linear-algebra

[onyx tundra]

thanks :)

[slender yarrow]

yeah 'propre' just means eigen stuff

[brittle juniper]

🥖

[onyx tundra]

yeah fortunately a lot of maths words in french are quite similar too

[gray glen]

🥖

[onyx tundra]

🥖

[gray glen]

[slender yarrow]

[onyx tundra]

💩 @gray glen

[gray glen]

[onyx tundra]

damn nitro

[honest marlin]

Hi, I was watching this video https://youtu.be/LyGKycYT2v0. At around 7:38 seconds, he starts to talk about how a vector can also be thought of as a 1 by 2 matrix. I don’t understand why this is the case and how it actually relates to the dot product.

[broken hawk]

write a vector as $$\begin{pmatrix} a \ b \end{pmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

flip it 90 degrees

there’s your 1x2 matrix

now calculate both $$\begin{pmatrix} a \ b \end{pmatrix} \cdot \begin{pmatrix} c \ d \end{pmatrix}$$ and $$\begin{pmatrix} a & b \end{pmatrix} \begin{pmatrix} c \ d \end{pmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

and note how they simply give you the same number back

and that’s all there is to it

(well, except for the whole dual space thing but I’m not gonna explain that here)

[honest marlin]

Well I get that they give the same answer but I’m wondering why it’s like that which I guess is the dual space thing.

[broken hawk]

well, when you carry out the computation of the dot product, you can recognize that you’re doing nothing else than that matrix-vector multiplication in the second row

and thus you can associate taking the dot product with (a,b) (as a column vector) as applying the linear map given by the matrix [a,b] (as a row matrix)

the dual space of ℝ² (which can be seen as 2×1 matrices) is just the 1×2 matrices

[honest marlin]

Ok, one more question. Why is it more useful for the dot product to be projection*length of a vector rather then just being projection?

[broken hawk]

I believe I’ve already explained this to you on reddit

:)

[honest marlin]

Ooh, was that you?

[broken hawk]

well I certainly have tried to give an answer to exactly that question about half a day ago

or so

or maybe it was yesterday actually idr

as I said there, the only reason is becasue you can do the exact same things as with your idea, except that you have way nicer properties for calculating with it

so why would you ever choose the one that has ugly properties (like not being symmetric)

also, even more pragmatically: the standard dot product is super easy to calculate. just multiply the components and add them up

measuring the length of the projected vector is harder to compute

you essentially have to calculate the standard dot prodct and then divide by the length of the vector you project onto

which will involve square roots n stuff

so basically you’re giving up both nice computational properties (symmetry, bilinearity) and easy computation… for what?

what do you hope to gain?

also, one thing that I’m not sure you misunderstood or are just communicating badly: the dot product of two vectors is a number, not a vector

[honest marlin]

I’m not pushing for all mathematicians to change what the dot product represents, I’m just asking so I can understand it better which I think I do now. Just to clarify, we happened to define the dot product as projection* direction which happened to be calculated by taking a one dimensional array of the vector and multiplying it with another? Also, is the idea of a vector also being a transformation just the idea of all 2d vectors also being able to represent two 1d vectors some 1d line? And those two vectors also represent the dot product when used to transform another vector?

[honest marlin]

<@&286206848099549185>

[placid oracle]

is a symmetric matrix one thats equal to its trnaspose? arent these all equal to their transposes?

[sonic osprey]

Check the signs in B

[placid oracle]

thats such a dumb q

ok thank u

[slow scroll]

a space of eigenvectors

[dusky epoch]

does your linear algebra book not have definitions

for an operator A and an eigenvalue λ of A, the eigenspace of A corresponding to λ is the set of all vectors v such that Av = λv

it would do you good to check that this does indeed form a subspace of K^n

where n is the size of A

[slow scroll]

do not use technology
lol

[dusky epoch]

ok so clearly 0 is an eigenvalue right off the bat

lemme see

gonna throw it into matlab rq to check

[ember pilot]

,w eigenvalues [[-1,0,1],[-3,0,1],[-4,0,3]]

[slow scroll]

i think it has to be {}

[dusky epoch]

alright

yeah

your eigenvalues are correct @robust swallow

now for each of those finding a basis for the eigenspace boils down to finding a basis for ker(A - λI)

[ember pilot]

@slow scroll [] works

[slow scroll]

oh ok

[dusky epoch]

is that for λ=1

you should say which is which

yeah, because that isn't a basis for R^3

it has too few vectors

you'd need another eigenvector for λ=1 to get an eigenbasis

A isn't diagonalizable

an* eigenbasis

but yes

[slow scroll]

I got $(\text{Ran}, A^T)^\perp = \text{Ker}, A$ and $(\text{Ran} , A)^\perp = \text{Ker} A^T$. Is that correct?

[stoic python]

kxrider:

[winter sage]

g

[slow scroll]

ood?

[oblique crag]

Anyone help@with #6?

[dusky epoch]

which part

showing W is a subspace or finding a basis for it?

@oblique crag

[oblique crag]

basis

[dusky epoch]

i guess the simplest way would be to convert those conditions into equations in the coefficients

[kind arrow]

Hi could anyone explain me the proof for the formula of the magnitude of a vector?

[brittle juniper]

What do you refer as "the formula for the magnitude of a vector"?

[broken hawk]

I’d e more interested to see the proof because it might as well be a definition

[limber sierra]

do you just mean like, for a vector (a,b) in a conventional vector space? because thats just the pythagorean theorem

[broken hawk]

I assume the intended thing is ‖v‖ = √(x² + y² + z²)

…

[kind arrow]

yes

[dusky epoch]

is that not a definition

[limber sierra]

yeah thats usually a definition

[broken hawk]

you can either see that as the definition of magnitude, or see the geometric intuition of length as the definition and then prove this formula from pythagoras

as like, a one-liner

[kind arrow]

my question is why do we use the Pythagorean theorem

[broken hawk]

if you wanna do it geometrically

[kind arrow]

yeah that's fine

[broken hawk]

then you first define length only for those vectors pointing along an axis

in which case it’ll just be the one nonzero component

and then for any other vector, it’ll be the sum of those

and you can apply pythagoras becasue the axes are at right angles

[dusky epoch]

in which case it’ll just be the one nonzero component

(taken with its absolute value)

[broken hawk]

is that satisfactory?

[kind arrow]

yes thank you

just didn't quite the intuition behind it and I didn't want to just memorize the formula and go plugging in numbers because that's what the textbook said

[broken hawk]

however, bear in mind that when you procede further into linear algebra, there’ll be more general definitions for magnitudes (usually called norms) and this one’s just one possibility

the one that’s most intuitive to us

but not the only thing that makes sense

[kind arrow]

ah ok well I am only hs rn and the lin alg we do is pretty basic

[broken hawk]

this one’s called the euclidean norm or 2-norm, it’s the one that coincides with our “flat” geometry

[kind arrow]

we don't even do matrices lol

[winter reef]

Where can I learn about Orthographic prjoections and stuff like that in euclidean spaces? I can't really imagine in, I've been reading through many sources, the drawings ive seen from my teacher were not helpful as well, are there any like a 3blue1brown animations vids out there?

[wintry steppe]

@winter reef i read the book linear algebra and its application 5th edition and really explains it really well

[winter reef]

thx I found it will read it, but why would you EVER use greek letters as scalaras and normal one for vectors smh

will have to rewrite

[dusky epoch]

thx I found it will read it, but why would you EVER use greek letters as scalaras and normal one for vectors smh

visual differentiation

also they're called roman, not "normal"

[limber sierra]

nah, god himself descended from the heavens and decreed

"the true language of humanity is urban latin"

speaking of, ban the letter U

[dusky epoch]

why

[wintry steppe]

because

greek for vektor

and normal for scalar

NORMAL

[dusky epoch]

no

[wintry steppe]

not ROMANL

[dusky epoch]

that's like

the fucking worst

[limber sierra]

nah nah nah

Cyrillic for vectors

chinese for scalars

[wintry steppe]

yes

[limber sierra]

or better yet

traditional chinese for vectors

and simplified for scalars

[dusky epoch]

bad

[grave plank]

what if you used numbers for scalars

[winter reef]

@wintry steppe the book actually explains it real;ly well with the drawings, thank you so much

[swift plinth]

i got matrix AxX = B I need to find matrix X

[broken hawk]

did you try following the hint

[swift plinth]

yep. but I am not sure how it should look like

[broken hawk]

can you solve equations of the form Ax = b, where A is a matrix and b is a vector?

(and x is also a vector)

cause your thing here is just three of those, with each x being a column of X and each b a column of B

[swift plinth]

ahh sure ! haha easy! thx buddy

[broken hawk]

(make sure you understand why that’s the case)

[dusky epoch]

A * [B, C] == [A*B, A*C] (matlab syntax)

[thin bloom]

Quick question, when we find the eigen values of a matrix lets say 2 x 2 shouldn't we get an eigen value of 0 everytime since the zero vector will always satisfy the solution?

Ax=\lambda * x

wait nvm

we specifically let det(A-\lambda * I)=0 so that we specify that x cannot be 0

[oblique crag]

Any1 help with 17?

[jagged pendant]

well, when you have known roots of a polynomial

you can factor it

[honest marlin]

I was watching this video:https://youtu.be/LyGKycYT2v0?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&t=152. At that part in the video, he explains why the order you do the dot product does not matter but I did not understand his explanation. Could someone please explain why order does not matter in a similair way to how he did it?

[slow scroll]

well the dot product of (a,b,c)dot(x,y,z) = ax + by + cz
and (x,y,z)dot(a,b,c) = xa + yb + zc

oh hold on im watching the video

[slow scroll]

3b1b made the point that its clear that the dot product commutes for vectors of unit length, so if we start with two normalized vectors, $\hat u$ and $\hat v$, we can say that $\hat u \cdot \hat v = \hat v \cdot \hat u$. Multiply both sides by $\lvert u \rvert$. Now we have $\lvert u \rvert\hat u \cdot \hat v = \lvert u \rvert\hat v \cdot \hat u = u\cdot \hat v $. This is the projection of $u$ onto $v$. Now multiply both sides by $\lvert v \rvert$. Then you have $\lvert v \rvert u \cdot \hat v = \lvert u \rvert \lvert v \rvert \hat v \cdot \hat u = \lvert u \rvert (v \cdot \hat u). \ \$ Whether you choose to think of the dot product as a projection of $v$ onto $u$ or a projection of $u$ onto $v$, you have the scaling factor of the magnitude of the other vector that makes everything equal, i.e. $$\lvert v \rvert (u \cdot \hat v) = \lvert u \rvert (v \cdot \hat u)$$

[stoic python]

kxrider:

[slow scroll]

@honest marlin

[honest marlin]

I'm confused on what you mean by its clear that the dot product commutes for vectors of unit length, so if we start with two normalized vectors and why multiplying one of the unit vectors by the magnitude of one of the vectors is still equal to the dot product.

[slow scroll]

3b1b made this argument for vectors of the "same length."

All unit vectors have length 1, and using unit vectors allows me to talk about things in terms of projections like 3b1b. All i did was manipulate the equation

u dot v = v dot u (with hats over the vectors)

[honest marlin]

So is u a different vector from u-hat in your explanation?

[slow scroll]

u hat is a vector that points in the same direction as u, but has a length of 1.
v dot u hat just means v projected onto u and
u dot v hat just means u projected onto v

maybe it would be easier to think of u hat and v hat as vectors with the same length, not necessarily 1, and |u| and |v| as just scaling factors on the first vector in each product.

whyyyyy texit whyyyyy

there is another definition of dot product that makes this easier to explain. $$u\cdot v = \lVert u \rVert \lVert v \rVert \cos \theta$$ where $\theta$ is the angle between the vectors. If you align $u$ on the x axis, then its easier to see that the projection of $v$ onto $u$ is $\lVert v \rVert \cos \theta$, where $\theta$ is the angle $v$ makes with the x axis. $ \ \ $ If you instead align $v$ on the x axis, then the projection of $u$ onto $v$ is \lVert u \rVert \cos \theta$. Now compare these projections along with this definition with 3b1b's...

[honest marlin]

I understand that algebraically, the explanation makes sense but I can’t seem to think of it visually

[slow scroll]

Well the thing is, projection is a visual thing, but dot product really isn't that i know of. Other than that it should be zero when the vectors are perpendicular

[honest marlin]

I guess what I really don’t understand is if the projection of one vector onto another is bigger then the vector being projected onto and why when he scales up vector v by 2 and projected it onto w in the video, then you have a larger projection and you are multiplying by the length of w. I understand that when w is normalized, it makes sense but if it wasn’t, then why would the projection of 2v onto w and multiplying it by the length of w be the same as projecting w onto 2v and multiplying it by 2v.

Sorry, I’ve got to go. Could anyone please DM me if they know the answer?

[slow scroll]

@gilded plover projecting w onto 2v is the same as projecting w onto v. The only thing we take from the vector we are projecting onto (v in this case) is the direction of v. In the dot product, the extra factor of 2 doesn't just go away, but it scales the whole product by two

[random trellis]

any hints for (a) (or (b)) implies (c) in qn 3? i’ve shown that a and b are equivalent, but cant seem to figure c out

i cant think of how to use qn 2 fully, because in the primary cyclic decomposition, there may be multiple subspaces with the same irreducible factor

alpha is normal and it’s a positive definite form, so there’s the fact that alpha and alpha* are simultaneously diagonalisble. i saw some stuff on stackexchange, but they said to use lagrange interpolation, which we didnt learn, so there has to be an easier way

[half ice]

Let me draw your attention to the snipping tool, good for copying parts of the screen

[random trellis]

okay sorry bout that!

[half ice]

Otherwise Isry I don't know how to answer, but there's smarties on the server

[random trellis]

i was guessing maybe the polynomial expression of alpha* is the same when the cyclic subspaces have the same irreducible factor, but am not sure how to argue that

[random trellis]

so, any hints?

[thin bloom]

Just want to know if this was correct

[broken hawk]

your algebra there is a bit off. I think in line 2 you accidentally wrote λλᵏx when you meant Aλᵏx

[dusky epoch]

bad proof.

[broken hawk]

it’s generally just written messily

the logic is sound but it’s hard to read

[dusky epoch]

major argumentation hole

[broken hawk]

and you never say what x is

which is probably the hole ann means?

[dusky epoch]

and also wording

[broken hawk]

you also need to show that λᵏx is an eigenvector of A

which is another one-liner

but it has to be done to be rigorous

[dusky epoch]

you attempted to do a proof by induction, but neglected to state the base case, however trivial it may appear

and to prove $A^{k+1}x = \lambda^{k+1}x$ you cannot start with $A^{k+1}x = \lambda^{k+1}x$

[stoic python]

Ann:

[dusky epoch]

you do not start with X to prove X.

[broken hawk]

here‘s how I’d write the same thing more cleanly:

Assume $x$ is an eigenvector of $A$ with eigenvalue $\lambda$. This gives us the base case $A^1x = \lambda^1x$. Now assume the statement holds for $A^n$. Then
$$A^{n+1}x = AA^{n}x = A\lambda^nx = \lambda^nAx = \lambda^n \lambda x = \lambda^{n+1}x,$$ which was to be shown.

[stoic python]

Sascha Baer:

[dusky epoch]

@thin bloom

[thin bloom]

Ok so for a proof you never start with the statement that you want to prove

@dusky epoch For the base can I use k=1?

Since k=1 is assumed true by the question

Also @broken hawk

[broken hawk]

the base case is trivial but for a proof by induction you at least have to state that

like “base case n=1 holds by assumption” is enough

[dusky epoch]

Ok so for a proof you never start with the statement that you want to prove

i mean honestly

how was that not blindingly obvious

[broken hawk]

eh, I messed that up a bunch. I usually intended to write sth like “I want to prove X=Y” and then just wrote X=Y

[thin bloom]

Can I start with what I want to prove and "convert it" using what I know is true to another statement that is true? @dusky epoch

[broken hawk]

it’s the algebra mindset

where you start with an equation

and then deform it until you get what you want

but really what you have here is sth like

A = B = … = C = D

so you should start from one side and try to get to the other

[dusky epoch]

an equality chain

[broken hawk]

or you can start in the middle and show that it’s both equal to A and to D

[thin bloom]

ok

[broken hawk]

this for me often ends up looking like
A = B … = C
‖
D = … = E

(where the goal was C=E)

but you can’t just write C=E because you don’t know that yet

also, one more thing

if C=E is wrong

then you can derive true things from it

[dusky epoch]

ex falso quodlibet

[broken hawk]

so starting somewhere and then showing a true thing is not a proof

[dusky epoch]

1 = 0
0 = 1
1+0 = 0+1
1 = 1

[broken hawk]

(however, starting somewhere and then showing a false thing is a counterproof)

[thin bloom]

ok so if they asked prove A=D I have to use A=B=...=D therefore A=D? Rather than A=D=C=B since A does = B A=D must be true?

[dusky epoch]

you cannot say A=D

[thin bloom]

even if I showed that A=D implies something true such as A=B

[dusky epoch]

you want to prove A=D, therefore you cannot say A=D

anything implies truth

[broken hawk]

read through ann’s example above

[dusky epoch]

just because X implies Y and Y is true does NOT mean that X is true.

[broken hawk]

where she shows that 0=1 implies 1=1

and yet we know 0 doesn’t aqually equal to 1

from true things you can only derive truth
from false things, you can derive everything
you don’t know whether what you wanna prove is true

so deriving anything true from it tells you nothing

deriving something false from it does however tell you that it’s wrong

you need to go either
true statement ⇒ … ⇒ thing you want to prove
or
thing you want to disprove ⇒ … ⇒ false statement

[thin bloom]

Alright I understand now.

[broken hawk]

here‘s another humorous false proof

assume 1=2
invite the pope into your room
there is now a pope in your room
there are two people in your room
since 2=1, there is now one person in your room, and in particular, as already stated, one pope
since you are in your room, and there is one person in your room, and one pope, it follows that you are the pope

so from a false statement, I derived something ridiculous

[thin bloom]

@dusky epoch @broken hawk Thank you.

[winter reef]

cool examples

[thin bloom]

@broken hawk So never start from what you want to prove since it can lead to false conclusions

[broken hawk]

never start from something you want to prove, because you will never be able to tell if it was true

[thin bloom]

what if I did start from the proof and ended up getting statements that I knew were true?

I wont do this again but what if this happened?

[broken hawk]

as we’ve said about five times now

if you conclude something true, this tells you nothing about the original statement

put another way

if you ever see a true statement, then everything after that will be true
if you ever see a false statement, then everything before that was false too

but it does not go the other way

(before/after within the same chain of logical argumentation)

[broken hawk]

black being “unknown”

[thin bloom]

whats the T thing

[broken hawk]

shorthand for the word next to it

[dusky epoch]

[thin bloom]

Im sorry I dont understand what the picture really represents.

[dusky epoch]

$\top$ and $\bot$ are the logical symbols for true and false, resp

[stoic python]

Ann:

[dusky epoch]

the squiggles are statements

[broken hawk]

the two columns are independent

[dusky epoch]

the columns are arguments

[broken hawk]

each is a sequence of statements, e.g. your calculations or what not

[dusky epoch]

in which every line follows from what came before

if one line is known to be true, then everything afterwards is guaranteed true as well

if one line is known to be false, then everything before it is guaranteed false as well

[broken hawk]

but you never know anything in the other direction (the black lines)

the statements there could be true or false

[thin bloom]

So we're reading from up to down

[broken hawk]

yes, same as the equations in your proof

[thin bloom]

So how does some statement we dont know lead to something true?

[broken hawk]

here in your ”proof”

anything that would go after that last line is true too

anything above, we can’t conclude just by looking at this

[thin bloom]

OHHH

[broken hawk]

every line follows from the one above by a logical implication, but this doesn’t help

[thin bloom]

So if I started from false statement why don't i know if the next one is false or unknown?

[broken hawk]

because 1=0 implies the world is a cube

false statement imply anything

[thin bloom]

which can be true so can I start off with a false statement and end up with something ture?

[broken hawk]

you can conclude whatever you want if you start with a false assumption

indeed

[thin bloom]

ok let me work on this proof again

[thin bloom]

I've fixed the logical flaw that I had

Have i said enough about the base case?

[proper crescent]

I'd reword, don't say "we don't need to calculate for the base case" so much as "the base case follows from the definition"

[thin bloom]

If I solved for the eigen vectors of a 2x2 matrix and got complex eigen vectors do these eigen vectors span the space ?

They are distinct eigen vectors

[broken hawk]

if they are linearly independent, then yes, because any two linearly independent vectors will span a 2D space

however, if x is an eigenvector, then 2x is too

so getting two eigenvectors doesn’t mean anything yet

[thin bloom]

distinct eigen vectors

wait

[broken hawk]

if they have different eigenvalues, then they’ll be linearly independent,a lways

(but you can have linearly independent eigenvectors with the same eigenvalue)

[thin bloom]

I got x1 = [5 3-i] and x2 = [5 3+i]

[broken hawk]

those are independent

those two span ℂ²

[thin bloom]

C^2 does also span the entirety of R^2 right?

@broken hawk

[broken hawk]

uh

that statement makes absolutely no sense

that’s like asking whether apples fit inside bananas

if you’re working in ℝ², then [5, 3+i] doesn’t exist

if those are the eigenvectors of your matrix, then your matrix simply doesn’t have eigenvectors over ℝ

with linear algebra, you always have to pick a field (e.g. ℝ or ℂ) and then you stick with it

[thin bloom]

So the original question was "Find the eigenvalues and corresponding eigenvectors for each matrix below. Do the eigenvectors form a basis for the space? " A=[1 -5; 2 -5]

[broken hawk]

and what is the space?

[thin bloom]

idk it seems like R^2

by the matrix

but the eigen vectors are complex

[broken hawk]

is this a matrix over ℂ that just so happens to have real entries, or one over ℝ

[thin bloom]

Codomain C but range of R?

[broken hawk]

what?

[dusky epoch]

that makes 0 sense

[thin bloom]

I don't understand your question

[broken hawk]

appearance bananas but looks apples?

[dusky epoch]

can you repost the problem you are doing once again

EXACTLY AS IT WAS STATED TO YOU

[thin bloom]

I did

[dusky epoch]

i said repost

[thin bloom]

Oh

[dusky epoch]

i don't want to scroll a mile up

[thin bloom]

Find the eigenvalues and corresponding eigenvectors for each matrix below. Do the eigenvectors form a basis for the space?

A=[1 -5; 2 -5]

[broken hawk]

it’s an ill-stated question if it doesn’t state what “the space” is anywhere

seen as a real matrix, A has no eigenvectors

seen as a complex matrix (which so happens to have real values as its entries), it has your two linearly independent eigenvectors and they span ℂ²

[thin bloom]

I think she means C^2 then

Therefore the two eigenvectors are linearly independent and are basis for the space

C^2

Hold on.

can the linear combination of the eigen vectors span R^2? I understand its complex but we can make the complex part be zero with the right linear combination of the eigen vectors

[broken hawk]

that’s not how it works

[dusky epoch]

R^2 isn't even a subspace of C^2

[broken hawk]

like, yes, if it spans ℂ², then ℝ² is a sub_set_ of that

[dusky epoch]

also, "complex part" isn't a real term

[broken hawk]

but it’s not, as ann just said, a subspace

and you can’t span something which isn’t a vector space

[thin bloom]

R^2 isn't a subspace of C^2?

hmm

but its a subset of C^2

yes sorry not the "complex part" I meant the imaginary part

[broken hawk]

it’s not a subset of ℂ² because the scalars in ℂ² are complex numbers, and so if x is a vector in a subspace, i*x is too

but i*(1,1) is not in ℝ²

[thin bloom]

right

So the question can't be answered?

Since we don't know what its really asking for?

Can I just assume that they're talking about C^2

[slow scroll]

@thin bloom i think complexification allows for vector spaces over R to have complex basis vectors

[thin bloom]

k so in conclusion these 2 complex vectors are basis for C^2 which will include the set of R^2

soo.

[slow scroll]

not a basis for C^2. We have two basis vectors over R, not C. It takes 2 basis vectors over C to span C^2.

your eigenbasis here just spans R^2

[broken hawk]

his eigenbasis has complex vectors m8

they don't even live in R^2

[slow scroll]

ok then ignore what i said. i don't quite know how to talk about this

[vagrant shale]

what are the elements of a curve?

[thin bluff]

anyone know any noob friendly vids on matrix transformation? Having a difficult time understanding how to go about it

[winter reef]

[honest marlin]

In the 3blue1brown video https://youtu.be/LyGKycYT2v0 he explains how the dot product can be though of as a 2x1 matrix. He then goes on to solve for what the numbers in it are. I know that u hat’s x and y position encode a transformation but how do we know it is the transformation that we want? In the video, he says he needs to find a 1x2 matrix that takes 2d vectors to numbers but how does he know it will be u hats x and y position?

[dusky epoch]

in order for a transformation to be what you want it to be, it needs to send the basis vectors where you want them to be sent

[honest marlin]

Could you elaborate?

[dusky epoch]

doesn't 3b1b say this pretty often in his linalg videos

a transformation is determined by what it does to a basis

[honest marlin]

Yeah but how do we know where the basis vectors should end up? He’s trying to explain why the dot product is also just a transformation but how do we know that the x and y positions should be as they are when they are projected?

[swift plaza]

He explains that as well at 9:00, since by symmetry the projection of î onto û is the same size as the projection of û onto î, which is just the u_x.

[honest marlin]

Right, I get that. What I do not understand is why we know that those are the coordinates that are the dot product/projection.

[honest marlin]

How do we know that the x and y coordinates are the ones we are looking for? At 8:31, he says we need to find the matrix that defines projction*lenght but how does he know that the x and y positions are the correct ones?

[placid oracle]

Consider thje link diagram : A ---> B. Ralph as a 2/3 chance of starting at page A and a 1/3 chance of starting at page B. Once per min, he click on a random link, if possible. What is the probability that ralph will be on page B after four minuteS?

[jagged pendant]

not familiar with link diagrams, or what that is supposed to represent.

[astral junco]

So this set of vectors can at most span R^4 because to span R^4 you need at least four vectors?

[native lodge]

it can span at most R^4 because they only have four components

right away we can tell one of the vectors is useless because you only need four linearly independent vectors to span R^4, you have 5 vectors, so one is a linear combo of the others

assuming that four are independent

[astral junco]

thank you. how can i tell which four are independent assuming that there are four linearly indpendent ones?

[native lodge]

you can form a matrix with them and do the RREF if you want

pretty sure you only have to go as far as getting to upper triangular form for you to see which one is redundant

[astral junco]

ok will do. thanks just making sure i understand.

[placid oracle]

its a little cut off but:

thats a graph of links where R(A)=R(C), R(B)=R(A), and R(C)=R(B)

It says the pagerank vector is, unsirpisngly 1/3,1/3,1/3

and i found the altered transition matrix to be

0 0 1

1 0 0

0 1 0

not sure how to do the next two parts though,,,

[half ice]

I'm not certain of the content, but I'm guessing Px is just a multiplication?

Then P²x, P³x...
There should be a pattern in the powers of P

@placid oracle

[placid oracle]

yes thats what i was notiicng

does my altered transition matrix seem correct though?

because i just used the normal matrix as there was no need to change it since it was already stochastic

also, its basically cyclying through the order of x_0 for every power

so how does that relate to the sequence approaching r in the last question

[half ice]

It doesn't, and that's your answer

It will never get anywhere near r

[placid oracle]

@half ice is my logic for the altered trans matrix correct as well?

also ty

[half ice]

Yes I like it

[random violet]

hi guys

[tacit sigil]

hi

[astral junco]

I'm trying to find a matrix to change basis from A to B. I'm pretty confused. Is it inv(A)*B ?

[quaint heart]

Think about it

[astral junco]

I mean inv(A) get me to standard basis and then applying B gets to that basis?

[quaint heart]

Given any 2 matrices, when can you transform them into each other by a change of basis?

Wait do you mean that those are your basis?

[astral junco]

yeah those are both bases

[quaint heart]

Oh ok

I think what you said is correct

It's either that or it's inverse

Ie B^-1 A

[astral junco]

I tried that and then tried to check it but maybe i made a mistake. didn't seem right.

[quaint heart]

Let me think for a moment

Is it A M A^-1 or A^-1 M A?

I always forget

[astral junco]

hmmm

i mean A*A^(-1) gives you I

?

[quaint heart]

X is the matrix your changing the basis of

[astral junco]

ah

[quaint heart]

Fixed

[astral junco]

eh i'm so confused. how can i even check which one is right?

[slow scroll]

i don't really understand what this question is asking, A and B aren't similar matrices, so there definitely isn't a change of basis between them

[quaint heart]

A and B are your basis

[astral junco]

ah crap. i was trying to create an example so i could practice.

what does similar mean?

[quaint heart]

@slow scroll you can definitely do a change of basis between those 2 basis

[slow scroll]

it means two matrices can be written as a change of basis of each other.
If $I_{\mathcal{AB}} is the transformation that sends a basis $\mathcal{B}$ to $\mathcal{A}$, then
$$T_{\mathcal{AA}} = I_{\mathcal{AB}}T_\mathcal{BB} I_{\mathcal{BA}}$$

cmon texit....

[astral junco]

wish i knew tex

can't tell what could be wrong

[slow scroll]

Anyway, similar matrices share properties like trace and determinant, thats why your example with A and B doesn't work (unless i just don't understand what ur asking).

[astral junco]

hmmm

[quaint heart]

You don't understand what he's asking

Given any 2 basis

You can do a change of basis between them

He's asking what the change of basis matrix.is

[slow scroll]

ohh....

[astral junco]

wouldn't they both have to be invertible?

[quaint heart]

Yes

To be a basis

The matrix must be invertible

[slow scroll]

so in this case, if the matrix $A$ is a transformation that sends the basis, $\mathcal{A}$ to the standard basis $\mathcal{E}$ to, call it $I_{\mathcal{E A}}$. And if $B$ is a transformation $I_{\mathcal{E B}}$, then the transformation that sends $\mathcal{A}$ to $\mathcal{B}$ is $I_{\mathcal{B E}} I_{\mathcal{E A}}$. You get $I_{\mathcal{B E}}$ by inverting the matrix that describes $I_{\mathcal{E B}}$.

[stoic python]

kxrider:

[astral junco]

hmmm. makes my brain hurt

does this make sense at all?

i'm just trying to find out how i can check if i've found the right matrix.

and also to see if i understand it.

[brittle juniper]

A change of basis matrix from $A$ to $B$\ is a matrix $P\in\GL_2(\bbR)$ such that $P^{-1}AP=B$ iirc

[stoic python]

Tuong:

[astral junco]

what does GL_2 mean?

[honest swift]

the invertible 2x2 matrices

GL stands for "general linear"

[brittle juniper]

Mhmm seems like it's not that xd

huh

it's more like Q¯¹AP=B

[honest swift]

?

[slow scroll]

err thats what i thought he was talking about at first too, but i think A and B are matrices that send a particular basis to the standard basis, and he wants to compose them so that one basis is sent to the other or somethin like that

[brittle juniper]

oh nvm

[honest swift]

it will be conjugation because it's inverse is the change of basis matrix for the inverse transformation.

[astral junco]

they're supposed to be matrices that send the standard basis to them A sends the standard basis to A which is the new basis.

same with B

[honest swift]

yeah

[astral junco]

this is supposed to be what A represents

[honest swift]

it's just what you conjugate by to get the matrix of a transformation in a different basis (as it's name would suggest)

[slow scroll]

im gonna suggest that you watch this video if you haven't already https://www.youtube.com/watch?v=P2LTAUO1TdA&t=520s

[astral junco]

i've watched it 10 times lol

i have a tiny brain 😉

but it is the best one i've come across. guess i should probably watch again

[slow scroll]

tl;dr $\begin{pmatrix} 1 & 5 \ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix}$ = \begin{pmatrix} 1 \ 3 \end{pmatrix}$. $ \ \ (1,0)^T$ is written in the basis of $A$, and the transformation you gave converts it into the standard basis, i.e. the wacky way we would think of this choice of basis when we see it.

[stoic python]

kxrider:
Compile Error! Click the reaction for details. (You may edit your message)

[astral junco]

yeah i think i get it up to that far. just can't seem to put the rest together. anyway i'm a little closer to getting it i think.

[slow scroll]

think about it: whenever you are even defining a basis, you are trivially using the standard basis to describe it, so in a sense, anytime you talk about a different basis, you are transforming it to the standard basis to make sense of it.

This is honestly one of the most trippy topics I've come across in LA so far, so ur certainly not alone if you are confused xd

[astral junco]

yeah it's like every time i think i get it i try it in a problem a week later and i'm lost. thanks for the help so far though. think i just need to rewatch that 3b1b vid.

[slow scroll]

aight np.

[sacred quest]

Typically, is a Intro to Linear Algebra class easier than, say, Diff Eq. or Calc IV?

[winter reef]

yes

[keen nexus]

Depends on how you lean. It's conceptually heavier than what you might be used to, whereas Intro to DEs is more methodology-centric.

[sacred quest]

Aye, taking 261 (intro to linear algebra), 254 (calculus IV), and 256 (diff eq) in my first year in upper division CS. Already taken the first three calc classes.

If 261 is easier, then I'll stack it in a term with a harder CS class

[chilly burrow]

Are there any good books for linear algebra by chance anyone would recommend

[broken hawk]

there’s two suggestions in #books-old but I’m sure if you have more specific wishes (in particular about what approach to linalg you need) people can find sth better

[chilly burrow]

I just remembered those channels and was gonna delete lmao

[proper crescent]

Hoffman and Kunze is a bit old school but basically the correct linear algebra book

At least for the old school treatment. I hear "Linear Algebra Done Wrong" is the modern correct linear algebra book but I haven't seen it myself

[chilly burrow]

I see thank you very much! I'll check them out

[broken hawk]

the name “linear algebra done wrong” is clearly a play on axler’s book (linear algebra done right), are they in some sense doing “opposite” approahces to the theory?

[proper crescent]

Axler's got this imo overdone to the point of stupidity anti-determinant thing going

[broken hawk]

in the preamble he never actually explains the title. he just says he first wants to get some other comments out and then never says the actual thing

to be fair we covered determinants only at the very end of linalg I and I didn’t feel like I ever missed them

[proper crescent]

What Axler doesn't like is, to be fair, valid. A lot of linear algebra courses do this thing where they're like "Okay a determinant is where you give me a matrix and I bash some stuff out"

[broken hawk]

we did a very matrix-lite course though

we introduced determinants as multilinear alernating maps with det(id) = 1

[proper crescent]

And then when it's time to prove that linear maps C^n -> C^n have eigenvalues, a geometric concept, everyone's like lmao tfw det(tI - A) is a polynomial so it has a root gg

Doing it that way is pretty shit, since like, the "input a matrix output a number through this algorithm" doesn't have much conceptual content in itself and then it's used to prove something of substance

[chilly burrow]

Is there a specific math background that would be helpful for linalg by chance? Or is it something I could probably jump into and be fairly alright? I'm just rebuilding my foundation for Calc, haven't had the chance to look at what linalg entails exactly quite yet as finals just finished up

[proper crescent]

Now there's a correct way to do determinants and that's what you talked about, it's what Hoffman-Kunze kinda does as well. Axler's solution is to avoid determinants altogether

[broken hawk]

linalg can easily be done as your first “rigorous math” class

however

depending on the book used it wuold help to get a bit of background in proving things

[proper crescent]

He waits until much later, after he even does stuff like characteristic and minimal polynomials, and then he's like okay we're always working over C so det = product of eigenvalues

[broken hawk]

and in stuff like set theory notation

other that that?

high school algebra suffices and is barely needed

it’s a very “basic” subject

[proper crescent]

And he does some other stuff that's not in my taste coming from algebra. Since he only thinks of R or C, he never distinguishes between polynomials and their functions, etc

[broken hawk]

one thing I can absolutely recommend is to watch 3blue1brown’s series on it just to get a bit of an intuition for how to imagine things

for the most basic definitions

[chilly burrow]

As for applications I'm going into pchem if that really matters at all (granted I really have no problem with just learning more for the sake of learning)

I love 3b1b

[broken hawk]

basically the reason I asked for that is because there’s, very broadly speaking, two different “linear algebra”s

there’s theoretical linalg, and there’s computational linalg

[chilly burrow]

I'm doing computational chemistry specifically

[broken hawk]

the latter is uh, boring

but extremely useful

in a lot of applications

the former explains what the hell the latter is doing

and is very rich in nice mathematics,imo

and quite accessible

[chilly burrow]

Yeah I wasn't sure if there was a different approach in learning that's better suited based on actual applications of it, but I feel like a nice conceptual understanding of it would help me a lot

[dapper hemlock]

chonk, supposing you already know proofs, maybe try lay's linear algebra book, its more of the applied type

[chilly burrow]

I'll give them all an overview thanks guys!

[half ice]

@chilly burrow
If we're throwing out suggestions, Zill's applied engineering mathematics covers applied linear algebra in a chapter

[astral junco]

if a 3*3 matrix has three distinct eigenvalues does that mean that there must be an eigenbasis?

[half ice]

@astral junco
Each eigenvalue represents a subspace of vectors. Each subspace has a basis.

It sounds like you have 3 one-dimensional subspaces

[astral junco]

ah maybe i'm not forming the question right due to a lack of understanding

i guess i don't really understand what this is saying

anyway if i find three eigenvectors that are linearly independent does that mean that there is an eigenbasis?

[broken hawk]

yes

[gusty crystal]

hey is this right
-5x < -45x + 200
-5x + 45x < 200
40x < 200
x < 5
? its linear inequalities.

[wintry steppe]

👌

It good

[gusty crystal]

Okay thank you

I might ask a couple more questions if thats okay with you?

[wintry steppe]

That’s fine 👍

[gusty crystal]

do you happen to know how to do compound inequalities?

[wintry steppe]

Yeh

[gusty crystal]

50<10(x+2)<180 this is one of my problems and I am not quite sure how I am supposed to solve it if you dont mind could you explain it to me

[wintry steppe]

So you can simplify the middle part

And then isolate x

[gusty crystal]

So am I removing the 2 and subtracting it from both sides?

[wintry steppe]

wut

[gusty crystal]

I am confused sorry

[wintry steppe]

XD it cool

So 10(x+2)

Can simplify to 10x + 20 straight off the bat

[gusty crystal]

okay and that removes the parentheses right?

[wintry steppe]

Yep

[gusty crystal]

Thats what I was having trouble at thank you

[wintry steppe]

No problem

[gusty crystal]

then can I divide everything by 10?

[wintry steppe]

Yep

[gusty crystal]

3<10x<16?\

[wintry steppe]

Well

[gusty crystal]

yeah remove 10

sorry

[wintry steppe]

👌👌👌

[gusty crystal]

3<x<16

Thanks alot.

[wintry steppe]

Ez

[gusty crystal]

Yeah its not so hard once you figure out what you are doing

[wintry steppe]

Basically math

😂

[gusty crystal]

haha

Okay I am sorry I know im helpless but this problem is confusing me. Yeast, a key ingredient in bread, thrives within the temperature range of 90°F to 95°F. Write and graph an inequality that represents the temperatures where yeast will NOT thrive.

[wintry steppe]

So it can survive from 90 to 95

But nothing bellow 90

Or above 95

[gusty crystal]

and above yea

[wintry steppe]

Yeh

[gusty crystal]

I got the graph I think but the inequality is the part confusing me

[wintry steppe]

So F greater or equal to 90

And f less or equal to 95

[gusty crystal]

https://gyazo.com/ae1b798df366c8e1438e4dd3a927fa2c is this right?

90<=f<=95?

[wintry steppe]

👌👌👌

[gusty crystal]

I guess I was supposed to put an open point because it was equal too

[wintry steppe]

Yep 😂

[gusty crystal]

That one is on me lol

[wintry steppe]

[gusty crystal]

How do I work a word problem lol.

[wintry steppe]

Oh boy word problems

[gusty crystal]

would it help if you saw the problem?

[wintry steppe]

Probably 😂

[gusty crystal]

3x+5=>-19 and -5x-3>-18

[wintry steppe]

So simplify them first

Getting x alone

[gusty crystal]

3x=>24?

[wintry steppe]

Yep

-24

[gusty crystal]

I guess my calculator is bad xD

[wintry steppe]

XD

[gusty crystal]

3x=>-24

[wintry steppe]

And then divide 3

To get x alone

[gusty crystal]

x=>-8

[wintry steppe]

Yep x=>-8

[gusty crystal]

I was just fixing it lol

[wintry steppe]

😂

And then if I remember correctly the second one is -5x-3>-18

[gusty crystal]

x=>-8 and -5x-3>-18

Yeah I was just resending it lol

[wintry steppe]

So simplifythe second one

Making x a loner

[gusty crystal]

-5x>21?

I am not confident it is 21

[wintry steppe]

Adding 3 to both sides

-18+3 😂

[gusty crystal]

so I add it to the 8 as well lol

-15 lmao

[wintry steppe]

Nah you don’t have to

Because they separate

[gusty crystal]

-5x>-15

[wintry steppe]

Yep

[gusty crystal]

then divide by -5

[wintry steppe]

Ez

[gusty crystal]

x>3

x=>-8 and x>-3

or did I fuck up the negative again lmao

[wintry steppe]

X>-3

[gusty crystal]

I knew them pesky negatives lol

[wintry steppe]

XD

[gusty crystal]

what it said I got it wrong

[wintry steppe]

x>3

And then you can connect them

I think

XD

[gusty crystal]

yeah which side is open left or right?

I thought it was left but it said right side had open circle.

[wintry steppe]

Yehhh

[gusty crystal]

https://gyazo.com/c87aaec1c6ad4a4f0e35a16e04b2f648

[wintry steppe]

So x>3 x=>-8

[gusty crystal]

https://gyazo.com/3bb079b67209320c1a57585767af9913

wtf it flipped?

[wintry steppe]

-8<=x<3

[gusty crystal]

So its not open circle on the left side?

fuck me

I am dumb

it just now hit me lol

[wintry steppe]

Hadeth we fucced up

[gusty crystal]

Open circle means it cant be there closed circle means it can I just got hella confused

[wintry steppe]

[gusty crystal]

I always fail on the easy stuff

[wintry steppe]

That’s confusing

Closed circle is filled in so that makes it included I guess

[gusty crystal]

yeah

Luckily my state is 50th in state education woo hoo

[wintry steppe]

😂

[gusty crystal]

-10<5x+5<25

-3<x<4

Did I do this one right I feel like I did

[wintry steppe]

👌👌👌

[gusty crystal]

Yeah I did one right lmao

[wintry steppe]

Ezzz

[gusty crystal]

watch it be wrong xD

it was right

[wintry steppe]

Ayyy

[gusty crystal]

Okay this one combines all my 3 worst things in compound inequalities

Kevin is baking bread for a family function. The initial temperature of the oven is twice the room temperature. He knows that yeast, a key ingredient, thrives within the temperature range of 90° F to 95° F. So to facilitate yeast growth, Kevin decreases the temperature of the oven by 44° F.

Which inequality represents the given situation?

[wintry steppe]

oh god

[gusty crystal]

Where do I even begin lmao

[wintry steppe]

what does that even mean

[gusty crystal]

IDEK

I want to die lmao

Would it help if I gave you the multiple choice anwsers?

answers?

[wintry steppe]

yes

[gusty crystal]

It will take a second because you cant copy them for some reason you have to manually write them down

[wintry steppe]

hard work

[gusty crystal]

A.90<=2x-44<=95
B.90<=2x+44<=95
C.90=>2x+44<=95
D.90=>2x-44<=95

That was a doozy lmao

I am pretty sure its not b or c

[wintry steppe]

Okay c and d don’t make sense

[gusty crystal]

yea so its a?

[wintry steppe]

I don’t know yet XD

wait

[gusty crystal]

it says decreased by 44 so they cant add 44

or am I dumb

[wintry steppe]

oh god try it

[gusty crystal]

It was right oh thank god lol

I cant get any more wrong or I get a 70 which means I will have to do this satan work again

[wintry steppe]

NO

[gusty crystal]

https://gyazo.com/c5379d9a02b1b7099db5e625549301ca

I should have done this from the start lol

I suck at these line graphs

I think its D

[wintry steppe]

Yeh

[gusty crystal]

IT WAS B

[wintry steppe]

WHAT

[gusty crystal]

Oh god I will have to do it again

[wintry steppe]

[gusty crystal]

Why god why

lol

[wintry steppe]

oh god it can’t be both

noooooooo

[gusty crystal]

this was the explaination On the graph, the numbers to the left of x = 2 and to the right
of x = 10 are shaded. Therefore, the compound inequality will be an "OR" inequality, or a disjunction.

[wintry steppe]

it tricked us

[gusty crystal]

They have done it again

they are out to get me the entire school system is out to get us man lol

[wintry steppe]

There is only one thing to do

Run

[gusty crystal]

commit sepuku?

or that

[wintry steppe]

blame it on the teacher

😂

[gusty crystal]

Exactly

I am taking this to the supreme court gosh dangit

[wintry steppe]

XD

[gusty crystal]

god I hate the noise of failure

Every answer I get wrong it makes a noise and a part of me dies with it lol

[wintry steppe]

the shame

[gusty crystal]

I know will you fall on your sword with me partner?

[wintry steppe]

yes

[gusty crystal]

Activate seppuku

[wintry steppe]

[gusty crystal]

I cant believe I am going to get a 50 man

Fml I am so dumb xD

[wintry steppe]

I blame the questions

[gusty crystal]

Same and the answers

I blame it all but me

and I guess you

[wintry steppe]

Exactly

Shizzzz

[gusty crystal]

No I am saying I don't blame you this is a huge honor to be bestowed upon

[wintry steppe]

It’s 1:39 am

[gusty crystal]

Damn do you live in asia?

[wintry steppe]

XD

Netherlands 🇳🇱

[gusty crystal]

Ohh red light district?

[wintry steppe]

Close enough

Oof

[gusty crystal]

I think the only people that live there are prostitues and homeless people so I just insulted you

And I don't apologize

[wintry steppe]

XD

I mean most people here

Are dicks

[gusty crystal]

It kinda makes sense

Your next to both france and germany you were fucked from the start lol

[wintry steppe]

XD too true

Imma sleep

[gusty crystal]

aight gn thanks for helping me

[wintry steppe]

have fun withsatan’s work

[gusty crystal]

You know I wont

[thin bloom]

Does Repeated eigenvalues Imply Linearly dependent eigenvectors ?

[native lodge]

no

[thin bloom]

Does a zero eigen vector imply anything?
I understand that now its not invertible

@native lodge

[quaint heart]

That's all it implies

[native lodge]

I forget what zero means

[quaint heart]

And you mean eigenvalue

Not eigenvectors

[thin bloom]

ys

yes @quaint heart

My mistake

[quaint heart]

Zero eigenvalue is actually equivalent to non invertible

[thin bloom]

yes

But what does this mean for the eigenvectors

[quaint heart]

Nothing

It gives you 0 information

[thin bloom]

ok

[quaint heart]

About the eigenvectors

Without knowing more info about the matrix you're working with

[thin bloom]

Question asks if the eigen values of a matrix A is 1,1,0,2 what conditions does the matrix A meet for it to be diagonalizable

[quaint heart]

Ok that's different

The 1 eigenvectors must be distinct

[thin bloom]

yes

[quaint heart]

That's it

[thin bloom]

so I answered using the eigenvalues you must be able to obtain 4 linearly independent eigen vectors

[quaint heart]

Um

You can do a bit better than that

[thin bloom]

Ok so like stating that the repeated eigenvalues must produce 2 linearly independent vectors?

[dusky epoch]

for each eigenvalue λ, there must be as many linearly independent eigenvectors as the multiplicity of λ when considered as a root of the charpoly

this is the condition for diagonalizability

so in your case, dim(ker(A - 1I)) must be 2

[thin bloom]

is dim(ker(A-1I) same as nullity of A-1I? I dont remember

[dusky epoch]

seems so

yes

nullity = dimension of kernel, definitionally

[thin bloom]

Ok thank you

[ember pilot]

<@&268886789983436800> um idk but off-topic/creepy also #calculus

[jagged pendant]

👍

[ember pilot]

[astral junco]

i'm trying to figure out if a vector is in the range of a linear transformation.

sry if this is stupid, but if the vector isn't in the kernel of the transformation then wouldn't it have to be in the range?

[dusky epoch]

no

[broken hawk]

the kernel is part of the domain, the range is part of the, uh, range

I guess you use range the way I use image?

[astral junco]

ok thanks. i'll have to read up on the difference between range/image think i'm confusing the two.

[broken hawk]

the way I use them (but they're not quite standardized) is that the image is precisely those that are actually hit by the function, while the range is more generally all the things it could hit, like the entire target space

[astral junco]

gotcha yeah. a lot of new terms to memorize. probably gonna have to retake lin alg from the beginning to get any sort of understanding of it lol

[slow scroll]

hmmm from every resource ive heard from, the range is the same as the image and the "co-domain" is the target space

[jagged pendant]

same for me.

[broken hawk]

yea, as said, not everyone uses them the same

image seems pretty clear tho

I think it’s only range that’s ambiguous

[honest marlin]

I’m linear algebra, I have learned that a 2x2 matrix represents a 2 dimensional transformation and that the columns can be interpreted as where they take I hat and j hat so when you do vector matrix multiplication it’s just scaling the new I hat and j hat and then adding them which is pretty much what you do when you have a vector. But now I’m confused with matrix multiplication. Both vectors can’t represent where I hat and j hat land because then the result would always be the second matrix applied so what is an intuitive way to think about it?

[broken hawk]

think of it as applying the rightmost matrix first. that’ll transform not just your basis vectors, but also the rest of the grid. so now there’s two new vectors that can be considered “i” and “j”. and then the second (left) matrix will move those to its columns

[honest marlin]

Ok, could it be thought that the transformed I hat and j hat are now just vectors in a new, normal I hat and j hat space and then that space also gets transformed?

So whenever all of space gets transformed, it is still inside another normal vector space?

[broken hawk]

what does “normal” mean

the image of a linear transformation will again be a vector space. if the function is surjective, it will be the entirety of the target space, otherwise some subspace

basically, if you apply the function encoded by $\begin{bmatrix} 1&2 \ 3&4 \end{bmatrix}$, then the vectors (1,0) and (0,1) will move to (1,2) and (3,4), respectively. and then you take (1,0) and (0,1) again, and transform those. and if you multiply the matrices, that’ll tell you where the original (1,0) and (0,1) land when you first apply the right and then the left transformation

[stoic python]

Sascha Baer:

[honest marlin]

Well I learned that a matrix could be thought of as transforming the entirety of vector space since changes to I hat and j hat change all vectors. By normal, I just mean j hat is (0,1) and I hat is (1,0). I’m imagining something like a bunch of vector spaces stacked on top of each other like paper. The first matrix transforms that one on top and then applying the second one transforms the one below it which also affects the top one.

[limber sierra]

Hard disagree

The only thing that sucks is me

:(

[jagged pendant]

[limber sierra]

Wait

Is it "me" or "I"

#linear-algebra help me with my grammar

[jagged pendant]

no

cool dude. that isn't linear algebra though

[feral crow]

I just need help with the last part of b

[broken hawk]

just divide through some norms

[feral crow]

I proved that each left eigenvector is orthogonal to a right eigenvector with a different eigenvalue

But how do I normalize each left eigenvector

What does that even mean

[broken hawk]

well, if you do $b_i \cdot a_i$, you’ll get some value

[stoic python]

Sascha Baer:

[broken hawk]

divide $b_i$ by that value

[stoic python]

Sascha Baer:

[feral crow]

But didn't I just show that $b_i\cdot a_j$ =0

always

because theyre orthogonal

[stoic python]

Cosack:

[broken hawk]

no, you showed that $b_i \cdot a_j = 0$ for $i \neq j$

[stoic python]

Sascha Baer:

[broken hawk]

for $i = j$ it should be nonzero, that was part of the exercise

[stoic python]

Sascha Baer:

[feral crow]

So if I normalize them

it works for even i = j

I still get nonzero answers

Whats the point of normalizing it then

[broken hawk]

your goal is $b_i \cdot a_j = \delta_{ij}$

[stoic python]

Sascha Baer:

[broken hawk]

not $= 0$

[stoic python]

Sascha Baer:

[feral crow]

But isnt $\delta{ij}$ = a number

[stoic python]

Cosack:

[broken hawk]

it’s 1 if $i=j$ and 0 otherwise

[stoic python]

Sascha Baer:

[feral crow]

I see that

What's the point tho

[broken hawk]

because orthonormality is just nicer to work with than only orthogonality

I can’t tell you what’s the point in this specific example

but in general, it’s nicer to only work with 0s and 1s, than with 0s and random numbers

[feral crow]

Probably helps with this

i guess

Looks like a2 is the same as a1

just opposite direction

Thats cool

[broken hawk]

uh, do you have distinct eigenvalues?

either way, that’s not a basis

you should have (1/2, 1) and (-1/2, 1)

[feral crow]

Yes theres different eigenvalues

for a1 and a2

yeah

[broken hawk]

yea that’s not what you said

[feral crow]

So if you start at origin

a1 goes positive x and y

same slope as a2

but opposite

[broken hawk]

“opposite direction” usually means just - on everything

[feral crow]

Oh

Opposite x

Well

I see what you mean

my bad XD

Also a question

I found that you can just find left eigenvectors of A

by doing right eigenvectors of A transposed

but when you do that

Doesn't work because matrix multiplication requires certain dimensions

you cant do (2x1) x (2x2)

[broken hawk]

2x1 would be a column vector

transposed wuold make it 1x2

[feral crow]

You have to transpose the eigenvector too

once youre done?

[broken hawk]

that’s what you just said, no?

[feral crow]

I said you just do A transpose

[broken hawk]

I’ve never worked with left eigenvectors

[feral crow]

and find its right eigenvectors

and those are the left eigenvectors of A

[broken hawk]

ah, yea that makes sense

[feral crow]

but then I can't satisfy

[broken hawk]

so you’ll get the following:

[feral crow]

So im not sure

[broken hawk]

yea, give me a sec I’ll write it up

[feral crow]

kik

kk

[broken hawk]

so let’s say $\hat b_i$ is a (right) eigenvector of $A^T$

[stoic python]

Sascha Baer:

[broken hawk]

then $(A^T \hat b_i) = \lambda \hat b_i$

[stoic python]

Sascha Baer:

[broken hawk]

now transpose both sides of the equation

and you’ll get $(A^T \hat b_i)^T = \lambda (\hat b_i)^T$

[stoic python]

Sascha Baer:

[broken hawk]

and $(A^T \hat b_i)^T = \hat b_i ^T A$

[stoic python]

Sascha Baer:

[broken hawk]

so, a left eigenvector of $A$ will be the transpose of a right eigenvector of $A^T$

[stoic python]

Sascha Baer:

[feral crow]

So I just have to write it as a row

instead of column

[broken hawk]

yes

[feral crow]

So if I have

and lets call it bi

bix = 2

biy = 0

if i transpose it

x and y are the same still yea

bix = 2, biy = 0

Cause for this part

I can assume that x = first value, y = second value

from left-> right or top->down

or is that wrong

[broken hawk]

that’s what is intended, yes

only thing that’d make sense

[feral crow]

Ok

So its only in row form because it satisfies the equation

[broken hawk]

here’s where you’ll need that normalization :P

[feral crow]

yeah

makes it nicer

I literally have no idea what this means

[broken hawk]

basically all you have to check is that B is diagonal in the sense shown below

it’s weirdly worded

[feral crow]

i swear this stuff is made to confuse me

XD

[jagged pendant]

yea. by fundamental theorem of algebra, char polynomial has at least 1 root, so u get at least 1 eigenvector. then u use properties of the hermitian-ness to show that its eigenvalue is equal to its conjugate, so it has no imaginary part

@feral crow

[feral crow]

Thanks!

[jagged pendant]

and probably some invariant subspace stuff so u can take quotients without messing anything up

to get the orthogonal basis

[feral crow]

But if the left eigenvectors are always equal to the right eigenvectors

How can they be orthogonal

nvm

The two eigenvectors with the same eigenvalue

are orthogonal

[fringe cave]

@woeful hawk I don't know what it is you need, but this channel is empty atm so good enough

[feral crow]

What's R(-theta)

is that just R times -1

full question for context

dont know how to do e

[fringe cave]

R(-theta) is like f(-x)

[modest jasper]

Hey.

Does anyone know how I can solve a question like this...?

[fringe cave]

So, what all have you covered?

Also lmao questions channels who needs those amirite

[modest jasper]

What all have I covered?

[woeful hawk]

@fringe cave a friend already helpe me thanks

[fringe cave]

aight

@modest jasper as in have you gone over gaussian elim?

[modest jasper]

Yeah I have.

But I kinda just have a basic knowledge.

I don't know if there are a lot of complex techniques and stuff.

[fringe cave]

Just do the gaussian stuff and get it so that you can more easily see what changing k does

[modest jasper]

What changing k does?

I'm confused, sorry...

[fringe cave]

Like, get it into a triangular form or something, you know?

[modest jasper]

Triangular form?

[fringe cave]

Alright, so, you know how you can add and subtract the equations from each other

[modest jasper]

You mean reduced echelon form?

[fringe cave]

ye that's even better

[modest jasper]

Oh, okiii.

[fringe cave]

triangular is just a slightly different type of the same thing

[modest jasper]

Oh, okay.

Then what?

[fringe cave]

Echelon form wouldn't work too well I fear since you can't exactly fully remove the kz, but if you manipulate it to find x or y or something from, say, the top two equations

it would help

[slender yarrow]

should work

[half ice]

U blu

[modest jasper]

So I should learn about the triangular form then?