linear-algebra

[fringe fjord]

Doesn't S contain vectors for all choices of a and b here?

[wintry steppe]

For example span({a, b, a+b}) = a(1, 0, 1,) + b(0, 1, 1).

a basis {(1, 0, 1), (0, 1, 1)}, dim = 2

[nocturne jewel]

Oh yeah duh

[fringe fjord]

In which case {(1,1,1,-1),(0,1,0,-1),(4,2,1,-2)} would be a basis.

[wintry steppe]

@fringe fjord sorry im confused... what are you trying to say?

[fringe fjord]

I'm suggesting a basis for your span.

[dusky epoch]

hold on what

@wintry steppe can you show the problem exactly as stated please

as a screenshot maybe

[wintry steppe]

Sorry not in English. :( But I have to find a basis if $V = span({a, b, \frac{a}{b^2}, -b}), a, b \in \bR, b \neq 0$.

[stoic python]

mate

[wintry steppe]

Basically I have a set of all 2x2 matrices ({a, b}, {c, d})

and a = -d, bc = -(a/d)

[dusky epoch]

can you send the screenshot no matter what language it's in? please?

i have a strong suspicion that you are messing up the notation severely

and span() does not mean what you think it means

[vivid crypt]

hi

[dusky epoch]

mate

do you have

a PICTURE

i want a PICTURE not a translation not a transcription just a PICTURE or a SCREENSHOT

[wintry steppe]

Sorry, Ann, I do not. And that actual problem does not exist (I just need it to solve something else).

It just a part of my problem.

[dusky epoch]

can you show the whole problem then

im extremely suspicious of all this

you're tasking us with finding a basis for something that either isn't a vector space at all or has a very obvious one-element basis

[wintry steppe]

Lets try with this.

$X$ is a set of all 2-by-2 matrices $A$ such that $A^2 = 0$. Find a basis for span(X).

[stoic python]

mate

[wintry steppe]

We defined a linear span as a set of all linear combinations of elements of that set.

[dusky epoch]

okay so you start with the set of all 2x2 matrices whose square is zero and you take the span of that

[tall gull]

hey guys

im only in calc 1 in school but i got a book on linear algebra and its rlly interesting

ok nvm this question was dumb

sorry

[wintry steppe]

I find out if I take any reals a, b(not equal 0) then for matrix

({a, b}, {(-a^2)/b, -a})

A^2 = 0.

[dusky epoch]

ok sure that describes a subset of X

i don't think that'll help you much in finding span(X)

[wintry steppe]

And I am stuck because I have never had to find a basis if I have squares or fractions.

That is the problem...

[dusky epoch]

you're confusing this desc of a subset of X with the span of X

i think span(X) might just be all of R^(2×2) but im not 100% sure about that

[wintry steppe]

We need to find a basis for a vector space, right? I am confused.

[dusky epoch]

you're confusing yourself on what said vector space even is

[wintry steppe]

Take all such matrices and make all linear combinations and find a basis for that vector space.

[dusky epoch]

and make all linear combinations

thats precisely what your desc doesnt capture

(but i would STRONGLY ADVISE AGAINST trying to make it do that)

i can say for sure [0 1; 0 0] and [0 0; 1 0] are in our space so its dimension is at least 2

[fringe fjord]

If Mate does indeed have all the nilpotent matrices there, then the span does not have dimension 4 because they all have trace 0.

[dusky epoch]

oh true

so span(X) ⊆ {A ∈ R^(2×2): tr(A) = 0}

hence its dimension is at most 3

[fringe fjord]

I think trace and determinant both 0 is a necessary and sufficient condition.

[wintry steppe]

Yes.

That is what I am using here.

[dusky epoch]

necessary and sufficient condition for what

[fringe fjord]

For being nilpotent, not being in the span.

[dusky epoch]

yeah my point exactly

ok, hold on

$\bmqty{1 & -1 \ 1 & -1}$

[stoic python]

Ann

[fringe fjord]

OTOH I found three lin indep members an hour ago, so we are indeed looking at just the trace-0 matrices.

As matrices we could just take ((0,1),(0,0)) and ((0,0),(1,0)) and ((1,-1),(1,-1)).

These are all easily nilpotent, independent, and since we agree every matrix in the span has trace 0, there can't be more than 3 matrices in a basis.

[wintry steppe]

@fringe fjord and @dusky epoch , thank you very much!!

[barren sandal]

What would you guys say are the prerequisites to self learning linear algebra?

[nocturne jewel]

Nothing really imo

[molten pilot]

What really is the cross product?

I mean, it seems awfully limiting to create an operation that would only work in R^3

[nocturne jewel]

Exterior Product

[molten pilot]

What?

[nocturne jewel]

but it's just a product that gives a mutually orthogonal vector to 2 other vectors

[molten pilot]

Hmmm

Ig a better way to frame the question is "why does the cross product exist?"

[wintry steppe]

one way to look at it is as the vector v x w with <v x w, -> = det(v, w, -). this lets you extend it to any R^n, but then you have to put in n - 1 vectors instead of two

[molten pilot]

I didn't quite get that

What does <v × w, -> mean?

[wintry steppe]

for any vector u, <v x w, u> = det(v, w, u)

[molten pilot]

Ah, you mean the dot product

[wintry steppe]

yes

[dapper gorge]

Does the generalized binomial theorem hold for 1/n, where n is a natural number in the case (I+N)^(1/n) where N is nilpotent? Wikipedia assumes the case for n=2. How would you go about proving something like this? N is nilpotent so the formula is just a finite sum, but rather complicated

Just curious, as I don't even feel too comfortable with power series with just ordinary real numbers

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

4.4. Jordan decomposition

[fringe fjord]

Another answer could be that the cross product is a prototypical example of an antisymmetric bilinear map. Now for each vector space V there's a systematic way to construct a universal antisymmetric bilinear map, meaning a vector space W and a antisymmetric bilinear map f: V×V -> W such that every other antisymmetric bilinear map V×V -> U for any U is the composition of f with a unique linear transformation W -> U.
In general, if V has dimension n, W ends up having dimension n(n-1)/2, so W is cannot be the same vector space as W -- except that n(n-1)/2=n iff n=3. So exacty for 3 dimensions we can choose W to be V itself, and the cross product then turns out to be a possible definition for a universal f: V×V -> V.

(There are many possible f: R³×R³ -> R³ with this universal property, though, so it's not the entire story. However, the cross product (together with its negative) additionally distinguishes itself by being the only one to satisfy T(v×w) = T(v)×T(w) whenever T is an orientation-preserving rotation, which is nice enough to prefer it. The fact that this is even possible is another low-dimensional coincidence, though).

[thorn nexus]

hey, can someone help me?

with a system of equations

10x+7y=263
4x+8y=230

i have to solve for x and y

bro wtf

is no one gonna snwerr

answer

[nocturne jewel]

set up the augmented matrix then invert.

[thorn nexus]

what

im in 8th grade

[nocturne jewel]

since you're asking here and not #prealg-and-algebra , I'm assuming you're learning Linear Algebra

[thorn nexus]

8th grade regents class

so should we chat there?

[nocturne jewel]

You can ask your question there, yes

[still lodge]

what material should be covered in college level lin alg class

for theoretical purposes let’s make it an honors class

im starting to feel like my lin alg class was lacking and would like to fix that

[gleaming knot]

I'm curious, what's lacking?

[deft apex]

what did you cover?

[nocturne jewel]

well what did you cover?

[still lodge]

i dont have my notebook with me to say exactly (there was no syllabus ) the matrix stuff youd expect, eigenstuffs, JCF

some very light inner product stuff

cayley hamilton (w/o proof) and some work with polynomials

[nocturne jewel]

I mean, sounds typical

[still lodge]

and in our last lecture he introduced RCF

that last bit is what feels lacking

[nocturne jewel]

but you can also just add the proof for Cayley-Hamilton in yourself, it's not that hard imo

[still lodge]

cuz i remember it looking cool/important and opening up to other stuff

perhaps im overthinking im just curious what other people might have done in their classes

[nocturne jewel]

R^n stuff, abstract vector spaces, inner products, least-squares, transformations/matrices, eigentheory

[still lodge]

wdym exactly by abstract vector spaces

the entire class was done in R2 R3 and polynomial space for me

[nocturne jewel]

polynomial space, matrix space, function space

vectors that aren't coordinate vectors

[still lodge]

ah

ok well he gave us a question involving matrix space on the final which was very cool and good

but thats it lol

i’ll probably just try and fill in some gaps then

[hushed hedge]

Is it possible to check whether two lines are intersecting with each other only using the affine form?

If i have this:
x+y+3z+6=0
2x+y-2z-10=0, can i check using the affine form or do i have to make a equationsystem?

[queen oxide]

anyone have any ideas for this

i tried the usual x+yi stuff but it becomes a huge mess and idk if its the right way of doing it

[dusky epoch]

consider: |z* + 2i| = |z - 2i|

[queen oxide]

we still have sum of surds that doesnt simplify nicely afaik? like idk how to get rid of the sqrt

or is there a way of doing this without converting to cartesian lol

[dusky epoch]

you'll have the equation of an ellipse.

[queen oxide]

from sqrt(x^2 + (2-y)^2) + sqrt(x^2 + y^2)?

[dusky epoch]

no you're overthinking it

you know the definition of an ellipse, right?

it's the set of all points for which the sum of the distances to two foci is constant

[wintry steppe]

@dusky epoch can I ask you a question

[dusky epoch]

why me specifically

[wintry steppe]

Cause it's related to LA and you seem knowledgeable

[dusky epoch]

you can certainly try but don't expect me to answer immediately

[wintry steppe]

How important are polynomials in LA? Thus far we have used them alot in class but from what I've seen in LA books they are not covered excessively

If you have a good resource on polynomials that'd be great

[dusky epoch]

i mean

i guess they serve as good examples of vector spaces to examine that arent just R^n

[wintry steppe]

I see

Are polynomials in LA used alot in cs?

[zinc timber]

(sometimes useful to find the eigen values)

[vapid lake]

Let V be the free F2 vector space over {a, b} and let U = {0, a + b} ⊆ V .
What does this mean?

V = {a, b, U} ?

[stable kindle]

is this just {0, a, b, a+b}?

[subtle walrus]

ye, all F2-linear combinations of a and b

[wintry steppe]

One quick question: i have a 2x2 matrix C, a 2x3 matrix A and a 3x2 matrix B such that AB = C, now i want to show that C is invertible, i was wondering if it suffices to show that both A and B have rank 2

I already forgot linear algebra lol

[zinc timber]

AB=C does not necessarily mean C is invertible, even if A and B have rank 2

$A = \m{1 & 0 & 0 \ 0 & 1 & 0}, B = \m{1 & 0 \ 0 & 0 \ 0 & 1}, C = AB = \m{1 & 0 \ 0 & 0 }$

[wintry steppe]

Ah damn you are right

Ye i think it was rank(AB) <= min(rank(A), rank(B)), not "="

Any advice what i could try here?

[zinc timber]

you need to be more specific about your problem

it's too vague

[wintry steppe]

Its a little bit much to explain, but i think i will just brutally calculate it. Thanks!

[haughty berry]

You could write it that way, or just rank(AB)<=rank(A),rank(B), but yes. Not equal

[wintry steppe]

why is 0 a vector and 1 a scalar ?; can't 1 be a vector too ?

[broken notch]

What's the best book for numerical linear algebra?

[zinc timber]

0 is taken as an element of the vector space V, whereas 1 is taken as an element of the field

[wintry steppe]

why is 0 not a scalar too then ?

[zinc timber]

if you take it as an element of F then yes

if it confuses you, take $0_V \in V $ and $0_F \in F$ to distinguish them

[wintry steppe]

So multiplying by the zero vector is the same as multiplying by the scalar 0 ?

[tranquil steeple]

Best in what sense? NLA by Trefethen and Bau is pretty popular, and ANLA by Demmel and Iterative methods by Saad

[broken notch]

best as in easy to understand well explained etc

[broken notch]

this one?

[tranquil steeple]

Yes, it is quite widely used as far as I know

here is the saad book, the other ones are on libgen https://www-users.cse.umn.edu/~saad/IterMethBook_2ndEd.pdf

[wintry steppe]

Note that there is no vector multiplication required for a vector space. You only have a vector addition and a scalar multiplication

[broken notch]

do you know a numerical linear alg book that applies python?

[tranquil steeple]

No (I only use Julia so have never looked)

[wintry steppe]

Yes

You can write 0 for the 0 scalar and the 0 vector, if v is a vector and you write v + 0, its clear that its the 0 vector here. If you write 0*v then its clear that its the 0 scalar here

now im confused

But v + 0 makes no sense if 0 is meant to be a scalar

Yes, vector addition and scalar multiplication

So you can add vectors together and you can scale them, but nowhere it says that you can multiply vectors

You know what i mean?

yeah

[wintry steppe]

why are there two Rs ?

C(R) makes sense

but F(R,R) doesnt

[teal grotto]

F(R,R) is the space of all functions from R to R

im not sure why they were inconsistent

[wintry steppe]

yeah if so why not write C(R,R) ?

[teal grotto]

typically, you will see C(X,Y) denote the space of all continuous functions from X to Y, and if X = Y, its shortened to C(X)

[teal grotto]

not sure, but either way it doesnt matter

[wintry steppe]

i see

thanks

[vapid lake]

@subtle walrus what would a + (a + b) equal?

[subtle walrus]

a + a = 1*a + 1*a = (1 + 1)*a = 0*a = 0

so you get b

[hearty rapids]

can someone eli5 what a linear transformation is to me

all i can understand is you're changing the dimensions of a vector

[wintry steppe]

do you know what a vector space over a field is?

or do you want to keep it to R^n

[hearty rapids]

ooh no idk what a vector space is

should i learn what a vector space is first?

the video series i was using didn't teach the vector space before they did the linear transformation

https://youtu.be/Zd7wCsUM4pg

for context this is the video

ik what a span of vectors is

a span is just the set of all linear combinations of v1,..., vp

[silver cradle]

im not familiar with the cauchy schwartz inequality, could someone explain how i would go about using it here?

[nocturne jewel]

$\abs{u\cdot v}\leq \norm{u}\norm{v}$

[stoic python]

Mosh

[nocturne jewel]

That's cauchy schwarz

[silver cradle]

ah thanks, i see it now @nocturne jewel

[nocturne jewel]

yeah, abs value CS-Ineq QED

[silver cradle]

would this be trivial if we assume the first property?

i was thinking maybe they did not want us to use it in that case then

[nocturne jewel]

trivial based on what you just proved tbh

[silver cradle]

gotcha

[static jasper]

can someone help me putting this matrix in superior triangular?

[drifting scarab]

basically a transformation T is linear if T(a * v1 + b * v2) = a * T(v1) + b * T(v2)

[vapid lake]

On transformation,
T : V → W be the unique linear transformation such that T(a) = c + d, T(b) = c + d. where V ={a , b} and W = {c , d}. Im trying to find kernel and image of this

[halcyon spindle]

Since V = {a, b} one of the elements has to be zero of V, so it must be the other element is also 0 for V to be a vector space. Same thing with W. So the kernel is the just the only element in V which is 0.

The image would also be the only element in W which is 0.

[halcyon spindle]

Try row reducing it maybe.

[wintry steppe]

norm

it explains this below

[spring oriole]

What's a good linear algebra textbook for someone that needs to relearn linear algebra

[dapper gorge]

In most cases, we say that a subset S of some inner product space is orthogonal if given any two elements x,y in S, x and y are orthogonal. What if there are no two distinct elements in a set? Do we consider the set {x}, where x is non zero, to be orthogonal? Although by definition ⟨x,x⟩>0, but I'm wondering if this is a special case

In a text I'm using I think this assumption is made

[wintry steppe]

post text

[dapper gorge]

ok

[wintry steppe]

S should be orthogonal if that condition holds for distinct elements

otherwise the only orthogonal set is {0}

[dapper gorge]

This is Stephen Friedberg (and some other authors) text

[wintry steppe]

note "distinct"

[dapper gorge]

yeah

I have pointed that

So what about the case {x}? i.e., a set where there are no distinct elements?

Again, if x is non zero (x,x)>0, but idk

[wintry steppe]

singleton is vacuously orthogonal

there's no other element in {x} to take the inner product with

[dapper gorge]

This is a question on definitions

I don't think the definition above adresses the singleton set

?

If it does I don't see it at least

[wintry steppe]

.

[dapper gorge]

According to what definition of orthogonality of sets?

[wintry steppe]

the one in the book

[dapper gorge]

this?

btw

I'm not questioning singleton set is not orthogonal, just a matter of definitions

[wintry steppe]

the book's definition doesn't assume there are actually two distinct vectors in S to take the inner product of. it says if there are, then their inner product is zero

[dapper gorge]

which I'd like to understand rigorously

gocha

thank you

[wintry steppe]

[vapid lake]

How many elements are in the free F2 vector space over S when S is a set of n elements? F2 should include the basis, im not sure how it affects the dimension

[zinc timber]

free F2 vs of dim n has 2ⁿ elements

[vapid lake]

why @zinc timber

[zinc timber]

basic fact about vector spaces

$F_2^n ={ (x_i, x_2, \cdots, x_n) | x_i \in F_2 }$ you have 2 choices for each $x_i$ so $2^n$

[thorn cypress]

is the dot product iv.iv = ||v| |^2 where i is the imaginary

and v is complex vector

[zinc timber]

yes

$\ip{iv, iv} = i\ip{v, iv} = i \bar{i} \ip{v, v} = \ip{v, v} = \norm{v} ^ 2$

[quartz compass]

I would have said no by how I interpreted the question, since I interpret the dot product to be distinct from the inner product

[zinc timber]

dot products are inner products, so component wise multiplication (that u r calling dot) is not an ip

so can't be a valid dot (ip)

[quartz compass]

I'm saying without asking the person who asked the question, it's ambiguous

[wintry steppe]

What does direction ratio/number mean geometrically in 3D??

[errant palm]

i cant figure out how they got this answer, can someone pls help me? ty

[wintry steppe]

They performed row operations and reduced the matrix to a lower triangular matrix

[errant palm]

oh hbooy

[gentle relic]

Can u pls help me understand augmented matrix

[hushed hedge]

determine all the vectors that are orthogonal to u = (2,0,4) and v = (-1,2,3), aren't you suppose to use the cross product for u x v? I get (-8,-10,4) but that's wrong

[dusky epoch]

you've determined just one vector orthogonal to both u and v

you're not trying to claim it's the only one in existence, are you?

@hushed hedge

[hushed hedge]

ah sorry

are there more? is v x u another vector orthogonal?

actually v x u is wrong, if u x v is one vector, how do i find the second one?

[dusky epoch]

there are not just two such vectors

theres infinitely many of them

consider: scaling u × v by 2 would still yield a vector orthogonal to both u and v

[hushed hedge]

right

[dusky epoch]

scaling it by 3 would yield yet another vector

and so on

[hushed hedge]

but how does my book get t(4,5,-2)?

i should make a equationsystem right? To figure it how they behave (?)

[dusky epoch]

your book knows that the vectors orthogonal to u and v form a line going through the origin

and u×v is just one vector on said line

[hushed hedge]

aha ok

i didn't look at it that way, thanks

[halcyon spindle]

What don’t you understand about it, it really just the coefficients matrix A joined together with rhs of Ax = b.

[gentle relic]

wow ok like i dont know anything about augmented matrix but like i know matrix are the two related ?

like jee main level if u know jee

[fickle citrus]

It's just vocabulary

[gentle relic]

like ... ? i did not get you

[dusky epoch]

"augmented matrix" is just a name

it's what we call the matrix A with the column b attached to it on the right

[fickle citrus]

It's specifically referring to attaching the coefficients to the original matrix

^

[dusky epoch]

(where A and b come from your linear system, Ax=b)

[gentle relic]

oh ok

so like its transformation of matrix

[fickle citrus]

no, it's a more compact way of writing everything

writing everything out explicitly

Since the x does not matter

[gentle relic]

ohh like just the way a person denotes it ok

o ok got it

thanks

[wintry steppe]

i think it's genuinely confusing when students are asked to try interpreting matrices, and then we say "oh yeah and this augmented matrix is just like, better for rote row reductions, lmao"

[severe nymph]

x²+(y-3√2x)²=1 solve this equation

[zinc timber]

no

[haughty berry]

^

[hearty rapids]

so all a linear transformation is

is checking if T( vector u + vector v) = T(vector u) + T(vector v)

which is the first condition

and then

T( c * vector u) = c (T * vector u)

[gray dust]

yes

[hearty rapids]

oh then that's simple

i got tons of practice to do lmao

[crystal plover]

Yo, is this legit?

wolfram gave me x=8 as the answer

[fringe fjord]

Looks good to me.

[crystal plover]

thx

[polar marsh]

i have a problem from my linear algebra exam that i couldn't solve

let B is an element of M_n(R) and we now that BB=B and that B^T * B=BB^T, we now have to show that this implies B=B^T

[golden ermine]

Try showing that B^T = (B^T)^2

[polar marsh]

i have shown this already

you can transpose both side

s

[nocturne jewel]

then dont ask anything

Yeah, we don't care for stupid questions. You're not obligated to ask anything.

[crystal plover]

ok another question, wolfram tells x=67 but i don't really know how to get it :/

[winter harbor]

$\cdot$ (Part 1 : The Idea)

If you have a any matrix $B \in \mathcal{M}_{n}(\mathbb{C})$, then it is always the case that:
$$
\langle Bx, y \rangle = \langle x, B^{\ast} y\rangle
$$
Where $B^{\ast}$ is the adjoint of B, i.e its conjugate transpose, $x,y \in \mathbb{C}^{n}$ and $\langle \cdot, \cdot \rangle$ is the hermitian inner product on $\mathbb{C}^{n}$.

Since we are over $\mathbb{R}^{n}$, then the adjoint is just the usual transpose and the hermitian inner product is just the euclidean inner product on $\mathbb{R}^{n}$. So in any case, we have that:
$$
\langle Bx, y \rangle = \langle x, B^{t} y \rangle
$$
Still holds. Moreover, we have that, by definition:
$$
\langle x, y \rangle = x^{t} y
$$
For any two vectors in $\mathbb{R}^{n}$.
\
\
With this is mind, we can now prove your result by showing that for any two given $x,y \in \mathbb{R}^{n}$, we have that:
$$
\langle Bx, y \rangle = \langle B^{t} x, y \rangle
$$

Is this good enough so far?

I will write down the proof now

I just wrote down a few definitions to see if we are on the same page

[stoic python]

MISTERSYSTEM

[wise oriole]

can someone explain this

[polar marsh]

yes

exactly what is in my textbook

[winter harbor]

Hmmm

[winter harbor]

I am not actually being able to prove this only using trivial stuff.

But there's a really simple proof that just got to my mind

Since B^2 = B, then unless B = Id or B = 0, then the minimial polynomial of B is p(x) = x(x-1). In the case B = Id, then the minimal polynomial is just p(x) = x - 1 and in the case B = 0 we have that p(x) = x.

In any case, the only eigenvalues of B are either 1 or 0, which are all real.

Moreover, since BB^t = B^t B, then B is a normal matrix. Since B is a normal matrix with real eigenvalues, then B is hermitian, and as a consequence of B being a real matrix it is also symmetric; i.e B = B^t (as a consequence of the spectral theorem)

If you still haven't seen the spectral theorem for normal matrices in your class

You probably can't approach the problem like this tho lmao

[zinc timber]

I think I have an different approach to this problem, we already know that $B$ is diagonalizable with eigen values $0, 1$. Now If we could show that $E_0$ and $E_1$ are mutually perpendicular, then we'll be able to find an orthonormal basis that makes $B$ into a diagonal one, which would imply $B^t = B$. \

Now we see that $B^tB = BB^t$ i.e. they commute implying they preserve each others eigen spaces, which can be seen by following, let $v\in E_0$ of $B$ then $B^TBv = 0 = B B^Tv \imply B^Tv \in N(B)$.

[stoic python]

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

[dusky epoch]

k0 - 2k1 = 0
k2 +3(k3) = 0

however there seems to be no solution.

how are you getting that there is no solution

you claim the system $\begin{cases} k_0 - 2k_1 = 0 \ k_2 + 3k_3 = 0 \end{cases}$ has no solutions at all?

[stoic python]

Ann

[dusky epoch]

im adhering to your notation here btw, even though it is a little wonky (though not in a way that seriously affects anything)

does all this resolve your question

there are much less wonky ways of saying that but yes

[polar marsh]

okay thank you

i can't find really something on normal matrices in my book

we have just seen that if a matrix is hermitian then its eigenvalues are real

[barren sentinel]

guys i need some help

i have got these matrices which apply transformation to the world coordinates to convert it into image coordinates

(for a camera)

how would i do the opposite?

[devout hedge]

Are there any websites with linear algebra lessons?

[zinc timber]

yes youtube

[devout hedge]

Any specific channel?

[lavish jewel]

gilbert strang's MIT OCW

[devout hedge]

👍

[barren sentinel]

how do u invert R4 to R3?

do i need to transpose or something?

or its just not possible?

[dusky epoch]

what do you mean by "invert R4 to R3?"?

[odd kite]

yes do you mean a linear function from R4 to R3 for example

[slim cedar]

Probably a very simple question, but given a d times d matrix A, and two vectors x,y in R^d, can i say anything about <Ax, Ay> if Det(A)=1?

[barren sentinel]

trying to make a backwards imaging mordel but i understand that i cant invert 4x3 matrix 🤔

[slim cedar]

surely the inner product is the same as <x,y> right?

[barren sentinel]

idk if this is correct

but i hope it is

this is the diagram btw

[versed rock]

So, if i have these 2 vector [2, -4] and [-1, 2]. Can i say that [x, y] = [2, -4] + t[-1, 2] , right?

[nocturne jewel]

sure, that doesn't mean much though

however they're parallel vectors

[odd kite]

@barren sentinel can you link the whole document or give some more context

[barren sentinel]

@odd kite

[versed rock]

Exercise 18 and 19 should be trivial right?

I Know it's true, but i don't think the way i do it is right

[odd kite]

18 is

[versed rock]

,rotate

[nocturne jewel]

18 is trivial as in obviously true, 19 less so

[versed rock]

How can i prove 20 and 21?

21 for me looks right, like, intuitively i know this

[nocturne jewel]

show that an element from span(S) is in span(T)

like any other subset proof

[versed rock]

Like this?

[nocturne jewel]

yeah, just make c_i = 0 for all i b/w k+1 and m

$v=\sum_{i=1}^k c_iu_i=\sum_{i=1}^k c_iu_i+\sum_{i=k+1}^m 0u_i$

[stoic python]

Mosh

[hushed hedge]

My book writes the basis for the zero vector space like this: (-1,5,3), is it equivalent to (1/3,-5/3,1) ?

[nocturne jewel]

yes, and zero vector space is something different

you mean nullspace or kernel

[hushed hedge]

Yeah, i mean nullspace. Thanks!

[forest wave]

what steps do I need to do this simplification?

[wintry steppe]

factor the quadratic

[forest wave]

hmm

go on

[wintry steppe]

that's it

all you have to do is factor the quadratic

[forest wave]

well how do I do that?

[wintry steppe]

that's a question for #prealg-and-algebra

[forest wave]

actually I figured it out

[vast bobcat]

https://media.discordapp.net/attachments/853515478470361089/930999801733058560/IMG_3684.png

I don't remember how to answer these

can anyone help with with steps for 8. a at least

[fringe fjord]

The first step would be to reread the section in your textbook (or lecture notes or whatever) that describes diagonalization of matrices.

(If you're looking for a mindless step-by-step procedure, you might end up doing a lot more work than if you exploit the specific features of each example anyway).

[vast bobcat]

we actually haven't learnt diagonalization

[keen sierra]

You know that if $A$ is diagonalizable, then the diagonal elements of $P^{-1}AP$ will be the eigenvalues of $A$. And you know the columns of $P$ will be eigenvectors associated with those eigenvalues. So if you don't see any obvious shortcuts, you may as well proceed by finding eigenvalues.

[stoic python]

OurBelovedBungo

[dull rose]

I don't understand how these vectors becomes a line, and I am not sure how to do the vector equation.

This is the answer.

[golden ermine]

Well: the vector < 2, 4 > is just 2 * < 1, 2 > and similarly, < -3, -6 > = -3 * < 1, 2 >

So these vectors also belong in the span of < 1, 2 >

[dull rose]

but then I could make the same argument for < 2,4 > since it equals 2 * < 1,2 > and (-2/3) * < -3, -6 > ? And therefore wouldn't this make the vectors belong in the span of < 2,4 >? Or is < 2,4 > acceptable too?

[golden ermine]

That’s fine too

We just prefer < 1, 2 > over < 2, 4 >

[dull rose]

Oh alright, thanks!

[compact tartan]

if i have to put one more matrix in reduced row echelon form I am going to combust

[exotic horizon]

is there a faster method to calculate the inverse of this method or do I need to calculate the minor matrixes every time ? I know how to calculate the determinant here already

[wintry steppe]

pairing it with the identity, row reducing it itself to the identity, and taking the new matrix seems especially feasible as this is a triangular matrix

[exotic horizon]

I thought that this could only be used to calculate the determinant? Isn't the new matrix just the identity of this one ?

[zinc timber]

wym?

you calculate the inverse using row reduction

[sharp shell]

the difference between A*A^T and A^T * A is that the first one has the dot product of the the coloms on the diagonal and the second the dot products of the rows correct?

[fringe fjord]

It's the other way around. (And it's not just the diagonal).

[exotic horizon]

sorry i'm really confused, won't that just give me the reduced matrix?

[zinc timber]

No you attach the identity matrix with it

$\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \ 3 & 4 & 0 & 1 \end{array}\right]$ for example

[exotic horizon]

oh! you mean the matrix i obtain after row reducing here will be the inverse ?

[lavish jewel]

the one on the right, where the identity matrix originally was

[pliant kayak]

can you do the LU decomposition with a non quadratic matrix?

[tranquil steeple]

The matrix does not have to be square for LU.

[pliant kayak]

thanks

[exotic horizon]

okay thanks!

[wintry steppe]

I were you i would use Laplace

[pliant kayak]

if you multiply a row with -1/3 for example by doing the LU decomposition, how do you change the matrix L? I.e. here the second line has everywhere -1, if I wanna take that away, what do I have to do in Mat L?

[lavish jewel]

you can put a -1 in the corresponding diagonal element

[pliant kayak]

but what if I multiply with -1/3, do I have to put a -3?

[lavish jewel]

right

idk if this makes it clearer, so that you don't mix the two up. you're really doing L (L^-1 A)

so if you multiplied by -1/3 and you want the matrix to still be A, you need to do the inverse transformation too

which is -3

[pliant kayak]

I'm asking cause I did it, but it didn't work 💔

so I guess I'm making a mistake

[lavish jewel]

do notice that doing this essentially does nothing

you have to keep the two matrices separate

[wintry steppe]

y is supossed to be 7/38 what did I do wrong?

[clever karma]

you added 27/38 + (-10/19). not the same denominators

[wintry steppe]

ooooh

Now it comes out right

[hearty rapids]

is #5 just about getting it into reduced row echelon form?

i tried swapping row 1 with row 4 to get row 4 on top so there is a 1 on the top left corner

actually no it seems like swapping the third row with the first row is the better approach

sigh

ok i figured it out

x4 has to be a free

[lucid rose]

What does one dimensional flat mean?

Is it one free parameter?

[lavish jewel]

it's a line that doesn't necessarily pass through the origin

[lucid rose]

For example say we have a linear system in the form of a matrix

When is the set of solutions a one dimensional flat

[lavish jewel]

when the n-1-dimensional flats defined by the rows of the matrix intersect in a line

that would indeed be the case when you have one free parameter

since then the solution would have a particular term, call it x0, and then a null space component, let's say xn

so that the solutions are of the form x0 + c xn for some scalar c

[lucid rose]

Ok great thanks

I’m guessing for two dimensional flat it would be 2 free parameters?

[lavish jewel]

sounds about right

[lucid rose]

What about when the set of solutions is a single vector

What does that mean

[lavish jewel]

there's a unique solution

just the point described by that vector

[lucid rose]

So say if it’s like R^4

If the solution is a single vector

Then x1-x4 all have one possible solution

[exotic horizon]

do any of you know a resource where I could learn LU decomposition? It's still very confusing to me and I struggle with what I'm finding online

[sleek spruce]

replace x1 with x and x2 with y to help conceptualize it

[sleek spruce]

even if you replace it with z, you get a plane

but in a different axis

just let the grader know that x1 = x

and x3 = z

the purpose of the homework is to practice sketching in 3d, so both answers would work

if you really wanted to, you could do it for xy, xz, yz

[soft moss]

can someone help with this one please?

[zinc timber]

#prealg-and-algebra

[zinc timber]

#calculus

[sharp shell]

https://cdn.discordapp.com/attachments/761602106129645589/931541691217899561/unknown.png

whats the cross thingy for?

alternative to + for psuedo inverse or somthing else

[nocturne jewel]

dagger

[zinc timber]

phys guys use it for adjoint

[sharp shell]

so what do I take it for if I see it in lin algebra context

[zinc timber]

idk check yr textbook what they have mentioned

[odd kite]

often it's conjugate transpose

[lavish jewel]

i would really suggest to check in the book, i've just as often seen * and H for conjugate transpose

[odd kite]

but usually when that's the case, no daggers appear I think

[lavish jewel]

in those cases the dagger can mean pseudo inverse or something like that

if abdos showed the RHS, it would help

[zinc timber]

physicist messed things up again

[dusky wadi]

what's an orthogonal projection

linalg done wrong asked for it without explanation

[dusky epoch]

context?

[dusky wadi]

an orthogonal projection matrix from R² to the line x_1 = -2x_2

[zinc timber]

means you project onto a subspace along it's orthogonal complement

[dusky wadi]

context: I'm in chapter 1
they've barely introduced matrix multiplication

[zinc timber]

[lavish jewel]

are you sure you're supposed to be seeing how to solve this yet? 😛 have you done more abstract linear algebra before?

[dusky wadi]

I was just trying to parse what an exercise meant

[lavish jewel]

have you seen orthogonality?

[dusky wadi]

no

[lavish jewel]

just leave the problem for later

[dusky wadi]

this may be a problem with the texbook

[zinc timber]

[lavish jewel]

sounds like it

[zinc timber]

LA done wrong for a reason

[dusky wadi]

lol

I'll come back to it eventually

[zinc timber]

yeah read ahead like me then come back

[dusky epoch]

can i have a pdf and page reference

[dusky wadi]

here

pg 33 ex 5.4

[lavish jewel]

a quick ctrl f search says that is the first occurrence of "projection" in the book

[dusky wadi]

oh oof

[lavish jewel]

this book is like, hint:

[dusky wadi]

it's just like

once in a while

treil forgets where in the texbook he is

[lavish jewel]

page 1: the end

[pastel brook]

In the preface:

The title of the book sounds a bit mysterious. Why should anyone read this
book if it presents the subject in a wrong way?
That's a really good question (note that it's never actually answered)

[dusky wadi]

it's pretty good otherwise

[halcyon spindle]

I did that problem by following his example for that section for reflection and using demos to figure out how the hell am suppose to do it. His hint and stackexchange help.

He later asks you in the next section to find the matrix of the rotation around (2,2,3)^T after introducing invertible linear transformation.

[unborn heart]

i think you can still solve this problem by thinking projection as like an "informal everyday word" without formally knowing what projection is

[halcyon spindle]

He hint was confusing, “imagine your on the tip of the vector looking down towards the tail” or something.

[unborn heart]

very informally, if you were to project the blue point onto the red line, it would go to the green point

[dusky wadi]

ohhhhh

thanks

[wintry steppe]

Guys can i show that something is true by logic and not by using math? I explain: If I am consider Grassmann's theorem can i prove that it's true by logic and not by mathematical process? because to me is more easy

[gleaming knot]

Cool question. Yes, a proof by logic is a mathematical proof. To think otherwise would be a pretty weird misconception

After all, math is logic and logic is math

[wintry steppe]

Yes, you are right.. and sorry for my english but i come from italy and i'm try doing my best

[agile plover]

Are there any tricks of linear algebra that would allow me to compute a multivariate Gaussian likelihood without a full matrix inversion?

$e^{-\frac{1}{2}\sum_i(x_i - \mu)^tS^{-1}(x_i - \mu)}$

[stoic python]

PT

[agile plover]

x, mu are d x 1, S is d x d, so this becomes 1x1 at the end. there are tricks for sampling from a multivariate gaussian, where you only have to do either a cholesky or LU decomposition, but those aren't used for calculating likelihood.

[waxen turret]

how do we solve this kind of problem??

[zinc timber]

#geometry-and-trigonometry

[little frigate]

Let A be a 3x3 matrix and consider the invariant that AD = 0

how could one possibly find all the matrices D that would fit that condition

[zinc timber]

AD =0 means cols of D are in null A

[little frigate]

null A being... what people often refer as null space?

[zinc timber]

ye

[little frigate]

mmhm, the course started one week ago...

unless the concept itself is simpler than I think it is

I don't think we're being expected to know that- I only personally know that because I've already heard about it outside of that class context. It will be seen later

Here's what the matrix A looks like, I think that helps since it's filled with essentially 1s and zeros for the most part, so perhaps some technique could be preferred

I think that the two following propositions could help although

For clarifications purposes, B_1 ... B_n corresponds to the columns of B

so there definitely seems to be something revolving around columns hmmm

Oh

[little frigate]

doesn't seem that crazy, I'll take it for now, thanks

[ripe birch]

any idea how to approach this? A is just some arbitrary operator and a_n is an eigenvalue of that operator (and apparently it's multiplied into an identity matrix that isn't shown so that it can be subtracted from A).

My first thought was come up with a formula for the nth degree polynomial of (A - a_n)^n and see what kinda magic I could do but I don't really don't see anything. I do have a solution manual to this but I don't really follow a key step, and it may not make any sense to math people since it's a physics problem written in physics form

[dusky epoch]

sounds vaguely like cayley-hamilton

[viral magnet]

Write an arbitrary vector in terms of the eigenvectors and apply the product. One of the terms in the product will zero out every eigenvector

[zinc timber]

what if n doesn't have enough eigen values? what if min poly has an irred factor?

is your field ℂ?

[viral magnet]

I guess it's also nice to prove that every term in the product commutes. So you can just move the term to the front and get the zero vector right away

[ripe birch]

Looking into Cayley-Hamilton theorem. Seems like it could be related.

@viral magnet so the solution does say to have it act on what's essentially some arbitrary state vector. Not sure where a 0 will come from yet, but I'll keep looking.

@zinc timber I wouldn't think too much into it for that first question. If that 2nd question is asking if it can be complex (im not all that familiar with math terms yet) then yes it can

why do you say one of the terms will zero out?

[viral magnet]

Take just a 2x2 matrix with distinct eigenvalues

[zinc timber]

(if F=R² the you cannot write $A=\m{\cos 1 & \sin 1 \ -\sin 1 & \cos 1}$ in that form)

[zinc timber]

that's why I asked whether the field is ℂ of not

[viral magnet]

@ripe birch
Write an arbitrary vector as a linear combination of eigenvectors. So for 2x2 you have a sum of two vectors say lam1 * v1 + lam2 * v2 where v1 is an eigenvector corresponding to a_1 and similarly for v2.

Calculate (A-a_1 I) * (lam1 * v1) and similarly for v2

[ripe birch]

trying that real quick

oh that's wild

they do each go to zero dont they

[weak needle]

the determinant polynomial is the polynomial you get from multiplying (x-a) for all eigen values a (with appropriate multiplicity) so this is true by cayley hamilton

[weak needle]

Not every vector can necessarily be written as the sum of eigenvectorrs for a matrix

[ripe birch]

seems so weird to work. Only thing im caught up on is explaining why our arbitrary vector can definitely be written as a linear combination of eigenvectors. I'm assuming not every vector can be but no basis for that??

@weak needle beat me to it

[weak needle]

yeah so cayley hamilton is the way to go

[zinc timber]

can be written as an LC of generalized evs

[viral magnet]

You learn the Cayley Hamilton theorem in physics class?

[zinc timber]

(given field is ALC)

[ripe birch]

bro we don't learn ANYTHING about formal math in physics class. They gave us 1 month of LA, 2 months of ODEs, 3 months of PDEs, and random other crap

i like me some math dont get me wrong, i just am not up to speed on the fields that we use as much as I'd like to be

[viral magnet]

When can't you?

[ripe birch]

like @zinc timber you speakin another language to me rn

[viral magnet]

It's a physics class, they throw words like eigenbasis around like it's given

[zinc timber]

(ignore)

(that's why I'm enclosing them in paren)

[weak needle]

here is a concrete example: the 2x2 matrix with all 0's excet the top right entry

and 1 in that entry

[ripe birch]

my favorite part was when my professor said "a Hilbert space, just so far as what you need to use it within this class, is some place where you can do calculus". Kind of a paraphrase but you get the gist. We're basically bullied

The internet and this discord is how I learn my math

[zinc timber]

hilbert spaces are good tho

most well behaved

[viral magnet]

I'm also gonna assume they assume the matrix has full rank. If not, then it's a separate case

[zinc timber]

like an obidient pet

[zinc timber]

even if of full rank it's true

[viral magnet]

Right, a simpler proof I believe

[ripe birch]

am I gonna need to know what a ring is to figure out this cayley-hamilton theorem?

[zinc timber]

no

absolutely not

[ripe birch]

riemann i like your showing a lot, i just don't know if i can justify the arbitrary vector consisting of eigenvectors

[viral magnet]

You don't need Caley Hamilton for a physics class

[ripe birch]

are any of you familiar with bra-ket notation from quantum mechanics? I think there's a math equivalent of some sort but I'm not sure

in other words. Does this look anything like something you'd be familiar with?

where that alpha ket is basically a vector

[lavish jewel]

sure, kets are vectors and bras are linear functionals

[viral magnet]

Yea I am

[viral magnet]

Bras are dual vectors I think

[ripe birch]

bless. So I follow up until the start of the 2nd line. Looks like operator A just turned into a'' from nothing i dont see it

[zinc timber]

what sorcery is this

[viral magnet]

Or vectors in the dual space

[ripe birch]

i'm almost positive that's the case. dual space sounds super familiar from my textbook

[lavish jewel]

that's the same thing

[ripe birch]

@zinc timber This is wizardry

[viral magnet]

a'' is the eigenvector and the c_a'' are the eigenvalues

[zinc timber]

if the space is reflexive then they are same, ex for hilbert spaces, they are the the same

[lavish jewel]

yep it looks like they're decomposing the vector into a basis of vectors a''

[viral magnet]

Linear functional return kets when applied to a ket.

[zinc timber]

yes but why are they assuming there is such a basis

[lavish jewel]

linear functionals map to the base field, not to kets

[zinc timber]

[viral magnet]

Bras return a complex number when applied to the kets

[lavish jewel]

they take kets and return elements of the field

linear functionals are linear forms, not just linear maps

[zinc timber]

wth is happening here

[ripe birch]

so back to what you said @viral magnet . I see that |a''> is the eigenvector. is the a'' that appears in (a''-a') the eigenvalue from having A apply onto our ket?

[lavish jewel]

idk, riemann saying linear functionals map vectors to vectors

[zinc timber]

lol

(technically the truth)

[lavish jewel]

if you treat the base field as a vector space, sure

[viral magnet]

Right, a bit messy on my part. Vectors are functions in Hilbert spaces

[zinc timber]

not always

[ripe birch]

AcHkTuAlLy

[viral magnet]

Oh shit in uncountable Hilbert spaces

[zinc timber]

(not all vs are hilbert spaces)

[viral magnet]

Or countable I forget

[zinc timber]

(nothing to do with the coutability)

[viral magnet]

Yes? Could you clarify

[ripe birch]

so my question was that A just turns into a''. You mentioned that a'' was an eigenvector. When A turns into a'' it's still an operator and not a vector, so I'm assuming the a'' you mentioned being an eigenvector was the other a'', not the one I inquired about. Hope that clarifies it. Kinda weird to word

[viral magnet]

Oh man this abuse of notation

[ripe birch]

🙂

[viral magnet]

Can you rephrase it and use kets where you mean the vector and no kets when you mean eigenvalue

[ripe birch]

Sure. So we have a line which says (A - a') | a'' >

the following line has (a'' - a') | a'' >

with seemingly no other changes. so operator A just turned into the operator a'' (times identity matrix ofc) and it seems random

ill repost the pic to have it less buried

[viral magnet]

All uses of little a are either eigenvalues or eigenvectors

So A|a> = a |a>

[zinc timber]

[lavish jewel]

this notation is super unfriendly :x but as riemann says, the transformation A | a'' > yields the eigenvalue corresponding to the eigenvector | a'' >

[ripe birch]

So this theorem only works on kets which are eigenkets of A?

[lavish jewel]

and here they have simply removed the | > to denote an element of the field instead of a vector

this would honestly look so much simpler if you were to use matrix notation lol

[viral magnet]

Wtf now eigenket's a word

[ripe birch]

😂

[viral magnet]

Next thing you know they'll make up eigenSchrodinger

[ripe birch]

listen we're physicists for a reason. we live to brutalize your field

[oblique prairie]

i’m just lurking but i was about to say what the heck another eigen thing

[ripe birch]

eigenket is just an eigenvector in whatever dual space that is

[lavish jewel]

i'm not even a mathematician, this is just bad notation right now overloading of the same symbol

[viral magnet]

You're gonna have to change your name to eigenQuantum

[ripe birch]

I'm gonna eigenFreakTheFuckOut this semester

[zinc timber]

sometimes they look nice

[lavish jewel]

i can give you an alternative explanation using matrices and vectors

[zinc timber]

but this is not one of those moments

[lavish jewel]

if you'd like. otherwise, i'll leave you two be

[ripe birch]

yeah that works. Does this theorem only work on eigenvectors? or does it produce null kets on ANY vector?

[zinc timber]

like wtf a'', a and a' all together

[viral magnet]

Someone take eigenBruh to be the dual of eigenKet

[lavish jewel]

this works for diagonalizable operators A

[ripe birch]

and it does not matter if the vector this operator combination acts on is an eigenvector of A or not?

[lavish jewel]

it doesn't, because if the operator is diagonalizable, its eigenvectors can be used to span any vector it acts on

[viral magnet]

"All operators are diagonalizable" --physicists

[zinc timber]

the statement is also true if A is not diagonalizable tho

[ripe birch]

@viral magnet you're not wrong 🤷

[lavish jewel]

then by linearity, it boils down to using A on several eigenvectors

[lavish jewel]

yeah but i wanna avoid thinking

[zinc timber]

all functions are analytic

[ripe birch]

all operators are also linear and hermitian and have real eigenvalues

🙂

[lavish jewel]

also, all series converge

[ripe birch]

and if they don't

we approximate them to be 0

[lavish jewel]

anyway. do you want the matrix vector explanation?

[ripe birch]

oooo plz

[lavish jewel]

-1/12

ok

[ripe birch]

LMAO

[lavish jewel]

so let's say there is a diagonalizable matrix A, which has eigenvectors v_i with eigenvalues a_i

[ripe birch]

physics notation be like

[zinc timber]

(i actually prefer dot notation over d/dt)

[lavish jewel]

and we make a product pi_n (A - I a_i), where I is an identity matrix

the matrix (A - I a_i) has the eigenvector v_i corresponding to a_i as an element of its null space

[dusky epoch]

$\mathit{\imath}$

[stoic python]

Ann

[dusky epoch]

charge

[lavish jewel]

one can additionally decompose any vector u into a linear combination of the eigenvectors e_i, which in the proof were also assumed to be orthogonal, apparently

[ripe birch]

oo we're gettin into slightly spooky territory with the jargon, but ill try to follow

i thought we said earlier that perhaps not every vector can be written as a l.c. of eigenvectors

[zinc timber]

(why not just use Cayley hamilton )

[lavish jewel]

you can do a change of basis into the eigenbasis by doing a transformation V, a matrix whose columns are the eigenvectors v_i, and doing V V^T u

[ripe birch]

oooo is this the gram-schmitty process?

or different

[lavish jewel]

sure

[ripe birch]

let's go

[zinc timber]

no they are different

you may not necessarily have orthogonal eigen vectors

[lavish jewel]

the proof they showed kinda assumed they were

they're just orthogonally projecting on the eigenvectors lmao

[ripe birch]

i believe in quantum we always form an orthonormal basis with our eigenvectors

[lavish jewel]

now the trick is that V V^T u is a sum of eigenvectors

so when we consider A - I a_n, the operator can be replaced by a_j - a_n, where a_n is indexed by the big Pi in front, and j is the index of an eigenvector's corresponding eigenvalue

p_n (a_j - a_n) V V^T u is then 0,

since each eigenvector will be multiplied by some term (a_j - a_n) such that n = j

for completion, it seems the assumption made was that A is self adjoint

diagonalizable in an orthogonal basis

since the proof used orthogonal projections and assumed decomposition in an eigenbasis was possible without ever saying anything about it

[ripe birch]

i can nearly assure you that assumption is correct. we jump to conclusions more than a clingy ex-gf trying to win you back

[lavish jewel]

the proof you shared definitely made that assumption

the first step they write is "decompose the ket into the eigenbasis through orthogonal projections"

which is what i did with the V V^T, where V's columns are eigenvectors and therefore V is orthogonal, so that V V^T = I and doesn't affect the result

then in the eigenbasis, linear transformations are simply scaling factors

anyway, yes. orthogonally diagonalizable A

if not, you need something different for the proof

[zinc timber]

ye like m(x) be mini poly so some power of that poly will be divisible by m(x) which makes it nilpotent

[ripe birch]

so let me see if i understand this correctly. We take some arbitrary vector, and we project it onto the eigenbasis. Doing that allows the operator A to extract those eigenvalues a_j and eventually filter for where j=n?

[lavish jewel]

almost

[ripe birch]

ill take an almost. that means im close. What isn't quite right there?

[lavish jewel]

once you have the vector in the eigenbasis, you replace A with the eigenvalue of the corresponding eigenvector

then if you want you can use pigeonhole principle for the product part lol

if you have n pigeons and drill n+1 holes into then, then at least one pigeon has 2 holes

[zinc timber]

lol

[viral magnet]

Pigeonhole? More like pigeontroll

[ripe birch]

so by simply putting our vector into the eigenbasis, A MUST be replaced with the eigenvalue of the eigenvector I projected onto?

oh i guess. because when A acts on it you get that eigenvalue

[lavish jewel]

right

you don't HAVE to

[ripe birch]

but you'll get it anyway

[lavish jewel]

but if you don't, then doing this doesn't help you prove the thing

[oblique prairie]

horrifying

[quartz compass]

lmao

[ripe birch]

ok talking about the projection thing really helped, i appreciate that. I could feel the answer on the tip of my tongue but something was missing I couldn't quite get. Projecting the arbitrary vector along an eigenvector makes sense tho

[zinc timber]

at least one pigeon has 2 assholes

[ripe birch]

thank you all for joining me on this adventure. This was 100x more productive than the physics server

[zinc timber]

ya phys server sucks

[ripe birch]

it really shouldn't. but it do

[viral magnet]

Physics server: how about we run a simulation with a mole of spheres

[lavish jewel]

i'm still butthurt about the linear functionals though

[ripe birch]

math people are just better to be around. I love physics as a field, but the people of math are just better.

[zinc timber]

which one?

[viral magnet]

Kolmogorov will visit you in your dreams and tell you to suck it up

[ripe birch]

I've met a lot of physicists. wouldn't party with em. Went to a party of grad-level mathematicians, that was a blast. Way more fun 😂

[lavish jewel]

well get bent, you got this explained to you by an engineer

[ripe birch]

engineers are also more fun, but also more psychotic

stop using i,j,k as basis vectors you psychopaths

[lavish jewel]

i don't do that :x

[zinc timber]

they said maths server not mathematicians

[ripe birch]

then maybe you're a step above your bretheren

[lavish jewel]

why would i use j as a basis vector, when it obviously is sqrt(-1)

...

[ripe birch]

then again i can't speak, my people try to use phi as the azimuthal angle in spherical coorinates

[zinc timber]

quaternions

[viral magnet]

Why use e as unit vector when that's clearly Euler's number

[ripe birch]

like we use theta in 2D, but flip to phi in 3D

[zinc timber]

R³ iso to a subspace of quaternions

[ripe birch]

@viral magnet also a thing i never understood. one of my texts did that and i almost cried

thankfully i haven't seen it since

[viral magnet]

🫂 I don't wish that on my worst enemies

[golden ermine]

Number theorists using pi as a function to count the number of primes less than n

[ripe birch]

EH?

[quartz compass]

ever1 stop trying to attach universal meaning to symbols

[viral magnet]

Also what a lazy name

"Prime counting function"

[lavish jewel]

i'm afraid this has gotten a little off topic, please move over to discussion-i or chill

[ripe birch]

ill see yall in off topic

wait i cant find off topic. chill it is?

[lavish jewel]

yeah

btw

as i said

[viral magnet]

We'll ill be damned

[dusky wadi]

broke: $e\$
woke: $\vec{e}$

[stoic python]

Alison40

[spring pasture]

@zinc timber

[zinc timber]

@spring pasture ?

[spring pasture]

Hoe is you?

[zinc timber]

the army sam?

[spring pasture]

Yes

[old swift]

I've met a lot of physicists. wouldn't party with em. Went to a party of grad-level mathematicians, that was a blast. Way more fun 😂

[hearty rapids]

i'm so confused

why is a arbritary?

because there is already a 1 in R1?

how is it reduced row echelon form when there is a 3 above the 1 in column 4?

[sharp shell]

u have to choose the values so it is row reduced

there can be a 3 above the 1 if u choose the c accordingly

[hearty rapids]

huh?

i thought reduced row echelon form meant if there is a 1 in any column

it has to be zeroes above and below

[dim epoch]

only if its the leading 1

[tulip basalt]

Is there a way to like solve (A+B+C)x=0? A is a Vandermond matrix and B and C look like each other so I think this will be an easier way to solve my system but I have no idea how.

[lavish jewel]

wdym by "look like each other"

[tulip basalt]

They're both the operators applied to the exponential function with different exponent and such

and A contains those exponentials

It's like if T and S are operatos B=T(A) and C=S(A) applied component-wise

Does anyone how I can solve a system like this one?

Can I just take out an operator like (I+T+S)Ax=0 then apply the inverse on both sides to get Ax=0?

[viral magnet]

Your question is still too vague to answer. Come up with a 2 or 3 dimensional example

[tulip basalt]

Let $A=\left[\begin{matrix}
e^{ikx_1\cdot y_1} & e^{ikx_2\cdot y_1} \
e^{ikx_1\cdot y_2} & e^{ikx_2\cdot y_2}
\end{matrix}\right]\in (L^2(\Omega))^2\times(L^2(\Omega))^2$, $T,S: L^2(\Omega)\to L^2(\Omega')$ be compact operators.
Then, let $B=T(A),C=S(A)$ applied component-wise. How can I solve (A+B+C)x=0? Thanks!

[stoic python]

emphatic_wax

[viral magnet]

Well that's beyond my linear algebra knowledge

[tulip basalt]

thanks for trying though

thanks for everyone. i solved it

[echo night]

Can some kind soul check my life dilemma about a Change of Basis on #help-3?

[viral magnet]

What did you do

[tulip basalt]

Well I figured I proved some weeks ago that my operators are injective so I just pull them out and take the inverse so I'll have Ax=0

And I know the answer to this from a paper I read

[ashen bane]

it's not a leading 1

[serene solstice]

How do I answers the next two questions?

[boreal spindle]

hey can someone explain me the solution for this excersize question?

[fringe fjord]

Please don't ping helpers after waiting just two minutes.

[boreal spindle]

Oh sorry!

[fringe fjord]

Also, when the solution you're asking about takes up half a page, please explain which parts of it your problems start at. Otherwise you're basically asking people to repeat what is already there, in the hope that they'll randomly guess the rewording you need.

[boreal spindle]

I didn't understand almost all of it unfortunately

it's the solution for part a) though

I understood the part where they explained that the matrix is supposed to be 2x3 and the rest I'm lost

<@&286206848099549185>

[viral magnet]

What's the subscript on R?

[boreal spindle]

In the original question?

[fringe fjord]

It would make sense if $\mathbb R_k[x]$ means the vector space of polynomials of degree at most $k$.

[stoic python]

Troposphere

[boreal spindle]

it's 2 and 1

[sand sinew]

Moment of ecstasy

[ashen bane]

guy pings after waiting 16 minutes lmao

[oblique prairie]

sometimes people ping after exactly 15

[exotic horizon]

how can I find a matrix when given the eigenvectors associated to it ?

[lavish jewel]

do you have the eigenvalues as well?#

[exotic horizon]

yes

[lavish jewel]

are the eigenvectors orthogonal?

well, in general you'd take the eig vecs as columns of a matrix Q and the eigvals along the diagonal of a diagonal matrix D

then the matrix is QDQ^-1

[quartz compass]

the way I like to think of it is, you know the eigenvalue equation $Av=v \lambda $ instead of writing one of these for each eigenvalue/eigenvector, you can throw all the eigenvectors in a matrix at once and write this as $AV=VD$, here the matrix $V$ has its columns the eigenvectors and $D$ is a diagonal matrix of eigenvalues.

[stoic python]

Merosity

[zinc timber]

what if I write it as Av = lambda v

[lavish jewel]

it's the same thing anyway

[quartz compass]

heh I was gonna explain why I wrote it that way but it was getting too long

[lavish jewel]

once you consider more than one vector at a time

[quartz compass]

I think of multiplying on the right as column operations and multiplying on the left as row operations

[zinc timber]

yes that interpretation is actually helpful

[quartz compass]

since D is putting a scalar on the entire column, that's why I wrote it with the scalars on the right yup

[zinc timber]

though I still need to pause for a moment

[lavish jewel]

you don't have to worry about it if you use einstein notation

[zinc timber]

$v=v^ie_i$

[zinc timber]

btw I ended up ordering a larger one, edd

[lavish jewel]

ah

did you already receive it?

[zinc timber]

no it'll take like a week

they also dropped the price after I ordered it, those scammers

[lavish jewel]

haha

[exotic horizon]

that's a good way to see it ! thank you and thanks everyone who answered

[quartz compass]

you're welcome 👍

[dusky wadi]

would the matrix from R² to R² for the orthogonal projection onto $x_1 = -2x_2$ be $\begin{bmatrix} \frac{1}{\sqrt{5}} && 0 \ \ -\frac{2}{\sqrt{5}} && 0\end{bmatrix}$

[stoic python]

Alison40

[zinc timber]

no check again

[quartz compass]

a nice way to project onto a space spanned by a vector v is to make the matrix $P=\frac{vv^T}{v^Tv}$

[stoic python]

Merosity

[quartz compass]

(it follows from the usual derivation with dot products, also fun to check P^2=P)

[dusky wadi]

ok I'll actually come back to this question

i thought i could do it with rotation matrices but i was wrong

[quartz compass]

the only part where you "need" rotation matrices is the fact that $x+2y=0$ can be seen as $(1,2)\cdot (x,y)=0$ so you can do a 90 degree rotation or just see by inspection $(x,y)=(2,-1)$ is a basis for this 1D subspace

[stoic python]

Merosity

[dusky wadi]

i just realised i can solve it using simultaneous equations

[thorn yacht]

Hey, could I just ask whether the following statement is true or false?:
If f is a linear operator over the finite-dimensional vector space V, then V has a basis that consists of eigenvectors for f: V->V.

(My TA said that this was true, but from what I can see, this isn't always the case, so if someone could perhaps clear this up, that would be very helpful, thanks.)

[zinc timber]

no that's not always true, (the special case when it is true, we call T diagonalizable)

[lavish jewel]

as ryu says

[zinc timber]

ex the matrix $\m{1 & 1 \ 0 & 1}$ do not have a basis consisting of evs

[lavish jewel]

that's a so-called "defective matrix"

the requirement for what you said to be true is for the operator to have n linearly independent eigenvectors, where n is the dimension of the vector space

[zinc timber]

(My TA said that this was true, but from what I can see, this isn't always the case, so if someone could perhaps clear this up, that would be very helpful, thanks.)

[thorn yacht]

Thanks for clearing it up 🙂 👍

[subtle walrus]

please read #❓how-to-get-help

[severe heron]

Let A and B be 4 × 4 matrices. Suppose that A has eigenvalues
x1, x2, x3, x4 and B has eigenvalues 1/x1, 1/x2, 1/x3, 1/x4, where each xi > 1. Prove that A + B has at least one eigenvalue greater than 2. Can someone help? Any hints on how to start?

[quartz compass]

idk maybe try to look at manipulating their characteristic equations to try to make an inequality

[zinc timber]

no other information given on A or B?

[severe heron]

nope

here's the entire Q

[zinc timber]

maybe you can show that B and A^-1 are similar

[quartz compass]

something I have in mind is f(x) is the characteristic polynomial of A, then x^4 f(1/x) is the characteristic polynomial of B, then try to do something with cayley hamilton by plugging in the matrix A+B and break it into an inequality in terms of A and B separately

[zinc timber]

I'm not super sure on this because xi's may not be distinct

[severe heron]

okay i'll try that out

[quartz compass]

I guess now that I think of it, their characteristic polynomials are the same, just with the coefficients reversed

[quartz compass]

that might play well with this

[zinc timber]

if confused on the multiplicity part, like they didn't mention distinct

[quartz compass]

oh, why does it matter

[zinc timber]

like if they are taking something like [1 1 \ 0 1] has ev 1, 1, then it might not work, idk

[quartz compass]

hmm

[zinc timber]

because say $A = \m{2 & 0 \ 0 & 2}$ and $B=\m{1/2 & 1 \ 0 & 1/2}$ fits the description

[severe heron]

4 by 4

[zinc timber]

(just any 2x2 block, that's not the point)

[severe heron]

okay

[severe heron]

i used the fact that sum of ev of (A+B)=sum of individual ev of A and B= x1+1/x1+x2+1/x2+x3+1/x3+x4+1/x4 > 8 using am gm

and now since there are 4 eigenvalues of (A+B) we can say by PHP that one of them must be greater than 2 i guess

[exotic horizon]

Hi! does anyone know how to find the dimension of an eigenspace (we are given a matrix and eigenvalue) ?

[zinc timber]

nullity (A-cI)

[exotic horizon]

c is a real number right ? What is nullity ?

[zinc timber]

c is the eigen value

nullity = dim of the null space

[exotic horizon]

thank you!

[magic light]

Need some reminders on what operations are allowed on determinants

I know you can add columns together like C1: C1 + C2 - C3

and that you can take a constant out of a row like
|aa ab|
|c d|

a * |a b |
|c d|

but what else? can I do row operations?

[haughty berry]

(The transpose of a matrix doesnt affect its determinant, so anything that works for rows goes for columns as well)
You can add rows by a constant without affecting the determinant. So like R1: R1 + 3R2 wont affect the determinant
Multiplying a row by a scalar multiplies the determinant by the scalar
Swapping two rows multiplies the determinant by -1

Determinants are also multilinear

[magic light]

wait so why does this not change the determinant?

also 1 is constant right?

R1: R1 + R2 => unchanged?

[haughty berry]

It doesnt really matter in the case that youre adding rows

So like

R1: R1 + 3R2 => unchanged as well

Lemme show you

[magic light]

this makes me feel like I can just do gaussian elimination on a determinant, no?

also a follow up - can I take a constant from a column?

[haughty berry]

\def\sline{\text{---}}
Suppose you have:
\[ \det\begin{pmatrix} \sline v_1 \sline \\\vdots\\\sline v_n \sline\end{pmatrix} \]
And say you add $\alpha v_i$ to $v_1$ ($i\neq 1$):
\[ \det\begin{pmatrix} \sline v_1 + \alpha v_i \sline \\\vdots\\\sline v_n \sline\end{pmatrix} \]
Since determinants are multilinear:
\[ = \det\begin{pmatrix} \sline v_1 \sline \\\vdots\\\sline v_n \sline\end{pmatrix} + \alpha\det\begin{pmatrix} \sline v_i \sline \\\vdots\\\sline v_n \sline\end{pmatrix} \]
The second determinant above has $v_i$ twice (once in the first row, the second in the $i$th row), so it is just $0$:
\[ = \det\begin{pmatrix} \sline v_1 \sline \\\vdots\\\sline v_n \sline\end{pmatrix} \]
Which is your original determinant

[haughty berry]

Just about. Multiplying by a row by a constant multiplies the whole determinant by that constant, and swapping rows multiplies the determinant by -1

[stoic python]

Slurp

[haughty berry]

Anything that is true for rows is true for columns since the transpose doesnt affect the determinant

[magic light]

if a row of 0s => determinant is 0?

[haughty berry]

Yeah, but that's not related to adding rows

[magic light]

I was just curious why that thing is 0

[haughty berry]

Oh

You remember how the determinant of a non invertible matrix is 0?

So if the rows are linearly dependent, then the determinant is 0

[magic light]

I see.

My professor likes giving us a bunch of tricks in the determinants. If we solve it by brute forcing we get a cubed equation.

No calculators in the test.

[haughty berry]

that sucks

i cant do math without my calculator lol

[magic light]

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

[haughty berry]

Characteristic polynomials?

[magic light]

the question in this case was to find an orthonormal base for R^3 using the eigenvectors of A^TA

when A was given to us

Soo yeah.