linear-algebra

[still turret]

alright thanks 🙏

[wintry steppe]

yeyeyeye and no problem!

[modest notch]

Can someone give me a hint on how to show that if A^2 is equal to the zero matrix, then the only eigenvalue of A is 0? I've already shown that 0 is an eigenvalue, but I'm having trouble showing that it's the only one

[teal grotto]

if Ax = kx then A^2x = k^2x = 0 so k^2 = 0 since x is assumed to be non-zero

[viscid lagoon]

A finite cycle group is a group with the elements $\left{1, a, a^{2}, a^{3}, \cdots, a^{n-1}\right}$ such that $a^{i} \cdot a^{j}=a^{(i+j) \bmod n} \text { and } a^{n}=1$. Let $A,B$ be finite cycle groups of finite order. Show that A is isomorphic to B if $#A = #B$.

[stoic python]

lewis

[still turret]

(found the error)

[tired fossil]

Hello guys, does anybody know how to do d?

[glass axle]

cofactor expansion

[tired fossil]

So det of each element of the matrix?

[half ice]

How did you do a-c?

[tired fossil]

i didnt, but i am only doing the ones that have answers in the back of the book

[still turret]

can anyone explain me how to do this

[half ice]

@tired fossil
If you've never taken a determinant before, check up on YouTube

[tired fossil]

I got 2 for d

i know how to do a determinant, but how do i know what c is?

[tired fossil]

for this one if I am not mistaken you do [3,4//4, -1]=[1//0] and [3,4//4, -1]=[0//1] for each case you use x and y and multiply them by the 3x2 matrices and subtract them, you should get two columns, that is you A matrix

@still turret

[still turret]

,rccw

[tired fossil]

Tahts just the question

[still turret]

yeah I was reading the solution

alright thanks bud

🙏

[wintry steppe]

my professor's being a little condescending about this question, so if anyone who's taken linear algebra before could help out, it'd be much appreciated

tldr; the question is “if we use the second biggest eigenvalue” if it says 1, 1, -1, do i pick 1 or -1

[half ice]

I'm not learning from that professor, and cannot intuit what exactly they mean any better than you can

[still turret]

your professor uses discord lol

[wintry steppe]

"largest means >"

fair, just based on the slide/concept it’s based on i’m trying to get advice from people who’ve studied this concept

[wintry steppe]

yeah lol he doesn’t respond v well tho

[half ice]

Honestly, if the question is "What is the second largest" then just list them all in order

You can't be wrong

[wintry steppe]

no like the issue is there’s a repeating eigenvalue

it’s 1 with algebraic multiplicity 2

and -1 with algebraic multiplicity 1

[half ice]

So write -1, 1, 1 kekw

I don't see why they'd ask "what is the second largest" specifically

Seems odd

[wintry steppe]

i’m not solving for eigenvalues - it’s asking me for the second largest one because that’s what you need for constrained optimization problems

it’s used for the maximum value of a constrained function on some surface plot

which looks like this

[tired fossil]

how do i start, ?

[granite kraken]

Idk

Try expanding along 1st row and see if anything simplifies

[tired fossil]

didnt work

[wintry steppe]

@tired fossil why not take the determinant of the second one and plug in -2 for ad - bc

[tired fossil]

will try ty

[wintry steppe]

i think thats probably it - lmk how it works out

[tired fossil]

I just did b and got the answer like taht so i think it is the same for a

[wintry steppe]

for sure

[tired fossil]

yea, i miscalculated the det the first time

I figured

[raven badger]

Is this phrase true:

All diagonalmatrices in R^nxn space will have at least one eigenvector associated?

[wintry steppe]

maybe

[winter harbor]

If you have a diagonal matrix A = diag(a_1, ..., a_n) and you take e_i = (0,...,1,...,0) the vector whose i-th coordinate is 1 and the others are 0, then:
A(e_i) = a_i * e_i

[stable kindle]

i don't understand what the trace actually is

what's the intuition

[lethal tangle]

Hi could someone help me with this question please?

[wintry steppe]

a geometric way of viewing it is that $$\left.\frac{d}{dt}\right|_{t=0} \det(I + tA) = \mathrm{trace}(A),$$ so the trace tells you how a matrix $A$ "close to $I$" changes volume

[stoic python]

TTerra

[wintry steppe]

or something like that

https://mathoverflow.net/questions/13526/geometric-interpretation-of-trace

lovely thread

[wintry steppe]

do you know how to show something's a subspace?

[lethal tangle]

I think so

?

[wintry steppe]

yes

so to show that $U^\perp$ is a subspace of $\bR^n$, you have to check that those conditions hold

[stoic python]

TTerra

[lethal tangle]

ok

I think I can do that😅

thanks!

[stable kindle]

tf

[wintry steppe]

the same thing's probably in the thread but with a better explanation

[stable kindle]

is that even true

[wintry steppe]

yes

[stable kindle]

i don't think i can prove it

[wintry steppe]

you could use one of the long formulas for the determinant involving permutations or cofactors or some messy thing like that

you could also try to relate det(I + tA) to the characteristic polynomial, which you can factor into a product of (t - eigenvalues)

[stable kindle]

oh, true

that's sorta approachable

[wintry steppe]

if you know some properties of the matrix exponential you may also be able to use those

since I + tA = exp(tA) + o(t)

but that might be a nuke

nuke: det(I + tA) = det(exp(tA) + o(t)) = det(exp(tA)) + o(t) = exp(t trace(A)) + o(t) = 1 + trace(A)t +o(t)

[stable kindle]

no clue what the matrix exponential is

out of my depth

[wintry steppe]

it's the best thing you'll ever learn

[stable kindle]

bold of you to assume i'll be able to learn it

[wintry steppe]

you will

it's literally just the taylor series for e^x but you stick a matrix in

[stable kindle]

huh

that's pretty awesome ngl

[wintry steppe]

you know how the one dimensional ODE x' = cx, x(0) = 1 has the solution x(t) = e^(ct)?

well if you go to n dimensions and replace c with a matrix

the solution is the matrix exponential of tA times ur initial data

it's cool

[stable kindle]

bruh

[wintry steppe]

im being a little imprecise

[stable kindle]

i get the gist of it

[wintry steppe]

but it's all very natural stuff

[vagrant hazel]

Hi, could someone point me in the right direction with this ? All I can figure out is the for all n it is = 1

Find a formula for the general term of the sequence a n where
a_(n+3 )= 5a_(n+2 ) − 8a_(n+1) + 4a_n ,and
a1 = a2 = a3 = 1.

[zinc timber]

you are posting this in LA means you want a matrix solution?

[vagrant hazel]

yes Indeed 😄

[zinc timber]

you can try similar thing like in the fibonacci seq

$$\mqty[F_{n+1} \ F_n] = \mqty[ 1 & 1 \ 1 & 0 ] \mqty[F_n \ F_{n-1}]$$

[stoic python]

Ryu?

[zinc timber]

but since you have 3 starting element, you'll have to form a 3x3 matrix

[vagrant hazel]

ok thanks ill have a go now 😄

[zinc timber]

you have to do some diagonalization shit

[worldly bear]

i love diagonalization shit

[zinc timber]

[silver heath]

Question.... If we have $PAP^T$ Where P is an orthagonal matrix

[stoic python]

Elonmosqito96

[silver heath]

Is this equal to just A?

because P is a rotation matrix

and then we apply the linear transformation A

and then undo the rotation?

Is something wrong with my logic?

[agile bronze]

For orthogonal matrices Q with orthonormal columns, transpose and inverse are equal

If PAP^T = A then PA = AP

You can definitely think of a counterexample with an easy A

[silver heath]

yeah ik this statement is false but what is wrong with my logic/geometrically speaking

[agile bronze]

What you said describes P^-1(PA) = A

The inverse on the right is not the exact same as undoing what P did on the left of A

You might be familiar with matrices acting "on the rows or columns" depending on whether it's on the left or right of a matrix

[silver heath]

Wait im confused, does AB apply transformation A and then transformation B? If so doesnt PAP^-1 apply rotation P then transformation A then undo rotation P????????

*edit... what does P^-1 do if its not undoing the rotation then?

[agile bronze]

ABC applies B to C and then A after (to BC)

[silver heath]

wait what

i thought it goes from left to right?

[agile bronze]

As long as you associate right, (AB)C = A(BC)

Mightve gave a confusing explanation

[silver heath]

(AB)C means that A applies to B and then AB applies to C?

[agile bronze]

To answer your question, the P^-1 does not completely reverse what P did on the other side

[silver heath]

sry to be a fly, but what does P^-1 do then?

[agile bronze]

Let's say P on the left did something to the columns of A. Then, P^-1 = P^T did something similarly to the columns (it would be rows since it's on the right, but it's the transpose) again of A.

If anything, they have a constructive effect instead of a reversing effect.

[lethal tangle]

Hi could someone help me with part c) and d)

[vocal isle]

hey folks can someone show me a proof for why the minimum of f(x) is the solution to Qx = c ?

[lavish jewel]

if Q is symmetric positive definite, then f(x) is a quadratic form, which is strictly convex

that means that, if it has a minimizer, it is unique, and at that point, the gradient of f(x) will be 0

the gradient of f(x) is precisely Qx + c, and the minimizer x* is found by setting the gradient equal to 0

so that Qx = -c, and x* = Q^-1 c

this isn't really linear algebra, though, in spite of there being matrices and vectors involved. it's part of multivar calc/optimization

you can easily google proofs of the uniqueness of the minimizer of a quadratic form, and also explanations on convex optimization

lots of fun with line segments/convex combinations and setting up nice inequalities based on the definition of convexity

the only real link to linalg is in the final statement that x needs to satisfy Qx = -c, since the problem may have one, infinitely many, or no sols depending on the rank of Q

[vocal isle]

thanks Edd, Ok maybe i'll use the multivariable calc for further questions. But all of these statements are trival except for the ones which involve linear algebra. For example, why does Q being positive definite mean that it' sstricly convex? Also how do you take the gradient when you've got transposes in there?

solving this in the 1d case is trivial. It's min 1/2 Q x^2 + cx ---> Qx + c = 0 ---> x = -c/Q

but it's the linear algebra part that gets me 😦

[lavish jewel]

for the gradients, that's still calc. what i can tell you is to use the matrix cookbook and/or look up matrix calculus on wikipedia. at the end of the day, what you want is to write the matrix-vector products as sums and then use your regular calculus rules on each element of the result

for example, x^TQx is a scalar that depends on all of the x_i in the vector x. the gradient is therefore a vector of the same size as x

and you can compute the gradient by differentiating every element in the (double) sum that defines the product x^TQx

you should only need to do it for a couple of elements to conclude that the gradient of x^TQx is 2Qx

as for the part of positive (semi) definiteness implying convexity, gimme a second to fetch some notes

that is a bit more involved

(and also not linalg)

ok, this is the direction convex -> pos semi def

and next, pos semi def -> convex

[vocal isle]

woah! Can you provide link plz?

i don't think this was in the cookbook

[lavish jewel]

it wasn't

these are notes i wrote for my students

[vocal isle]

thanks Edd

I've also been looking at the following source: I can understand how this is true:

and now i need t somehow use the product rule to show tha tthis is true:

i think d/dx(xTAx) = d/dx(xT b) where b = Ax

then d/dx (xT b) = bT

[lavish jewel]

sure, seems ok

i think there's a factor 2 missing

[vocal isle]

yes there is, but i'm breaking it into parts

so i'm just considering the xT A x part first

[lavish jewel]

here are the full notes. lemme know if you catch any mistakes

[vocal isle]

thank you so much! Do you made these, is hat right?

so you made this, is that right **

[lavish jewel]

yeah

[vocal isle]

wow impressive, thanks for sharing!

[lavish jewel]

i don't have any matrix calc stuff in there though, i only use its results in the framework of convex opt

[vocal isle]

do you have any other notes in convex optimization? At the end of the day thats what i want to really learn 🙂

[lavish jewel]

i think the references in the pdf will do well

i link to some slides used by other unis, and also to a book and a smaller manuscript by stephen boyd

stephen boyd is pretty up there in convex opt

regarding the derivatives you're looking at rn, i just found this on stack exchange

[warm frost]

https://i.stack.imgur.com/782AU.png ♥

[lavish jewel]

depending on the equation and the original system, either nothing, the system becomes inconsistent, or there is now only one unique solution

[vocal isle]

thanks Edd, I'll look at all these resources! But there's one thing, in this image, I believe the ansewr is wrong. the answer should be a, not a^T, right?

[lavish jewel]

ah, that's the shitty part

there is no universal convention for this

it could be either

from the linalg perspective, a makes sense because we wanna see what happens to the vectors in the original vector space

thinking about linear transformations instead of elements of the original vector space though, the transpose can make more sense because it is used to define differentials

i.e. the gradient is also a linear transformation, and particularly, it is involved in determining a first order linear approximation to a function

in these cases were we have f:R^n -> R, the line between jacobian and gradient is blurry, i would say 😛 (i might be wildly wrong, i'm not a mathematician)

at any rate, in the real case, this is only a transpose

in the complex case, CR calc time, usually

[vocal isle]

so are you saying df/dx (where x = nx1 vector) is not well defined?

I would have thought df/dx = [df/dx11; df/dx2; df/dx3 ...]

which is also a nx1 vector

[lavish jewel]

it is well defined, the notation is not unique

[vocal isle]

also df/dx1 is the PARTIAL derivative

[lavish jewel]

and the deal is that the dual space of R^n can be obtained by transposing vectors in it

so the question is, do you want vectors in R^n, or linear maps R^n -> R

and ofc it depends on what you're doing

if you wanna do gradient descent, for example, you care about the gradient as a vector in R^n

in taylor approximations, you care about it as a linear form

so what you have to do is simply make your notation clear

at the beginning of what you're writing, you say something like "we use numerator/denominator notation" or "let the gradient be ... "

[teal grotto]

the map f(A) = AA^T - I_n from the space of real n by n matrices to the space of real symmetric matrices should be surjective right?

[lavish jewel]

can't confirm nor deny :x

[teal grotto]

cry

like, its directional derivatives are, Df_A(V) = AV^T + A^TV

wait

that might be a lie, i need to check something

well, my whole goal was to show that the map from Mat_n(R) --> Sym_n(R) given by AA^T - I_n has zero set O(n)

and that this map was smooth

[lavish jewel]

what is O(n)?

[teal grotto]

group of real orthogonal matrices

[lavish jewel]

aight yeah

for square matrices, makes sense

[teal grotto]

because this map basically shows you that O(n) is a compact Lie group and an n manifold without boundary as a subset of Mat_n(R) (that last part i also need some help with)

working with manifolds is just super not natural/intuitive for me rn

[lavish jewel]

i know nothing about them :x let's wait for someone else to materialize

[zinc timber]

any non brute force method to see why this is true?

$$
\det(AB-BA) =\frac{\tr[(AB-BA)^3]}{3}$$

[stoic python]

Ryu?

[zinc timber]

feels like surjective but can't say 100%

[teal grotto]

i dont think i really need it anyway

i just have no idea how to show that something is a manifold

[zinc timber]

hmm like you can solve AA^T for any any B+I, feels like it dhould have a solution

[teal grotto]

yea

been playing around with a bunch of diff expressions. nothing is really working

but again, i dont think i require it. i was confusing something with something else

[zinc timber]

if B+I is +ve def then cholesky would work

[teal grotto]

gonna look up what that is later

[teal grotto]

is this just true for all square matrices?

[zinc timber]

it's a T/F question and answer is given true

so yeah

[teal grotto]

that seems crazy to me

[zinc timber]

IKR

but again, AB-BA has a lot of degeneracy

wait A^TA is always +ve semidefinite but not all symmetric matrix are, don't think it's surjective now @teal grotto

[lavish jewel]

it does have that -I term that mangles the eigenvalues though

[zinc timber]

don't think it can manage all -ve s

like -1000

[lavish jewel]

ok, that much is true

the eigvals can only be as negative as -1

[zinc timber]

hmm

[lavish jewel]

since AA^T - I is diagonalizable as Q (D-I) Q^-1, and D has diag entries that are >= 0

[teal grotto]

https://math.stackexchange.com/questions/3505997/prove-that-detab-ba-frac13-left-mathrmtraceab-ba3-right

these are kinda just answers, but i assumed u wanted one since u were asking here anyways, lol

[zinc timber]

yeah should have googled first

[teal grotto]

its more fun asking here tho

[vocal isle]

So I've been trying to find a linear algebra way to differentiate xTAx where x is a nx1 vector and A is a nxn matrix. I was previously told that the answer to this is 2Ax. But this is not true! The general answer is (A+AT)x which only = 2Ax if A is symmetric (because AT=A)

https://www.sfu.ca/~haiyunc/notes/matrix_calculus.pdf

i've found these lecture notes, but they only prove the result by breaking it down into its constituent components

is there a "linear algebra way" to find d/dx (xTAx) ?

[lavish jewel]

ah, i assumed A was symmetric, since you were looking at a quadratic form earlier

[zinc timber]

yu cn find it componentwise and reconstruct the matrix

[lavish jewel]

this is calculus, there is no "linalg way" of calculating a derivative

[vocal isle]

why does quadratic form imply that A must be symmetric? I thought quadratic form just meant 1/2 xTAx + xTb

[lavish jewel]

all of the properties you see in the matrix cookbook come from doing what ryu says

no, it's only a quadratic form if A is pos semi def

[vocal isle]

oh

[lavish jewel]

at any rate, you'd have to split it componentwise and see what you get

say x^T (Ax), then use the chain rule

[zinc timber]

also A is a quadratic form then you can write A equivalently with a symmetric matrix B, , say B=½(A+A^T)

[lavish jewel]

you'd get Ax + (x^TA)^T = Ax + A^Tx

(that easily if you already know how to differentiate forms c^T x and Ax, which is done... by looking at it componentwise and learning the result from the pain)

or transpose the result if you like row vectors better

the answer will always be "LoOk aT tHe SuM!"

[zinc timber]

yeah, I do find the derivatives little confusing in my optimization class

wHeRe ShOuLd thE ^T gO

[lavish jewel]

[zinc timber]

that cook book came in handy tho

[dark brook]

If I wanted to check a set of 3 4x1 vectors if they are linear independent, I would just do RREF as one way to do it?

[lavish jewel]

sure, that would work

[dark brook]

Would you prefer any other ways?

[lavish jewel]

on paper, not really

unless the vectors are so simple that you can just look at it and go "ah!", RREF is pretty safe

[dark brook]

Neat, it is just an assignment, but the exam is going to be on paper. Thanks 🙂

[outer goblet]

u_1 and u_2 are ortonormal

i have to show that u_1q=0=u_2q

[wintry steppe]

is there a way to generalize tensors to fractional ranks? or as the methmeticians would say "extending the definition"

[outer goblet]

but isnt q here just 0?

[dusky epoch]

no, it's not 0

it's only 0 if w happened to be in span{u_1, u_2}

[outer goblet]

so only if w=(1,1)

[dusky epoch]

what no

[outer goblet]

hm

[dusky epoch]

do you think span{u_1, u_2} consists of u_1, u_2, u_1+u_2 and nothing else

if so then you have a huge misunderstanding of what span is and should read up on it later

[outer goblet]

hm

[dusky epoch]

anyway, you should remember that $\mathrm{proj}{\bd{u}}(\bd{w}) \cdot \bd{u} = \bd{w} \cdot \bd{u}$ and also that $\mathrm{proj}{\bd{u}}(w)$ is parallel to $\bd{u}$

[stoic python]

Ann

[outer goblet]

aaa projection on a vector can be shorter/longer than that vector

right?

[dusky epoch]

of course it can???

the length of the projected-onto vector is literally immaterial

[dark brook]

When you want to find an orthonomalbasis for V, you'll use the Gram-Schmidt process for finding it, right?

[dim epoch]

gram schmidt gives you an orthogonal basis but you can define every element of that basis multiplied with 1/its norm and get an orthonormal basis

[dark brook]

Yea, I was just not sure if I had to use another gram Schmidt process or the regular one

[zinc timber]

pretty sure gram-schmidt also includes the normalization

[nocturne jewel]

Gram Schmidt orthonormalizes and maintains span iirc

[zinc timber]

ig it doesn't include normalization then,

oh wait

[dim epoch]

?

[teal grotto]

bruh i’m misreading things my bad holmes

[dim epoch]

LOL

all g

uh in my course we just considered them different algorithms

gram-schmidt orthogonalizing algorithm and gram-schmidt orthonormalizing algorithm (idk if that translation is correct)

since the second just includes an extra step after every vector

[zinc timber]

so there are some inconsistencies

[dim epoch]

yeah

i just checked wikipedia

the german article covers both and considers them different processes but the english one only talks about orthonormalizing

[ebon river]

Is this correct? If so then is the result not true for $M_n(\mathbb{R})$?

[stoic python]

MrFlaze

[zinc timber]

just describe a little more why E_\lambda = C^n means A = \lambda I

[ebon river]

The maps corresponding to A and $\lambda I$ match on entire domain, so they must be the same right?

[stoic python]

MrFlaze

[zinc timber]

yeah

but just write another line showing that

$A-\lambda I = 0 ,\forall v\in V \implies A=\lambda I$

[stoic python]

Ryu?

[ebon river]

Ohk

But so this doesn't work on $\mathbb{R}^n$ right?

[stoic python]

MrFlaze

[ebon river]

Atleast this proof

[zinc timber]

the proof doesn't but the statement is true

in abstract terms we say that the center of M_n is scalar matrices

[ebon river]

ohh...so can you give a hint as to prove that for that case also?

Since the characteristic polynomial may not have a root, idk how to proceed exactly

[zinc timber]

I had one approach in mind but I don't know if you'll be able to understand it

let me think of a better one

[gleaming glacier]

Hello everyone, im writing a paper on exploring the use of eigenvectors and eigenvalues to solve systems of linear differential equations, and was wondering how to formally explain why each eigenvalue of a matrix will correspond to a unique solution to a linear ODE, could anyone please help? Thank you!

[zinc timber]

what do you mean by "each eigen value of a matrix will corresponding to a unique solution ..."?

[gleaming glacier]

systems of linear DEs come in the form y' = Ay, where A is a matrix. Apparently, each eigenvalue of A will correspond to a linearly independent solution to the system of ODEs, but why does this occur? is essentially what I am asking

hope this is a bit clearer!

[zinc timber]

by unique do you happen to mean Lineary Independent solution?

[random axle]

What does it mean 'forms a group'

[gleaming glacier]

yep sorry

[zinc timber]

anything that contains 'matrix' isn't linear algebra, try #groups-rings-fields

[dusky epoch]

ryu, did you mean "not everything that contains matrices is linear algebra"?

[zinc timber]

hm..

[dusky epoch]

you said $$\forall x (x \text{ contains matrices} \to x \text{ isn't linear algebra})$$ when you meant $$\neg\forall x(x \text{ contains matrices} \to x \text{ is linear algebra})$$

[stoic python]

Ann

[zinc timber]

lol

[random axle]

[dusky epoch]

anyway

[zinc timber]

[dusky epoch]

azeem, do you know what a group is

[random axle]

a group of what

[zinc timber]

I'm not gonna edit that, let others enjoy the humor

[dusky epoch]

a group in the group-theoretic sense

i.e. a set of elements with a binary operation upon it which satisfies a bunch of axioms

[random axle]

nope. dont think i have learned it. is it hard to pick up?

[dusky epoch]

have you encountered this before

oh

what

can you look in your text for the defn of a group

[random axle]

im using old a-level textbook, it has nothing here on groups

this question is under matrix algebra is there no way to solve the question with just knowledge of matrices?

[dusky epoch]

well this question requires you to know what a group is

no way around it

[random axle]

oki, thanks anyway

[zinc timber]

have you solved any system with eigen values/vectors before?

[gleaming glacier]

yeah ive done some questions n stuff, but intuitively im not immediately sure why this occurs

[zinc timber]

do you know that distinct eigen vectors corresponding to distinct eigen values are linearly independent ?

it has to do something with this fact

[gleaming glacier]

ah okay, yeah i seem to remember learning that a bit back

if I may ask, what is the direct relationship between this fact and solns to systems of odes?

[zinc timber]

say you have 2 distinct eigen values $m_1, m_2$ and the corresponding eigen vector $v_1, v_2$. you write the general solution as $\vb{x} = c_1 v_1 e^{m_1t} + c_2v_2 e^{m_2 t}$

[stoic python]

Ryu?

[zinc timber]

so sol corresponding to $m_1$ is $v_1e^{m_1t}$ and for $m_2$ we have $v_2e^{m_2t}$

[stoic python]

Ryu?

[zinc timber]

what can you say about them given v1, v2 are linearly independent?

[gleaming glacier]

im not very sure.... sorry!

[zinc timber]

ok, how do you solve a system of ODE given eigen values m1, m2?

[gleaming glacier]

wouldnt the solutions of the system come in the form similar to what you said, such as c1e^m1t and c2e^m2t

[random axle]

2 more weeks of learning this matrix algebra and then i can leave this channel for good

[zinc timber]

yeah so if $m_1 \neq m_2$ can you write them $e^{m_1t}$ and $e^{m_2t}$ as a constant multiple of other?

[stoic python]

Ryu?

[gleaming glacier]

are you asking whether we can write e^(m1)t as c1e^(m2)t or something like that?

[zinc timber]

yes

if yes, they are linearly dependent, if not, they are independent

[gleaming glacier]

oh okay, id presume no then right, as they are linearly independent if i am correct (im really not sure, im struggling w lin algebra, thanks for being patient)

[zinc timber]

have you encountered wronskian before?

you learn it in your ODE class, not LA class

[gleaming glacier]

yep ive learnt it while doing variation of parameters i think

[zinc timber]

hmm it's basically the same

to show $y_1, y_2$ are linearly independent, you show that $$\left| \mqty{y_1 & y_2 \ y_1' & y_2'}\right| \neq 0$$

[stoic python]

Ryu?

[zinc timber]

try to do the same for e^m1t, e^m2t

m1 ~= m2

[gleaming glacier]

okay sure

okay, so i calculated and since m1 and m2 are different, e^m1t and e^m2t are linearly independent, so then how do i relate it back to the eigenvalues thing?

sorry once again for being a bit slow lol

[zinc timber]

well yeah

if eigen values are different, the so is the exponential part

so they are LI

isn't that what you were asking?

[gleaming glacier]

i was more asking why each eigenvalue corresponds to a solution of the system

[zinc timber]

so you were asking why each eigen vector corresponding to a LI solution

if eigen values are different, then you just showed that $e^{m_1t}$ and $e^{m_2t}$ are LI, so the solution corresponding to them are $\vb{v}_1e^{m_1t}$ and $\vb{v}_2e^{m_2t}$ are also LI

[stoic python]

Ryu?

[zinc timber]

so each eigen values corresponding to LI solution of the system

[gleaming glacier]

ahhh okay, makes sense, thanks a lot for your time haha

[outer goblet]

if ||u1| |=1

is the dot product

u1 . u1 =1

?

ah its 1^2 i google

[nocturne jewel]

$\langle v,v\rangle=\norm{v}^2$

[stoic python]

Mosh

[nocturne jewel]

picking <> as the dot product gives v.v=|v|^2

[silk scroll]

maybe just let B = Eij, then we can get aii=ajj, aij =0 , i not equal j. then we can get A= a11 I. R and C seem no change.

[winged prairie]

hey guys i have a question about the dual map. My understanding of this is that the LHS of the equation goes from W'---> V' but the RHS goes from V --> F, so how can they be equal

[north hedge]

T' takes in a map phi in W' and gives you back a map in V', yes

and that given map is phi composed T which as you said is from V to F so is indeed an element of V'

T' maps a map to another map so its a little bit of a tongue twister at first

@winged prairie

[winged prairie]

I understand this part

[winged prairie]

its this part i am having trobule undertanding

[north hedge]

T' maps phi which is in W' to phi composed T which is in V'

[winged prairie]

why is phi composed T which is in V'

isn't it in F?

[north hedge]

phi composed T is an element of V'

but you can also consider it as a map from V to F

[winged prairie]

and an element of V' is an element which takes an element in V to an element to F

[north hedge]

elements of V' are maps from V to F yes

[winged prairie]

ok thanks

imma defo need some time to wrap my head around this

[north hedge]

its more useful to think of the map V to F as an object thats being mapped onto rather than a map itself in the context

[winged prairie]

i don't get this

[north hedge]

phi composed T is the result of applying T' to phi

it's what you get out

[winged prairie]

ye it makes sense when i see it like that

[north hedge]

maybe this was phrased badly ignore it lol

[winged prairie]

dw

ty

ok

the RHS would represent matrix multiplcaiton?

[north hedge]

yes

you can write phi and T as matrices

[winged prairie]

phi would be a 1*n matrix right

[north hedge]

yes it looks like a row vector

[winged prairie]

ye helps to look at it like that

[winged prairie]

sorry another theory question, why do we define the annihalitor on a subspace rather than a vector space

[tropic pebble]

Anyone knows can geometric multiplicity be zero?

[zinc timber]

no

GM of an eigen values is at least one, since an eigen value exists mean there is a eigen vector

[zinc timber]

you can define it on the whole space, but it'll just be the {0} functional

so not much interesting

@tropic pebble@winged prairie

[wintry steppe]

annihilator of a subspace is just the kernel of the dual map to its inclusion

very natural thing to consider

[winged prairie]

ok ty

[outer goblet]

is asking for base for eigenspace the same as asking for eigenvectors?

[wintry steppe]

the word "base" should give it away

[outer goblet]

so yes or no???

[lavish jewel]

in general no, because a basis needs to have lin indep elements

[zinc timber]

wth is a 'base'?

[lavish jewel]

and the eigenvectors need not be lin indep for defective matrices

i assumed they meant basis

[zinc timber]

oh

[outer goblet]

how did i find the eigenvectors then

[lavish jewel]

anyway, you do need to find the eigenvectors to find if they're lin indep

[outer goblet]

all the book has is base for eigenspace

[tropic pebble]

thank you!

[wintry steppe]

If M is a nontrivial subspace of V, what is the linear span of V \ M?

[teal grotto]

is this V mod M or V cut M

[dark brook]

This is about the least square method. I've got a function
$f(t) = \alpha_1 (t^2 - 1) + \alpha_2 t + \alpha_3$ and I've put in the values for 4 coordinates (-1,0), (0,4), (1,-2) and (2,2). My question, is it true that I only get two values out for x? I try to do $A^T A x = A^T b$
I get $x = \begin{pmatrix} \frac{1}{3} \ 0 \end{pmatrix}$

[stoic python]

HrJonas

[wintry steppe]

set difference I believe

[teal grotto]

no

rank of an m by n matrix is less than or equal to the minimum of m and n

[stoic python]

Phi

[dull pilot]

the text says that these are easy consequences and thus doesn't bother to prove them

but I don't intuitively see the connection

could someone help explain?

[crystal oracle]

@dull pilot What is a_λ and what is g_λ?

[dull pilot]

the algebraic and geometric multiplicity of an eigenvalue

respectively

@crystal oracle

[crystal oracle]

@dull pilot About Corollary 42.10. For any n×n matrix, the sum of its algebraic multiplicities is never greater than n, and the sum of its geometric multiplicities is never greater than the sum of its algebraic multiplicities. Having a basis consisting of eigenvectors is equivalent to the sum of geometric multiplicities being n.

About corollary 42.11. If an n×n matrix has n distinct eigenvalues, then it has n distinct eigenvectors, so there is a basis of its eigenvectors, so it's diagonalizable by Theorem 42.9.

[zinc timber]

what's T and T_i, i'm confused

[zinc timber]

hmm So $T_i$ is the row of the matrix, i.e.
$$ T_{ij} = f^{(i)}(x_j)$$?

[zinc timber]

I thought you are talking about a sequence of T_i's

[crystal oracle]

Oh shit, I made some typos there. One moment.

[zinc timber]

T \in ?

[dusky epoch]

no ryu don't you know

set theory bad

types good

[zinc timber]

[ebon river]

Does this work as a proof or is induction necessary?

[crystal oracle]

Rewrote a bit:
Consider the space $\mathcal{P}{m-1}$ of real polynomials of degree at most $m-1$. Suppose $T \in \mathcal{L}(\mathcal{P}{m-1}, \mathbb{R}^m)$ is a linear function s.t. the i-th row of its matrix (w.r.t the basis $1, x, x^2, x^3, \dots, x^{m-1}$) evaluates the $k_i$-th derivative of the polynomial at a point $x_i$. So, $T_{i,j}=(j-1)(j-2)\cdots(j-k_i)x_i^{j-1-k_i}$. Suppose $T$ has the property that, if some row of $T$ evaluates the $(k+1)-th$ derivative at a point, then another row of $T$ evaluates the $k$-th derivative at the same point (we consider the zeroth derivative to be the identity operator). I want to prove that $T$ is invertible.

@zinc timber @limber sierra Sorry for the confusion and for typos. Btw, I no longer think that a simple proof exists because looking up such things on the interwebz gives me blah blah chinese remainder theorem in the best case and huge unreadable formulas in the worst case.

[stoic python]

Phi

[random axle]

I am trying to make sense of this proof. What does the first summation represent?

[dusky epoch]

the (i,j) entry in AB

[zinc timber]

try to compute the matrix for the given basis and you'll see why this is invertible

[random axle]

But why is that a summation?

[limber sierra]

do you know how to multiply matrices?

[random axle]

please spell it out for me if you can, it's early morning again

[limber sierra]

thats just a way to represent a given entry in matrix multiplication

[dusky epoch]

do you know how to multiply matrices

[limber sierra]

you multiply every member of the i'th row of A with a corresponding entry in the j'th column of B

and add them together

[dusky epoch]

this is the definition of matrix multiplication

[random axle]

Oh, I see

thanks

Just wasn't used to this notation

[ebon river]

Anyone?

[crystal oracle]

@zinc timber Alright, but even for the case of two points and one derivative (so a 4×4 matrix) the matrix doesn't look nice.

[zinc timber]

k+1 th derivative of x^k should be zero

should look nice

[stoic python]

Phi

[zinc timber]

btw try to think of rref now

Not gonna ask again how you got it

[crystal oracle]

Performing Gaussian elimination on this particular matrix didn't give me any insight on why it would always work in the general case 😦 Alright, I think I'm gonna give up because this is too difficult and not important enough for me. If I ever need this in the future, I think I'll read https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Hermite_interpolation (where I don't understand why a polynomial's remainder after division over (X-x_i)^{r_i} must be its Taylor polynomial at x_i) and https://artofproblemsolving.com/community/c1157h990758

[random axle]

Ideas?

@zinc timber

[stoic python]

c squared

c squared

[wintry steppe]

or u can just directly use the definition of transpose

[random axle]

Show me this method pls

[wintry steppe]

no

[random axle]

oh

[wintry steppe]

(im on my phone and don't wanna type indices)

[teal grotto]

are u talking about the dual space def?

[random axle]

By definition of transpose, what do you mean? I was taught that the transpose is just the rows and columns interchanged. I don't get that fancy double summation thing above

[wintry steppe]

no just the matrix transpose lol

[wintry steppe]

rows and columns interchanged, yes

[lavish jewel]

if your matrix is originally indexed as A_ij, the transpose is A_ji

i had never seen such cursed notation for an identity matrix, what the fuck

[teal grotto]

that’s where i was gonna go with this

[teal grotto]

wait what lol

[lavish jewel]

why E_ij, what

[teal grotto]

that’s not an identity matrix

[lavish jewel]

just call it I like normal people

oh you were gonna do a permutation

[teal grotto]

no

lol

[lavish jewel]

what do you wanna do with that then

[random axle]

im so confused

[teal grotto]

E_{ij} is literally the matrix with a 1 in its ith row and nth column, zero everywhere else

[lavish jewel]

oh shit you wanted to split it into one matrix per element

[teal grotto]

yea lol. the thing that bases are used for

[lavish jewel]

denoting scalars AND matrices as M_ij is cursed anyway

[zinc timber]

,, (AB)^T = B^T A^T

[teal grotto]

lmao how else do u do it

[lavish jewel]

why not just use the definition of the transpose

[teal grotto]

what is this definition that everybody is talking about 💀

[lavish jewel]

i would use boldface caps for matrices, boldface lower for vectors, and no bold for mats

A^T_ij = A_ji

[teal grotto]

oh sheet

[lavish jewel]

what the fuck lol

[teal grotto]

dying over here rn

[lavish jewel]

worst part is you were probably gonna use that anyway

but applying it on your cursed E_ij matrices

[teal grotto]

yea i was holy fawk dude

[lavish jewel]

c squared like

[zinc timber]

lmao what did I just witness

[lavish jewel]

[sudden narwhal]

[lavish jewel]

first time that emoji is suitable for a situation

[teal grotto]

yw

[zinc timber]

[zinc timber]

I have to show that it's false, approach I am trying is assuming the EVs are $\lambda_i;i=1, 2, \cdots r$ and we have show that $2^{\sqrt{n}} \leq |\tr(A^n)| \leq 2020\cdot 2^{\sqrt{n}}$

[zinc timber]

i.e. $2^{\sqrt{n}} \leq \sum_{i=1}^r \lambda_i^n \leq 2020\cdot 2^{\sqrt{n}}$ is not possible

[wintry steppe]

M = {(a, b, c, d) : a + b + c + d = 0, b - c + d = 0}

I must find a basis for this subspace. I did the following: a + b + c = b - c, which means a = -2c. So our subspace is M = {(-2c, b, c, d)}. The answer is (for example) B = {(-2, 0, 1, 0), (0, 1, 0, 0), (0, 0 ,0, 1)}.

Could someone tell if this is correct?

[zinc timber]

no it's not

you have only used one condition, use both and you'll get a 2 dimensional subspace, not 3 dimensional

[wintry steppe]

Thank you, @zinc timber

[teal grotto]

i feel like something with cauchy schwarz helps here when you square everything

[zinc timber]

there's a general form of CS which can be used but I don't see how it'll be useful

[bold sun]

hey basically i am trying to do this question.....''Prove that a triangular matrix is invertible if and only if its diagonal entries are all non-zero.''

this is how ive started but would it be valid like as a proof for =>

[zinc timber]

think about the rank

[bold sun]

wdym rank?

[zinc timber]

also you said you are doing for triangular matrix, then why have you written diagonal?

[bold sun]

cuz i have to prove that diagonals entries are non zero

[zinc timber]

showing that diagonal entries are non-zero and actually assuming your matrix is diagonal are totally different thing

[bold sun]

also it didn't specify upper or lower triangular so technically diagonal matrix is a triangular 1 right?

[zinc timber]

not all triangular matrix can be turned into a diagonal one

[wintry steppe]

I have to find intersection of two 3D subspaces in R^4. Is there a better way than solving 4 equations system?

[bold sun]

hmm i thought that diagnal matrix are triangular like both upper and lower?

[zinc timber]

honestly use a computer

[zinc timber]

diagonal are triangular, but not the converse

[wintry steppe]

thats my homework unfortunately

[bold sun]

hmm so how would i start then......

[zinc timber]

use gaussian elimination then

[zinc timber]

think about the rank

[bold sun]

wdym rank?

[zinc timber]

rank of the matrix

[bold sun]

sorry never heard of that before ngl

[zinc timber]

[bold sun]

i did quick find in notes and the term rank is used in chapter 8 were only on 3 lol....

[zinc timber]

I would have suggested you to look at the characteristic polynomial, but since you haven't reached that far yet

[bold sun]

hmm nope aint done that either

[tranquil steeple]

the eigenvalues of the triangular matrix are given by the diagonal. can you invert a matrix with a 0 eigenvalue?

[bold sun]

whats eigenvalues??

[tranquil steeple]

😛

[bold sun]

lol haven't heard of that either- i only began uni and lin algebra 2 months ago now lol so havent done alott

[nocturne jewel]

Assume A_kk is 0, then Ax=0 has a 0 row, hence a free variable

[bold sun]

basically we have a test next week so im just trying to revise for that and do as many qs for revison as possible cuz i have no idea what uni tests r like....

[nocturne jewel]

They're like regular tests.. just typically shorter

[bold sun]

isnt that kind of what i did?

[bold sun]

ooh really its meant to be like 1hr 15 mins or something but the room is booked fr 2 hrs or smthing....he sed revise proofs multiple choice qs and normal calculation qs so thats basically evrything weve learnt so far...but im doing content propely and im awful at proofs so im focusing on that too

[bold sun]

oh so here can i say it haz zero column too so its not invertible too...which is a contradiction?

[nocturne jewel]

Yeah, you just prove a bigger case, ie Ax=b has a unique solution for all sensible b iff A is invertible

You're proving one of the equivalences in FTIM

[bold sun]

oh is that what the q is asking?

[nocturne jewel]

It's asking for the 3rd statement of FTIM (in typical taught order)

[bold sun]

FTIM?

[nocturne jewel]

fundamental theorem of invertible matrices

https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

[bold sun]

ooh okk i think it was second fr us

wow okk yeah we only learnt 4 properties out of them

so how would i prove this way then <= assume that it has a uniqe solution prove its invertible?

[nocturne jewel]

If it has a unique solution, RREF(A) must be I

[bold sun]

oh ok and if it is an identity matrix then it must have inverse as a(a^-1) is identity matrix

[outer goblet]

for a set to be a subspace of Rn, its span has to be the same or lower than Rn? right?

like only sets in R1,R2,R3 can be subspaces of R3?

[wintry steppe]

If basis for A is (a1, a2, a3) and for B is (b1, b2, b3) and (a1, a2, a3, b1, b2, b3) is a lineary independent set, does that mean A intersection B is {0}?

[wintry steppe]

R^1 and R^2 are not subspaces of R^3.

[outer goblet]

sets thru origins in R_1 and R_2 are subspaces of R_3

vectors and planes thru origin are subspaces of R_3

[wintry steppe]

subspaces of R3 are

a) {(0,0,0)}
b) {t(x,y,z)} (lines)
c) {t(x,y,z) + s(x, y, z)} (planes)
d) R3

[outer goblet]

yeah thats what i said

[wintry steppe]

subspaces of R2 are

a) {(0, 0)}
b) {t(x, y)}
c) R2

[wintry steppe]

But R2 is (first, second). R3 is (first, second, third). Right?

[outer goblet]

i guess so

[spring pasture]

Not enough info

Are a1,a2,a3,b1,b2,b3 independent or you deducting from A and B basis

[wintry steppe]

no vector from B can be expressed as a linear combination of vectors in A.

[spring pasture]

So like b1= k1a1 + k2b2 + k3b3

Not all k1,k2,k3 being zeros

[wintry steppe]

yeah

[spring pasture]

Then A intersction B won't be {0}

[wintry steppe]

thanks. why?

[spring pasture]

For any k in R , kb1 belongs to both A and B

[wintry steppe]

thank you

[faint dune]

Why does this inequality work?

[stoic python]

Toge

[faint dune]

do we assume these norms to be compatible with vectornorm

[zinc timber]

do you know why $\norm{A^{-1}\cdot\delta b} \leq \norm{A^{-1}}\cdot\norm{\delta b}$ is true?

[zinc timber]

,texconfig colour white

[stoic python]

You have switched to the white colourscheme.

[faint dune]

I can say this is true only if matrix norm is compatible with the vector norm.

[zinc timber]

yeah that's how you define norms for operators, operator norm are induced norm of the spaces involved

Now $$\norm{b} = \norm{A\cdot A^{-1}, b} \leq \norm{A}\cdot \norm{A^{-1}, b}$$

[faint dune]

you mean delta b.

ok

[zinc timber]

so this gives you $$\frac{1}{\norm{A^{-1}b}} \leq \frac{\norm{A}}{\norm{b}}$$

[zinc timber]

combine it with previous result and you have the what you need

[faint dune]

ahh thank you.

[zinc timber]

you should have tried proving this on your own, it's a good exercise

[faint dune]

it was not a proof task, but yes. The prof just describes steps where I think I am missing an obvious step...

[wintry steppe]

a = (1, -2, -1), b = (-1, 2, X), c = (1, -Y, 0)

For which X, Y is span{a+b, c}= span{a, b}?

Could someone help me?

[outer goblet]

how do i know in which order to put the eigenvector bases for a diagonalizing matrix, is it from biggest eigenvalue to smallest?

[zinc timber]

doesn't matter

[tranquil steeple]

Depends. most numerical solvers return them in non-decreasing order. But you might want them in some other order.

[outer goblet]

so it doesnt matter?

[tranquil steeple]

it matters sometimes. but if you know, you know.

[outer goblet]

i dont know tho

[tranquil steeple]

then it won't matter for you. 🙂

[outer goblet]

ait thx

[tranquil steeple]

so, the most standard way would be to sort them in non-decreasing order (real part first if complex). And most svd solvers will return the singular values in the opposite order (non-increasing)

[outer goblet]

ill just do what the book does cuz im not advanced yet to understand what you mean lol 👍

[tranquil steeple]

smallest first 🙂

[lavish jewel]

hmm i'm curious why you'd call non-decreasing the most standard

[nocturne jewel]

What would you call most standard Edd?

[spring pasture]

Why would you call Edd?

[zinc timber]

[outer goblet]

If im finding basis for column space, i reduce the matrix to row echelon form, note which column number has leading ones, and the base will be the columns that had that number, but in non reduced matrix? right?

or do i have to transpose it first?

[silver heath]

a

[outer goblet]

so no transpose?

[silver heath]

transposing first would get you the row space

which is the same thing as the column space but just transposed

row A^T = col A

[outer goblet]

thats what i thought too

but the exam solution does this like this

where this is the matrix

[silver heath]

could you translate the german bit at the bottom right?

[outer goblet]

its norwegian, but basicaily says that column vectors are (1,2,-2) and (0,1,-3/5)

[silver heath]

oh sry xD

hmm... not sure. Are we trying to get the row space or the column space here?

[outer goblet]

column space lol

i mean i checked the span and its the same with those vectors and the vectors i found

im just trying to understand why to solve it weirdly like that

[silver heath]

Oh wait hahaha...

Elementary row operations don't change the row space

so the pivot rows of the rref matrix is the basis for the row space

[outer goblet]

so basicailyl they overcomplicate it for no reason at all

[silver heath]

yep

[outer goblet]

like using quadratic formula on x^2+2x+1 lol

[silver heath]

well as you can see its quite obvious this x^2+2x+1 equation is homomorphic to the blah blah blah blah maximums and minimums of the quadratic form in r^6

[outer goblet]

haha

[tranquil steeple]

Most numerical solvers do that (at least for Hermitian matrices. For non-Hermitian, for example, Matlab can return very strange orderings)

Julia is better in that regard.

[wintry steppe]

I have to find basis for vector space that contains polynomials p degree 3 or less such that p(2) = 2p(1). I just found polynomials of degree 1, 2 and 3 that satisfy that condition. Is there any other way of doing it?

[lavish jewel]

i would've thought finding them in the reverse order is easier, using something like arnoldi

[wintry steppe]

so this is the only way for a beginner? i have just started learning linear algebra

[zinc timber]

?

[wintry steppe]

sorry could you explain what is confusing you

[zinc timber]

oh nvm I was thinking of char polynomial

[faint dune]

Let | |B| | < 1. Then the matrix I+B is regular. part of proof: | |(I+B)x| | >= (1 - | |B| |) | |x| |. Since 1-| |B| |>0, I+B is injective so regular.

I dont get the last part 😞

[wintry steppe]

if x is non-zero then |(I + B)x| >= (1 - |B|)|x| > 0, which means you cannot have (I + B)x = 0

or is it the regular matrix part

[faint dune]

I get ur first line. I have to think about the second. 🦢

[wintry steppe]

if you had (I + B)x = 0 for a non-zero x (which is precisely what non-injectivity is), then the computation gives 0 > 0

so you gotta be injective

[warm frost]

Hi guys, i have a question, if i know that A=A^-1, so A=+=I?

(A unit matrix)

@wintry steppe Can you help me please?

[wintry steppe]

i don't understand your question

what does A = + = I even mean

[warm frost]

• = I (unit matrix)

[wintry steppe]

are you asking if A^2 = I implies A = I?

i don't get what + = I means

if you're reading something can you post a screenshot?

can you type it precisely using the latex bot?

[marble lance]

😵‍💫

[warm frost]

@wintry steppe

that what i meant

[wintry steppe]

1 0
0 -1

0 1
1 0

two examples

[tranquil steeple]

For example the discrete sine transform $A_{ij}=\sqrt{2/(n+1)}\sin(ij\pi/(n+1))$

[stoic python]

Sven-Erik

[wintry steppe]

cool

[tranquil steeple]

julia> n=5; A=sqrt(2/(n+1))*[sin.(i*j*pi/(n+1)) for i in 1:n, j in 1:n]
5×5 Matrix{Float64}:
 0.288675   0.5          0.57735      0.5          0.288675
 0.5        0.5          7.0705e-17  -0.5         -0.5
 0.57735    7.0705e-17  -0.57735     -1.4141e-16   0.57735
 0.5       -0.5         -1.4141e-16   0.5         -0.5
 0.288675  -0.5          0.57735     -0.5          0.288675

julia> norm(A*A-I)
8.446231779064342e-16

[surreal river]

how do u find the dimension of kernel and image without calculating kernel and image

[wintry steppe]

specific example in mind?

A=span{(1,-1,0,2), (3,-1,-3,1), (2,-1,-1, 2)}, B=span{(1,2,-1,-2), (1,-2,-3,4),(-1,-1,3,-1)}

Guys, how do I find basis for A + B? I am confused.

[surreal river]

this matrix

[agile bronze]

The columns of A are able to be written of a linear combination of the vectors in the first span (and columns of B for second span)

Regardless of what the new columns are in A + B, you should consider how the spans together will contribute

[wintry steppe]

thanks, @agile bronze

but I didnt understand... are you talking about matrices?

[agile bronze]

I am

Would you prefer it in terms of a vector space

[wintry steppe]

sorry I am a beginner and havent studied matrices yet

[agile bronze]

Ok, so are A and B vector spaces?

[wintry steppe]

yes, subspaces of R4

span is a subspace, right

[agile bronze]

The set of all vectors expressed with the span is a subspace

Might be more accurate ^

So the span tells you that every vector in R^4 that's in A is written as a linear combination of the 3 vectors you see in that span

Same logic with B's vectors with the span of B

Let's call the vectors in A's span v_1, v_2, v_3, and B's span w_1, w_2, w_3 (the order we label these does not matter). So generally, for any $a\in{A}, a = c_1v_1 + c_2v_2 + c_3v_3 \text{and for any} b\in{B}, b = d_1w_1 + d_2w_2 + d_3w_3$.

[stoic python]

Len
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[agile bronze]

Ok whatever i'll do it in two steps

$a\in{A}, a = c_1v_1 + c_2v_2 + c_3v_3$

[stoic python]

Len

[agile bronze]

$b\in{B}, b = d_1w_1 + d_2w_2 + d_3w_3$

[stoic python]

Len

[agile bronze]

The c_i's and d_i's are just the coefficient's that let the vectors a and b be some linear combination of the things in the span

A + B is the vector space of all vectors e = a + b, where a comes from A, b comes from B

So what can you say about vectors e with respect to the above stuff?

[wintry steppe]

vectors e can be expressed as a linear combination of some vectors from basis of A and of B?

[agile bronze]

Yep

So we know every vector e can be expressed as a combination of v_1, v_2, v_3, w_1, w_2, and w_3

But the question asked for a basis, which means we want a set of linearly independent vectors

[wintry steppe]

so i pick any four from basis A and B that are linearly independent? for example v1 v2 v3 and w2?

[agile bronze]

If v1, v2, v3, and w2 are linearly independent then that's right. You have to check, though. Of course, our basis at least 3 (because the span of A and B should be linearly independent already) and at most 4 (because we always have at most n in R^n).

So picking v1, v2, and v3 and finding which of those w's are independent if any will be your answer.

[wintry steppe]

I understand. Thank you very much, Len!

[agile bronze]

no problem

[wintry steppe]

@agile bronze if you have one more min

And to find a basis for intersection of A and B I have to write some v as a linear combination of vectors in basis of A and then as a linear combination of vectors in basis of B and find coefficients for which equality holds? is there any shortcut?

[gray dust]

pretty much that

[wintry steppe]

thanks!

[wintry steppe]

{b1, b2...bn} is a basis for B. L is a subspace of B which contains only linear combinations of b1...bn such that sum of coefficients is 0.

is dim L = n - 1?

x from L

x = -A2(b1 - b2) -A3(b1 - b3) -... -An(b1 - bn)

and basis for L {b1 - b2, b1 - b3... b1 - bn}?

[stable kindle]

yes

[wintry steppe]

thanks :)

[bold sun]

hey sorry if this is a silly question i just wannt this clarifying in maths what does notion mean like for example ''define the notion of a determinant? - would that just be det(a)??

or like define notion of like elementary matrices that would just be e1...ek?

like is it symbols or does it mean to like actually write the defintion of it?

[nocturne jewel]

notion = idea

Like what's the idea of a determinant? It's how a unit of "volume" changes under a matrix transformation, it's a function that maps a matrix to a number with certain properties (f(1)=1, multilinear, alternating)

@bold sun

[bold sun]

ooooh okk that makes sense lol i acc thought it was just the symbols whoops lol

so elementary matrices would be the operations you use to get to identity matrix right?

[worldly bear]

correct

if it takes 3 row operations to reduce to the identity matrix then you will have 3 elementary matrices

each of which has the same respective row operation used to reduce the matrix A to the identity

[nocturne jewel]

Elementary matrices induce an ERO when left multiplied to a matrix

So it's the "matrix form" of row reduction

[bold sun]

oooh okk i think i get that now

Thank you so muchh!!

[wintry steppe]

If w1, w2 and w3 could be expressed as a linear combination of v1, v2 and v3, would that mean that B is a subspace of A and A + B = A?

[bold sun]

hey ok i need some help on this - how would i know where to start from and when and where to stop

in the lect notes in the e.g they did theve started at random places and randomly stopped it dont make sense....

like what is the aim obv to find the determinant but yeah idk what im doing with row reduction here..bit confused if im making sense

[nocturne jewel]

you pick the place that makes it easiest to evaluate

[bold sun]

so that would be the -1?

[nocturne jewel]

for example, you'd want to use the 2nd row in A

cause you have 2 0's

[bold sun]

hm so what would the row reduction would be to get it 0?

[nocturne jewel]

??

oh using row reductions my bad

[bold sun]

yeah...idm using the formula its easier then this but the q dont alow it

[nocturne jewel]

When you find det with EROs, you aim to get the matrix into upper triangular

cause it's known the determinant of an upper triangular matrix is the product of the diagonal entries (Prove it yourself)

[bold sun]

oh ok so make the 2 and 5 equal 1

[nocturne jewel]

Not necessarily

[bold sun]

hm then

[nocturne jewel]

You want it upper triangular

[bold sun]

also it has a 0 in the diagonal is that even allowed?

[nocturne jewel]

yep

[bold sun]

oh so make 2 and -1 equal 0

[nocturne jewel]

yes

[bold sun]

oh ok didnt know that i thought all diagonal had to be non 0

[nocturne jewel]

Then of course keep track of the EROs you do, because ERO affects determinant

[bold sun]

ooh okk yeah

let me try it out then .....gimme a min

[nocturne jewel]

(alternatively you just use elementary matrices and then apply det(AB)=det(A)det(B) a bunch)

[bold sun]

okk i got determinant is 10 for first

[bold sun]

okk yeah tht makes sense

[nocturne jewel]

I think it's suppose to be -10

[bold sun]

hmm okk lemme check my workings....

[bold sun]

still get 10 what did i do/am i doing wrong?

ohh wait its okay my bad i figured it out i should have swapped rows duhhh

that makes sense and works now- im gnna try the rest then see if i need to find more qs like this for practice too....

thanks tho for explaining what to do 🙂

[vapid pine]

if someone can help me in the help channels

its related to graphs

[barren sentinel]

i got null space basis as a vec <-9, 5, 1, 0>

im not sure if u is in it?

it shouldnt right?

[teal grotto]

can really help if we don’t know what A is

[zinc timber]

yeah

[barren sentinel]

sec

u dont really need a cuz i already told u what the basis for null space is

but here u go

lemme also know a and b in this case, just to double check cuz im unsure

a) k = 3, b) k = 4, and then c)i) and ii) im not sure

i found 2 eigenvectors

so dimension is 2?

[wintry steppe]

perhaps

assuming you mean you found a basis for the eigenspace and it had two vectors, that is

[wintry steppe]

a - d - 2e - f = 0
-a -d - 3e - 2f = 0
b + d - f = 0
c - d - e = 0

whats the best strategy for solving such systems (no matrices)? if every equation had 6 unknowns, that would be easy, but now i am stuck

y no matrix

the first thing i'd do is try to add some equations together to try and eliminate some variables

[dusky epoch]

a + 0b + 0c - d - 2e - f = 0
-a + 0b + 0c - d - 3e - 2f = 0
0a + b + 0c + d + 0e - f = 0
0a + 0b + c - d - e + 0f = 0

here now every equation has six unknowns

[lavish jewel]

dont forget 2 equations are missing tho

0a + 0b + 0c + 0d + 0e + 0f = 0

twice

[dusky epoch]

oh yes how could i forget

[wintry steppe]

sorry but i am a beginner and we only told that matrices are vectors and how we add them

[wintry steppe]

thanks!

matrices can also help you solve systems of linear equations

but if you don't know how yet that's ok

[wintry steppe]

thanks Ann! this is brilliant

[dusky epoch]

what

[wintry steppe]

it's brilliant

[dusky epoch]

are you being serious rn

i was almost joking there

[wintry steppe]

nah

[dusky epoch]

okay so you were being sarcastic.

[wintry steppe]

i like your and Edds comment, didnt mean to be rude

[dusky epoch]

but you are admitting to not being serious, therefore i must have been of no help to you.

[wintry steppe]

you have!!!

wait

i tried to say you did help me

[dusky epoch]

yeah, and then i asked you if you were serious and you said no

[lavish jewel]

in their defense, i'm always serious about my sarcasm

[barren sentinel]

yeah it does

https://cdn.discordapp.com/attachments/540211747613704221/914432282276794388/unknown.png

[wintry steppe]

i am going to try to solve the system with 0s added

[barren sentinel]

can someone help me with d here pls?

[lavish jewel]

for part i, do you know what the definition of the null space is?

[barren sentinel]

yeah i got null space to be span of -9 5 1 0

but im unsure of u being in it

i feel like its not

but online it says the two are linearly dependent so im not sure

[lavish jewel]

well, what's the span of that vector?

[barren sentinel]

linear span?

like a line

[lavish jewel]

what do the vectors in the span of {[-9 5 1 0]} look like

[barren sentinel]

idk

points/arrows?

passing through origin

[lavish jewel]

mathematically, not visually

what's the definition of the span of a set of vectors

[barren sentinel]

a times that vector

[lavish jewel]

right

[barren sentinel]

where a belongs to R

[lavish jewel]

so, does there exist an a such that a[-9 5 1 0] = [3 -2 -1 0]?

that'd be a no

alternatively, you could just multiply Au and see if you get zero, since that's the definition of the null space

the set of all vectors x such that Ax = 0

[barren sentinel]

yeah so answer is no?

cool thnx, i thought that was right but was not sure, thnx

[lavish jewel]

well idk if what you found as a basis of the null space is correct

i just told you how to double check though

[wintry steppe]

@dusky epoch no success

[barren sentinel]

how do i show this?

i mean b first part

actually whole b part

i got reduced form as
1 1
0 -1
0 0

[lavish jewel]

first, show that A has rank 2, meaning it is injective

[barren sentinel]

bruh this injective shit is so confusing cuz everything has like 3 names

our class calls them onto or something? idfk tbh

[barren sentinel]

ok then?

[lavish jewel]

injective is one-to-one, not onto

that's surjective

[barren sentinel]

ok

[lavish jewel]

then you show that the columns of the matrix A are both orthogonal to the normal of the given plane

[barren sentinel]

wait

why does that prove?

and how do i write range space?

oh its same as column space 🤦‍♂️

[lavish jewel]

because then you note that the image of the matrix A is of the form c a_1 + d a_2, where a_1 and a_2 are columns of the matrix and c and d are scalars in R

[barren sentinel]

i dont quite get it but im in a rush so ill just write it out thnx 😂

[lavish jewel]

and so the vectors in the image, call them, u, are of the form u= c a_1 + d a_2

and if both a_2 and a_2 are ortho to the normal of that plane, then c n^T a_1 + d n^T a_2 = n^T u = 0, satisfying the equation of the given plane

is this a quiz/test?

[barren sentinel]

bro what u r writing makes no sense

whats c a_1 + d a_2?

[lavish jewel]

please read what i wrote, otherwise my answer can only be "what the fuck"

[lavish jewel]

here

[zinc timber]

just finished a horrible test, ngl it was a disaster

[barren sentinel]

i did read

[zinc timber]

[barren sentinel]

oh that

ok

[teal grotto]

feel that bro. what was it on

[lavish jewel]

what the fuck.

[zinc timber]

wanna hear my sad story

[barren sentinel]

whats this?

[barren sentinel]

nah bro

homework

lots of fucking homework

[zinc timber]

look the questions

[lavish jewel]

ryu

plz tell me this is not by hand

plz

[teal grotto]

i’m gonna puke actually if it’s not

[zinc timber]

it was edd it was

[lavish jewel]

[teal grotto]

[zinc timber]

1hr for the whole fucking test

1 hr

[teal grotto]

[barren sentinel]

thats nothing

[teal grotto]

part A???

[zinc timber]

even if I copy it, I still won't be able to finish

[teal grotto]

what was part B lmao

[zinc timber]

oh yeah, another part

[barren sentinel]

i had to write like 10 pages of this shit by hand

• an entire report around it

[zinc timber]

taht one assignment,

[teal grotto]

bro that feels like busy work