, don't think upon it.

[wintry steppe]

if it's wrong then think upon why it's wrong

[round coral]

yeah, read the previous message too

[wintry steppe]

you could try doing something that effectively ignores the last three coordinates of F^5 and gives you something nice on the first two

@errant cedar

then the kernels of your transformations could just be {(0,0,x,y,z) in F^5 : x,y,z in F}

just some food for thought

one example || K(x,y,z,w,t) = (x,y) and L(x,y,z,w,t) = (y,x) ||

this is usually how things go wrong when mapping from a higher dimensional space onto a lower dimensional one

[errant cedar]

i did not know that i can ignore

[wintry steppe]

well of course you don't have to

but it's nice to have simpler examples

i could have taken it to be projection onto the last two coordinates, for example

[hollow finch]

making good choices like these is how you can make these kind of exercises easier to digest/understand

[errant cedar]

than here is null(K) = null(L) and null(K) = (0,0)

[wintry steppe]

i don't know why i thought clicking on the spoiler text in the message preview would unspoil it

[wintry steppe]

no, that's not the correct kernel

you should be more explicit

let me make myself more explicit to help (previous message edited)

[steady fiber]

I did the same

[wintry steppe]

i should be checking over my homework right now

6 pages of complex analysis

[hollow finch]

if a vector in F5 (x,y,z,w,t) is mapped by K to the zero vector 0 in F2 (which is (0,0)) then

K(x,y,z,w,t)=(0,0)
(x,y)=(0,0)
so for an element in F5 to be in the kernel then x=0 and y=0

to find a kernel you can always set the entries of the transform of a generic vector to zero

[steady fiber]

lol, is it all residue calculations

[wintry steppe]

ill post it in #real-complex-analysis

[hollow finch]

if its a more complex transformation then you may have to set up a system of equations to solve for the kernel

[hollow finch]

i wish i could do 7 pages of complex analysis instead of being 7 labs behind in my physics lab

[wintry steppe]

complex analysis is just busywork that i have to get done so i can get to the classes of mine that are actually interesting

at least, that's how it's been the past while

[errant cedar]

can you give an example

[hollow finch]

K(x,y,z,w,t)=(0,0)
(x,y)=(0,0)
so for an element in F5 to be in the kernel then x=0 and y=0

[errant cedar]

K(x,y,z,w,t) = (x-y+z-w+t,0) L(x,y,z,w,t) = (-x+y-z+w+t,0)

when i say like tihs

is null(K) equal to null(L)

?

[wintry steppe]

read the pinned message, this is not linear algebra

[wintry steppe]

My apologies

[copper stratus]

What is the process for solving this?

<@&286206848099549185>

A ping would be nice if someone can help 👍

[bitter hornet]

I don't know how to find the linear combination for a single number when three equations are involved but basically I have to solve:
0 = a(2-x+x^2) + b(2-2x+x^2) +c(-3+3x-2x^2)
1 = a(2-x+x^2) + b(2-2x+x^2) +c(-3+3x-2x^2)
2x = a(2-x+x^2) + b(2-2x+x^2) +c(-3+3x-2x^2)
3x^2 = a(2-x+x^2) + b(2-2x+x^2) +c(-3+3x-2x^2)

I got the first two by checking and guessing but there has to be a better way to do this right?

[wintry steppe]

it'd make your life a lot easier to write whatever that is as a matrix equation, for one

[bitter hornet]

This is what the problem actually looks like

But the process is essentially the same. I'm stuck finding a linear combination that involves three variables

Edit: I understand it now

[errant cedar]

Give two linear maps L_1, L_2: F ^ 2 → F ^ 2 such that L1 ◦ L2 not equel to L2 ◦ L1
Do you have an idea about it

[wintry steppe]

think of it in terms of matrix multiplication

after that think of some simple examples, matrices with 1s and 0s

[hollow finch]

if that vector is a steady state vector then it is an eigenvector with eigenvalue 1

assuming they want the columns to add up to one then there is something else that we know

A^T (1,1)=(1,1) so (1,1) is an eigenvector of A^T with eigenvalue 1 as well

notice the fact that the columns sum to 1 implies (1,-1) is an eigenvector of A

$\begin{bmatrix}a&b\1-a&1-b\end{bmatrix}\begin{bmatrix}1\-1\end{bmatrix}=\begin{bmatrix}a-b\b-a\end{bmatrix}$

[stoic python]

nix

[hollow finch]

so we know two eigenvectors of a 2x2 matrix

we can then do a diagonalization

$A=\begin{bmatrix}\frac{15}{29}&1\\frac{14}{29}&-1\end{bmatrix}\begin{bmatrix}1&0\0&\lambda_2\end{bmatrix}\begin{bmatrix}\frac{15}{29}&1\\frac{14}{29}&-1\end{bmatrix}^{-1}$

[stoic python]

nix

[hollow finch]

then you just have to pick some lambda2 with absolute value less than 1.

@copper stratus thats how id do it anyway

[copper stratus]

@hollow finch Is this for me?

[hollow finch]

yes

[copper stratus]

Ah, I was confused for a min'

[hollow finch]

sorry i thought jh was trying to help you

[copper stratus]

I thought you credited someone's problem's solution to me

XD

[hollow finch]

you could also do

$A=\begin{bmatrix}15&-1\14&1\end{bmatrix}\begin{bmatrix}1&0\0&\lambda_2\end{bmatrix}\begin{bmatrix}15&-1\14&1\end{bmatrix}^{-1}$

[stoic python]

nix

[hollow finch]

that may make the calculations slightly easier but it doesnt make a big difference

some other neat things to note as long as |lambda2|<1:
the bigger lambda2 is, the slower the system will converge to the steady state vector
if lambda2 is positive then the sequence will steadily approach the steady state vector
if lambda2 is negative then itll bounce around the steady state vector like an alternating series

just in case you want to test out how the matrix looks for different values/situations

its definitely the kind of thing i like to do

[hollow finch]

its far harder to find two linear maps where L1 ◦ L2=L2 ◦ L1. if you just make them up randomly the chances are high that they will satisfy that.

[wintry steppe]

far harder
just take L1 = L2

[willow marten]

<@&286206848099549185>

[odd kite]

if $y_i=\sum_j A_{ij}x_j$ then $\frac{\partial y_i}{\partial x_j} = A_{ij}$

[stoic python]

Timon

[odd kite]

all the terms on the rhs for $\bar{h} =\dots$ are equivalent to one matrix multiplying $\bar {x}$

[stoic python]

Timon

[willow marten]

ok, so i need to move for each value and derive?, but what does having both the uper and under index mean in -W

[odd kite]

it means the matrix squared that is $W^2_2 = W_2W_2$

[stoic python]

Timon

[acoustic path]

dimension=rank+nullity

[bleak ginkgo]

Anyone up to help me finding a linear transformation from this:

-R3 to R2
-T(1, −1, 1) = (1, 0)
-T(1, 1, 1) = (0, 1)

[willow marten]

ah ok, thank you

[hollow finch]

Try turning it into a matrix equation

[shrewd wharf]

actually I am supposed to use the fact that (A^2+B^2) = (A+iB)(A-iB) to proceed
as I am unaware of diagonalization (I am in HS)

[hollow finch]

Would you mind posting a picture of the question

[half ice]

Consider (A + B)², which has a positive determinant

[native rampart]

I don't think that's valid considering we are talking about Mn(R) and not Mn(C)

[shrewd wharf]

why it's not valid

[native rampart]

det's domain is Mn(R)

[half ice]

It's possible to get going, but will need more work.

Plus, probably not the easiest way to go about it

[native rampart]

det is a function from Mn(R) to R.

[half ice]

Huh. I'll have to think on that

[shrewd wharf]

okay thx
btw I think since n=2, using the fact that det(A+B) + det(A-B) = 2(det(A) + det(B)) might be useful

[native rampart]

That's not true

[shrewd wharf]

for second order square matrices?

[native rampart]

Wait,did you check it manually?

Because it's generally not true

[shrewd wharf]

yea generally it's not true, but it is true for second order square matrices

[half ice]

Oh haha, yeah M2(R) may be making all the difference here

As the first question asks about n×n matricies

[dusky epoch]

det(A+iB) and det(A-iB) are conjugates

[shrewd mortar]

ye

you can think about det as sum of diagonal products for this

[dusky epoch]

[round coral]

I am very confused on how to prove this if I have two sets X and Y and a field F, then I can't able to make an isomorphism between F^(X ⊔ Y ) and F^X ⊕ F^Y . I am trying to show they are isomorphic

[brisk sparrow]

can someone explain how this transformation matrix was made? https://i.imgur.com/3zHQYcX.png

[dusky epoch]

P_3(R) comes with the basis {1, x, x^2, x^3} presumably

so they just, yknow, constructed the transformation matrix as you normally would

applying the transformation to each basis vector and writing those as the cols

[vocal isle]

does anyone know how to find a solution to the following nonlinear matrix equation

$$[A]X^2 + [B]X + f = 0$$

Where $[A]$ and $[B]$ are 2x2 constant matrices and $X = [x_1, x_2]$ and $X^2 = [x_1^2, x_2^2]$. Also, $f$ is a 2x1 vector. I could do this easily using simultaneous equations, but I wanted to know if there was a way to solve this using matrices.

[stoic python]

The Smelly Cherry

[timber magnet]

any thoughts?

[simple hornet]

ah just a question about notation

if we denote a function f : S to F where F is a field, when we talk about f(x) is that an element of the field?

[marble lance]

Yes

[simple hornet]

And when we just talk about f, that's a more abstract thing

that's a member of the set of functions from S to F, while f(x) is not

[marble lance]

Yes. f is the function, f(x) is the output of the function which comes from F

[simple hornet]

thank you

[dense wyvern]

1. Differentiate with respect to x: 2sin 2x cos 3x
   can some help me with this

[acoustic path]

yeah

think of differentiation as a linear transformatio

[dense wyvern]

then....

[acoustic path]

and then go and ask in #calculus

[wintry steppe]

[thorn lichen]

is there such thing as a stochastic matrix that does not enable x_k to approach some steady state vector q?

theres a theorem stating every regular stochastic matrix must enable to this to be the case

but the inverse doesnt seem true, as the identity matrix appears to be an irregular stochastic matrix, but still has a steady state vector

in fact every vector is a steady state vector for the identity matrix

ooooo

nvm

i found one

for 2x2 for instance, r1: 0 1 r2: 1 0

would work

😈

[little frigate]

Does anyone have some steps in order to find the neutral element for addition, addition being □ depending on the following conditions
Let V = (2,∞). For u and v that belongs in V [...] we can define the internal addition operation such as :

u □ v = uv - 2(u + v) + 6

i don't know how to manage to cancel u on both sides of the equation to just be left with v

[native rampart]

Neutral element=3

It's mostly trial and error

[little frigate]

you're not talking to me hopefully aren't you o_o

[native rampart]

Yes,I am talking to you

[little frigate]

dang

how come is it trial and error?

doesn't sound very formal to me

[native rampart]

Also,This belongs in #proofs-and-logic

[little frigate]

?

uh... i forgot to mention but those are vector spaces..

ah im big dumb nvm nvm

[marsh veldt]

anyone wanna join the VC and help me and my boi out with a basic ass linear algebra question

we struglging haha

[foggy nebula]

plz

si vous plait

[hollow finch]

$\int_0^t e^{(t-u)A}\vec{v}(u), du=\int_0^t e^{uA}\vec{v}(t-u), du$

[stoic python]

nix

[hollow finch]

is this a valid change of variables where A is a matrix?

or does the fact that A is a matrix make it wonky

wait is this a convolution? o.o

[fast stag]

idk if I'm in the correct section but how do I solve this?

[hollow finch]

That is not linear algebra

[fast stag]

@hollow finch what is this then?

[hollow finch]

#prealg-and-algebra

[fast stag]

ok thank you

[rose coral]

I believe the change of variables is still valid. I think pretty much all properties of integrals generalize to these settings with vector spaces

[lime dust]

Hey cud someone pls explain why only that option is right?

[hollow finch]

Okay thank you :)

[rose coral]

No problem

[rose coral]

The explanation for b and d being incorrect is similar. Are you confused by one of those or a?

[lime dust]

I just don't get why a b and d are not valid :(

[hollow finch]

Yeah b and d only require a counterexample

Starting with b, can you think of a vector which is orthogonal to every vector in Rn?

[lime dust]

Not really no...

[hollow finch]

It's the one everyone forgets about

[lime dust]

0?

[hollow finch]

Yeah

[lime dust]

Ahh I thought 0 vectors are always diacarded

[hollow finch]

No you still have to invite the zero vector to the vector space party even if it's a buzzkill

[lime dust]

Oh kk that makes sense, n what about a then🤔

Cos if U is a subspace of Rn shouldn't U'contain everything else in Rn

[gray dust]

a is false

in R^2 the orthogonal complement of the x axis is the y axis. their union clearly isn't R^2

[lime dust]

How though? 🤔

[gray dust]

you might've conflated union & sum of subspaces

[hollow finch]

Oh I see. But the span of the union would be Rn right?

[lime dust]

I haven't exactly learnt to sum subspaces😅

[gray dust]

sum of subspaces U & W is defined U+W:={u+w: u in U, w in W}

span of the union would be Rn
yes

[lime dust]

Ah ok, so what would be an example of a vector that lies in Rn but not in this union?

[gray dust]

(1,1)

[lime dust]

Ahhh so the in between regions are not covered

[gray dust]

visualize the union of the x & y axes as the fence between quadrants. there are many pts off the fence

[lime dust]

Riiight I think it's kinda complicated to think abt it for over 2 dimensions but alright I'll give it a think

Thanks so much both of you!

[gray dust]

you're welcome. also a nice exercise is to prove c is true

[acoustic zodiac]

Can someone explain to me simply what a dual basis is? I don't really get it

[wintry steppe]

the kth dual basis element just specifies the kth coordinate of a given vector in the basis

that statement is very concisely summed up by what you'll often see written as $e^i(e_j) = \delta^i_j$, where the $\delta$ is the kronecker delta, the $e_j$'s your basis, and the $e^i$'s the dual basis

[stoic python]

TTerra

[acoustic zodiac]

so for example the 3rd element of (1,1,1) in the canonical basis would be the third element, also a 1, in the dual basis?

[wintry steppe]

each dual basis element, where the dual basis is dual to the canonical basis (im assuming this is in R^3), will indeed send that to 1

in a few minutes i can type something that expands on the first thing i said

[acoustic zodiac]

thank you

[wintry steppe]

wow

fuck you too texit bot

[stoic python]

TTerra

[wintry steppe]

i should point out two things

1. these are really ordered bases, so if you tend to make the distinction, note that i really do mean ordered bases here (else, how would any of this make sense)
2. using upper indices to denote dual space elements is a differential geometry thing, i don't know how common it is in LA books (also at this point i should have used upper indices on the coefficients of v too )

[acoustic zodiac]

wow thank you man

[wintry steppe]

that's a bit long but i hope it can clarify the definition a little

[mild igloo]

the process for getting the eigenvalues of a 3x3 matrix is the same as a 2x2 right?

just gets more complicated because taking the determinant is harder?

[wintry steppe]

I have two (possibly infinite dimentional) linear maps T and S on some space H. I know that T = T* and S = S*, and I want to show that if <T(v), v> = <S(v), v> for all v in H, then T = S.

Since f is infinite dimentional I can't just plug in orthonormal basis vectors and show that the entries must be equal, unfortunately

[shut falcon]

Bro I have no idea how to do this problem, I and some of my other classmates think it is not possible can anyone help?

[stoic python]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

Please give me something to evaluate.
See ,help calc for usage details.

Dec 1, 2020, 12:53 AM

[wintry sphinx]

@shut falcon it is possible, but read the pinned message; this is not linear algebra

[shut falcon]

I know that i couldn't find any other chat.

[steady fiber]

this is not the chat though

[wintry steppe]

maybe one way to get started is by re-writing the given condition as <Tv - Sv, v> = 0; i think that'd be fruitful

well it's not much

but i feel like you can do some funny cauchy-schwarz thing

[steady fiber]

cauchy schwarz?

oh wait nvm

[wintry steppe]

||especially since (x,y) \mapsto <(T-S)x, y> is an inner product sometimes (always?)||

[steady fiber]

complex has infected my brain

[wintry steppe]

hehe

schwarz has quite a few things in that field

schwarz lemma, schwarz reflection principle

[steady fiber]

cauchy/schwarz/christoffel/reimann

those names

have shown up too much

[wintry steppe]

i'm thinking that ||if the thing i described is an inner product then |<(T-S)x, y>| \leq <(T-S)x, x><(T-S)y, y> = 0 so by whatever argument makes sure this is actually an inner product, i.e. nondegenerate, you get T = S||

just throwing ideas out there

[wintry steppe]

i only know christoffel for christoffel symbols and i'm not sure anything can top those

[steady fiber]

I love the schwarz-christoffel mappings

they're very cool

there's also the tensor named for him

I forget the other guy

oh lol, it's reimann

reimann-christoffel tensor

[mild igloo]

could someone explain how these solutions were reached?

im not quite understanding

[steady fiber]

last row affects nothing

middle row is v2 = 0

top row gives that v1 = v3

and they set it to t

[mild igloo]

so were allowed to make v3=t because its all 0

yea I still cant apply that reasoning

dont get it

[nocturne jewel]

v_3 is a free variable since the 3rd row is all 0's

so we let v_3 = t

then you have v_1 - v_3 = 0 from top row

so v_1 = t

and the only way 2nd row is true is if v_2 = 0

Thus the v vector is t[1,0,1]

[mild igloo]

alright

what math am I doing wrong here then?

every calc I give this matrix gives me diff eigen vectors

and my own math hasnt produced the right answer either

[mild igloo]

pls

[shrewd mortar]

@mild igloo so if A(v) = -9v = -9(a, b, c), from the matrix we get

A(a, b, c) = (9c, -5b, 9a)

[mild igloo]

it says its wrong

maybe this question is broken?

[wintry steppe]

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

Why is (x,x,y,y) + (x,x,x,y) = (x,x,y,z)?

what?

that's an addition of sets there

I know

it's absolutely not the case in general that (x,x,y,y) + (x,x,x,y) = (x,x,y,z), you should try proving that equality by showing each set is a subset of the other

Why did the third coordinate become a y instead of an x, for example?

well you can think of it like this: when you take any element of U, say (a,a,b,b), and any element of W, say (x,x,x,y), then when you add you get (a+x, a+x, b+x, b+y). the first two coordinates are the same, which is represented in set builder notation using the dummy variable x by writing the first two coordinates as the same. the third coordinate, represented in set builder notation using the dummy variable y, could be anything

these are all dummy variables

i strongly suggest writing out a proof as to why that equality of sets holds

so you can see that the x's, y's, z's are just placeholders

Ok. Would it be too much to ask for help writing that proof?

i can't help out with that right now, sorry

i really think you should try it yourself

generally it's a good idea to verify or think about the examples you come across, and get proficient at doing these "mini exercises"

[half forge]

can someone explain this problem too me

[hollow finch]

well how do you find the jordan form of a matrix

[round coral]

@half forge I think the question is quite clear, do you know how to find the Jordan form

[hollow finch]

im assuming your teacher gave a lecture on jordan forms if you are given that kind of homework problem

do you have notes?

[half forge]

her lecture was confusing

im not quite sure

[hollow finch]

have you looked at your textbook?

[half forge]

yes, i understand up to here in yellow

after that i am lost

[round coral]

that's just matrices, you won't be able to understand the algorithm with that

to figure it out, you need to understand the theory behind it as well

that image looks more like a calculator!

[hollow finch]

the jordan form of a matrix is given in a block pattern based on how defective each eigenvalue is

theyre called jordan blocks

so lets start with the eigenvalues. theres only one of them so it wont take that long

whats the geometric and algebraic multiplicity

[tame mural]

Is there such thing as a matrix for which repeated application permutes through the whole space?

[wintry steppe]

what do you mean by permutes through the whole space?

as in, you can reach any vector by applying the matrix a certain number of times to some vector?

or such that you can reach any matrix of the same size by taking powers?

both of those aren't possible, at least over the reals (uncountable spaces)

[tame mural]

like a matrix M when applied repeatedly to [1, 2, 3] gets you something like: [123] -> [231] -> [312] -> [321] -> [213] -> [231]

Which are all the permutations of [1, 2, 3]

I'm thinking it's not possible because 123 -> 321 is a different kind of matrix then 123 -> 231

So you always need two different matrices

[round coral]

I can see what you mean. For that you can define a vector space as a set X={1,2,3} on the field F . Then you can define isomorphism from F(X) to F(X)

there will be total 6 such matrices

[wintry steppe]

ah yes just take the 3 element field

[round coral]

and these 6 matrices will form a group under matrix multiplication

[wintry steppe]

iso to S_3?

ok i stop interjecting now

[round coral]

@wintry steppe Am I saying wrong?

[stoic python]

Kanishk

[wintry steppe]

i'm not quite sure what you mean by "on the field F" and by F(X)

but i'm liking this idea

@tame mural if you're comfortable with basic abstract algebra maybe you can try looking into connections between finite matrix groups and symmetric groups

i don't know much algebra though so i don't know if you can say anything significant

kanishk may be on to something though

[native rampart]

For this case,M doesn't exist. As M is a permutation of cycle length 1,2 or 3,M^k(v) can cover a max of 3 vectors

If you can find integers $a_1,a_2...a_k$ such that $a_1+a_2+a_3...a_k=n$ and $lcm(a_1,a_2,...a_k)=n!$ you have your required matrix

[stoic python]

DrunkenDrake

[native rampart]

S_n is not cyclic for n>2 so no(
(12) and
(13) are 2 different subgroups of order 2 for n>=3.
n=1 and n=2 are trivially cyclic)

[tame mural]

I see, thanks

So the number of "generators" I need = the number of matrices

and the symmetric group isn't 1-generated

sad

[dreamy iron]

https://cdn.discordapp.com/attachments/488104216678760469/783260598896033802/162017091996745728.png

What else do I know about the basis vectors???

......
i guess i can conclude that there also live inside V.:

[crystal oracle]

Every basis of null(T) can be extended to a basis of V

[dreamy iron]

Okay. Yeah. I was hoping to get there....

I just don’t know how to do that.

i know the basis of null(T) is linearly INDEPENDENT in V?

That makes intuitive sense....but I don’t know if I can prove it directly....no matter.

How do I know I can start adding vectors to the list of basis vectors?

Like, okay, how?

Is this by like a set theory axiom where im allowed to form set unions?

[crystal oracle]

@dreamy iron If we are talking only about finite dimensional vector spaces, then look here

[dreamy iron]

That feels right: I take a new vector in V and that that as a singleton set and Union it to the basis set.

LOL, Im in AXLER TOO

[dreamy iron]

Is induction in the under belly of all of these proofs?

(That’s the thing about Axler, he does induction but doesn’t declare that out right.)

[crystal oracle]

I think the proof of 2.31 uses induction. Namely, I think you can prove it by proving by induciton that for all j, there exists a subset of {v_1, ..., v_j} which is a basis of span({v_1, ..., v_j}).

[dreamy iron]

Oh!

This might be my first encounter using subset is induction.

[cosmic pawn]

can someone help with my question in #help-6

[brittle orchid]

Hi, could someone help me understand a bit of notation?
The Er, Dr, Dc

Please @ me if responding, thank you

[wintry steppe]

@brittle orchid just from looking at the picture it looks like it's describing the row or column operations going on

for example

in the first line try comparing the row operation going on with the thing under the arrow

the notation is kind of pointless, it just seems confusing and cluttering

[brittle orchid]

I understand that the notation may not be ideal but I believe that we are required to present our answers in a particular way for university therefore I think I have to understand what this means

[wintry steppe]

like maybe the letter is the type of elementary operation, the subscript whether it's a row or column operations, and the arguments determining what happens to what row/column

why not just ask your instructor then

[brittle orchid]

well, I've sent him an email requesting for an appointment

[steady fiber]

what's with lin alg profs and using wacky notation

I knew a lin alg prof in first year who made up all his notation

[wintry steppe]

today we're gonna denote change of basis matrices using the clunkiest notation possible

[steady fiber]

literally 0 notation used in his course was widely accepted notation

it was extremely idiotic

[wintry steppe]

i cringe a little every time i see the change of basis matrix notation with like, a backwards arrow between capital letters under a [I]

[brittle orchid]

I mean I wish they would at least topics of Algebra at the same university with relatively consistent notation

[wintry steppe]

it's so.... clunky

[steady fiber]

oh lol, I've seen that

[brittle orchid]

but it also doesn't help that last year I was slacking 😄

[wintry steppe]

consistent notation is a myth, but there's definitely such a thing as bad notation

idk, i feel as if just the arrow with R2 -> (whatever) is perfectly descriptive

it's concise and captures 100% of what's happening

[steady fiber]

$\bR^{m \times n}$ is how people normally denote the space of m x n matrices, but this prof had to use some garbage like $M_{m,n}$

[stoic python]

PorosInMyAshe

[wintry steppe]

🤢

[steady fiber]

and that was some of the best notation he used

[wintry steppe]

well yeah then you get the students who are like but isn't that technically R^(mn)

[brittle orchid]

I think the Er (i, j, k) means multiply k times row j and add it row i or something

[steady fiber]

E_x (i, j; k) seems to mean [x] i = [x] i + k * [x] j

where x is r or c

[x] is row or column

based on what x is

D just seems to be multipying a row/column by some constant

[wintry steppe]

use D for multiplication

[steady fiber]

I've always just seen E used

for all the elementary row operation matrices

[brittle orchid]

ngl E almost reminded me of elementary matrices

[wintry steppe]

almost like elementary row and column operations correspond to elementary matrices

[brittle orchid]

pff

Btw does this fall under linear algebra: real canonical form and complex canonical form?

[wintry steppe]

no? canonical form means something different than this

row/column reduction usually falls under "reduced (...) echelon form" i guess

[brittle orchid]

this isn't RRE though

if you saw the image

it was 1 or 2 steps from real canonical

[steady fiber]

it seems to be going to rref

would love to see the rest of the image

[wintry steppe]

im a little hesitant to say for certain since ive never actually seen row or column operations used simultaneously

[brittle orchid]

They're fairly similar, they've just got double operations instead of single

[steady fiber]

same to ttera

[brittle orchid]

real canonical form is almost identical (visually, at first glance) to RRE with 2 operations happening simultaneously

at least to a noob like me

😄

[wintry steppe]

in my mind "canonical form" usually refers to something obtained by and related to the characteristic polynomial

but "canonical" is also one of math's many overused words

[brittle orchid]

idek what it means it's just algorithmic imho

[wintry steppe]

you can study all of this from a non-algorithmic point of view; the theory behind linear algebra is very nice 😌

[steady fiber]

if only that were the norm

[brittle orchid]

tbh I have hated my Maths department for a long time and as a result I wanna get done with my degree asap 😅

[dire thunder]

sorry occupied

lmao

ok

so can we prove this

using that every odd degree polynom has root in R

[hollow finch]

thats how id do it

when it says "has an eigenvalue" does it mean a real eigenvalue?

[steady fiber]

"real vector space"

so any complex eigenvalue isn't an eigenvalue

in that space

[hollow finch]

complex eigenvalues are people too 😢

no but i get you

[steady fiber]

yes, you can intuitively think about it as an odd dim matrix, which has a characterstic polynomial that is odd, so it must have a real root

[hollow finch]

IVT ftw

[prime delta]

f(6) = 9 , f(-5)=0 can someone explain how to solve this please

[dire thunder]

no we can't

read #❓how-to-get-help

[hollow finch]

that is not linear algebra

[stoic python]

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

[dire thunder]

(i need to formalize it but still)

[steady fiber]

the "better way" to prove it is via induction

however

you can trivially prove it for dim V = 1

[dire thunder]

well yes

[steady fiber]

and you can do induction on every "next" odd dim space

[dire thunder]

but like is it really much needed

oh that's how axler does the proof

[steady fiber]

it doesn't say the operator is a matrix

which is why I would steer away from the characteristic polynomial proof

it's still possible to do it that way

but it's not as clean?

[dire thunder]

we already showed one two one corresponednce between linear maps and matrices tho

[steady fiber]

ya, I mean I know it's easy to do, it's just not the "cleanest" of proofs, for lack of better words

it's not wrong

[dire thunder]

and like he does not really use char polynomial

[steady fiber]

but it's better as a way to think about it rather than to prove it

[hollow finch]

characteristic polynomial makes it intuitively obvious but yea induction is probably cleaner

[dire thunder]

i mean he does not derive char poly itself

he just says "we can use polynomials with operators"

[wintry steppe]

axler avoiding characteristic polynomials

[dire thunder]

privyet ttera

kak dela

[wintry steppe]

privyet MVT

[dire thunder]

wotsup ttera

btw @wintry steppe

today we had quiz on series (and a bit of double integrals)
everyone: doing stuff through pain
me: ok calculus
10 mins later: i am done

[magic light]

Looking for some guidance.
Let S,T:V->V be linear transformations s.t Ker(T) = Ker(S) and Im(T) = Im(S)
prove S=T

My incomplete proof: if T(v) = 0 then v in KerT, therefore v in kerS, therefore S(v) = 0 and vice versa
so T=S when the result is 0

but I'm struggling to use Im(T) = Im(S) to get another equality like this, any guidance?

[tame mural]

If the domain and image are the same, then the function is the same, isn't it -_-a?

[magic light]

well I need to prove it

lol

ideally using actual linear algebra and not just general sets

[half ice]

No that's not true in general. Consider R → R, S = |x| and T = x²

Have same domain and image but aren't the same

[magic light]

Right

[half ice]

Now, yours is different because the nullspace is also the same

And these are linear operators

[magic light]

I know S=T, I'm just struggling to prove it

Basically for any v in V if T(v) = 0 then S=T for these v in V

that's easy

but T(v) = w and S(v) = w'

proving w=w' is hard

or alternatively
T(v) = w, S(v')= w => v=v'? not sure if that's correct

[hollow finch]

no

what if w is the zero vector?

v and v' could be different vectors in the kernel

[magic light]

w != 0 because I've dealt with 0 in my incomplete proof but yeah

[hollow finch]

maybe proof by contradiction would work

[magic light]

I have tried, but just came up with an idea

[tame mural]

Wait, the proposition isn't wrong, is it?

[magic light]

what proposition?

[native rampart]

Are you sure it's correct?

[tame mural]

For A = [1 0; 0 1] and B = [2 0; 0 2]

Don't they have the same domain and image?

And nullspace?

[magic light]

I'm not sure it's correct

[tame mural]

I'm a noob too so don't take my ideas too seriously

[magic light]

I think so but 🤷‍♀️

[tame mural]

but A and B are clealry different functions...

but if you're sure your teacher isn't going to throw you a trick question

then I'm not sure

[magic light]

the functions can be different in implementation what matters is that f(x) = g(x) no matter what x you throw at it I guess

but let me check your example

[half ice]

Good counter example, the prop doesn't seem to be true

[magic light]

Why is that example false?

S[x,y] = [x, y], T[x,y] = [2x, 2y]
KerT = {0} = KerS

ImT = R^2 = ImS no?

[half ice]

A is the identity
B is the operator that doubles a vector's distance from the origin

Clearly null is trivial
And range is all of R²

For both

[magic light]

yeah edited

ah

I see

so for [1, 1]

S[1,1] = [1, 1] and T[1, 1] = [2, 2]

man

I've been sitting on this for an hour trying to prove it instead of thinking of a counter example

[tame mural]

I see now, that's why counterexamples are useful

then you'll save work if you're doing the wrong thing

😄

[magic light]

sometimes you get stuck with a stupid idea and just can't see you're wasting time

[half ice]

Oh it was a true/false wasn't it?

[magic light]

I was sure because I used the KerT incomplete proof there must be a way for ImT

[magic light]

honestly, I didn't notice it was a T/F... I just saw "prove"

[half ice]

Haha

Good mental exercise

[magic light]

good at mental exhaustion

50 minutes wasted XD

thanks guys @tame mural

[wintry steppe]

[tame mural]

Does the cayley graph of a symmetric group always have a hamiltonian path?

Given the 2 generators you need

such as an n-cycle and a transposition

And is it easy to find?

[hollow finch]

you should probably @ the helpers if you still need help with your question

[tall thunder]

Is anyone able to explain this to me ? Or link a YouTube video that shows me how to answer it ?

I know what an orthonormal basis is but I don’t really know what this question is asking and how to do it

[tame mural]

an orthonormal transformation is one that doesn't change the "angles" or the "lengths" of your world

So if you had a unit cube in your world, that cube wouldn't get stretched or tilted

They want you to use a combination of the vectors in B

to reach w

[tall thunder]

Oh@got you thank you for deciphering that lmao

[tame mural]

B is called an orthonormal transformation because every vector inside is length 1 long

And because every vector is "orthogonal" to every other

[tall thunder]

Yeah I got that part, so how do I go about going from B to W ?

[tame mural]

You can imagine this as a systems of equations

[tall thunder]

Okay

[tame mural]

B applied to an answer vector X = w

you are trying to find X

[tall thunder]

Soooo like should I do b1 = w

And b2 = w all the way down etc

[tame mural]

RREF should get you the answer for problems like this

[tall thunder]

Ah ok

Thanks

[tame mural]

RREF(B augmented with w)

[tall thunder]

Got it

@tame mural like this ?

[tame mural]

Yeah you look like you're on the way to getting the answer

[tall thunder]

And just rref it ?

[tame mural]

yup

you're almost done

keep doing the RREF

[tall thunder]

It’s because i have a classmate and I’m looking at his work and he did something different so I’m just double checking

[tame mural]

Well one thing I'd check is if your classmate did anythign with orthognoal projection

that was imo the one slight ambiguity in the question

if your classmate did, then I'd be suspicious of misreading the question

[tall thunder]

Ohh hold on lemme show you his stuff

This is just a study guide so it’s not a test or anything cheating

[tame mural]

but otherwise you haven't finished your RREF

I'm not near my normal computer

so I can't check it

[tall thunder]

Yeah but I mean he diddnt do rref at all

[tame mural]

There are often many algorithms to the same answer

[tall thunder]

He did some like c1(xxxx) + c2(xxxx) or something

Ohhh ok ok

[tame mural]

If you got the same answer as him

then it's all good

[tall thunder]

Ok cool

[tall thunder]

Ahh it’s not coming out the same 😭

Nvm dude

[alpine sequoia]

You got it?

[tall thunder]

You were right hahaha

Yeah my dumbass just wrote the augmented matrix sideways

[alpine sequoia]

I felt that lol

[tall thunder]

Coincidentally the numbers are similar both ways

[tame mural]

I'd check out why your friend did it different

that could be the way your class wants

[tall thunder]

Yeah I have no idea but your way was way easier and we both got the same as the teachers answer

[wintry steppe]

I'm trying to find a linear map T on l^2(R) -> l^2(R) such that T* (the adjoint) doesn't exist. Is there some well known example of this, because I'm really struggling on this

My thoughts were that clearly T would have to be unbounded, so there's probably some thing with picking a countable set out of the basis and defining a linear map using that.

[half forge]

can someone help me with this one

its a true or false question but im not understanding it

if someone can walk me through it since im not sure how to start it

[tame mural]

My intuition is the answer is false, but I'm not very good at this

Basically they're saying I have two elements from the same vector space

[half forge]

yes it false i forgot to mention it

[tame mural]

And they each generate cylclic subspaces

so they're saying if I know they generate the same subspace

then they are the same

but no, of course not

there are many choices for generators that might not be equal

within the same subspace

[half forge]

ohh, wait so what would a counterexample be

[tame mural]

Well my example is gonna be kinda messy...

but my intuition came from doing stuff with error correction codes

[half forge]

oh can i see your messy example

[dreamy iron]

@brittle orchid and I are just talking and this is exactly what is going on.

https://cdn.discordapp.com/attachments/540211747613704221/783322190367948810/unknown.png

[brittle orchid]

I'm trying to speedrun videos on RRE as we speak 😂

[deft gull]

Bro when do you even take this math

this looks so hard

[tame mural]

@half forge Ah I just thought of an easy one. Rotation 90 degrees either forward or backward.

[half forge]

oh that would be an example

[brittle orchid]

Tbh it's not actually that bad

If I actually paid attention and studied 2 hours daily I would actually come out with a 1st lol

But of course no one has ever done that

[dreamy iron]

$${\begin{bmatrix}1 & 0 \ -3 & 1\end{bmatrix} \cdot \begin{bmatrix}3 & 9 \ 9 & 2\end{bmatrix} = \begin{bmatrix}3 & 9 \ 0 & -25 \end{bmatrix}$$

[stoic python]

ninnymonger is a physics main.
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tame mural]

lol

[dreamy iron]

cannot get that over brace to cooperate

[tame mural]

What r u trying to figure out or show

[dreamy iron]

the left most matrix is E_r (1,2;3)

that's the name according to his instructor

[brittle orchid]

what does this mean?

[tame mural]

It might be easier to take a picture of the book

[dreamy iron]

$$ \overbrace{\begin{bmatrix}1 & 0 \ -3 & 1\end{bmatrix} } ^{E_r(1,2;-3)} \cdot \begin{bmatrix}3 & 9 \ 9 & 2\end{bmatrix} = \begin{bmatrix}3 & 9 \ 0 & -25 \end{bmatrix}$$

[brittle orchid]

I came across this in some online notes (Google search) but I'm not sure how this relates to the row operation described in the earlier image

[stoic python]

ninnymonger is a physics main.

[dreamy iron]

the matrix encodes the row operation.

[brittle orchid]

Do you think it would be a bad idea to just ask my professor for help with notation when he calls me tomorrow most likely?

Stupid questions like "what does E_r(1,2;-3) mean"

[tame mural]

Ah I see

well the notation is defined above

but it's such an ugly notation lol

[dreamy iron]

$$ \overbrace{\begin{bmatrix}1 & 0 \ -3 & 1\end{bmatrix} } ^{E_r(1,2;-3)} \cdot \begin{bmatrix}3 & 9 \ 9 & 2\end{bmatrix} \cdot \underbrace{\begin{bmatrix} 1 & 0 \ -3 & 0 \end{bmatrix}}{E_c(1,2;-3)} = \begin{bmatrix}3 & 9 \ 0 & -25 \end{bmatrix} \cdot \underbrace{\begin{bmatrix} 1 & 0 \ -3 & 0 \end{bmatrix}}{E_c(1,2;-3)} = \begin{bmatrix} \sqrt{3} & 0 \ 0 &-25 \end{bmatrix} $$

[half forge]

@tame mural can you explain more of what you said

[stoic python]

ninnymonger is a physics main.

[dreamy iron]

@brittle orchid

[brittle orchid]

So what does the E mean then?

[tame mural]

E represents all the operations you've been doing so far

If the matrix is reversible, you'll ultimately end up with an inverse matrix

assuming this is rref

[brittle orchid]

it isn't necessarily ^

[dreamy iron]

okay, so piece by piece:

E_r(1,2;-3)

the E means it's an elementary row operation where there's no row switching

the r means it's a row operation so you have to multiply on the left side

the (1,2) means the value to consider happens in column 1 and row 2

the (1,2;-3) means the value to put in 1,2 is -3

[brittle orchid]

From my understanding, this is the -3 times row 1 of I_2? added to row 2 of I_2?

[dreamy iron]

hold up

hold up

[brittle orchid]

okay

[dreamy iron]

https://cdn.discordapp.com/attachments/540211747613704221/783322190367948810/unknown.png

okay so your professor is applying two "functions" that do the same thing

the top of the row is what you said. R_2 -> R_2 - 3R_1

and bottom of the arrow is saying what i just said, with the matrices.

https://cdn.discordapp.com/attachments/540211747613704221/783486104665325618/162017091996745728.png

[brittle orchid]

Edited

[dreamy iron]

the E_r(1,2;3) is just telling you what specific elementary matrix to "hit" the original matrix with.

[brittle orchid]

and then that's just matrix multiplication from there?

[dreamy iron]

yes

[brittle orchid]

Hmm I see

[dreamy iron]

step two, using his same notation:

https://cdn.discordapp.com/attachments/540211747613704221/783487443335774248/162017091996745728.png

[brittle orchid]

So, I have a question about the elementary matrices

[dreamy iron]

(it is a guy prof right. i just assumed. )

[brittle orchid]

yes

I would have corrected you

I don't really understand the E_r and E c

[dreamy iron]

if you can do elementary row operations using row switches and and row additions, then stick to that.

this E_r and E_C stuff is just the equivalent matrix representation of what it is you're actually doing, but using matrices to do the row switching for you.

[brittle orchid]

So, do you necessarily need to apply the "elementary" matrices or can you simply apply the row operations?

[dreamy iron]

okay, so say you want to do row switching in excell.....or google sheets.

and i've wanted to do this....

i can't tell google sheets to "do this row operation". the spreadsheet doesn't understand that language.

what the spreadsheet does understand is elementary matrix operations, so i use that instead.

[dreamy iron]

both, one or the other.

if youre confused, and you seem to be, stick to row operations

but remember that "whenever you do a row operation" that's just a matrix multiplication in disguise.

[tame mural]

@half forge How about this potentially similar and bad example; let S = {0, 1, 2}, or the integers mod 3. S can be generated through repeated application of n + 1 or n + 2.

[dreamy iron]

@brittle orchid one last question so youre not feeling defeated:

what is the most basic row operation?

[tame mural]

IMO a good way to encapsulate what you've been doing with matrices and all that E_r stuff is to simply put your matrix next to an identity one, and perform the same row operations on both

You are slowly creating that matrix

If your matrix is invertible, then you'll turn your original matrix into identity, and your agumented matrix into the inverse one

For many systems of equations problems, that inverse matrix is the "answer" matrix

[dreamy iron]

what that image is doing, is exactly what we're doing.

but that image is working with a larger elementary row matrices. we only need 2x2 matrices in our case.

if you look at E_1, E_2, and E_3, they're all simple looking matrices.....they are very close to the identity matrix....that's a big clue as to what is going on.

[brittle orchid]

Hi, apologies for the late response, give me a minute to catch up to this, was just sending an email

[brittle orchid]

The most basic row operation is simply a row multiplied by a scalar, idk?

[dreamy iron]

the most basic row operation is to do nothing.

the do nothing operation.

[brittle orchid]

so that would be the same as multiplying by a scalar, wouldn't it?

[dreamy iron]

no, it would the same as multiplying the entire matrix with the identity.

okay, let's be a bit more atomic with this:

$$\begin{bmatrix}3 & 9 \ 9 & 2\end{bmatrix}$$

[stoic python]

ninnymonger is a physics main.

[dreamy iron]

what's the first step to row reduce that matrix?

[brittle orchid]

I mean, assuming these matrices are coefficients of equations surely if you multiply a row by a scalar it "does nothing"

[dreamy iron]

i mean, i'm not looking at an individual equation here. we need to take entries in the matrix whole-cloth.

not piece-meal

[tame mural]

you both mostly understand each other

but ninny is more precise

and reasonably more precise

[brittle orchid]

well, the first step is to get rid of the 9 in the second column

[tame mural]

in programmer terms, ninny means to feel out the "API" of the matrix and don't get lazy with it

because that's what you'd want to come away with

as opposed to the ability to eyeball the problem

[dreamy iron]

are you able to do the reduction on a paper and send a picture?

[brittle orchid]

tbh, I've never done a reduction before because I never had to sit an algebra exam 😂 but yeah, time to try

imagine, doing algebra 3 without knowing 1 and 2

[dreamy iron]

yeah, this is a finger exercise. we gotta get our hands dirty.

[brittle orchid]

yeah, I really really appreciate the help, btw

also, it sounds like you got some programming experience just from the way you describe this

[brittle orchid]

the last person who described a problem as a "finger exercise" was my Computing teacher in school

[tame mural]

where r u from whiteberry?

[brittle orchid]

UK

[tame mural]

ohhh

no wonder I Didn't recognize what u were saying

algebra 3, computation teacher

I was like huh

[wintry steppe]

these are not really a question i can solve with numbers

anyone know where I should start with these questions?

its homework problems

[brittle orchid]

Oh I study (the module) Algebra 3 at university, I meant Computing teacher while I was at school ("high school"). I study Mathematical Computation at university (UCL, to be specific, and I'm in my 2nd year)

[tame mural]

They're basically asking you for 6 and 7, what's the definition of a vector space?

Some people memorize it as a list of properties

Use those properties as a test

[dreamy iron]

7. is a vector space

im not sure about 6.

[wintry steppe]

properties?

[tame mural]

Yes, exactly.

[wintry steppe]

I tried to use this to figure it out

[brittle orchid]

Btw ninny here we go. The reason I have no notation is because I don't know how to present it... 😅

[dreamy iron]

omg, lol

the arrows are hiding so many steps. haha

[brittle orchid]

I just saw this Chinese kid do it once

Next to me

[dreamy iron]

okay, so we need to start from the tippy-top

[brittle orchid]

and he did it with no notation >.>

[tame mural]

@wintry steppe how about this, you agree what they're basically saying is can you have vectors with integer "entries"

those aren't closed under scalar multiplication

because you can multiply by 1/2

[wintry steppe]

i think I am just confused with the wording on number 6

[tame mural]

Then you end up with fractions

[wintry steppe]

so you say scalar multiplication is not an due to fractions

[dreamy iron]

what if the scalars are only integers.

[tame mural]

Yup, because we need at LEAST fractions to qualify as scalars

"least"

[dreamy iron]

yeah, i guess the scalars do have to come from a field

what meow said

[wintry steppe]

without numbers how were you able to figure out

i am just confused I understand where you are coming from tho

[tame mural]

I imagine a scenario where I can do a valid operation on polynomials

but end up "outside" of the set

then I know I broke "closure"

[dreamy iron]

@brittle orchid https://www.youtube.com/watch?v=9LYVi-n-6Jw

[brittle orchid]

Thank you, will watch

Right away

[dreamy iron]

i hope i'm not talking down to you. and that this is of your required level.

[tame mural]

Yo, just do the augmented thing already lol

RREF([M | I ])

Do this

[wintry steppe]

I have two more questions on my hw

[brittle orchid]

Not at all, too much explanation really doesn't hurt anyone

[brittle orchid]

in a worst case scenario I'd be reviewing some concepts

[wintry steppe]

number e and g

i did from A to J but E and G just don't make sense

[tame mural]

Mmm...

not understanding e is dangerous, IMO

very very dangerous

[wintry steppe]

its true and false

[tame mural]

you are apparently in a hard class, but you need to get your foundations straight

[wintry steppe]

I think its false

but i can't prove it

[tame mural]

This one is worth understanding.

[dreamy iron]

column space of A is just the range of the linear transformation communicated by A?

^i think thats right.

[cursive narwhal]

Let $V$ be a vector space with dimension $n$ and let $v$ be one of the basis vectors in a specific basis. Then, define a set:

$${v,2v,3v,\ldots,(n+1)v}$$

This is a set of $n+1$ vectors that does not span $V$

[gray dust]

A is a matrix representation of a linear map T. indeed col(A)=im(T)

[wintry steppe]

@tame mural you are right tho I don't hace my foundation down all I know is to solve problems.. so when it comes down to true and false or anything without numbers I don't know how to start it

[stoic python]

Abhijeet Vats

[gray dust]

also try proving e

[brittle orchid]

lockettojampuu!!

omg

[gray dust]

hi

[brittle orchid]

hii

[cursive narwhal]

@wintry steppe For g, is what I wrote reasonable or no?

[wintry steppe]

oh i'm sorry let me check real fast

so you are saying it does not spann

[cursive narwhal]

Indeed. I'm assuming that V is a real vector space over here, of course

[wintry steppe]

i'm not sure the how you are able to define a set $${v,2v,3v,\ldots,(n+1)v}$$

[stoic python]

HUMBA CHUMBA

[wintry steppe]

I don't think I took a high enough class to understand that part

i have not taken math proof

MATH 300 yet

[cursive narwhal]

Well, to explain why you can construct a set like that, you do need to study a bit more set theory. On the other hand, if you are willing to accept that I can take subsets of a given set, then all I'm doing above is taking a specific vector $v$, looking at $1v, 2v,3v, \ldots, (n+1)v$ and then putting all of these different vectors in a single set.

[stoic python]

Abhijeet Vats

[gray dust]

assume every linalg student has only seen naive set theory, abhi

[cursive narwhal]

haha well that's probably a fair assumption to make

@wintry steppe Does that sound reasonable, though?

[dreamy iron]

i defo don't have any axiomatic set theory under my belt

[gray dust]

most undergrad classes don't care

[cursive narwhal]

Well, it's not necessary to have axiomatic set theory to get started

[wintry steppe]

yes it makes sense

thank you so much it took me a couple minutes to understand

sorry lol i'm not the brightest

[cursive narwhal]

You're welcome

Well

[brittle orchid]

@dreamy iron I watched part 1 & part 2 of Row Reducing a Matrix - Systems of Linear Equations by PatrickJMT as you recommended. So I can really just use simple notation like that? I don't have to use lambda (i, j), delta (i, j), epsilon (i, j, k), idek

[cursive narwhal]

you're still learning the subject so don't worry about it too much

But you should try to learn a bit of set theory to get yourself acquainted with dealing with different kinds of sets and so on

[wintry steppe]

@cursive narwhal can i show you my calculator real fast

[cursive narwhal]

Sure

[wintry steppe]

i hope its not out of your comfort zone

[cursive narwhal]

Why would it be?

[wintry steppe]

i would say its linearly dependent because

the row of 000 on the bottom

[dreamy iron]

your instructor uses the same notation

[brittle orchid]

what is this though

[dreamy iron]

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

[wintry steppe]

so S is not a basis for r^2 since S is not linearly independent

[brittle orchid]

So is that the same as for instance R_2 -> 5R_2 to represent row 2 being replaced with 5 times row 2

[dreamy iron]

yes

[brittle orchid]

I see

[dreamy iron]

that's the general case for matrices of any size

the r means it comes from the left, okay.

and your instructor is saying $(1;\sqrt{3})$ just so that the first entry is to be replaced by $\sqrt{3}$

[brittle orchid]

How do you read that verbally, btw?

[stoic python]

ninnymonger is a physics main.

[brittle orchid]

This, I mean

[dreamy iron]

i have no idea, lol. i would write it on the board and point and say, do this thing.

maybe "multiply on the left by a diagonal matrix where the first entry is square root of 3"

that's prolly how i would "read that verbally"

[brittle orchid]

or perhaps multiply each element in row 1 by 1 over root 3

[dreamy iron]

that i would not say

definitely not.

[brittle orchid]

I understand that it does not capture the purpose of the operation

but it does capture what the end result it

does it not?

[dreamy iron]

yes, it captures the end result but it's not good math behavior.

there's no mention or appeal to a matrix multiplication.

so if there's not appeal to matrix multiplication then why use diagonal matrices.

[brittle orchid]

You know, I didn't quite know why my professor used the D symbol till I clocked that the way to scalar multiply each element would be to use a diagonal matrix

[dreamy iron]

just to clarify, you asked "how do you read that verbally" which is not the same question as "what do you get when you do this"

[brittle orchid]

since, in our Algebra 1 module, we were taught the notations lambda (i,j), P(i,j), amongst others, when performing row operations

and the diagonal thing was a subtle reference which was never made

perhaps they expected us to figure it out but I wasn't that sharp 😂

[dreamy iron]

you run into it explicitly eventually, which is now, it would seem. so congratz is in order

[brittle orchid]

😂 thank you?

[dreamy iron]

[brittle orchid]

ahh I wish I paid a little more attention during algebra 1

But thank you so much, I must go to sleep now

@dreamy iron I hope you have an amazing day 🙂

[wintry steppe]

would s be is a basis for r^2

because even if there is 000 on the bottom there are still 2 rows

1 0 0

0 1 0

0 0 0

there are still R^2

[tame mural]

Yes, so you could be asked to solve for Mx = [0 0 1]

And that would be outside of the span of your matrix

And that's when your systems of equations is "inconsistent"

[wintry steppe]

thank you

[tame mural]

Is there an easy way to find the hamiltonian path for the cayley graph of a finite symmetric group?

Given two generators, a transposition and an n-cycle

[wintry steppe]

How do I show that if $U ={(x,0,0) \in \mathbb{F}^3 : x \in \mathbb{F} }$ and $W = {(0,y,0) \in \mathbb{F}^3 : y \in \mathbb{F} }$ then $U + W = { (x,y,0) : x, y \in \mathbb{F} }$ by showing that the sets are equal?

[stoic python]

LINEAR_ALGEBRA_GUY

[wintry steppe]

I need to show that U + W is a subset of the right hand side and the other way around

[dire thunder]

oh you multiposted haven't you

[rose grotto]

how do I add demand function

[wintry steppe]

are you sure this is linear algebra?

check the pinned message

[hallow lodge]

what's the difference between an ordered basis and a standard ordered basis

[native rampart]

Standard ordered basis is an ordered basis {v1=(1,0,0,..),v2=(0,1,0,0..)...vn=(0,0,0....,1)}

[dusky epoch]

a standard basis is one that your space "comes with"

like

the default basically

if there is one

[hallow lodge]

ty's

[dense wyvern]

https://cdn.discordapp.com/attachments/496784996510466057/783686192075505664/1.png

hey can any1 say wht do i get the final answer of the chain rule
for this math

[round coral]

@dense wyvern this is not a linear algebra question, wrong channel. Also if you want to check your answer just differentiate it online on symbolab or wolfram. Better than asking here

[faint lintel]

So I'm taking a proof based linear algebra course this next semester.

This is the textbook: Peter Lax, Linear Algebra and its applications, second edition.

I have literally 0 linear algebra experience. I know how to do determinants and basically vector operation (dot and cross) and that's it. What's a good way to go about self studying some intro linear algebra at a high conceptual level over break?

[shrewd mortar]

if you're just looking for intuition or something, people seem to like the 3b1b series

i'd recommend you read ahead though

just like get acquainted with proof-based LA

to make the transition to it easier, assuming this is your first time

[quartz compass]

I'd recommend learning eigenvectors/eigenvalues and diagonalization as being an important goal

[faint lintel]

ok

I'm taking intro to proofs rn but we'll see how different it is to LA

[half ice]

Honestly Axler is a good book

It has a weird treatment of determinants though

But you already know about that!

[round coral]

Yeah Axler doesn't give much importance to determinants in most of the book except the last chapter. But overall it is really good for theory. Other good books maybe are Linear Algebra, by Hoffman and Kunze or Linear Algebra, by Friedberg, Insel and Spence

[uncut wave]

Hello. I am trying to understand this solution on problem:
A linear isometry on R^n is an orthogonal matrix And the solution I found is as shown in the image. However I am unable to understand on how the first equality came?
<Ax, Ay > = 1/4 {|Ax....} (this term) Can someone please help

[brave pier]

If matrix A is invertible, how would you determine if the power of said matrix is invertible?

[native rampart]

Note (A^-1)^n is the inverse of (A^n)

So,A^n is invertible

[brave pier]

Oh.. of course lol

I am dumb

[native rampart]

Use properties of inner product

|A(x+y)|^2=<A(x+y),A(x+y)>

[uncut wave]

Ok, I am trying to simplify using properties

[tame mural]

Does a metric on a set mean there is an identity element inside?

[dusky epoch]

no

[tame mural]

I see, thanks

[tame mural]

Does a distance/metric on a vector space mean you have a normed vector space now?

Because you have the 0 vector now

[dusky epoch]

no, not necessarily

you can put a very stupid metric on your vector space

theres no guarantee that the metric comes from a norm unless you require it to

[tame mural]

I see.....

thanks, much clarify

[uncut wave]

Hello, I am trying to prove something, but not sure if my argument is correct or not, can anyone please comment if there is any mistake or lack of clarity.

The problem is:
Suppose V is a finite dimensional vector space, and T : V -> V, Prove that  T is a scalar multiple of identity iff ST = TS for every S : V -> V. 
==> If T is a scalar multiple of identity then T = c I for some scalar c. 
TS = c I S = c S = S c = S c I = S T 
<== If TS = ST 
Then T v = c v, for some eigenvector v for an eigenvalue c. (But I can only write this when V is finite dimensional ( WHY ? = Because for a finite dimensional vector space, the smallest Vector space of dimension 1, there will be at least 1 eigen vector of dimension 1 spanning it or a 'dimension' of it, so we can write this), but what happens in infinite dimensional case?
So we have T v = c v, left multiplying by some S 
S T v  = S c v
T S v = c S v 
which means Sv is an eigenvector of T for eigenvalue c, since S was arbitrary, this holds for any S
So this means T x = c x for all x, so T is a scalar multiple of identity, i.e. T = c I 

Here, since S was arbitrary, but 'v' was fixed, why can I say this for all 'x'? Does this means every x belongs to V is of the form Sv for a fixed v, but some linear operator S will ALWAYS exists which will map 'v' to 'x'? 

[wintry steppe]

this is kind of hard to read and might get a quicker answer if you typed it in latex

[round coral]

Not always this may be the case, I think. You can have a T and S operators on finite. dim vector space, and T may not be a scalar multiple of the identity matrix but you may still have TS= ST , you can take S=T or you can take S as the zero operator or S as the identity operator

Correct me if I am wrong somewhere though

[native rampart]

The person is asking if such a S exists

[round coral]

Didn't he say TS = ST iff T is a scalar multiple of Identity

[native rampart]

i.e.,can you always find a map S such that S(v)=x

T satisfying those conditions,for any choice of S

It's an axler question,I believe

Yea,You can always find a linear map S: S(x)=v for any choice of x and v

Just consider a basis,one of whose elements is x

A linear map is defined by how it acts on the basis elements

[uncut wave]

The iff condition holds here if TS = ST for every such S, but yes, it may not be true if we are talking of a fixed S which commutes with T.

[uncut wave]

Very well, this makes sense! So I can choose a basis such that the linear operator maps one vector to another. So it will depend on the basis. This made sense. Thank you!

[wintry sphinx]

my interpretation is that T is a scalar multiple of the identity if T commutes with every linear operator S from V -> V

[uncut wave]

Yes. with "EVERY" such S

[wintry sphinx]

that's not what you wrote though

[uncut wave]

TS = ST iff T is a scalar multiple of Identity for every such S from V -> V

[wintry sphinx]

yeah that's very confusing wording that can be interpreted multiple ways

for one, you've redefined S

and for two, you could read it as TS = ST if (T is a scalar multiple of identity for every such S from V -> V) which is technically always true

[uncut wave]

The if part is trivial, so it will always be true, the converse is where I had questions

[wintry sphinx]

I'm not making a statement about the thing you have to prove

I'm just saying that the wording is off

[uncut wave]

Ok, I will take care from next time. Thanks for pointing it out.

[wintry sphinx]

well, you should take care of it this time and rewrite your proof in LaTeX so it's easier to read

[uncut wave]

I have taken care of the wording. I cannot change it to LaTeX as I need to figure out how to do that in here. So that I will learn and do next time.

[calm hamlet]

If your matrix T is a rotation in V with real scalars, you cannot find eigenvalues (unless particular cases)

[wintry sphinx]

complex eigenvalues if you take the complex numbers

[calm hamlet]

that's why I said real

[native rampart]

You can easily show T has an eigenvalue by considering S such that Se1=e1 and S maps everything else to 0

[uncut wave]

But can I comment on the existence?

[rose umbra]

is group called vector space if it contains objects that can express every point in the space?

[calm hamlet]

In a C-vector space, you always have eigenvalues

But not in a R-vector space

[native rampart]

Benefits of C being algebraically closed

[wintry sphinx]

real vector spaces are trash

[native rampart]

What about real dot product?

[round coral]

@calm hamlet The rotation matrix can have real eigenvalues too when the angle of rotation is k \pi. This may not be a good answer though. But you are right apart from these, it does not have eigenvalues in real vector space

[rose umbra]

@calm hamlet what are c and R

[native rampart]

C is field of complex numbers

R is field of real numbers

[rose umbra]

@native rampart so group called R- vector space if it contains objects that can express every R point in the space?

[native rampart]

What's a R point

[rose umbra]

i mean every point of real numbers

[calm hamlet]

@round coral that's why I said "unless particular cases" haha

If it works for any S, it necessarily works with well chosen matrices S Zenquiorra

[uncut wave]

So when I assumed this in the proof
Then T v = c v, for some eigenvector v for an eigenvalue c.
I cannot always write this if V is finite dimensional but NOT algebraically closed. Because such an eigenvalue may not exist?
Please correct me if I am wrong.

@native rampart @calm hamlet Please comment, I am so confused.

[native rampart]

You can't say that in general

But with the info given in the question, you can

[native rampart]

This

[errant mist]

If a matrix is irreducible is it also primitive and vice versa?

[round coral]

Primitive matrices are always irreducible but not the other way around always

[uncut wave]

Ok.

[shrewd mortar]

@half ice @round coral yeah i actually quite like axler

yes, determinant treatment is weird

[errant mist]

@round coral Thanks)

[shrewd mortar]

but you can just supplement it with another source

but i think axler's exposition and the way the book is structured and the proofs given etc etc are all good

[copper stratus]

Does anyone know how to prove this?

<@&286206848099549185>

[half ice]

@copper stratus
So watch out for the helper ping, it is expected to be used 15 mins after not being answered.

[magic light]

haha my question is very similar

if you want we can share answers

[copper stratus]

cool

[acoustic path]

cool

[wintry steppe]

or you can just prove it

[copper stratus]

Anyone wanna help with that 🤔

[wintry steppe]

how do you usually show something is linearly independent

[calm hamlet]

Well I was about to give an answer but yeah it's useless

[copper stratus]

don't stop......: 🤣

[wintry steppe]

if you just want an answer then use google

[acoustic path]

is it if the determinant is nonzero @wintry steppe