linear-algebra

[wintry steppe]

you might as well. if $q(x) = x^TAx$ for any matrix $A$, and if the characteristic of the field is $\neq 2$, then $$x^T\left(\frac{A+A^T}{2}\right)x = \frac{1}{2}x^TAx + \frac{1}{2}x^TA^Tx = x^TAx,$$ and the matrix $(A+A^T)/2$ is symmetric, so you can always assume when working with things of the form $x \mapsto x^TAx$ that $A$ is symmetric

[stoic python]

MOTHMAN

[wintry steppe]

sometimes the "symmetric" is baked right into the definition and you don't need to worry about this

[drowsy flower]

Yeah I see thx a lot

[hard drum]

Of course characteristic 2 comes up

[teal grotto]

take U = span(1,0,0) and V = span((1,0,0), (0,1,0))

then the direct sum of U and V is still well defined, but U intersect V is equal to U.

[glacial terrace]

would this be an upper triangular matrix?

[teal grotto]

no

it’s not even a square matrix

[glacial terrace]

what would be the diagonal in a 3x4 matrix?

I know it doesn't make sense but I don't see why not according to this question

[wintry steppe]

the diagonal is just the entries A_{ii}, this always makes sense

[glacial terrace]

but then this is an upper triangular matrix

[wintry steppe]

yeah it is according to the definition you just posted

[teal grotto]

looks like a trapezoid now instead of a triangle

[glacial terrace]

ikr lol

[drowsy flower]

Hey, If a complex matrix is skew symmetric, can I say that the matrix is also hermitian?

My textbook definition of hermitian matrix is A^H = A => A is hermitian matrix

unsure if A^H = -A also implies A is hermitian

[tranquil steeple]

No it is not Hermitian, but it is a Hermitian times i (Hermitian has real eigenvalues, Skew symmetric has imaginary eigenvalues)

[drowsy flower]

sorry, confused with what you mean by hermitian times i,
So lets say A is matrix such that A^H = -A, then iA is hermitian?

[tranquil steeple]

yes

[drowsy flower]

interesting, and I assume the sign does not matter here, that is it -iA is also hermitian

yup thx a lot

[hard drum]

if A^H = -A we call A^H skew hermitian btw

[haughty dust]

Does anyone know what epsilon means in this case

[limpid vine]

where did you get this picture @haughty dust

what is it in reference to?

[haughty dust]

from google

delta function

wolfram alfa

https://www.google.com/search?q=dirac+delta+function&client=safari&rls=en&sxsrf=ALeKk03zpAEGbSsYKp2hnHI7SALiPwchzQ:1628967303127&source=lnms&tbm=isch&sa=X&ved=2ahUKEwiD_M-LmLHyAhW7ZzABHb1kBaQQ_AUoAXoECAEQAw&biw=1361&bih=735#imgrc=ovP9jQf1GRpAkM

[limpid vine]

you could start by reading the page its from

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

[silver heath]

Is the answer false?

Because it just has infinitely many solutions?

any x in its respective vector space is a least squares solution?

[lucid glacier]

If A is the zero matrix, what is Ax going to look like?

[silver heath]

0

[lucid glacier]

So if b is not 0, can any x be a solution?

[silver heath]

key word normal equations (least squares)

[lucid glacier]

Ah

My bad, I assumed it was some awkward terminology

[silver heath]

xD yeah i should have provided context mb

[nocturne jewel]

should be false, yeah

cause A*Ax will be the 0 vector for all x, and A*b will be the 0 vector as well

[silver heath]

Uh... I have another question xD

The explanation for the answer to this problem makes sense

but

lest say col A spans a plane in r^3

maybe z=0

wait

nvm im stooopid

xD

I was thnking we oculd have a vector (x,y,1) that would be perp 😆 🤦‍♂️ '

[wintry steppe]

[inner lagoon]

let $\mbb{K}$ be commutative field and $E$ be a $\mbb{K}$-vector space.
let $k \ge 2$, $(Vi){1\le{i}\le{k}}$ is a finite family of $k$ subspaces of $E$ ( $V_i \not = {0}$ and $V_i \not= E$). if $E = V_1\cup{V_2}\cup{...}\cup{V_k}$ show that $\mbb{K}$ is a finite and $k \ge Card(\mbb{K}) + 1$.

[stoic python]

cos/ ev

[hard drum]

(aren't all fields commutative by definition?)

[inner lagoon]

Emmm no but the commutative ones are useful

Just to say i'm not talking about theorem of Wedderburn

Because i generalize if the field is finite or not

Anyway it isn't our topic in the problem

@hard drum are you familiar with the field of quaternions?

[nocturne jewel]

why was this question posted by 2 people?

[inner lagoon]

Which one?

Ah i saw it my friend does it in the wrong channel

Sorry

[hard drum]

from the wikipedia article

but obviously this is a bit of topic lol

[inner lagoon]

@hard drum people haven't same idea about quaternions

(the quaternions almost form a field. They have the basic operations of addition and multiplication, and these operations satisfy the associative laws, (p + q) + r = p + (q + r), (pq)r = p(qr). ... The only thing missing is the commutative law for the multiplication.) from google

[hard drum]

yup

[inner lagoon]

But it's more logical to say field and commutative field

Because the other rules still the same

But most of fields are commutative

[hard drum]

isn't that going against the normal definition of a field? what's wrong with saying division algebra here?

[inner lagoon]

Normal definition is different from countries

I Think

It's like convention

It's like bourbaki stuffs

[hard drum]

oh, interesting, sure

yeah so France ye

corps commutatif vs gauche apparently, sure

[inner lagoon]

xD

But exactly what i mean

[hard drum]

ah okie sure, sorry for getting this off topic aha

[inner lagoon]

No worries

[austere moss]

Why does the transpose move from the t to X after taking a partial derivative?

[sudden nacelle]

I have a question that requires me to find the inequalities of a convex hull

given a set of points

can anyone help?

[wintry steppe]

just post it

and someone might help

[wintry steppe]

I think this is correct

did i make any mistakes?

[teal grotto]

mind elaborating on what the question is ?

[sudden nacelle]

i posted the question in the picture

[teal grotto]

it should be all of them

[wintry steppe]

i thought the rows only span 1xn

[teal grotto]

all of the n linearly independent rows of A

[teal grotto]

i do not see a question in the picture you posted

[wintry steppe]

wouldn't the rows of A spanning R^n imply that rank(A)=n?

[teal grotto]

yea, that is true in this case, since A is invertible

[wintry steppe]

OHhhhh

[teal grotto]

lol

[wintry steppe]

wait im still confused about the rows spanning 1xn

[silver heath]

@wintry steppe another idea

[wintry steppe]

i used this

[sudden nacelle]

F

[silver heath]

if A is invertible then dont we know A^T is invertible?

and if A^T is in vertible the col A^t span R^n a.k.a row A span R^n

[wintry steppe]

yes

[wintry steppe]

oh okay

[teal grotto]

oh bruh, does your class make a distinction between R^n as a vector space of column vectors versus a vector space of row vectors?

[silver heath]

Sorta what i was thinkin' but maybe the problem is just weirdly worded

[wintry steppe]

Uhh

[silver heath]

well i mean they are isomorphic....

[teal grotto]

because why else would anybody write "span the set of all 1 x n row matrices"

[wintry steppe]

Yeah

it must make some distinction

[teal grotto]

yea but they're not the same tisk tisk

[wintry steppe]

fuck fuck

ok

[teal grotto]

that is very annoying

[wintry steppe]

lemme ask the prof

[silver heath]

does the inner product space <0, v> = 0 ??

for all inner produc space's and v in V

[teal grotto]

huh?

[silver heath]

its part of a question

all of the other "requirements" of the linear transformation are satisfied

due to properties of inner product spaces

but i am not sure about the T(0) = 0

[teal grotto]

yes <0,v> = 0 for all v in V

[limber sierra]

note that 0 = 0 * 1

[silver heath]

RIGhhhhhhhhhhhhhhhhttttt

[limber sierra]

apply linearity in the first argument

[sudden nacelle]

does anyone know how to do my question

[silver heath]

Waiitt whaaaa???

i expanded

and got

$0 \leq 2\langle u, v \rangle + 2\sqrt{\langle u, v \rangle}$

And if $\langle u, v \rangle$

[stoic python]

Elonmosqito96

[silver heath]

is negative we get a negative complex number?????

[stoic python]

Elonmosqito96

[teal grotto]

this follows from cauchy shwarz

[silver heath]

uhh

[silver heath]

how does this apply here?

[teal grotto]

directly

lol jk, but there are proofs of this online

i kind of dont feel like retyping and explaining them

[silver heath]

what do i search to find the proof?

[teal grotto]

triangle inequality
here is a link: https://en.wikipedia.org/wiki/Triangle_inequality#Normed_vector_space

go to the section about normed vector spaces

you should see a proof of it there

[silver heath]

also another dumb question is <x, y> = 0 an inner product?

[limber sierra]

its an equation involving an inner product

[silver heath]

i mean like

would it be possible to define an inner product of u and v to be 0

or would thta not meet one of the reqs?

[teal grotto]

this is a perfectly valid definition for an inner product

you are defining <x,y> to be 0

[silver heath]

wait what about the

[limber sierra]

oh

i see what you mean

[silver heath]

<u, u> = 0 if and only if u=0?

[limber sierra]

you mean <x, y> = 0 for all x, y

[teal grotto]

yea

[limber sierra]

if so, then yeah, thats an inner product

[teal grotto]

not a very interesting inner product tho

everything is the same point

[silver heath]

wouldt that not meet <u, u> = 0 if an only if u =0 tho?

[limber sierra]

er

wait

whats your definition

[silver heath]

uh

[limber sierra]

if you require that, then yeah, its not an inner product

[silver heath]

1 sec

[teal grotto]

ah wait. i mispoke.

[teal grotto]

it is a perfectly valid bilinear form, but it does not define an inner product

[limber sierra]

okay yeah

positive definiteness disallows your product

[silver heath]

so this would be false?

[limber sierra]

from being an inner product

[limber sierra]

except over the trivial vector space

[silver heath]

*true

[teal grotto]

well thats a different question

[limber sierra]

?

[silver heath]

i was thinking it could be false if <u,v> = 0 was a valid inner product

[wintry steppe]

i think thats true

[limber sierra]

it clearly follows

[silver heath]

but thtas the only way i can think of

[limber sierra]

precisely

[wintry steppe]

gonna pull a theorem from my textbook

[limber sierra]

no

thats not necessary

x = x + 0

so <x, y> = <x, y> + <0, y>

but then <0, y> = 0

hence y = 0

[teal grotto]

this says, given an inner product <.,.>, given a fixed y, it has the property that <x,y> = 0 for all x

[limber sierra]

by definiteness

[silver heath]

oh...

[torn hornet]

huh

[silver heath]

ty

[teal grotto]

it is a different question than what this is asking

[limber sierra]

er wait

hold on

that argument is wrong

sorry im being a dumbass

[silver heath]

its asking is it true that y must be zero if <x,y> = 0 for all x?

[torn hornet]

i'd do <y,y> =0

[limber sierra]

yeah thats better

and still works

x = x + y - y

[teal grotto]

why this? <y,y> = 0 suffices, since it implies y = 0

[limber sierra]

oh misunderstood

thats faster yes

and cleaner

man idk what im thinking

im gonna check out lmao

being an idiot

[teal grotto]

lol same im so tired

[limber sierra]

its immediate from taking x = y

bleurgh

[silver heath]

its too late for math...

[sudden nacelle]

for this question i tried finding the line through each two points but i think that's too much work

[haughty dust]

in the collaz conjecture

what does iterations mean

does it mean the times it takes to get to 1

[limber sierra]

please don't multipost.

[radiant yarrow]

I don't understand how a direct sum of upper triangular matrix and a symmetric matrix makes the set of all square matrices

[dusky epoch]

every n by n matrix can be written uniquely as the sum of an upper triangular matrix and a symmetric matrix

take $A \in F^{n \times n}$ and let $A = S + U$ with $S$ symmetric and $U$ upper-triangular

[stoic python]

Ann

[dusky epoch]

...wait is it actually unique hold on

ah9

U is not just upper triangular but STRICTLY upper triangular

no

wait

i got myself confused

@radiant yarrow W1 is the subspace of all strictly lower-triangular matrices

not upper-triangular

in any case

the decomposition is as follows: ||take the contents of A on and above the diagonal and copy them into S. fill in the subdiagonal portion of S to make it symmetric. then S is the symmetric component of your S+L decomposition, and L is what remains, i.e. L = A - S. it should not be hard to see that it cannot be any other way||

[radiant yarrow]

sub diagonal meaning when i<j?

[dusky epoch]

subdiagonal meaning below the diagonal

which actually means i > j

[radiant yarrow]

Ok, but don't we already have a lower triangular matrix for subdiagonal place

[dusky epoch]

what do you mean

[radiant yarrow]

You said A is lower triangular matrix

[dusky epoch]

no, i did not

[radiant yarrow]

Wait

[dusky epoch]

i never said that

[radiant yarrow]

But wait

[dusky epoch]

don't put words in my mouth

i really don't like having words put in my mouth and my statements misinterpreted so wildly

[radiant yarrow]

Ok

[dusky epoch]

how am i expected to account for such misinterpretations when explaining things?

[radiant yarrow]

W1 is A

It's my fault

you're not accountable

[dusky epoch]

swing and a miss..

no, W1 is a subspace, not a matrix

you're confusing matrices with sets of matrices

[radiant yarrow]

But A belongs to W1

So A is a lower triangular matrix right?

My words

Not yours. Sorry

[dusky epoch]

no, i said A was arbitrary

[radiant yarrow]

Oh

[dusky epoch]

we're trying to decompose an arbitrary n by n matrix A into the sum of a symmetric matrix (which i'm calling S) and a lower-triangular matrix (which i'm calling L)

[radiant yarrow]

Okay A is an arbitrary matrix with S symmetric and U as upper triangular

[dusky epoch]

no

[radiant yarrow]

Oh

[dusky epoch]

the only reason i said anything about upper-triangular matrices before is because i was mislead by your misidentification of W1 as the subspace of all upper-triangular matrices,

instead of the subspace of all strictly lower-triangular matrices that it is

[radiant yarrow]

ok sorry

[dusky epoch]

and apparently my utterly algorithmic description of the decomposition i have in mind is too complex to understand

which strikes me as a huge surprise

[radiant yarrow]

Am I missing something?

[dusky epoch]

you're missing a lot of things

we wish to find S symmetric and L strictly-lower-triangular such that A = S + L.

do you understand that this is our sub-goal or not

[radiant yarrow]

yes

[dusky epoch]

okay

now let's look at my description again

1. take the contents of A on and above the diagonal and copy them into S
2. fill in the subdiagonal portion of S to make it symmetric
3. take L = A - S

[radiant yarrow]

holy shit

[radiant yarrow]

Can I just say it won't be a direct sum if a vector doesn't belong to either W1 or W2 because the vector space isn't closed in the sum of two subspaces?

[dusky epoch]

sum != union...

also what's your definition of direct sum?

[wintry steppe]

most likely W1 + W2 = V and W1 \cap W2= empty

@radiant yarrow homie you there?

[radiant yarrow]

combinations of all the x+y such that x belongs in W1 and y belongs in W2

[radiant yarrow]

W1 intersection W2 = {0}

[wintry steppe]

ye ofc

so like try to show both impliactions

assume each v in V can be written uniquely as x1 + x2, so just need to show that w1 cap w2 ={0}

[radiant yarrow]

Oh

[wintry steppe]

if you show this, then you have to assume w1 and w2 make direct sum of V, need to show the uniqueness, which also should be quite easy

[radiant yarrow]

Okay

[wintry steppe]

let me know if ure stuck

[lucid glacier]

={0}, not empty

[wintry steppe]

yes acn u read

[radiant yarrow]

To show uniqueness you need to make a contradiction of one of the vectors not belonging to either subspaces right?

[wintry steppe]

assume its not unique and find a contradiction.

assume there are x1+y1 = x2+y2 for x_i in W1, y_i in W2

then you have x1-x2 = y2 - y1

BUT left side is in W1 and right side is in W2 (because those are subspaces, so closed under addition)

but they are the same thing

and the only same thing in W1 and W2 is 0

therefore x1-x2=0 => x1=x2 and y2-y1=0 => y2=y1

That shows uniqueness

[lucid glacier]

There ARE vectors not belonging to either subspace, but this a sum, not a union, so any vector will be of the form w1+w2 with wi in Wi.

[radiant yarrow]

Oh

[radiant yarrow]

Got it

thanks a lot @wintry steppe

Sorry for late reply

[silver heath]

Uh

I am a little confused

with symmetric diagonalization

why are we allowed to orthagonalize the eigenvectors

and still treat them as such?

[hard drum]

wdym 'still treat them as such'?

the important thing with symmetric matrices in this connection is that eigenvectors corresponding to different eigenvalues are orthogonal, and even if two eigenvectors correspond to the same eigenvalue, we can find an orthogonal basis for the corresponding eigenspace by Gram-Schmidt

[silver heath]

i am a little confused about the eigenspace part

if we use gram-schmidt the orthagonal vectors will not be eigen vectors so whats the point of starting with eigenvectors in the first place?

[hard drum]

We're creating an orthogonal basis of the eigenspace, so all vectors in that space are eigenvectors, in particular the elements of the basis

[silver heath]

Righhhhhhhhhttttt

[silver heath]

Is this true?

I mean a seems correct

but i am not sure

[wintry steppe]

why are you unsure?

[silver heath]

idk

[wintry steppe]

well if you know A = PDP^T what happens when you invert both sides

do note that the inverse of P is P^T

[silver heath]

PD^-1P^T

[wintry steppe]

yes

so that means that A^(-1) is orthogonally diagonalizable, right?

[silver heath]

yeah

[wintry steppe]

so it's a

[fleet orbit]

in an orthogonal basis defined by Span{v1,v2,v3} for the column space of a matrix that is linearly independent, does the dot product between v1 · v2 · v3 = 0? or is this true only for the dot product of v1 · v2, v2 · v3, v1 · v3?

wait if v1 · v2 = 0, then its just 0 · v3

is the dot product even defined for a scalar and vector

[dusky epoch]

v1 · v2 · v3 = 0
what would that even mean?

[limber sierra]

it is not

so youre right, v1 · v2 · v3 is nonsense

but yes, each of v1 · v2, v2 · v3, and v1 · v3 is 0

[fleet orbit]

so in order to check my work all I have to show is that the dot product between two of the vectors are 0

[limber sierra]

right, between every pair of vectors

[fleet orbit]

Okay thank you very much!

[silver heath]

is this true?

i mean it can be a projection on to the identity matrix??

because proj is

u*u^t y

and y can just be identity matrix?

[grim bridge]

it can be written as PDP^-1 where P is diagonal, and a diagonal matrix can be written easily as a linear combination of matrices with zero everywhere except for a 1 on the diagonal

and then you develop the sum and it becomes a sum of orthogonal projections since P is orthogonal

[hollow finch]

have you discussed spectral decomposition in your class?

[tribal moss]

I'm a high school student and have a bit of confusion with respect to Euclidean space, so if you have a vector space together with an inner product then we get a euclidean space which is your basic geometric space used in calculus. But how can you construct such a geometric space if there is no order relation with respect to the inner product. Or am I understanding something wrong, any help would be really kind 🙂

[hollow finch]

Having an inner product isnt a necessary axiom of a vector space. Most vector spaces have a standard inner product defined for them, but it's not required for something to be a vector space. You may want to look into the 10 axioms of a vector space

[silver heath]

ye

[hollow finch]

alright then. you know that uu^T is a projection matrix. how does that relate to spectral decomposition?

[silver heath]

ye i figured it out 🙂

is there an easier way to do this?

what came into my mind was making x = [x_1, x_2]

and parametize the boundry x_1^2+x_2^2 = 1 with a sin x, cos x

or use some sort of lagrangian multipliers which is super overkill

but is there a method thats obsurdely easier

as i am pretty sure thats not the intended method

[hybrid portal]

It is related to definition of matrix norms. Calculating 2nd norm of M is enough.

[silver heath]

wait what?

[hybrid portal]

Largest singular value is the answer.

[silver heath]

hmm.

.i havent learned singular value decomposition yet

[wintry steppe]

the largest value is the highest eigenvalue of M i think

[hybrid portal]

Yes since it is symmetric.

and positive definite.

[wintry steppe]

^

i think it follows regardless if the quadratic form is positive def or not

but M needs to be sym

[hybrid portal]

It needs to be pos def. For example, consider -M, it is still symmetric singular values doesnt change, but its eigenvalues are negative now.

[wintry steppe]

err

tbh im just basing my stuff of my textbook

heres the theorem im using

[hybrid portal]

my bad, i read sentence like it says "largest singular value is the ...", I was replying to that

[wintry steppe]

ah its ok

but thats true

largest singular val has to be pos

[hollow finch]

it doesnt need to be pos def. if the two eigenvalues are 4 and -2 then the largest value is 4 and the smallest is -2.

for example

its just the eigenvalues

for example the matrix $\begin{bmatrix}0&1\1&0\end{bmatrix}$. the quadratic form is $$Q(\vec{x})=2xy$$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

$$Q\left(\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)=-1$$
$$Q\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right)=1$$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

and -1 and 1 are the smallest and largest inputs possible for a unit vector

[hybrid portal]

Yes, it is totally true. In the original question, it is not a problem to use matrix norms because matrix is already pos def.

[wintry steppe]

Do the columns of an mxn matrix A with rank(A)=n span R^m?

[teal grotto]

rank(A) is always less than or equal to m and n

[wintry steppe]

Yes

but what if its less than m

[hard drum]

I mean if rank(A) = n then we've said the colspace is n dimensional

And it's a map to R^m, so the answer depends on whether m > n or m = n

[wintry steppe]

yeah

so its not gauranteed

this was the q

[hard drum]

I think I agree

But I am tired so I may be overlooking smth lol

[manic summit]

Hey guys how do you do domain/range/function linear graphs?

[nocturne jewel]

General notation I saw, if V is a vector space what's End(V)?

[gray dust]

End(V) is the set of linear maps V->V

[ember cedar]

Err hello can i ask if identity map is the same as zero translation??

[ocean sequoia]

do you mean zero transformation?

if so no the zero transformation takes a T(v) = 0 for all v in V

the identity map is f(x) = x for all x

[ember cedar]

Ah yes i meant transformation

Thanks

[ocean sequoia]

yep

they are the same for the zero vector

[ember cedar]

Ah ok i see

[ocean sequoia]

which maybe was your example

[teal grotto]

if you meant zero transformation, then they are the same if and only if your vector space is the trivial vector space, i.e., the vector space containing only the zero vector

[ember cedar]

Alright understood thanks c^2 and obedience🙏 that was the answer i was looking for

[silver heath]

wait

this is a stupid question but is

*what is min {-1,0.5,-0.5,1}

NEVERMIND IM BIG BRAIN DUMB

[wintry steppe]

[silver heath]

thats just sad on my part

[silver heath]

Uhhh what does it mean for a matrix to be negative semidefinite??

my textbook says " Other terms, such as positive semidefinite matrix. are defined analogously"

Which is.... awfully discriptive

[wintry steppe]

negative semidefinite means x^TAx <= 0 for all x

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

[silver heath]

so wait

negative semidefinite just means its never positive?

[wintry steppe]

yes

[silver heath]

does this imply that none of the eigenvalues are positive?

[wintry steppe]

u tell me

[silver heath]

yes...

[wintry steppe]

yea

[silver heath]

ok wtf... am i mising braincells or do i not know how to read this question

whats the question?

[wintry steppe]

is this supposed to be a multiple choice

it's kind of strangely worded

[silver heath]

its a true false

question

but i dont understand what its asking?

[wintry steppe]

i guess it's trying to ask whether or not you can go from y^TDy to x^TAx by setting y = P^Tx

[sudden ferry]

anyone willing to help me with eigenvectors

it would be very much appreciated

:)

it involves checking something for me

[dusky epoch]

don't ask to ask

just post the problem & your work

[sudden ferry]

ok thanks ann

[sudden ferry]

I got the negative sign for the j component of v_1 instead of the i component

can someone do this and tell me if they get the same?

I'm not sure where I went wrong

This is my working out:

[gray dust]

@sudden ferry scalar multiples of eigenvectors are eigenvectors

your answer is -(their answer) so it’s fine

[forest quiver]

Does anyone have a good Linear Algebra problem that tests multiple different skills?

I need to use a problem for an essay

and I want to find a problem that has multiple components to it so I can break it up into different sections/body paragraphs

[gray dust]

here’s one i typed sometime ago, see if it’s worth writing an essay on

Take $\bR^n$ as a $\bR$-vector space under usual adding and scaling. For this problem let’s write all vectors in $\bR^n$ as columns.\

\begin{itemize}
\item Let $\mbf x,\mbf y\in\bR^n$ be non-zero vectors. Show that $\mbf x^T\mbf y$ is the dot product $\mbf x\cdot\mbf y$ and that $\mbf{xy}^T$ is an $n$ by $n$ matrix.
\item Viewing $\mbf{xy}^T$ as the standard matrix of an operator on $\bR^n$, find its image and kernel, then find the dimensions of these spaces.
\item Let $\mbf x_1,\mbf x_2,\mbf y_1,\mbf y_2\in\bR^n$ where $\brc{\mbf x_1,\mbf x_2}$ and $\brc{\mbf y_1,\mbf y_2}$ are linearly independent. Viewing $\mbf x_1\mbf y_1^T+\mbf x_2\mbf y_2^T$ as the standard matrix of an operator on $\bR^n$, find its image and kernel, then find the dimensions of these spaces.
\end{itemize}

[stoic python]

The Singing Sands of Shangdu

[sudden ferry]

indeed

[forest quiver]

I appreciate this

But I would prefer ones with actual applications

because I feel like I have a better understanding of those

but yeah this is a good example it has 3 components to it

maybe I can find a way to combine previous ones I have done

something like this could be interesting

Maybe instead of arbitrary vectors/matrices I can use actual values

and do some sort of polynomial interpolation thing

idk

[silver heath]

Is there a quadratic form Q on r^2 such that Q(x) = 1 is an eclipse

[hard drum]

yes

assuming you mean ellipse

what is the standard equation of an ellipse?

[silver heath]

yeah

[silver heath]

$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

[stoic python]

Elonmosqito96

[silver heath]

@hard drum ?

[scenic fulcrum]

Its almost a quadratic form

[hard drum]

well how about an ellipse about origin?

[silver heath]

x^2/a^2+y^2/b^2=1

🤦‍♂️ I meant to ask "Is there a quadratic form Q on r^2 such that Q(x) is an eclipse and is indefinite

[hard drum]

oh lol fair

[limber sierra]

this isnt linear algebra, try #prealg-and-algebra or an open questions-_ channel

[silver heath]

oof

[forest quiver]

How are Lagrange Multipliers related to Linear Algebra?

[sudden ferry]

I don't understand equation 4.68, what is the significance?

(Topic: diagonal matrices)

[limber sierra]

this is very broad

[forest quiver]

I guess they are related because of determinants?

What I am thinking is horribly wrong

[limber sierra]

the lagrange multiplier theorem itself is a statement that, given certain conditions and under certain constraints, the derivative of a function is a linear combination of the derivative of the given constraints with your lagrange multipliers being coefficients

[forest quiver]

hmm

[limber sierra]

er, gradient

not derivative

[forest quiver]

Oh right ok that kinda maeks sense

I haven't taken calc 3 yet, but from my understanding

a Lagrange Multiplier is factor multiplied to one side of an equation of the gradients of two multivariate functions, essentially matching up the arbitrary gradient vectors

does that work ?

but then I don't have the connection to Lin Alg

I will have to understand the linear combination part

oh wait yeah that is linear

[limber sierra]

take calc 3

[nocturne jewel]

Wronskian's time to shine?

[glacial mango]

Why doesn't (III) hold?

[wintry steppe]

1 0
0 0

[glacial mango]

Does that only hold in integral domains and the ring of matrices is not one?

[wintry steppe]

yeah M_n is generally not an integral domain

unless n = 1 and your matrices have entries in an integral domain im pretty sure the example i gave (appropriately generalized) shows that it's never an integral domain

[hard drum]

note as well that matrices with P^2 = P are (by definition) projections, which can help give some intuition

[silver heath]

Get outa here

[nocturne jewel]

[silver heath]

[wintry steppe]

[glacial mango]

If I have a set of vectors, I can find out if they're linearly independent by creating a matrix with them and seeing if the matrix reduces to the identity or not. If it doesn't, shouldn't the solution (interpreting the row vectors as equations) give me the coefficients that show that one is a linear combination of the others?

[nocturne jewel]

I dont automatically see anything wrong with that statement

[teal grotto]

yea. won’t necessarily be the identity tho

[stable kindle]

can someone check my proof pls

essentially what needs to be shown is that there exists a vector in V that maps to either w1 or 0, since you could just form a basis from that vector?

so let that be v1

dim V = m

so dim range T =< m

now, if w1 is not in range T, then by the condition Tv1 = 0

and that fulfils all the criteria easily?

if w1 is in range T, there exists v1 such that Tv1 = w1

and then just extend that to a basis??

have i done something wrong? it feels wrong

i am very tired

[stable kindle]

<@&286206848099549185>

[ocean sequoia]

I’m not sure you need the dim V = m

If t(v1) = 0 = 0w1

[stable kindle]

yeah, that's extraneous

it just seems so simple

well...

hmm

no

[ocean sequoia]

I think this is a “simple” question don’t over think it

Honestly I’m not sure if the second part is enough

Why does that show that there is a 1 in 1,1 and 0 everywhere else

[stable kindle]

uhhhh

ok i think that's what i was missing then ig

i didn't say anything about v2, ..., vm, but ofc i should have

kdsjf

[ocean sequoia]

I don’t think you need to mention v2…vm

Because we are extending w1 to a basis right like you said is def correct

[stable kindle]

ok what if

we find a space U such that range T is the direct sum of U and the space generated by w1 alone

then...

ok no first what are we actually doing i'm confused, lemme think

start over

[ocean sequoia]

Hey I have to go fuck

I’m sorry

[stable kindle]

np

ty for trying

[ocean sequoia]

I think what you had at first is basically right

I was trying to think if we could make it cleaner but that seems close

[stable kindle]

i've flipped my position, i think it's insufficient, it doesn't show that there's a 0 everywhere else

i'll think about it some more on my own

[ocean sequoia]

Yea but I think you were close

[weak needle]

Isn’t this problem just the equivalent of subtracting the first column of the matrix times some appropriate scalar from all the other columns, and then scaling the first row appropriately? I think if you phrase this in terms of basis vectors it gets a little messy but should still be doable. Since your soln doesn’t talk about any of the other basis vectors, I feel like it can’t be correct. Atleast I don’t see why the first row has 0 in all other places

[stable kindle]

oh, yeah, i think i figured this one out but didn't post it here

essentially if we start with some random basis where Tv1 = w1, then if there's some wi that doesn't work, so wi (where i > 1) = a1Tv1 + ... + amTvm, we can just adjust the original basis?

so for example we could let vm* = vm + (a1/am)v1

and then replace vm with vm* in the original basis

and cancel out the issue

except we wouldn't want to be replacing vm, we'd want to be replacing vi, whoops

so for each wi that doesn't work, we adjust vi to vi* to fit, and it should still be a basis

[weak needle]

This seems a bit backwards, you should see how Tvi is expressed in terms of wi’s not how a wi is expressed in terms of Tvi’s. Indeed it could be outside of the range

My point was, start with some random basis {vi}

Then look at Tvi’s, if none have w1 in the expansion, we are done

If one of them does, wlog Tv1=a1w1+smth

Scale v1 by 1/a1 to get Tv1=w1

+smth

[stable kindle]

well i was just ignoring the fact that it had to be 1 because it's so trivial to scale it

but ok

[weak needle]

Then if w1 appears in the expansion of Tvi=aw1+smth, replace vi with vi-av1

Now these new vi’s will be the exact basis we need

[stable kindle]

that's basically what i was describing?

[weak needle]

You had wi=sumation of Tvi’s

That seemed backwards

[stable kindle]

i don't see the problem

i think it works either way round?

[weak needle]

Why does wi have that expansion? I think I may have missed something you said

[stable kindle]

well that's what the matrix actually describes, right?

[glacial mango]

Shouldn't (5,-1,1) be a solution to this? Let z = t, then x = 5t, and y = -t?

[stable kindle]

wi = ai,1*v1 + ... ai,m*vm, that's what the matrix is, right?

M * v = w

thinking about wi as a summation of vi is the forwards direction of the matrix multiplication

surely?

[weak needle]

Why is Mv=w?

[stable kindle]

have i fundamentally misunderstood matrices somehow

ok so

[weak needle]

I think maybe we are using different matrix conventions or something

[stable kindle]

possibly

or maybe i've just fundamentally misunderstood them

that would be funny

[weak needle]

Ok so my understanding is that the colums of the matrix M, say the first column, is the coefficients appearing in expansion of Tv1 under the {wi} basis

[stable kindle]

ok

yeah, it's on my end

i've just gotten it all the exact wrong way round, lmao

well that's shit

[weak needle]

I see, yeah we all have brainfart moments

[stable kindle]

i guess i've been working with M^-1 the whole time? essentially?? idek

it seemed to work out, lol

[weak needle]

The problem is that M may not be invertible, but if it is something like what you say would probably work

[stable kindle]

yeah

ok i just need to start over again and approach it with the correct convention

this is pretty clearly a valid approach, i think, so at least i know where i'm going now

ty for the patience lol

[weak needle]

Np

[wintry steppe]

it is a solution to Ax=0 where A is that matrix

[fleet orbit]

how do I find the determinant of this using the properties of determinants and how can you get just a number out of this???

pls ping if you have any insight

[dusky epoch]

@fleet orbit first thing i'd do here is subtract col 3 from col 1 to zero out some entries

also i think you may need to know the determinant of [a b c d; e f g h; i j k l; m n o p]

but the first col is clearly almost a copy of the third

[tranquil steeple]

>> det([3 2 1 0 0;b a b c d; f e f g h; j i j k l; n m n o p])

ans =

2*a*f*k*p - 2*a*f*l*o - 2*a*g*j*p + 2*a*g*l*n + 2*a*h*j*o - 2*a*h*k*n - 2*b*e*k*p + 2*b*e*l*o + 2*b*g*i*p - 2*b*g*l*m - 2*b*h*i*o + 2*b*h*k*m + 2*c*e*j*p - 2*c*e*l*n - 2*c*f*i*p + 2*c*f*l*m + 2*c*h*i*n - 2*c*h*j*m - 2*d*e*j*o + 2*d*e*k*n + 2*d*f*i*o - 2*d*f*k*m - 2*d*g*i*n + 2*d*g*j*m

[wintry steppe]

Not helpful

[teal grotto]

nope

[teal grotto]

i think by laplace expansion, you really don’t care about expanding on the second and last two entries in the first row, since they will be zero

[lucid glacier]

I think doing expansion for anything over a 3x3 matrix is torture. I think bringing it to upper triangular form might be better

[teal grotto]

no but a bunch of stuff cancels here

focus on what happens when u expand on the first and third entries in the first row

you would just need to know the determinant of [a b c d; e f g h; i j k l; m n o p] like ann said

[lucid glacier]

tru

still a bit of a pain tho but 🤷‍♂️

[teal grotto]

det(G) = 2det [a b c d; e f g h; i j k l; m n o p]

[dusky epoch]

first thing i'd do here is subtract col 3 from col 1 to zero out some entries

yeah

i think lytei is withholding some info from us

[teal grotto]

spill the beans @fleet orbit

[wintry steppe]

😂

[tranquil steeple]

?

[tranquil steeple]

this is the same as I posted.

[teal grotto]

i just meant like “nope, nope, nope, not taking the time to read and understand that”

[half ice]

Yikes multiple 4×4 determinants.

[lean acorn]

Hello I was wondering if someone could tell me if this is true. If I have a finite dim vector space V and define K = V \directsum V' (V' is the dual of V). Then take the dual of K as K' = (V\directsum V')' . Would the dual of K just be V' \directsum V''?

[sonic osprey]

yes this is true

[lean acorn]

amazing! tyvm

[wintry steppe]

Is it 95?

[sinful acorn]

You need to find the amount of vectors that aren't 0 in Both W1 and W2

there are 20 0's in W2

and 25 in W1

[wintry steppe]

X€(W1 intersection w2) if xi st i is divisible by 20

[sinful acorn]

I don't realize what do you mean

[wintry steppe]

W1 intersection w2 ={(x1,x2,.....,x100) : xi=0 if i is divisible by 20}

[sinful acorn]

no

[wintry steppe]

Why

[sinful acorn]

W1 intersection w2 = {(x1,x2,.....,x100)} : xi = 0 if i is divisible by 4 OR by 5

at least by one of them

not by both

[wintry steppe]

Intersection is logical AND

[teal grotto]

(x1,…,x100) is in W1 intersect W2 means that xi = 0 if i is divisible by 4 and i is divisible by 5.

so how many numbers are there of the form
1<= 4^k * 5^m <= 100 with m,k>= 1?

[sinful acorn]

I know, another way to look at it is W1 intersection w2 ={(x1,x2,.....,x100) : for each i, xi not equals to 0 in W1 and in W2}

its the same thing tho

[teal grotto]

huh?

[sinful acorn]

you need the vector to not be 0 in both the subspaces

[wintry steppe]

3

[teal grotto]

wot? lol. i count at least 5

20, 40, 60, 80, 100

[wintry steppe]

I didn't get 40 60

[sinful acorn]

@teal grotto what is answer that you get

because I think one of us is wrong

of the whole question

[teal grotto]

wait wtf am i stupid

[wintry steppe]

20(k=1,m=1),100(k=1,m=4),80(k=2,m=1)

[sinful acorn]

I will write here the vectors that are 0 in at elast one of the groups: {x4, x5, x8, x10, x12, x15, x16,....}

these vectors

AREN'T in the intersection

do you get that?

[teal grotto]

40 is divisible by both 4 and 5, so x40 should be zero. i set something up wrong

[wintry steppe]

5×4 is enough?

[teal grotto]

this bottom part is wrong

don’t know why i said 4^k 5^m

it should be the number of integers k between 1 and 100 such that k mod 4 = 0 and k mod 5 = 0

my bad

[sinful acorn]

its all wrong

you need the vector to be not 0 in BOTH groups

and then it goes into the intersection

[teal grotto]

what?

[sinful acorn]

its not a Union

its an Intersection

[teal grotto]

(x1,…,x100) is in W1 intersect W2 provided that xi = 0 whenever i is divisible by 4 AND i is divisible by 5

[wintry steppe]

K=4,m=5

[sinful acorn]

so do you say that the answer is 95? @teal grotto

because the only numbers are 20 40 60 80 100

these are divisivble by 4 and by 5

[wintry steppe]

yes.

Numbers which are divisible by 5 & 4 are the numbers which are divisible by 20

[teal grotto]

yes. that should be right based on this

[sinful acorn]

its not tho

[teal grotto]

nani?

[sinful acorn]

the correct answer is 60

[wintry steppe]

Its right

[sinful acorn]

like the dimension

[teal grotto]

through what line of reasoning

[sinful acorn]

.

and thats happens when i is divisible by 4 OR by 5

there are 40 numbers like that

[wintry steppe]

Not OR ITS AND

[sinful acorn]

and 100 - 40 is 60

its not

[teal grotto]

bruh

it’s and

[sinful acorn]

its AND here

but in order to solve this

you change it to OR

[sinful acorn]

This is correct....

[teal grotto]

so what your telling me is that (1,1,1,0,1,0,…,0) is in both W1 and W2

[sinful acorn]

we are using X's not numbers

what is 1 1 1 1 0 0 0 0

x1 is in the base of the intersection

as well as x2 x3 x6 x7 x9

[teal grotto]

a vector

[sinful acorn]

but x4 x5 x8 isnt

[teal grotto]

and now that i’m thinking about it, it should be in both lol

bad example hold on

there we go

[teal grotto]

this vector is not in the intersection of both W1 and W2 since the fifth coordinate is not 0

[sinful acorn]

true

[wintry steppe]

Yeah

[teal grotto]

so point proven

it’s and

[wintry steppe]

Yeah

[sinful acorn]

its AND HERE:

but in order to find how much i there are

you can do it anyway you want

[wintry steppe]

And it has to be we are intersecting two sets and intersection means both the conditions has to be satisfied for the elements belonging to intersection

[sinful acorn]

the easiest way is to do it like I did

and use OR

[teal grotto]

u got the wrong answer lol

[sinful acorn]

what is the correct answer?

[teal grotto]

95

[sinful acorn]

its not

[wintry steppe]

95 make sense right ?

[sinful acorn]

its not true tho

[teal grotto]

ur argument is not convincing me because of this

[sinful acorn]

I didnt say that the vector is

it ISNT

[teal grotto]

but according to u it would be in the intersection

[sinful acorn]

no it wont

[teal grotto]

because it satisfies the or condition

yes it would

[sinful acorn]

because i = 5 is divisible by 5

[wintry steppe]

@sinful acorn what kind of OR are you considering here INCLUSIVE / EXCLUSIVE?

[teal grotto]

yes but because i = 4 is divisible by 4 and the fourth component is 0, then it would be in the intersection, according to u

[sinful acorn]

inclusive

[teal grotto]

that is irrelevant here

[sinful acorn]

no it wont

[teal grotto]

explain this

[sinful acorn]

the vector that has 15 zeros, then 1 one and the rest is zeros

[wintry steppe]

Then there might be elements in intersection such that they are only divisible by 5

[sinful acorn]

isnt in the intersection

is that correct?

[wintry steppe]

And same goes with 4

[sinful acorn]

true

and

if they are divisible by only 5

[wintry steppe]

So its wrong

[teal grotto]

yes because it’s not in W1

[sinful acorn]

they AREN'T in the intersection

[wintry steppe]

We want it to be divisible by both 5 & 4 at the same time

[sinful acorn]

no

we want them to be divisible by at least on of them

if a number is divisible by at least on

he isnt in the intersection

the correct bese is: {1 2 3 6 7 9 11 13 14 17 18 19 21 22 23 26 27 29 30 31 .........}

and I already took down more than 5 numbers

[teal grotto]

wait sanity check. the dimension of W1 intersect W2 has to be less than the dimensions of W1 and W2, so it’s gotta be less than 80 and 75

welp

i’m fundamentally misunderstanding something here

[sinful acorn]

due to this...

[teal grotto]

well, just because W1 intersect W2 is a subspace of both W1 and W2

but yea that too

[sinful acorn]

Do you realize this?

[teal grotto]

no i’m not seeing it just yet. i’m having trouble seeing where i went wrong

[sinful acorn]

you found the union of bot subspaces

not the intersection

[stable kindle]

so it'll repeat every 20

1, 2, 3, 6, 7, 9, 11, 13, 14, 17, 18, 19

[sinful acorn]

yes

[stable kindle]

and similarly for the next 20, and the next 20, ie. 21, 22, 23, 26, 27, 29, 31, 33, 34, 37, 38, 39

5 20s

and that's 12 numbers, 12 basis vectors

12 * 5 = 60

[teal grotto]

where did 12 come from dude lol

[stable kindle]

len(1, 2, 3, 6, 7, 9, 11, 13, 14, 17, 18, 19) = 12

[sinful acorn]

@wintry steppeyou still here?

lol

[wintry steppe]

Yep

[teal grotto]

\begin{align*}
W_1\cap W_2&={(x_1,\dots,x_{100}):i\mod 4 = 0\Longrightarrow x_i = 0}\cap{(x_1,\dots,x_{100}): i\mod 5 = 0\Longrightarrow x_i=0}\&={(x_1,\dots,x_{100}):(i\mod 4 = 0\Longrightarrow x_i = 0)\wedge(i\mod 5 = 0\Longrightarrow x_i = 0)}
\end{align*}
is $(i\mod 4 = 0\Longrightarrow x_i = 0)\wedge(i\mod 5 = 0\Longrightarrow x_i = 0)$ is true whenever $i$ is divisible by either $4$ or $5$?

[stoic python]

c squared

[wintry steppe]

Okay a implies b is false only when a is true and b is false then we will not consider that i . But if a is false and b is false still implication is true we have to consider i

Okay got it

[stable kindle]

i think this is wrong from the first line

the full condition for the intersect is that whenever i mod 4 = 0 or i mod 5 = 0, xi = 0

that means that i mod 4 => 0 and i mod 5 => 0

afaict?

[teal grotto]

i messed up on the first line???

[teal grotto]

explain

[sinful acorn]

the first line is correct

[wintry steppe]

I think this is the explanation for OR like behaviour of AND in this case

[stable kindle]

what do you want me to explain about it, it seems simple to me

[sinful acorn]

thats the definition

[stable kindle]

no offence

[teal grotto]

dude ik. im not getting it

[stable kindle]

what about it don't you get

i don't know what to explain

[sinful acorn]

.

[stable kindle]

it's not

[teal grotto]

bruh am i just dumb?

[stable kindle]

wait

i might be braindead

[teal grotto]

same?

[stable kindle]

no, it's me, sorry

stupid question, did you at any point edit an or symbol into an and symbol

[teal grotto]

no

[stable kindle]

i could have sworn one of them was an or

ok it's just me

i apologise

[teal grotto]

no worries

[stable kindle]

ok this looks right then

[wintry steppe]

Guys is this correct?

[teal grotto]

im struggling to see why the condition from the second line turns into the condition you say it does.
is (p-->r) and (q-->r) equivalent to (p v q) --> r?

[stable kindle]

i have no idea what you're talking about, it's incoherent

[wintry steppe]

Okay wait

[stable kindle]

so it's basically what you said in the last part of the first line

except i've shortened it

[stable kindle]

so if i mod 4 = 0, xi = 0; and if i mod 5 = 0, xi = 0
therefore if i mod 4 = 0 or i mod 5 = 0, xi = 0

right?

[teal grotto]

oh. my. god.

i am a child with a small brain

[stable kindle]

f

[teal grotto]

f indeed

@stable kindle @sinful acorn thanks for being patient lmao

[stable kindle]

np

[wintry steppe]

Is this explanation correct?

[dire thunder]

it is not really clear

what are you trying to prove

[wintry steppe]

We wanted intersection of
(a implies b1) AND (c implies b2)
i such that i is divisible by 5 and 4 . So someone will consider i which is divisible by both 5 and 4 if take such i ans turns out to be wrong and if we consider same
(a implies b1) OR (c implies b2) then answer turns out to be right so i wanted to prove that there is no or between those two statement that and in statement just acts similarly like or coz of implication that i mentioned above

[teal grotto]

they are trying to prove the above in the case where
p is i mod 4 = 0
q is i mod 5 = 0
r is x_i = 0

but @wintry steppe it’s true regardless of what p, q, and r actually are since

(p v q) —>r = ~(p v q) v r
= (~p ^ ~q) v r
= (~p v r) ^ (~q v r)
= (p—>r) ^ (q—>r)

[wintry steppe]

Yeah it is. so thats why @sinful acorn was saying OR ig

[teal grotto]

yup. i just didn’t realize that until kaisheng pointed it out for me

no need to ping ariel either lol

[wintry steppe]

I realised it after you mentioned that logical equation

[obsidian axle]

How do you know whether to pre-multiply or post-multiply a matrix by a transformation. I know how to do the pre and post multiply but how do I determine which one will give me the transformation I expect such as a horizontal shift or a clockwise rotation?

[tranquil steeple]

what do you mean? you could either learn it or just try on a model problem?

[limber sierra]

when you think of matrices as representing linear functions, they are usually written as left-multiplication

to match up notationally:

$T(\vec{v}) = M\vec{v}$

[stoic python]

Namington

[limber sierra]

that said, in theory you could find a matrix that does the same thing, but upon right-multiplication instead

that is to say: there's no general rule, and you just have to compute what it does to a vector to see if it has the desired effect

(or memorize specific forms)

[wintry steppe]

if anyone can help with this, i'd be really thankful

the question is how to prove (2)

So, how to prove that A** = A?

[glacial mango]

But 5(1)-2=3, not -3?

[rugged echo]

Wait uploading

Can you help me pls

[plain saffron]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[rugged echo]

<@&286206848099549185>

[limber sierra]

you delete your ping and then ping again, still ignoring the 15 minute rule that i just posted

wtf

[rugged echo]

Sorry

[wintry steppe]

What's the point of C here

It seems entirely useless to define C only to immediately unfold the definition

But the proof itself is fine

[rugged echo]

Oh ok so i can put sum(a+b) already

[wintry steppe]

Sure

[rugged echo]

Thanks bro

[limber sierra]

recall properties of a hermitian form.

in particular, $\langle A^*v, u\rangle = \overline{\langle u, A^*v\rangle}$

[stoic python]

Namington

[limber sierra]

can you complete the argument?

@wintry steppe

(sorry for late response, you mightve already figured it out)

[wintry steppe]

???

What

5(1)-5=0

Also

5+(-2)+(-3)=0

[wintry steppe]

.

[wintry steppe]

Yes, I can complete it now, thanks a lot!

[glacial mango]

I'm an idiot. I was setting it equal to the third column.

[glacial mango]

Thanks for that

[wintry steppe]

Sorry I wasn't trying to be mean

But yeah no problem

[fleet orbit]

@dusky epoch @teal grotto @tranquil steeple my bad guys, im notorious for not reading the directions, it was like a 10 part question with different scenarios so I forgot they gave me det|a,b,c,d;...]=5, but otherwise I solved it after realizing this, I did a cofactor expansion along the 1st column

thanks though!

[wraith patio]

so if you do $QR$ decomposition, $A=QR$ then will $A^TA=R^TR$?

[stoic python]

taxminion

[teal grotto]

yup

[tranquil steeple]

Yes because $A^T=R^TQ^T$ and $Q^T=Q^{-1}$ so $A^TA=R^TQ^TQR=R^TR$

[stoic python]

Sven-Erik

[wraith patio]

neato

[torn stag]

Suppose $N \in L(\mathbb{C}^n)$ is nilpotent and $c > 0$. Can we prove that $N$ is similar to $cN$ without using the Jordan decomposition?

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Maybe by using Schur upper triangularization?

[reef nebula]

Hi everyone! Hoping I could get some confirmation on this

When we're dealing with homogeneous coordinates, is the reasoning behind adding the third coordinate just so that we can convert a 2D point into a 3D point? I understand that this makes transforming a bit easier, i'm just wanting some confirmation on this

I'm currently getting into Computer Vision and so theres some math concepts that I want to make sure I understand before I continue on

[sonic osprey]

I'm not sure what you mean by this. Are you asking why three homogeneous coordinates is like 2 regular coordinates?

[reef nebula]

basically 🙂

[sonic osprey]

One reason is that there are copies of 2D within the three homogeneous coordinate grid

So you can send (x,y) to [x:y:1]

[reef nebula]

what do you mean by copies?

I'm also trying to understand what the one represents

I've graphed it to give me a visual of both 2D and 3D but even though it's still not there for me

when it comes to understanding the relation of the 1 in the homogenous coordinate

[sonic osprey]

If you look at the set of all [x:y:1] inside of 3 homogeneous coordinates, it looks exactly like the usual 2D plane

1 isn't super special here, you can replace it with any non-zero number

[reef nebula]

i thought having the 1 there was special since in order to convert back to a simple (x,y) values we the homogenous coordinates by whatever the 3rd value is

[sonic osprey]

Well I mean

[x:y:2] is the same as [x/2 : y/2: 1] so it doesn't really matter since you can always scale to 1

[reef nebula]

I think im starting to understand now

so we have it set to 1 because no matter what values x and y are, if we see it 3-D line (1,1,1) top down from the z-axis we will essentially see the 2D representation of it

the 3rd 1 makes it to where we can go from 2D to 3D and vice versa without changing the characteristic of the line

[wise basin]

When solving for eigenvalues and eigenvectors, What implications arise when the eigenvectors are not all linearly independent.

[torn stag]

@wise basin Eigenvectors corresponding to distinct eigenvalues are always linearly independent. More generally, the same holds for generalized eigenvectors.

[hard drum]

and even if you have multiple eigenvectors corresponding to a single value, we can always turn that into a linearly independent set anyway

[wise basin]

What can you do when you the eigenvectors are not all independent and the eigenvalues are degenerate? The symmetric matrices are used all for their independence properties, but what can we do when they are not all independent?

I guess what I am asking is what can we say about a system when the eigenvectors are not all independent.

[weak needle]

Have you heard of the jordan canonical form for a matrix? When you can’t find independent eigenvectors, you can still find independent “generalized eigenvectors”. That is pretty much the best you can do in case of an arbitrary matrix I suppose.

[fleet orbit]

How do I approach this problem?

all I have in my notes is that {area of T(S)}=|detA| {area of S}, which isn't very helpful

[rugged echo]

I need help in this

<@&286206848099549185>

[weak needle]

What is the question?

[rugged echo]

Whether (A+B)(A-B)=A^2-B^2 is a valid matrix equation

And

(A+B)(A+B) = A^2 +2AB +B^2 is a valid matrix equation

[half ice]

All of that is correct @rugged echo

[rugged echo]

Thank you 💝

[wintry steppe]

@teal grotto what book did you use to learn LA?

[teal grotto]

the book that my class followed loosely was Hoffman and Kunze

[wintry steppe]

did you take two classes or one?

[teal grotto]

just one so far. it wasn't explicitly called a linear algebra class per se. there was a lot of linear algebra tho.

[wintry steppe]

Oh

Was it called Algebra?

[fleet orbit]

anyone able to help with this one?

nvm figured it out!

[zinc copper]

It’s a little unclear if they’re referring to the internal or external direct sum here: in the book the external direct sum was notated with squares instead of circles for the finite case, but the internal one was not defined as an operation on arbitrary vector spaces but as a means to recognize a space as a “direct sum” of some of its sub spaces

If S and T are sub spaces of V and we can write V as a direct product of both on the two orders then clearly it’s commutative (internal) but in the external sense they are not the same set wise though they have the same structure (I haven’t reached the notion of an isomorphism for vector spaces but I assume it’s similar as for groups:in this case the sums are isomorphic)

Then there’s also the fact that the book hints to the fact that external and internal direct products are in fact in a sense isomorphic

[subtle walrus]

probably skip this exercise until you know of the relationship between external and internal direct sum (they are in fact just different points of view of the same thing) and the notion of isomorphism (the questions are to be understood up to isomorphism)

your assessment is correct though

[zinc copper]

Alright good to know, thanks

[zealous junco]

Not that it matters, but I feel like the hermitian in second picture for the h in the tilde{A}_j should be a transpose? Unless I missed something

nvm nvm ignore

[tardy vault]

I have a math question

Let X be a full rank matrix so that at least one pair of columns are highly correlated (say X1 and X2 are correlated)

Let A be the Grahm Matrix X'X. Can we relate the smallest eigenvalue of A to the correlation between X1 and X2?

[north narwhal]

is linear algebra algebra 2?

[ocean sequoia]

hm @tardy vault are you familiar with the SVD? The short answer to this should be yes because X^T X will always be positive semi definite so we can find the eigendecomposition of the matrix Q /lamda Q and the eigenvalues of the lamda(the trace) will be the total amount of variance in the matrix

[nocturne jewel]

No

[shell sierra]

Nope, linear algebra comes after algebra 2.

[nocturne jewel]

LinAl is a typically first year course in uni

[shell sierra]

It's about the same level as calculus as far as when it's taken.

[nocturne jewel]

Linear algebra is the study of vectors, matrices, vector spaces (including inner product spaces, the geometry of vectors, etc.), and linear transformations

as well as the pinned message will give further info

[wintry steppe]

yes, linear algebra can be algebra 2, i.e., a course that comes after your basic course in group, ring, and field theory

module theory is just linear algebra but it smoked some weed beforehand

[mild crypt]

[dusky epoch]

very far from it

[teal grotto]

given a vector space V over some field F, is it possible to create a non-zero linear functional in V* without choosing a basis for V?

i have tried for a while, but no ideas really come to mind.

[dusky epoch]

if F = R i think you may be able to do it with hahn-banach

(or C)