linear-algebra

[fringe fjord]

(FTA wants the polynomial to have degree at least 1 :-)

[graceful mauve]

GOOD point, constant polynomials need not have zeroes

[wary girder]

I am trying to derive the Fourier Transform from Fourier Series and I want to do it in a way that emphasize that the Fourier transform gives a function of frequency

so i attempted to derive an expression for the coefficient of the Fourier Series such that it can be turned into the Fourier transform

but I have been struggling for quite a few days

[hushed hedge]

Find the line that:
• goes through the point (3,2,1)
• Is perpendicular to the line (x,y,z) = (1,2,3) + t(1,-1,2)
• Is parallel with the plane pi: 2x+2y+3z= 5

Is my drawing correct of what they are asking for?

Since w is parallel with the normal of pi and it is perpendicular with v

[dull kestrel]

I know Im missing something stupid here, but if addition is not well defined in the projective plane P2, what operation are we using to define an automorphism of P2

[torpid moat]

like if we just take the normal matrix multiplication $m(e_{ij} \otimes e_{kl})=\delta_{jk}e_{il}$ how would we define its adjoint $m^$? so what would $m^(e_{ij})$ be?

[stoic python]

thestonethatrolled

[dusky epoch]

do you mean adjoint as in dual

with some identification of M_n(K) with its own dual (the two being isomorphic as vector spaces but not naturally so)

[torpid moat]

I mean its Hilbert space adjoint

[dusky epoch]

do you have a definition handy?

[torpid moat]

so like we need $\innerproduct{m(e_{ij}\otimes{e_{kl}})}{e_{pq}}=\innerproduct{e_{ij}\otimes{e_{kl}}}{m^*(e_{pq})}$

[stoic python]

thestonethatrolled

[strange elm]

Hey guys can someone help me with this one?

I've somewhat got part a)

for part b) id guess that we use the fact that dim(im(b) \cap ker(a)) = 0 given rank(AB) = rank(B) but i dont know where to go from there

[slim totem]

Hi, This is example 3 from Stephen, Friedberg, Arnold, Insel Linear Algebra

What do the square brackets mean and how do they differ from the round brackets?

[dusky epoch]

they do not differ from round brackets here.

the only reason they are written here instead of round brackets is aesthetics

[languid sphinx]

it's not just 'for looks'

You pair square bracket to square brackets

You can also use all (()), but then it becomes easier to get confused

Since clearly there is no marker to show which bracket pairs with which other bracket

[dusky epoch]

i mean it is unambiguous

[wintry steppe]

You can also interpret it as saying that $\mathcal{F}(S,F)$ which is the family of functions from $S$ into $F$ forms an equivalence relation where $[f(s)]$ is the equivalence class of functions equal to itself at each point. Then you can argue the author is saying an equivalence class can be operated on by scalar multiplication if we view it as scaling the function before evaluating at $s$. Similar stuff could be said for addition

[stoic python]

engineersanonymous

[wintry steppe]

f(s) is an element of F, so [f(s)] isn't going to be the kind of equivalence class you want. this is not at all what the author means, and it seem like a super clunky way to interpret the (in my opinion, already clear) author's notation

"scaling the function" is precisely what the author is defining here, so this is circular anyways

it doesn't make sense

@wintry steppe

[obsidian hearth]

Can someone provide me an intuitive (preferably geometrical) explanation to when given an orthonormal matrix, why does the transpose of that said matrix's columns are still orthonormal, when their elements are from separate (but I suppose mutual) vectors?

Stems from $A^{\top}A = AA^{\top} = I$. Algebraically it's easy to prove this, but on a visual level I can't wrap my head around it.

[stoic python]

kura

[quartz compass]

I'd say the issue is not just orthonormal matrices, but inverse matrices in general that we have $A^{-1}A=AA^{-1}=I$, the rows of $A^{-1}$ dotted with the columns of $A$ give you either 1 or 0, and this is true even when you do it "backwards" which is surprising

[stoic python]

Merosity

[humble token]

i dont really understand what negative determinant means

i tried making some graphs what i pulled out is that determinant is negative

if i hat is below the line formed by j hat

or j hat is above the line formed by i hat

is there another meaning?

or intuition

i watched a 3b1b video now and i undertsand now

[humble token]

wait so

would the determinant of the inverse of a matrix be the inverse of its determinant

lemme write what i mean

also det(kA) = k^2det(A)?

yup alright

[gusty axle]

Yeah this is correct

[gusty axle]

I believe this is only true for 2x2 matrices

[humble token]

thanks

[humble token]

oop yeah i forgot to specify im working with 2x2 matirces

[gusty axle]

For an nxn matrix, det(kA) = k^n * det(A)

[humble token]

imma tryna derive for 1/A

for 2x2 matrices

[willow crag]

Question about dot product:
I'm currently trying to adjust a vehicle path using line detection, and it goes a bit wonky (pic included). To get the average path deviation I'm currently averaging all lines, already a bad idea, and calculating angle between that vector and the heading (0, -1):

arccos( v_n . (0.0, -1.0) )

Path vector is p1 - p0 (light - dark red dot)
Heading has y = -1 because Y is positive down since it is pixel coords, but my assumption is that as long as we have identical direction between the vectors the inner angle will be calculated

However this doesn't seem like the right value for the angle, is my assumption not correct? I would expect a value of about 45deg here

In python, if it matters:

# lines is returned from cv2.HoughLinesP. It contains uint16 rows of [x_0, y_0, x_1, y_1]
x_0, y_0, x_1, y_1 = np.mean(lines[:,0], axis=0, dtype=np.uint16)
# Treat point with most pos y-value as the "bottom" point
# ... because of y axis being positive down.
if y_0 >= y_1:
  p0 = np.int32((x_0, y_0))
  p1 = np.int32((x_1, y_1))
else:
  p0 = np.int32((x_1, y_1))
  p1 = np.int32((x_0, y_0))

vec = p0 - p1
vec_n = vec / np.linalg.norm(vec)

# get the deviation angle between heading and averaged line:
head_line_deviation = np.arccos(np.clip(np.dot(vec_n, (0.0, -1.0)), -1.0, 1.0))

[gusty axle]

You shouldn't need the sqrt

Wait I'm rusty let me think more xd

Ok so $\vec{x} \cdot \vec{y} = |\vec{x}||\vec{y}|cos\theta$

[stoic python]

hello2248

[willow crag]

Oh yea! The sqrt is done for the norm, thanks, I'll correct that.

• But regardless, the norm is found inside np.linalg.norm(vec) so the normalised vector in the code, should be alright, I think

[gusty axle]

$\theta = arccos(\frac{\vec{x}\cdot\vec{y}}{|\vec{x}||\vec{y}|}$

[stoic python]

hello2248

[gusty axle]

)

So in your case just divided by magnitude of v

[willow crag]

Yea, but that's actually already done implicitly in the code.
You made me think if if the norm function perhaps did something else than the 2-norm by default, but it's correctly doing the 2-norm, getting the same normalised vector value through wolfram

And a quick wolfram again gives me an angle of 63deg, this baffles me a bit, probably code related, should ask somewhere python-related probably

[gusty axle]

I mean the triangle you have drawn looks like a 30-60-90 triangle to me if it's 2D

[willow crag]

Ah you mean the angles? Yea, would've guessed 45 tbh, but python apparently wants it to be ~1 <.< Could be something weird going on with the data types

[gusty axle]

It's in radians probably

[willow crag]

That would be pretty close tbh, but isn't theta directly related to deg?

Omg if it's that

I think you're right fuk me lol

[gusty axle]

That's funny lmao

Happens to me a lot too with my coding

[willow crag]

This is why I wanna make video games and not important engineer stuff

[gusty axle]

Hahaha yeah I feel that

[willow crag]

Anyway thanks mate. Saved me a nights headache haha

[gusty axle]

Yeah no problem gl with the code

[willow crag]

Ty!

[humble token]

how would i solf for matrix efgh

[obsidian hearth]

Yeah but I just can't visually wrap my head around it. Algebraically it's so obvious. But I don't understand why

[fringe fjord]

Ah, so a, b, c, d are known constants, and e, f, g, h are unknowns?

[humble token]

yup

[fringe fjord]

Then you just need to invert the abcd matrix and scale it appropriately so its determinant becomes 1. This is impossible if the original determinant was negative.

[humble token]

well see this is the problem

im using this equation

to derive how to invert a matrix

after a bit of math i found out that the inverse of matrix a b c d

is e f gh times 1/sqrt(det(abcd))

where efgh * abcd = [x,0,0,x] x is any real number

i might be totally wrong though

i just started linear algebra

[fringe fjord]

Looks mostly right; I get $$\begin{pmatrix}e & f \ g & h\end{pmatrix} = \frac{1}{\sqrt{ad-bc}}\begin{pmatrix}d & -b \ -c & a\end{pmatrix}$$

[stoic python]

Troposphere

[humble token]

damn i was actually right for once in my life

ok so in my equation the soltuion for efgh was d -b -c a

you used different meaning for efgh

i put efgh as equal to d -b -c a

and was going to find abcd inverse by multiplying efgh by 1/sqrt(det...))

but it same calculationd

alright

thanks!

[fringe fjord]

I used the meaning of e,f,g,h here.

[humble token]

ahhhh ok

well thanks i understand now

[fringe fjord]

And if ad-bc is not a perfect square, there'll be no solution where x is in Z.

[humble token]

wait that was my bad i somehow mixed up Z with R

this is the worst mistake i ever made

imma go cry now

[fringe fjord]

Ah, never mind then :-)

[humble token]

alr, thanks for the help

[mossy geyser]

Hi could anyone kindly explain what this solution is trying to say with the line in the red box?

I understand the part before that for A to have independent cols, Ax = 0 only when x = 0 since the nullspace of A only contains the 0 vector for A's col vectors to be independent

[gray dust]

if A has dependent cols then a nonzero sol to Ax=0 exists, giving a nonzero sol to A^T*Ax=0

[obsidian hearth]

Is this statement true?

[wintry steppe]

the transpose of an orthonormal matrix is orthonormal

yes

without writing out any entries, "A is orthonormal" means the same thing as "A^{-1} = A^T", and by taking the inverse of both sides you get what you want

[obsidian hearth]

On a more visual level, would saying the transpose of an orthonormal matrix A remains orthonormal because transpose insinuates simply a rotation, which would keep its perpendicular property?

[wintry steppe]

it's a little imprecise, but sure

it feels like you could turn that into a precise argument

"orthonormal" means "rows and columns are orthonormal", and transposing only switches up the rows with the columns

[velvet moss]

wait would it still be orthonormal when u transpose or just orthogonal

[obsidian hearth]

So I've ran into an issue of diag(-1, 1, 1), would mean that it's not only just rotation to fit with the linear transformation (as det = -1), while still retaining its orthogonality. On a "linear transformation mindset" mindset I suppose, would there be any other situations of this? (besides rotaiton, and this case)

[wintry steppe]

"orthogonal matrix" means the exact same thing as "orthonormal matrix"

[velvet moss]

oh I never knew that

I always thought it has something to do with unit vectors and orthogonal matrices

[obsidian hearth]

If it were not orthonormal, then it would probably break the rules of $A^{\top}A = AA^{\top} = I$

[stoic python]

kura

[obsidian hearth]

Or ig the assumption

[wintry steppe]

this is literally what it means for a matrix to be orthonormal/orthogonal, so of course it would

[wintry steppe]

okay, read my initial message as if i interpreted "rotation" as literally rotating the matrix

i don't see what transposing matrices has to do with rotations in the linear transformation sense

[obsidian hearth]

An orthonormal matrix, as a linear transformation, would simply just be a rotation or reflection
I was asking if the transpose of said linear transformation, maintains its orthonormal traits because it's simply just a rotation (or a reflection, given that determinant is negative).

My issue with this topic is mostly my incapability to visualize these things. I get it algebraically but like yeah

[wintry steppe]

An orthonormal matrix, as a linear transformation, would simply just be a rotation.
this is false and you gave a counterexample above

[obsidian hearth]

Ya

Oh wait

Yeah

Would there be any other counterexamples past rotation and reflection

[fringe fjord]

In 3D those are the only possibillities.

Wait, no, that's a lie.

-I is orthogonal, but is neither a rotation nor a reflection in a plane.

[gray dust]

in general if A is orthogonal there exists a set of coordinate axes where for some axes A does nothing, for some A reflects, and for some axis pairs A rotates by some angle

[obsidian hearth]

You guys are life saviors. Thank you for the help ❤️

[wintry steppe]

Hii Guys

Need some help with vector spaces

We know that C over C is vector space

But what will be the multiplication operation?

[gray dust]

@wintry steppe a scaling b is defined as the product of a,b when considered as complex numbers

in general this is how scaling is defined when considering a field as a vector space over itself

[stoic python]

meitar5674
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[silent aspen]

anyone knows how to get the span out of this one? W={p(x) /into R_3[X] | p(0) = P(1)}

[zinc timber]

what's R_3[x]? by any chance do you mean polynomials of degree at most 3 over lR?

@silent aspen

[silent aspen]

Yes

[zinc timber]

you don't you pick a general form a+bx+cx²+dx³?

[stoic python]

kura

[obsidian hearth]

$A \neq A^{-1}$

[gusty axle]

This isn't always true

[obsidian hearth]

^

[quartz compass]

show that its determinant is not equal to 1 or -1, that would mean it can't have its power the identity

[gusty axle]

Probably with determinants

[quartz compass]

that's a shame lol

[dim epoch]

[quartz compass]

then reason off the eigenvalues

$A=\mqty[0&1\1&0]=A^{-1}$

[stoic python]

Merosity

[uneven raptor]

This is necessary but not sufficient for what you want

[quartz compass]

any matrix with $A=A^-1$ satisfies $A^2=I$, so to make a bunch of examples, any kind of permutation matrix that permutes pairs of things would work. More trivially just -I works too

[stoic python]

Merosity

[uneven raptor]

Maybe you could share it

It might be possible to guess A^n in general then prove it by induction

[quartz compass]

why are they too advanced

you can still use them just be more verbose about it

teach them eigenvalues haha

you odn't have to take the trace

the eigenvalues of I are 1, so you'd need L1^n=1 and L2^n=1

you can demonstrate it with an eigenvector directly, multiply the matrix by it and show that the vector scales

and since it's an eigenvector the multiplication will just continue to scale it, so it will never be the identity

sort of, it's necessary but I don't know if it's sufficient

you're welcome

[quiet salmon]

can somebody help me understand this part of the proof:

To show that $U \cap W = {0}$, suppose $v \in U \cap W$, then there exists scalars...

I kinda see why it can make sense but I also don't see it

[stoic python]

texaspb :rainbow:

[wintry steppe]

remember what the basis of U and the definition of W are

[quiet salmon]

ooooh

right

makes sense.

okay thanks!

[uneven raptor]

strictly speaking $U$ is not a subspace of $U\oplus W$ but rather $U' = {(\bf{u},\bf{0}_W): \bf{u} \in U}$ is a subspace of $U\oplus W$. is this correct?

[stoic python]

giannis_money

[wintry steppe]

correct

[old yacht]

how do I find the two imaginary roots

@wintry steppe

[wintry steppe]

you pinged me out of nowhere for a question that belongs in #precalculus and not here, so i won't help

[old yacht]

my bad

I just read your role and it says honorable

So I assumed you were really smart

This question has been eating me alive for the past hour so just looking for some help

[old yacht]

Can you help me

I'm sorry

[wintry steppe]

go away

please stop pinging me

[old yacht]

I only pinged you once

Bruh

[old yacht]

Ohhhh

Huh

Replying counts as a ping?

[gray dust]

yep

[edgy shale]

How valid is this?

[hard drum]

No, you are essentially assuming the conclusion in saying B and B^-1 'cancel out' - playing around with it like this without actually using the definition of trace isn't really a viable strategy

(Moreover, B needn't be invertible)

You'll want to actually write what tr(AB) is in terms of the entries of A and B, or something similar

[young plinth]

I don't understand how to do this. I watched the lecture and didn't grasp the concept.

Would I multiply T(e_1) by (1,0) and T(e_2) by (0,1)

I don't know if I did that right xD

I was looking at the video and I guess this is how I would do it. not sure if it's correct tho

Oh that was correct 🤣

[wintry steppe]

there is no "vector multiplication" going on

how did your lecture define "standard matrix"?

you should copy that

your final answer should be a 2 x 2 matrix, not a vector

[young plinth]

It would be (2,0,0,4) right? I was asked for the first entry for row one and the second entry for row two.

[wintry steppe]

that matrix is not correct. if you're asked for those entries specifically (which you're not), then you would be correct that they are 2 and 4, respectively

[young plinth]

It's not this?

[wintry steppe]

it is not

[young plinth]

Is it flipped?

[wintry steppe]

no

[young plinth]

Like (0,2,4,0)

[wintry steppe]

if you didn't grasp the concept from your lecture, you should rewatch your lecture instead of trying random things

or google "standard matrix of linear transformation" and learn what it is

[young plinth]

I plan to do that later but right now I just need to complete my homework.

Which I already got the answer right for that question.

[wintry steppe]

that is definitely not the right answer lol

[young plinth]

I got it correct see

[wintry steppe]

thank you for now posting the full problem

[young plinth]

I just did what he did in the lecture lol

[wintry steppe]

whatever you did isn't correct, so that's probably not what your lecturer did

but you got the right answer so

if you're actually asked to compute the entire matrix instead of just some of its entries, you're going to have to learn what the standard matrix actually is

[young plinth]

Well he did something different in the lecture but I followed the steps for this problem kind of and end up getting it right xD

I plan to actually fully understand the concepts later whenever I'm not pressured on deadlines. It's important for me to get it all and understand it.

But I'm just trying to do my homework.

So I'm getting my grade so I don't have time to just research it all right now. Once my homework is complete though I definitely gonna watch some khan academy videos

Since he explain stuff well

[wintry steppe]

Geometric Question: If eigenvectors are meant to write an operator $A$ in terms of one basis as $Ax=\lambda x$, then supposing we have a full rank Eigenbasis, did we essentially linearize the operator $A$ in terms of new axes so that nonlinear things like rotation etc can be expressed as scaling?

[stoic python]

engineersanonymous

[wet stratus]

rotation is a linear operation. but yes in the new basis we can now apply the operator just by scaling the new axes

[vestal magnet]

Hey guys, some Jordan form related question -

Minimal polynomial - gives each of the eigenvalues Jordan max size.
Ker(T-idV) - how many Jordan blocks for each eigenvalue ?

Is this right? I might be using the wrong terms, no idea about the translation, sorry about that

[zinc timber]

should be dim ker(T-cI)

[vestal magnet]

where c is a scalar?

[zinc timber]

an eigenvalue

[humble elbow]

Any scalar works, if the scalar is not an eigenvalues then dim ker (T-cI) is 0.

[winter pond]

Could anyone help me with this?

[slender yarrow]

let's go in DM

[ripe compass]

why do they note matrix combinations AB when they first do the B transformation and then A transformation

[hard drum]

It's the same as with functions in general, doubt there's any good reason outside history

[restive raft]

Arabic style

[ripe compass]

wdym the same as functions

where in functions do they right to left

but ig I'll just have to accept it

and wiggle my brain around reading reverse

[leaden tide]

@ripe compass f(g(x)) means applying g, then f

ABx means applying B, then A

same thing

[ripe compass]

oh right

but that makes more sense in my head ig

[wet stratus]

for matrices it also works way better with the normal matrix multiplication when you want to apply them. when you apply AB to a vector x you first multiply y=Bx and then multiply Ay. so (AB)x=A(Bx)

compare that to AB meaning A first, so (AB)x=B(Ax)

[ripe compass]

thanks that helps

[hard drum]

If you do a quick sketch, it is possible to work out the vectors explicitly in coordinates

Or equivalently to find the angle between them

[whole thicket]

ok i will try thanks

but is my friends says right?

I will learn about how to make angle x axis and y axis

thanks anyway see ya

[hard drum]

Yeah 50 root 2 seems correct

[humble token]

Is this true?

Oops forgot a close parenthesis after -2abcd

[hard drum]

What was your working? I'm a bit confused where that's from

You can just write down the formula for the determinant and put absolute values about it

[humble token]

Well yes but I tried like solving for the area of the parallelogram (aka absolute value of determinant) by using vector lengths and fit products, wait I’ll send my reasoning

[whole thicket]

Many many thanks

[hard drum]

np

[humble token]

Bh being the area of the parallelogram

[quartz compass]

I'd try to expand that out and see if I could factor it into a perfect square - specifically you can expand (ad-bc)^2 and compare to it to save yourself the misery of trying to factor

[humble token]

Imma see what it’d give for a=d=1 and b=c=0

[quartz compass]

I just did b=c=0 leaving a,d unchanged and it gets you |ad| like you'd expect, so that's a good sign

[humble token]

Ah ok

So good so far

Imma try expanding the inside of the sqrt to see what I get

Imma simplify now

Hmm I don’t think it works when b and c are not 0 but I’m not quite sure

[quiet salmon]

what does an overdetermined system (most likely to have no solutions) says in regards to linear (in)dependency

does that mean it's linearly independent?

bc like either way I could choose coefficients equal 0 and it would be fine

ok

that should be obvious

of course it's LI

[hard drum]

Well needn't be linearly independent ig

[wet stratus]

well you could for example repeat a column multiple times and still get an overdetermined system

[hard drum]

(I assume you mean like the columns in the corresponding matrix?)

Yeah ^

[quiet salmon]

more equations than unknowns

if that helps

[wet stratus]

yeah. so ?

[quiet salmon]

it's always LI

[wet stratus]

if I put two columns of length 50 next to each other

[quiet salmon]

also I'm responding to potato

[wet stratus]

then I have 50 equations and 2 unknowns. and the columns are linearly dependent

[quiet salmon]

wait

how do you mean

[hard drum]

I meant what are you referring to as "linearly independent"

[wet stratus]

https://i.imgur.com/PEn1hIA.png

[hard drum]

But ye

[wet stratus]

paint skills ftw

[quiet salmon]

oh

hm

wait a minute

ok

but assume I'm choosing different coefficients for each vector

then if I have more equations than unknowns, this set of vectors is linearly independent

?

maybe that was written poorly

you just give the example

[wet stratus]

if you chose linearly independent columns then they are linearly independent, yes

[quiet salmon]

$S = {(4, -1, 2), (-4,10,2)}$

[stoic python]

texaspb :rainbow:

[quiet salmon]

show that they are linearly dependent (or not)

linearly dependent means that there is a nontrivial choice of coefficients a1, a1 such that

a1v1 + a2v2 = 0

correct?

[wet stratus]

yes

[quiet salmon]

ok then having the vectors be column vectors

it'd look something like this \

$a_{1}\begin{pmatrix} 4 \ -1 \ 2\end{pmatrix} + a_{2}\begin{pmatrix} -4 \ 10 \ 2\end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0\end{pmatrix}$

right?

[stoic python]

texaspb :rainbow:

[wet stratus]

yes

[quiet salmon]

then I'll have more equations than unknowns

so in this case

where every coefficient is different

I will always have a linearly independent set of vectors?

[wet stratus]

no

every coefficient being different is not enough

https://i.imgur.com/zeURX0K.png

whoops wrong pic

so every coefficient is different. but clearly the columns are linearly dependent

[quiet salmon]

true

ok I see it

uuuh

I feel like I'm forgetting something obvious

[quiet salmon]

in my example here

I came to the conclusion that the only choice for a_1 and a_2 that makes the expression equal 0 is if they are both equal 0 too

just solving the system

the thing is I was tryign to find a way to tell if it's linearly independent or not by just looking

[wet stratus]

yeah that's pretty hard

best way to check for linear independence is to solve the system

[quiet salmon]

yup

aight thanks

[copper hinge]

Sorry but not a lot people can’t tell what you need help with if you just post a screenshot with no context.

[serene solstice]

That's because I procrastinate at even writing a simple message. Duh.

Anyways....

[copper hinge]

Oh I didn’t know sorry.

[serene solstice]

No probs.
I still don't get the $X(tX - I_2) = t(tX - I_2) - X$ part.

[stoic python]

Jonathan Phan

[serene solstice]

Good people? A little help?

[dusky epoch]

aint this just elementary matrix algebra

wait, hold on

ah yes

$X^2 = tX - I$, so $X(tX - I) = tX^2 - X = t(tX - I) - X$

[stoic python]

Ann

[serene solstice]

"Ooooooohhhhhhhhh... I'm a fuckin' idiot!"

• Loud green-haired Irish man, c.2013

So is anyone else doing Titu Andresscu's Elementary Linear Algebra: A Problem-Solving Approach?

[serene solstice]

Please check if I did the following exercises correctly.

Haven't figured out 3c, though.

[wintry steppe]

latex your solutions if you want anyone to read them (at the very least, write them out cleanly and take clear, focused pictures)

it's also not clear what page of working corresponds to what problem

you also posted four problems at once. one at a time

[ripe compass]

what should I imagine with a linear image?

in my book it says "an image f: V->W is called a linear image if the addition and scalar multiplication is preserved. With this we mean f(u+v) = f(u) + f(v) and f(lamda * v) = lambda*f(v) for every u, v an element of V and all lambda an element of R"

I do not know what to imagine with this

[wet stratus]

is it called image or is that the word you chose to translate it as?

[ripe compass]

yea litteral translation

[wet stratus]

generally it's called linear mapping or linear transformation or linear function or stuff like that

[ripe compass]

one sec

this is the example they provide

[dusky epoch]

what language are you translating from

[ripe compass]

dutch

[dusky epoch]

and what's the word in dutch?

[ripe compass]

afbeelding

[wet stratus]

there is a video by 3b1b that gives some nice intuition about what linear maps are. check out his playlist about linear algebra. or just check out his channel in general

[hard drum]

ah function or map then Ruffy

[ripe compass]

what's the title

I can't find any with linear maps in teh title

[wet stratus]

he may have called it linear transformations. or maybe matrices. or linear functions. don't remember. don't wanna look it up

[ripe compass]

oh ty

linear transformation then

[dusky epoch]

yeah he called it linear transformations

[ripe compass]

but it looks like an interesting series guess I'll check it all out before starting to actually study

[real bone]

Anyone have an idea to prove that the sum is direct?

[wintry steppe]

i do, by appealing to the definition of "direct sum"

you probably want to use the second condition on the operators to show this, given that you (hopefully) used the first to show that there is a sum in the first place

[real bone]

i've already shown that this is a sum

i can't quite get to them having pairwise intersection 0

i think the idea is to somehow use $\pi_i\pi_j=0$ for $i\neq j$

[stoic python]

monkeman

[real bone]

so i get $\text{im}(\pi_j)\subseteq \ker(\pi_i),$

[stoic python]

monkeman

[real bone]

@wintry steppe i don't know how exactly to go further

i can apply this to show that $\text{im}(\pi_i)\cap \text{im}(\pi_j)\subseteq \text{im}(\pi_i)\cap \ker(\pi_i)$

[stoic python]

monkeman

[real bone]

but thats about it

[wintry steppe]

hmm

ignore what i deleted

the thing i was worried about was not a missing assumption, but follows from the ones you had

@real bone you should now try to show that im(\pi_i) \cap \ker(\pi_i) vanishes. to do this, try to show that \pi_i is a projection

you will have to use the \pi_i \pi_j = 0 condition and the \pi_1 + ... + \pi_k = id condition, somehow

hint: ||apply the first to the second||

thank god texit let me delete that

in the old days i would have used my message delete privileges

[dim epoch]

Isnt this already enough to argue that the images are disjoint

If the whole Image of some pi_i is contained in the kernels of all other p_j for j neq i then the intersection of that image with the images of pi_j can only be 0

And the intersections being 0 is whats missing for the sum to be direct

or am i missing some elementary fact here

[wintry steppe]

elaborate on why this follows

[dim epoch]

Found my mistake

nvm

[wintry steppe]

we both goofed

happens

[dim epoch]

ah wait lol now i see it how you get the projection property and then the statement

took me longer than it should have

[wintry steppe]

same

[real bone]

I don't think it actually does vanish right? Like there might be some v!=0 in the image that maps to 0 and therefore in the kernel

[wintry steppe]

\pi_i being a projection guarantees that this vanishes

for if v = \pi_i(w) is in the kernel of \pi_i, then 0 = \pi_i(v) = \pi_i^2(w) = \pi_i(w) = v

so now you must show that \pi_i^2 = \pi_i

[glacial terrace]

Why can we conclude the red line from the preceding statement?

[wintry steppe]

change of basis matrix. for you, P = [id]_C^B

this is exactly what they do

[real bone]

Wait so how would I go about proving it is a projection

[wintry steppe]

follow the hint i wrote

[wintry steppe]

.

[real bone]

Ohhh

I see

[wintry steppe]

the two conditions together give you what you want

[glacial terrace]

yeah I knew it

thaks

[glacial terrace]

For the (->) direction: Take B an ordered basis for F_n. Then [L_A]_B = A which is diagonalizable. So L_A is diagonalizable

[wintry steppe]

not just any B will work

you need to pick a specific B that makes [L_A]_B = A

but once you have done that, your argument works

and really, once you have [L_A]_B = A, the corollary follows immediately from what you just posted

[glacial terrace]

hum

[quiet salmon]

guys quick question

what is the dimension of a polynomial vector space

like for example P_2

will it be 3 or 2

the dimension

for the polynomials of the form $a_0 + a_1x + a_2x^{2}$

[stoic python]

texaspb :rainbow:

[wintry steppe]

what do you think it should be?

write down a basis of P_2 and tell me how many elements it has

[quiet salmon]

well it would have 3 elements

so yeah I guess the dim is 3

[wintry steppe]

there you go

[quiet salmon]

I was just confusing with the degree of it ig

[wintry steppe]

P_n will always have dimension n + 1, since you need to account for the constant term

[quiet salmon]

okay nice

ty

[brittle mortar]

can i ask someone to check my proof ? im not sure about the highlighted part, whether its a valid argument or not

[wet stratus]

yup that's valid

[hard drum]

Yup nice

[brittle mortar]

ok, thank you!

[glacial terrace]

TTerra, regarding the base for F^n. It needs to be B = ((1,0,..,0), (0,1,0,...,0),...,(0,0,...,0,1))

Then L_A(e1) = A for every element in the base

no

I mean, we will get the ith column for each action on the base

[wintry steppe]

this is correct (you could just say "standard basis")

[glacial terrace]

oh ok nice, thanks!

[brittle mortar]

btw, for the other side of this proof, can i just say that since U1 is contained in U2, their union is just equal to U2, which is a subspace so we're already done? (instead of taking the union and proving the properties of a subspace one by one the long way?)

[real bone]

lets say i have T,S: V -> V such that ker(T)=im(S). how do i construct S in terms of T?

TS=0 right?

hmm

[wintry steppe]

hmm

[real bone]

basically i'm trying to do part b

@wintry steppe you recall part a right?

[wintry steppe]

you just posted part (a)

but yes

[real bone]

yeah

well not just but like yeah

lol

sry

[wintry steppe]

hint/suggestion: ||T(T - id)(T + id) = 0||

[real bone]

yeah but am i not trying to find pi_1,pi_2,pi_3 such that im(pi_1)=ker(T)

im(pi_2)=ker(T-id)

and im(pi_3)=ker(T+id)?

because V_0=ker(T), V_1=ker(T-id) and V_2=ker(T+id)

@wintry steppe am i missing something here? because thats what it seems to imply

OOOOH

IM TROLLING

i'm so sorry

i legit didn't get ur hint for a hot second

wait no

hmm

[wintry steppe]

project onto ker(T)

few ways to do this. easiest is to pick a basis of ker T, extend to a basis of V, and chop off the components coming from the basis elements not in ker T

or pick an inner product and do orthogonal projection (admittedly, same as previous method with orthonormal bases)

[real bone]

wait so i'm assuming ur hint implied that the construction is something like pi_1=(T-id)(T+id)

pi_2=T(T+id)

and pi_3=T(T-id)

?

[wintry steppe]

i never said or tried to imply anything about what the \pi_i's are

[real bone]

oh

hmm

[wintry steppe]

it was just an observation that could be useful

[real bone]

oh

i see

do you have an idea of how to construct it

[rapid ivy]

what exactly does this

R^m and R^n mean?

doesnt that mean R is a vector of size m and n respectivly?

[real bone]

yes

[rapid ivy]

ok, thanks

[real bone]

wait

no

hold on no

[rapid ivy]

for future reference, how should i know that these are vectors?

oh ok

[real bone]

R^m is the set of all vectors of size m

and R^n is the set of all vectors of size n

[rapid ivy]

so its some m columns by some unknown

and n columns by some unknown

[real bone]

?

i didn't understand

[rapid ivy]

like there are m columns

since 1 column = a vector

well at least the trashy way im explaninng

[wintry steppe]

"R^n" means the set of vectors with n entries, each from R

$$\bR^n = {(x_1, \dots, x_n) : x_i \in \bR \text{ for each } i = 1, \dots, n}$$

[stoic python]

TTerra

[wintry steppe]

curly brackets are non-standard notation for vectors

[real bone]

oh oops

[rapid ivy]

not quite following

so its a collection of vectors?

[real bone]

ok just ignore my clumsy latexing

[wintry steppe]

it is. R^n is a set of vectors just as i wrote

it's a vector space, so you can speak of a linear transformation defined on it or to it

[rapid ivy]

got it

thank you

[rapid ivy]

can you explain this in simpiler terms, im still wrapping my head around all this

[mellow rampart]

but a vector of R^n is one that's cartesian product of n real numbers

[wintry steppe]

that's exactly what i wrote

[wintry steppe]

the "it's a vector space" part or "linear transformation..." part?

[mellow rampart]

like a vector in $\mathbb{R}^2$ might be $(1, \pi)$

[rapid ivy]

got it

[wintry steppe]

i am choosing not to distinguish between row and column vectors

[wintry steppe]

?

[stoic python]

_gmz

[rapid ivy]

i ask because im trying to wrap this around my head, but i dont think im grasping it correctly

if it perserves length, that means dot produt is same

[wintry steppe]

?

[rapid ivy]

so the kernel is the same from Rm to Rn?

[mellow rampart]

the kernel is the same?

no, the kernel is defined for a linear transformation

[wintry steppe]

what do you think the kernel of such a linear transformation is?

i will phrase my question to you precisely: "if L: R^m -> R^n is length preserving, what do you think ker(L) is?"

[rapid ivy]

hmm

[wintry steppe]

think about it a bit

if a vector x in R^m satisfies L(x) = 0, see if you can use the "length preserving" assumption to infer anything about x

[mellow rampart]

to clarify: this does not mean that for all x, L(x) = 0

[wintry steppe]

that's true, i need to be more precise with people like this

[rapid ivy]

my brain is mush, sorry

but hmm let me think some more

[mellow rampart]

@rapid ivy it's just that if the language isn't precise, it's hard to tell what the other person is saying

[wintry steppe]

i have edited it for more precision

[mellow rampart]

i'm sure they didn't mean it in a patronizing manner

[wintry steppe]

i didn't mean it in a patronizing way, but i can see how it came off that way

[wintry steppe]

if you can't figure it out in general, try finding some specific examples of length preserving linear transformations. compute their kernels, and see if you can make a guess for the general situation

it's always good to work with examples, and only then go to the theory, when you're stuck proving something general

[mellow rampart]

R^2 -> R^2 captures the essence of the problem already

[wintry steppe]

("like this" meant to refer to _gmz's message, not to belittle in any way. i apologize if it came off that way)

[rapid ivy]

jesus christ, i dont even know where to start

I think i can use the trasnpose is = the inverse

[wintry steppe]

this is where you can start, then

[mellow rampart]

do you have a geometric idea of what linear transformations from R^2 -> R^2 look like?

[rapid ivy]

like a rotation or flip across a line

[mellow rampart]

well, yes

now consider kernels

[rapid ivy]

ok

[real bone]

ok @wintry steppe for my problem can i write $\pi_1=1-T^2,$ $\pi_2=\frac{1}{2} T^2- \frac{1}{2}T$ and $\pi_3=\frac{1}{2} T^2+\frac{1}{2}T$?

the problem is i only get $\text{im}(\pi_1)\subseteq \ker(T)$ for example

[stoic python]

monkeman

[wintry steppe]

their sum isn't the identity map

[real bone]

oh i messed that up

1 sec

[stoic python]

monkeman

[real bone]

now the sum is identity map

[wintry steppe]

their sum is the identity and the pairwise products are zero, so you can apply part (a)

[real bone]

not quite

because the images aren't necessarily the complete kernel right?

[wintry steppe]

i said you can apply part (a), i didn't say it's the application of part (a) you need

that's something you'll have to check

[real bone]

oh

rip

how do i do this damn problem 💀

[wintry steppe]

[dim epoch]

It's a nice problem tho

[torn stag]

@rapid ivy Preserves length means $|Lx| = |x|$ for all $x \in \mathbb{R}^m$. It is a theorem that this implies $(Lx, Ly) = (x, y)$ for all $x, y \in \mathbb{R}^m$.

[stoic python]

IlIIllIIIlllIIIIllll

[mellow rampart]

are those intended to be angle brackets?

[torn stag]

the () mean dot product

[rapid ivy]

A^T A = I ?

[mellow rampart]

strange notation imo

[torn stag]

A^TA = I is equivalent to (Ax, Ay) = (x, y) for all x, y

[mellow rampart]

because V × V, defined correctly, is also a vector space

[rapid ivy]

so that means it orthogonal

[torn stag]

yes. The equivalence between length preserving and orthogonal is a theorem

[rapid ivy]

wait, so what step would i take after that once i recognize this?

im not seeing what this changes

[wintry steppe]

it's common enough and shouldn't cause confusion

[rapid ivy]

i learned what orthogonal matrices, but im unsure how i can see its kernel from that?

especially when its jut like R^m and R^n

my brain is truly pea

[wintry steppe]

suppose that x is in the kernel of A, which means that Ax = 0. can you use any properties of A to say anything about x?

[rapid ivy]

🥲

[rapid ivy]

that everything would map to 0?

wait

not necessarily

[wintry steppe]

im assuming that x is some vector which gets mapped to zero, im not saying that every vector gets mapped to zero

[rapid ivy]

hmm

since A is length perserving

then anything else would map to 0?

[wintry steppe]

i don't know what that means

[rapid ivy]

wait

[wintry steppe]

i want to know what the equation Ax = 0 would imply about x, if it were true for a given vector x

assume that x is some vector which A maps to zero (Ax = 0). what do you get to infer about x?

you must use the fact that A is length preserving at this step (or orthogonal)

[rapid ivy]

A is length preserving since its orthogonal

so the length of x is 0 as well

[wintry steppe]

YES

[rapid ivy]

🤯

[wintry steppe]

so x is zero, right?

do you see what this says about the kernel of A?

[rapid ivy]

ohhhhhhh

lmfao im an idiot

the kernel of A is 0

[wintry steppe]

exactly

[rapid ivy]

😹

im such an idiot

so just to clarify, with this question i should recognize that since orthogonal == length perserving, the only possible kernel is one where a vector maps to 0?

so im struggling to see how the idea of A^T A == (Ax,Ay) = (x,y) is useful in this case?

maybe im misunderstanding what this means?

[wintry steppe]

id rephrase that a little. instead of "orthogonal = length preserving" (which is true, but unnecessary), i just want to see how "length preserving" implies ker A = {0}

orthogonality comes later in the problem

(when you're asked to look at AA^T)

[rapid ivy]

thank you

[rapid ivy]

question, how would i find a transformation matrix for something like R^2 -> R^3

[gray dust]

google how to write a matrix of a linear map between finite dimension spaces (it requires a choice of basis of each space)

[mossy geyser]

Hi for this qn, my answer to a subspace S for V2 is : A line that lies in V2 which passes through the origin.

I'm wondering why the solution provided does not need to explicitly say that the line passes through the origin, and able to immediately point out that (0, 0, 0) is in the line through (1, -1, 1)?

[wintry steppe]

it's implied by the word "subspace" that it passes through the origin

the only line through a given point which is also a subspace is the one through that point and the origin

if you allow any line through your point, then you

1. will not have a subspace unless you pass through the origin
2. will have infinitely many lines; a unique one through each other point

it's just bad writing

[rapid ivy]

hi how tf do i get a vandermonde determinant

i have a 5x5 matrix and the number im getting is super large

[wintry steppe]

by using the formula for the determinant of a vandermonde matrix

in your example, the determinant you want should not be too large

[rapid ivy]

i tried a calculator and im getting like 288

but when i wrote it out

which is a very large number

so im so confused

[wintry steppe]

20321280 is a little larger than 288

[rapid ivy]

just a little

😭

[wintry steppe]

you're using the formula incorrectly

it should look like (5 - 4)(5 - 3)(5 - 2)(5 - 1) and so on

the determinant of this guy is $$\prod_{1 \leq i < j \leq n}(x_j - x_i),$$ and for your example, $n = 5$ and $x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4, x_5 = 5$

[stoic python]

TTerra

[rapid ivy]

not quite following

[wintry steppe]

(technically your matrix is the transpose of the one i've copypasted here, but that doesnt change the determinant)

[rapid ivy]

so (5-4) (5-3) (5-2) (5-1) (4-3) etc

?

[wintry steppe]

yes

[rapid ivy]

i thought it was by row?

[wintry steppe]

you're probably interpreting the formula incorrectly

the matrix you posted is a "column vandermonde" matrix, so to speak

transpose it to get a "row vandermonde" matrix, a matrix of the form i posted above

then you can apply the formula i wrote

[rapid ivy]

wait youre supposed to transpose first?

🤯

also 1 row is going directly down right?

or am i switching the naming

oof

[wintry steppe]

maybe i should rephrase

what is your definition of a vandermonde matrix, and what is your formula for its determinant?

just in case we're thinking of different things

[fierce relic]

Hi I don't want to interrupt but I can't tell if you guys are done so i'll just post it

[wintry steppe]

just posting it would be interrupting, but i don't particularly care

they seem to have disappeared, so go ahead

[fierce relic]

So I'm trying to learn about the pentagram map

but first i'm just trying to check if its a linear transformation

So i've started by trying to define a matrix equation by the intersection of the lines passing through every other vertex

and I know that every matrix defines a linear transformation

but does it follow that this matrix equation particularly defines a mapping from the point where the line originates to the intersection

I'm kind of confused

and I don't know how to go about looking at it other than this way

Everything I've found looks at it as a projective transformation

is that the same in this specific context?

And if it is then should i be looking at different matrices because it really feels that i should be

[dusky epoch]

the pentagram map?

[fierce relic]

so like

If you have a pentagon

and you connect all the vertices

the pentagon in the center is created by the intersections

and if you map the vertices from the outside one to the inside one

I mean it should be a linear transformation

I just have no idea on how to show that it is

Nvm I think you need more than lin alg for this lmfao

[dusty walrus]

Can someone describe to me what a 3x3 matrix transform is. I need to transform some data I have, but literally know nothing about matrix transforms. Can it be explained in layman's terms please?

[limber sierra]

a 3x3 matrix transform is a certain type of "well-behaved" function that maps ordered triplets to ordered triplets

[dusty walrus]

These values will apparently how me transform one color metric to another.

The values I have are the following

Red: 0.4686329572

[limber sierra]

by "ordered triplets" i mean lists of numbers of the form (a, b, c)

[dusty walrus]

Green: 0.4921898868

Blue: 0.4709272159

[limber sierra]

it probably just wants you to do matrix multiplication then?

[dusty walrus]

Yeah i'm not sure

It just says these values will help me transform my units to a different color metric.

[limber sierra]

,w {{0.9806, -0.0348, -0.0073}, {0.0065, 1.1191, -0.0601}, {-0.0132, 0.0017, 1.0297}} * {{0.4686329572}, {0.4921898868}, {0.4709272159}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Namington#1983

[dusty walrus]

Did you multiple the sum for each dimension per channel

being Red = SUM(0.9806, -0.0348, -0.0073) * 0.4686329572

[limber sierra]

not quite

so for computing the first entry

,calc (0.9806 * 0.4686329572) + (-0.0384 * 0.4921898868) + (-0.0073 * 0.4709272159)

[stoic python]

Result:

0.43720361750113

[limber sierra]

the second entry uses coefficients from the second row of the 3x3 matrix instead of the first row

the idea is that this "weights" each parameter

the first entry (red) of the final vector is based on a "weighted sum" of the entries of the input vector

you'll note that the red entry has the highest weight

0.9806

because presumably "red" in your different colour metrics looks fairly similar

the blue and green values have very low weights relative to red's

-0.0384 and -0.0073

[dusty walrus]

Oh, so you're basically going Transformed Red = (0.9806 * Red value) + (-0.0384 * Green value) + (-0.0073 * Blue value)

[limber sierra]

so red's weight ends up contributing "the most" to the final sum

yes.

the second row represents the weighting for computing the green value, and the third row represents the weighting for computing the blue value

same process, but now instead of red having the largest number, it's green or blue (respectively)

it's presumably just a matter of different standards of what colours are associated with what numeric values

[dusty walrus]

Interesting okay

Specifically this is to transform density metrics

optical density

[limber sierra]

this is just my assumption off of little context though, but it's the only thing i can see really making sense

and it produces sensible results

[dusty walrus]

Is this the same thing as a linear transform?

[limber sierra]

yes

matrix multiplication is notation for representing linear transformations

(between finite dimensional vector spaces, blah blah blah a bunch of asterisks that dont matter)

[dusty walrus]

I see, yeah I know they stated it was for a linear transform so you're probably on the money with it

I appreciate your help man, thank you

[mossy geyser]

Thank you for the detailed expln. Will my answer that: A line that lies in V2 which passes through the origin being a subspace S of v2 suffice then? Since any combination of vectors in that line will still lie in V2 ?

[serene solstice]

The eigenvalues of A are the sol'ns of the eq'n $t^2 - 2\cos(x) + \cos(x)^2$. How do I continue?

[stoic python]

Jonathan Phan

[dusky epoch]

i think you forgot some things in the equation

i think maybe you meant t^2 - 2cos(x)t + cos^2(x) = 0?

[quiet salmon]

any set of vectors that is linearly independent on R^n is a basis for R^n?

saw a guy say this in a stack overflow post

is it true?

[tribal willow]

yes

[quartz compass]

not any set, the set has to have n elements

[tribal willow]

oh wait yes oops

i made that assumption

[quartz compass]

I nearly did too

[tribal willow]

a set of n vectors which are lin indep span Rn

[quiet salmon]

cool

[tribal willow]

assuming vec space defined over R

idk if this applies for inf dim vec spaces tho

well i guess not cos we cant really say n vectors lol

[willow crag]

If i make a line with x = y*m + b, so that the intercept bis in the x-axis, and the slope m = (x1 - x0)/(y1 -y0) should the slope then not be equal to the angle between the vector (x1 - x0, y1 - y0) (green line) and a vector (0,1) (red line)?

I'm asking since I get some decimals off, but too much to be floating point precision

[wet stratus]

then it would only be possible to have slopes between -pi/2 and pi/2

does that seem reasonable?

or -90 and 90 degrees if you prefer that

[willow crag]

Well the slop can still be negative in value, that would allow it to be full radians, right?

slope*

[wet stratus]

well but still you couldn't for example have a slope of 1000

that should seem wrong

[willow crag]

Ah, I see what you mean, the slope could be any number?

But isn't it always a ratio

Ooh no

It's not rip

Right! Okay, so arctan(m) gives me theta

[serene solstice]

Ah yes. Thanks for the correction.
So we just solve this thing normally, right?

[wet stratus]

yes

you should probably notice that this has a very nice form

before just blindly applying the quadratic formula

[grave garden]

Hii guys

When they say "eigenvalues with multiplicity" generally this is algebraic multiplicity right ?

Sth like so

Also for this one

I can understand that we should have blocks of matrix on the diagonal

But I dunno how we get the following result

[wintry steppe]

teacher says the answer is -1 but I think it doesn't have any limit
because the limit of floor(sin(3π/2)) + ε is 0 and there is no limit for sin(3π/2) - ε (as sin is never less than -1) so the right and the left limit differ from eachother so it doesn't have limit

which is the correct answer?

this isn't linear algebra, go to #real-complex-analysis

[wild rapids]

while doing matrix elimination(like making upper trangular/lower trangular matrix), I mostly gets worst case scenario.. Can someone suggest me a proper way to do elimination.

[dapper jolt]

https://cdn.discordapp.com/attachments/879498465510572082/996857975153098934/image0.jpg

https://cdn.discordapp.com/attachments/879498465510572082/996858066953842781/image0.jpg

having issues with part d

if i’m understanding this correctly, all this matrix ‘P’ does is take the coordinate vector (a1, a2, a3) with respect to the ordered basis a to its equivalent coordinate vector with respect to the ordered basis a’

but like, X’ is apparently the coordinate matrix relative to a’? so wouldn’t X = PX’ mean that we are going from a’ to a?

i just don’t really understand how this works (textbook is hoffman & kunze btw)

[odd turtle]

is the z and y switched in the sum?

or does it not matter>

[dapper jolt]

variables dont matter, all its saying is that the first two coordinates are the same

[gray dust]

its important to read the form not the exact letters used

[glacial terrace]

is there any reason why we compute the eigenvalues with det(A-xI) = 0 and the polynomial characteristic is defined as det(xI - A) ?

Is it just because det(xI - A) makes the principal coefficient positive?

[gray dust]

det(A-xI) & det(xI-A) are off by a factor of (-1)^n

so u can actually define the charpoly as either one

hence use either to find eigenvalues

if u like monic polynomials then u can use det(xI-A)

[glacial terrace]

indeed

[polar marsh]

can two different bases of a vectorspace have the same linear combination for a vector?

[wintry steppe]

sure, the zero vector looks the same in every basis of a vector space

[polar marsh]

and except de zero vector?

[wintry steppe]

sure, if the two bases share a vector, that vector is gonna look the same in each

[quartz compass]

you can write it as an equation B_1 v = B_2 v and this means (B_1 - B_2)v = 0 so either v=0 or the columns of B_1-B_2 are linearly dependent

[polar marsh]

How can i prove that for a given n×n matrix we can find a linear map f wherfore this given matrix is the matrix representation of this map f, that is dependant on the basis we chose, or in other words why isn't it enough if i have a given matrix to find only one unique linear map?

I am sorry for my english, it is not my native language

[wintry steppe]

if $A$ is an $n \times n$ matrix and $\beta = (v_1,\dots,v_n)$ is an ordered basis of a vector space $V$, then you can define a map $L_{A, \beta}\colon V \to V$ by mapping $V$ isomorphically to $F^n$ using $\beta$, performing matrix multiplication, and then going back to $V$ using the first map's inverse.

[stoic python]

TTerra

[dapper jolt]

what does the T_v_k stuff mean here? the notation here seems confusing

[wintry steppe]

in more precise terms, using the notation of the previous post, let $\varphi\colon V \to F^n$ be the map $$\varphi(a_1v_1 + \cdots + a_nv_n) = (a_1, \dots, a_n), \qquad a_1,\dots,a_n\in F.$$ this is an isomorphism. set $$L_{A, \beta}(v) = \varphi^{-1}(A \cdot \varphi(v)), \qquad v \in V.,$$ where $A \cdot \varphi(v)$ is ordinary matrix multiplication. with respect to the basis $\beta$, the linear map $L_{A, \beta}$ has matrix representation exactly what you'd expect: $[L_{A, \beta}]_\beta = A$.

that's how to precisely construct it

on to the second part of the question, if you pick a different basis, the map \varphi is gonna change

(i probably should have used some notation that emphasizes that \varphi depends on the choice of basis, but whatever)

it should become clear if you think about this construction a little bit that picking a different basis will (in general) give you a different linear map

[wintry steppe]

Tv_k means T applied to v_k

[dapper jolt]

right

so A_1, k is just the row vector of the matrix A that the v1 basis vector is multiplied by in the linear transformation?

[wintry steppe]

A_{1, k} is not a row vector

it is a scalar, defined exactly as you say

[dapper jolt]

so it’s the scalar that’s multiplied by the v1 basis vector?

[wintry steppe]

sure

[dapper jolt]

is there a better way to describe it…

[wintry steppe]

i think that the way that the author's described it here is already pretty clear

maybe i can elaborate on it

[dapper jolt]

yea i’d appreciate it

[polar marsh]

Thx for the answer

[wintry steppe]

$Tv_k$ is an element of $V$, so it's going to be a linear combination of $v_1,\dots,v_n$, i.e., there are unique scalars $A_{1,k},\dots,A_{n,k}$ such that $$Tv_k = A_{1,k}v_1 + \cdots + A_{n,k}v_n.$$ that's how they're defined

[stoic python]

TTerra

[zealous mural]

I just join this server and when I saw the ''linear algebra'' I was hype. But, can someone tell me : how can we find RREF (reduced row echelon form).

[wintry steppe]

my latex vanished

in more precise terms, using the notation of the previous post, let $\varphi\colon V \to F^n$ be the map $$\varphi(a_1v_1 + \cdots + a_nv_n) = (a_1, \dots, a_n), \qquad a_1,\dots,a_n\in F.$$ this is an isomorphism. set $$L_{A, \beta}(v) = \varphi^{-1}(A \cdot \varphi(v)), \qquad v \in V.,$$ where $A \cdot \varphi(v)$ is ordinary matrix multiplication. with respect to the basis $\beta$, the linear map $L_{A, \beta}$ has matrix representation exactly what you'd expect: $[L_{A, \beta}]_\beta = A$.

[stoic python]

TTerra

[wintry steppe]

@polar marsh this disappeared for some reason, i might have accidentally deleted it

putting it back up if you need to look

[wintry steppe]

google

i don't think anyone wants to walk through the steps of gaussian elimination

[zealous mural]

Ok

[dapper jolt]

i see this, gonna try and understand it - ty!

[wintry steppe]

it's not that complicated

all that's happening is "vectors in V get representations in the basis"

[zealous mural]

and by the way, to find the Null space of matrix, do we need to make, in a way or another, a substraction ?

[ripe totem]

Someone can correct me if I'm wrong, but I think solving the homogeneous system AX=0 of matrix A should give you its Null space. That is, reducing the augmented matrix A|0 to RREF to solve.

[mellow rampart]

you're correct

[dusky epoch]

you are definitionally correct

[mellow rampart]

ofc you don't need to reduce to rref to know null space, but that is a way

[lucid glacier]

You also don't need to consider an augmented matrix. Just rref

Row operations preserve the kernel

[mellow rampart]

ofc for Ax = 0 the augmented part of the augmented matrix doesn't matter anyways

[polar marsh]

I don't really get this

[dusky epoch]

the highlighted paragraph?

[polar marsh]

Do they mean if i chose the same basis for in input as output? Because if not, then i can choose a linear map that takes b_i as an input with (b_1...b_n) a basis in V and then an out put c_i from another basis (c1.... cn) without having the identity function

[polar marsh]

Yes

[dusky epoch]

yes, they mean the same basis for input and output

[polar marsh]

Oh okay

[dusky epoch]

the identity is a map from R^n to itself after all

[polar marsh]

So this would mean that if i have the identity matrix , i don't necessarily have the identity function

?

[dusky epoch]

??

if you "have" the identity matrix then you have the identity matrix

[polar marsh]

The identity matrix as a representation of a linear map

Between two vectorspaces

[dusky epoch]

between two DIFFERENT vector spaces?

yeah, it won't represent the identity

[polar marsh]

No the same

[dusky epoch]

so from a vector space to itself

[polar marsh]

Yes

[dusky epoch]

why'd you say "between two vector spaces" then...

but anyway

UNLESS STATED OTHERWISE, the input basis and output basis are understood to be the same

[polar marsh]

Yes i get this so thats why i asked if i would choose two different bases in V i would have another linear map

[polar marsh]

My question was that if i give you the identity matrix as a representation of a linear map f, then f isn't necessarily the identity function since i can chose two different bases in V.

[dusky epoch]

if you want to be a smartass

then sure

[polar marsh]

Okay thank you

[little vigil]

Applied linear algebra: The decoupling principle Anybody familiar with the book? Will i be fine with just a basic linear algebra course before attempting this book?

[dapper crypt]

Anyone got a good problem sheet of questions where you factorise matrices? Such as $X_1(X_1-X_2)+(X_1-X_2)X_2$ is competing the square

[stoic python]

Max..

[serene solstice]

This one hurt my brain to calculate its eigenvalues. I know that it's a rotation, but a rotation of what?

[quartz compass]

I suppose in the eigenbasis you can think of it as a rotation in the complex plane in the components by x radians, idk that's sort of like 4 dimensional to think about

[dapper jolt]

can i get some help on 2 and 3b?

i was able to prove the reverse implication for 3b but i can't actually solve the problem

[wet stratus]

for 2 first find any basis oh H. then apply gram schmidt

[dusky epoch]

@dapper jolt would you like to have (b) be broken down into somewhat more elementary steps?

[dapper jolt]

yeah that'd be nice

[dusky epoch]

Let v ∈ V \ {0}.
(i) Prove that Jv ≠ 0.
(ii) Prove that Jv is not a scalar multiple of v.
(iii) Hence prove that {v, Jv} is a basis of V.
(iv) Hence prove that (v, Jv) is exactly the ordered basis you are looking for.

[prisma socket]

if A and B are singular matrices and A+B is also singular and AB=BA then does it mean A^2+B^2 is singular and what about A^3+B^3 and what about A^4 + B^4 ? (A and B are matrices in size of nxn )

[molten pilot]

regarding the proof of (b)

I don't get why we need row and column operations

can't we just put P=[B] and Q=[C] since multiplication by a matix can be viewed as a linear transformation? (B and C are from the proof of (a))

[mellow rampart]

i think so as well

[molten pilot]

nvm

the book acknowledges this as well

[mellow rampart]

but they probably want the proof of (b) to be from an entirely different angle, that doesn't rely on the direct connection b/w linear transformations and matrices

[molten pilot]

sigh

[mellow rampart]

anyways the vector space proof is a lot cleaner anyways so :/

[molten pilot]

haha

indeed

[molten pilot]

in (b)

is the las one a typo?

wait no

lol

[wintry steppe]

[molten pilot]

the 0 vector of two different vector spaces is technically a different object, right?

[slender yarrow]

sounds like T is V -> V here

@molten pilot

[molten pilot]

no, I know

I just wanted to make sure lel

[slender yarrow]

well yea the 0 depends on the vector space we're talking about

[lucid glacier]

Well for subspaces the 0 has to be the same

By uniqueness of the additive identity

[wintry steppe]

Hi, just seeking clarification. after checking out Gilbert Strang's linear algebra course for a little bit, is it right to assume that matrices and vectors can be assumed to anything and requires context?
like for eg```
| 1 5 |
| 2 4 |

[dusky epoch]

i think you're overthinking it

a matrix is a matrix

if it represents something then it'll be made clear in context

[wet stratus]

$$\begin{pmatrix} 1 & 5 \ 2 & 4 \end{pmatrix}$$ is a matrix. $$\begin{pmatrix} 1 \ 2 \end{pmatrix}, \begin{pmatrix} 5 \ 4 \end{pmatrix}$$ are two vectors. \begin{align*} 1x+5y&= 0 \ 2x+4y&=0\end{align*} is a linear system of equations with two equations and variables. these are all different things

there is an obvious way to convert between them. but at first they are all different things

[stoic python]

Denascite

[wet stratus]

sometimes we don't make a huge distinction between any of those because there is such an obvious way to translate between all them. but they are different

[wintry steppe]

ahh alright, thanks for the answer. I will think in context

[wintry steppe]

ahhh I see, I guess I got confused when we look at matrix as a column space but then proceed with row elimination.

btw, is that regex up there? for the math drawings, thats so cool

[wet stratus]

That's latex

[wintry steppe]

ahhh, looks so complex but the result is really cool

[wet stratus]

Yeah it has a steep learning curve

[glacial terrace]

I don't understand why we get (4,-3)

[wet stratus]

do you see why (4,-3) is a solution to the system?

[glacial terrace]

yeah but suppose there's something more abstract

suppose we have an eigenvalue t for some linear operator T and want to find the eigenvector

after doing the matrix representation with the basis

we find v the respective eigenvector for t by Tv = tv

what's the algorithm then?

[wet stratus]

well we solve the linear system (T-tI)v=0

if we already know t

and that's just a linear system which we solve in the usual way

for example by row reduction

[glacial terrace]

oh ok I see

thank you

[young plinth]

How would I solve this? I just need a start.