linear-algebra

[spare widget]

with the isomorphism 1 <-> (1,0,0), x <-> (0,1,0), x^2 <-> (0,0,1) you can actually write this as:

$\begin{bmatrix} 1 & -1 & 2 \ 0 & 1 & -3 \ 0 &0 & 1 \end{bmatrix}$

[stoic python]

criver

[twilit minnow]

ok but how do u connect this w the coordinates of P

[spare widget]

The above matrix transforms the coefs of a poly from B2 to B1

I derived it from these

So to go in thereverse directikn you have to compute the inverse

then do the samefor B3

[twilit minnow]

B1 would just be P right

[spare widget]

Finally you can compute (B1->B3) * (B2->B1)

[spare widget]

no

you have

f(x) = a + b * x + c * x^2

And you want f(1) =4, f(2) = 3, f(3) = 0

You can make linear system from that for a,b,c

Then (a,b,c) would be the coords of P in B1

[twilit minnow]

a = 3

b = 2

c = -1

in that case

[spare widget]

provided the above is correct

You have to transform (a,b,c) to B2

[spare widget]

You have to multiply it by the inverse of this matrix (the B2->B1 matrix)

Then derive the B3->B1 matrix,invert it and do the same

Finally compute the B2 ->B3 matrix as (B1 -> B3) * (B2 ->B1)

You will have all of those matrices from the previous steps

[spare widget]

This is the B2->B1 one

If you have

$\vec{f}i = \sum_j S{ji} \vec{e}_j$

[stoic python]

criver

[twilit minnow]

im unsure of how you got this matrix

[spare widget]

Where fi are the basis vectors in one of the bases and ej in the other, then S is the change of basis matrix from F to E

[spare widget]

#linear-algebra message

[spare widget]

From this

It's the S matrix

In this specific case f1 = 1, f2 = x-1, f3 = (x-1)(x-2)

e1 = 1, e2 = x, e3 = x^2

e.g. f3 = (x-1)(x-2) = 2 -3x + x^2 = 2 * e1 + (-3) * e2 + 1 * e3

That's how I derived the equations and the corresponding coefficients of S

[twilit minnow]

couldnt we compute (B3 -> B1) * (B1 -> B2)

[spare widget]

no

You want something that goes from B2 to B3

What you wrote maps from B1 to B2 first

And then maps from B3 to B1, but this makes no sense, since coords are in B2 after the first step, not in B3

You want B2 -> B3

So one such path is B2 -> B1 ->B3

Then (B1->B3) * (B2->B1) = (B3->B1)^{-1} * (B2->B1)

[twilit minnow]

MAKES SENSE

thank you criver

[lusty pumice]

Can someone please explain why this is true

[spice storm]

your steve jobs, you should think

[lusty pumice]

You're*

[tacit pelican]

i'm trying to prove this

and i'm a bit unsure

of whether i need to prove the alpha = 1 case

like

do i really need to do it?

or is what i did up to the Suppose alpha=1

enough

also, if it isn't, how would i go about this?

i've tried doing some stuff but none of them worked

the D(v_1, ..) stuff is just the determinant of a system of vectors btw

[fleet sun]

you need to handle alpha = 1 separately

because in that case, in (alpha - 1) D(v1, ..., vn) = 0, it could just be (alpha - 1) making the whole LHS zero

you have to show that in fact D(v1, ..., vn) = 0 itself

[tacit pelican]

i see

[tacit pelican]

could you give me a tiny hint on how to do this?

i have kinda no idea rn

these r the properties i have rn

that i can use

[fleet sun]

since v_j is the 0 vector, v_j + v_j = v_j is true

in fact, you could bypass the earlier part about alpha =/= 1 this way

$D(v_1, \ldots, v_j, \ldots, v_n) = D(v_1, \ldots, v_j + v_j, \ldots, v_n)$

[stoic python]

ManifoldCuriosity

[oblique prairie]

you could just let the jth column be zero and apply linearity, so you’d get detA = detA+detA

[oblique prairie]

dang it

[tacit pelican]

oh shit

bruh my whole proof was kinda useless then

lol

thanks anyway i got it now

[fleet sun]

it happens

[oblique prairie]

[fleet sun]

you basically found the right idea in the earlier part

[tacit pelican]

yeah

[vital drum]

i feel like any linear map and its double transpose should have the same ranks and nullities but i cant prove it

of fin-dim vector spaces

[slow scroll]

rank(A) = rank(A^T) so rank(A^T) = rank((A^T)^T) = rank(A)

[fleet sun]

nice

what about it used finite dimensionality?

rank(A) = rank(A^T) I assume

[slow scroll]

maybe. I haven't thought about the proof of that in a long time

[vital drum]

yea, i'm trying to not use that

but i found another way

[fleet sun]

for fin-dim, $(V^{})^{}$ is canonically isomorphic to $V$

[stoic python]

ManifoldCuriosity

[fleet sun]

if $M$ is a matrix, $L_M : V \rightarrow V$ its corresponding linear transformation

[stoic python]

ManifoldCuriosity

[fleet sun]

its transpose is a map $L_M^T : V^{} \rightarrow V^{}$

[stoic python]

ManifoldCuriosity

[fleet sun]

wait

I probably shouldn't be assuming the matrix is square

and I probably shouldn't even be starting with a matrix

idk what I'm doing, I gotta go grade

[vital drum]

good grading

[slow scroll]

what is your way?

[vital drum]

i used this exercise in friedberg

i didn't wanna use the fact about ranks of matrices that you mentioned above because ranks of matrices haven't been defined at the point of an exercise i'm trying do

and i thought i needed the fact a linear map and its double transpose have the same ranks and nullities to solve it

[slow scroll]

yeah this is the right way to do it

[void path]

trying to calculate the eignvalues of the matrix. I got to here (finding the determinate and subtracting lambda).

From here, i am foiling, getting a quadratic, then im doing the quadratic formula which results in a NASTY 2 solutions. Is this right? The khanacademy guy said something about "sum and product" and seemed to do his much easier and i feel im missing something (my problem isnt identical to his so maybe his has a special property)

[wintry steppe]

the results should be pretty simple

you're probably solving the quadratic wrong

[void path]

ill continue (on my picture) if that helps

[wintry steppe]

"sum and product" probably refers to the method of solving a quadratic x^2 + bx + c by writing it as (x - p)(x - q) = x^2 - (p + q)x + pq and finding p, q by solving p + q = -b (sum) and pq = c (product)

[void path]

im getting very strange root numbers, and the khanacademy guy got some very easy numbers. This is question 1 on my homework and it just feels weird that they wouldnt give me one that produces a nice easy result so im checking if its correct

those would be my eigenvalues?

[wintry steppe]

you should get 3 - 3λ - λ + λ^2 - 8

[void path]

i see 2 major errors

my bad

its 3:30am here lol

[wintry steppe]

and you wrote 3 - 8 = -11

go to sleep

[void path]

Errors aside, that is how you calculate that though, right? Like, thats the method?

[wintry steppe]

to find the eigenvalues of a matrix? yeah

[void path]

last dumb question before i go, this always messes me up bc i dont work with actual math a lot. (-x)*(-x) = -x^2? or x^2 (because negative times negative).

[wintry steppe]

x^2

[fringe burrow]

Are we agree that the following statement is false?: Real symmetric matrix has exactly n linearly indépendant eigenvectors

[lone flume]

A real symmetric matrix is diagonalizable

[languid sphinx]

I don't think symmetry has anything to do with full rank

[lone flume]

I'm always thinking about spd matrices

[languid sphinx]

psd?

Never seen people call them spd

[fringe burrow]

I mean real Sym matrix has n linearly independent eigenvectors but the key question is in “exactness”

[lone flume]

symmetric positive definite

[languid sphinx]

oh

[wintry steppe]

good luck finding more than n of them lmao

assuming the matrix is n x n

[fringe burrow]

If v is eigenvector a*v as well no? a real number

[lavish jewel]

is the matrix size n x n?

if so, it's true

[languid sphinx]

I don't understand

[fringe burrow]

Yes nxn

[wintry steppe]

oh, i see

[languid sphinx]

$$\begin{bmatrix}1& 1\1 & 1\end{bmatrix}$$ is real and symmetric

[stoic python]

pepper

[languid sphinx]

It's clear how many independent eigenvectors it has

[lavish jewel]

yeah, 2

[languid sphinx]

wait

[fringe burrow]

Yes but if family {v_1,v_2} is linearly independent av_1 is eigenvector bv_2 as well and {av_1,bv_2} is still linearly independent set

[lone flume]

Wait, rank doesn't matter

[lavish jewel]

it doesn't indeed

[languid sphinx]

oops

[lavish jewel]

what definition of exactness are you looking for

[wintry steppe]

now it feels more like a question about language

[fringe burrow]

Like there are no exactly n linearly independent eigenvectors no?

[languid sphinx]

Why or why not

[fringe burrow]

I can take multiplication by scalar and I still will have a lin indépendant family

Of eigenvectors

[languid sphinx]

While I did a bad digression, someone else has already said why the answer is true

[lavish jewel]

we'd need to see the definition of eigenvectors in your book, but i'd still be included to say it's true

i mean which convention to take regarding the scaling

[fringe burrow]

We defined it in class as v is eigenvector if there is some scalar lambda such that f(v)=lambda v

[lavish jewel]

but one usually considers scaled versions of the same eigenvector as being equivalent

[languid sphinx]

Your question already asks for independent vectors

[fringe burrow]

Juste knowing how tricky lin alg exam will be in preparing myself for this such of questions xd

[languid sphinx]

no need here since linear independence should take care of that aspect

[fringe burrow]

Alright thank you for your answers

[lavish jewel]

anyway yeah, i would say the point is to establish that real sym matrices are diagonalizable

[fringe burrow]

Yeah that spectral theorem we proved it as well

[lavish jewel]

as for the earlier discussion, i usually see symmetric positive denitive, not positive symmetric definite

[languid sphinx]

I was referring to positive semidefinite

[lavish jewel]

one usually has to specify symmetry as well, but that's a matter for another day

[scenic glen]

How would I approach this? If [1 2]_T was in the null space I'd understand how to approach it, since I'd set up this equation:

| a c | | 1 | = | 0 |
| b d | | 2 |   | 0 |

but since it's in the column space I'd have to do:

| a c | | x | = | 1 |
| b d | | y |   | 2 |

and not sure where to go from there.

[lavish jewel]

consider the reason why the dimension of W is 2

[prime snow]

Are all matrices in the Special Linear Group diagonalisable?

[dusky epoch]

no

$\bmqty{1 & 1 \ 0 & 1} \in \mathrm{SL}_2$ yet it is not diagonalizable

[stoic python]

Ann

[dusky epoch]

@prime snow

[prime snow]

Brilliant thanks

[pure crypt]

Can someone explain to me why this statment is true? Given that A is an n by n matrix. I thought the signs are supposed to alternate for determinants.

[quartz compass]

sure you could write it that way

the signs are inside the a_i

[lavish jewel]

as written here, all your image says is that the determinant of A - lambda I is a polynomial of degree n. the specifics of how the polynomial is obtained are to be found elsewhere, so the coefficients a_i could be anything, as merosity says

[twilit minnow]

i have one left inverse of a matrix

how would i find a second

for the same matrix

could i use qr factorization

[placid snow]

Looking for a proof check for the following, please:

One small mistake: “Clearly $s_i \in S_i$…” should be “Clearly $s_i \in span(S_i)$…”

[stoic python]

LosAngeles

[slate hatch]

u . v = | | u | | | | v | | cos(v)

Anybody that can tell where this formula comes from?

[placid snow]

It’s got a nice geometric interpretation; look up dot product as it pertains to vectors (as arrows)

[slate hatch]

Got it. Thanks 🙂

[placid snow]

Perhaps more to answer “where it came from,” sometimes a measure of how “aligned” two vectors are is of use. That’s one thing a dot product can indicate (and why that definition was adopted and works well)

[sage stirrup]

What is the second part of 1a referring to by uv plane?

im assuming just rewrite A where u and v equal values such that detA=0 ?

<@&286206848099549185>

[gray dust]

pls wait 15min before pinging helpers

its a 2d plane, u as the x axis, v as the y axis

[placid snow]

Oh I didn’t know I could ping helpers

<@&286206848099549185> I have a proof posted above, can anyone confirm please?

[sage stirrup]

sorry

[sage stirrup]

right, thanks

[gray pond]

I might not know what I’m talking about but my first guess is that V is over a finite field (assuming V finite dim)? Suppose not. Let ${w_1, \ldots, w_n }$ be a basis of $W$. The for all $\alpha \neq 0 \in \mathbb{F}$ we can form a distinct basis ${ w_1, \ldots , \alpha w_n }$.

[stoic python]

casualpolemic

[placid snow]

Wow there we go

[sleek helm]

also you should ping me bc i have this channel muted lol

[placid snow]

Ah ok, thank you

[ember pagoda]

I got to know that the eigenvectors with the eigenvalues zero are the ones that are present in the null space. And, we know every subspace is a null space of some matrix so, can I say that every subspace is an eigenvector with eigenvalue 0?

[undone fable]

Hi

I'm taking linear algebra and had a question

I found AtA and AAt and neither were identity matrices

so for part b

would none of the columns be orthogonal?

AtA was

4 4 0 2
4 3 0 2
0 0 2 2
2 2 2 4

AAt was

2 1 0 0
1 1 -1 0
1 1 1 2

also, looking at the dot products from each column pair for either result, i couldnt find any combo where the dot product would be zero

[wintry steppe]

to be more precise, say "every subspace consists of eigenvectors of some matrix with eigenvalue 0"

not "every subspace is an eigenvector"

[gray pond]

Look for zeros in $A^TA$ these would correspond to orthogonal column vectors. keep in mind there will be duplicates

[stoic python]

casualpolemic

[ember pagoda]

cool tnx

[undone fable]

thank you thank you

[spare widget]

I am not sure whether this fits in this channel, but here's my problem. I want to find an integer vector (n1, ..., nm) such that ni >=1, for which the following energy is minimized

$\min_{\vec{n} \in \mathbb{N}^m} \min_{c\in \mathbb{R}}|\vec{n} - c\vec{k}|_{\infty}$

[stoic python]

criver

[dusky epoch]

is the first min meant to be taken over k rather than n?

[spare widget]

it's over n, k is fixed

k is in R^m

[dusky epoch]

right

so let's see here

so geometrically this is drawing a line in R^n with direction k that passes thru a positive integer lattice point, such that the infty-norm distance of closest approach to the origin is minimized

[spare widget]

It arose from a rounding problem, where I need to allocate an integer number per cell >=1, with proportionality determined by k.

[spare widget]

yes, c however rescales k to fit as best as possible

the only apporach I could think of was to brute force it

and I forgot a constraint n1+...+nm = N

[spare widget]

by brute force it I mean set n1=...=nm = 1 initially since this is the lower constraint

then go over all n1,...,nm

and add a 1 to a single one

and then evaluate the energies for each such choice

then pick the best one

and repeat this process until n1+...+nm becomes equal to N

I thought there may be a simpler approach though, that would require the brute forcing only at the very end

[zinc timber]

is this some Integer Linear Programming?

[spare widget]

looks like an integer linear programming problem yeah

ultimately I just brute forced it so I would not have to think about it

[languid sphinx]

You could try to see if the convex hull will make it better

But rounding problem sounds in general like no

[zinc timber]

so you are allowed to vary n and c?

[spare widget]

well not c

c is computed to be the optimal

what I was looking for was n such that it is integer, ni >=1, and \sum_i ni = N

[languid sphinx]

The fact that your decision space is discrete is a bad sign

[spare widget]

I know 😂

I've read papers from hell on graph matching problems

it's ok, I posted it in case someone could easily come up with a better idea

don't think too hard on it, I already implemented that using brute force

[rugged nacelle]

Given some matrix A, is there a general name for doing this to it: B^(-1) * A * B where B is just some other matrix with no special restrictions.

[wintry steppe]

conjugation

[rugged nacelle]

that's what I was taught as well, but I literally couldn't find any mentions of the term, so I assumed it might have been domain specific

maybe my google-fu is just not up to speed

[wintry steppe]

conjugation is more commonly written as BAB^{-1}

so what you wrote would be "conjugation by B^{-1}"

[rugged nacelle]

thanks. I'll keep using that term then, just wanted to make sure I wasn't spitting some bogus

[quartz compass]

it's not uncommon to see it that way in linear algebra, if you're diagonalizing a matrix you have AE=ED and so E^-1 A E = D diagonalizes it

[tame pond]

It can also be called a similarity transformation

[spare widget]

this B is restricted, since B^{-1} has to exist

[rugged nacelle]

true, add that

[spare widget]

then you can interpret it as a change of basis

Let [v]_E = B [v]_F -> [A]_F = B^{-1} [A]_E B

[west saddle]

'm not sure if this is the right place, but i'm working with Anhilators in Linear algebra and differential equations. I noticed my book only wants me to solve for the coeffectent of the particular soltuion and not the complementary solution. Why is that?

[wintry steppe]

more context would help you get an answer

[hardy inlet]

My guess is that its used in the reverse direction, but I never used invertibility (that I know of) in the forward proof; is this correct? Why do i need (or not need) invertiblity?

[languid wigeon]

@spare widget you may try wrapping your math expression with a pair of double dollar sign.

Let $$[v]_E = B [v]_F -> [A]_F = B^{-1} [A]_E B$$

[stoic python]

vin100

[languid wigeon]

Let $$[v]_E = B [v]_F \to [A]_F = B^{-1} [A]_E B$$

[stoic python]

vin100

[wintry steppe]

just because v is non-zero, why should Rv be non-zero?

[hardy inlet]

i was just thinking that since positive isn't positive

its >= 0

OH

Invertibility means null space is 0

right

[wintry steppe]

for operators, yes

[hardy inlet]

so then Rv is strictly positive

and hence <Rv,Rv> is strictly positive

right?

[wintry steppe]

"Rv is positive" doesn't make sense

but you're aiming to show that <Rv, Rv> is strictly positive using the positive-definiteness property of the inner product

and for this you somehow want Rv to be non-zero

[hardy inlet]

well R is a positive operator (<Rv,Rv> geq 0), and since null is only 0, <Rv,Rv> = 0 ONLY when v = 0, so <Rv, Rv> > 0 (since we assume v != 0)

[wintry steppe]

yes

you may want to explicitly mention that it is R that is invertible

but you are right

this is where invertibility of T is used

[hardy inlet]

R is invertible bc T is invertible?

[wintry steppe]

yes

if you're unsure, prove it

[hardy inlet]

hmm. it makes sense (although i haven't proven it) that R is also invertible, but I dont see anything about that in axler

[wintry steppe]

you don't need to look in axler to find it, you can look in your brain and prove it yourself

it should take one line at most

[hardy inlet]

ye but its axlers question in the section so idk if that means he intended a different way of solving it

hmm. So suppose T is invertible, we know that R² = T, and T is invertible. uhh

R²=T iff R²T^{-1} = TT^{-1} iff R²T^{-1} = I not sure if that gets anywhere

[wintry steppe]

and this means R is invertible, because...

[hardy inlet]

idk. if R² = I then idek if thats enough, and i have a T inverse in there

aren't there ways to get I without inverses

[wintry steppe]

R(RT^{-1}) = I

[hardy inlet]

ahh

[wintry steppe]

alternatively, det(R) is a square root of det(T) which is non-zero, so it itself can't be zero. this would be fine if axler wasn't so vehemently objected to determinants

[hardy inlet]

is that last step with ur parenthesis and \implies good?

[wintry steppe]

honestly, you can fit this in a sentence instead of its own lemma

"Since T is invertible, R(RT^{-1}) = I, so R is also invertible."

words are easier to read than \implies

[hardy inlet]

i'll just add the words to the lemma 😬

[oblique prairie]

it doesn’t really matter lol

i prefer symbols as long as it’s not ALL symbols

[hardy inlet]

cause just doing the last jump skips the left or right multipication

[wintry steppe]

having a right inverse is enough for invertibility in this case

[hardy inlet]

yeah ik that one 😎

(although i wasn't supposed to)

[wintry steppe]

right inverse means you're surjective, and since you're an operator on a finite dimensional space, you get injectivity as a result

[hardy inlet]

i dont remember what homework i got like a point off xD

[wintry steppe]

it does if you're ever writing something for others to read

[oblique prairie]

yeah i thought this was just their notes

[wintry steppe]

i just assume everything here is homework

[oblique prairie]

makes sense

[hardy inlet]

yeah the main proof is part of HW problems

after words:

[wintry steppe]

i would exercise caution saying "the" square root, and i'd end it with "R(RT^{-1}) = I, hence R is invertible."

it's like saying "the inverse is RT^{-1}, so it's invertible"

having an inverse is part of being invertible, so the phrasing's backwards

[hardy inlet]

so like this instead?

[wintry steppe]

that's fine

if you'll let me nitpick even more, you really shouldn't be including this as a separate lemma. it's nowhere near important enough to warrant being put separately from the main proof, and it's short enough that you can include it as a single sentence

[hardy inlet]

ok

here are my modifications and hopefully finishing the forward

[wintry steppe]

this is correct

[hardy inlet]

do u think the whole That is, $Rv = 0 \iff v = 0$ is childish?

[wintry steppe]

"childish" isn't the word i'd use

it's just unnecessary

you and your grader both know exactly that it means the same thing as "null R = 0"

[hardy inlet]

ok so i made it

[wintry steppe]

good

[hardy inlet]

ty

The red underline is false right? It could just be that Tv and v are orthogonal?

[wintry steppe]

no, it's correct

if Tv were zero, <Tv, v> would be zero too

they can't be orthgonal because you assumed their inner product is non-zero

[hardy inlet]

right. I guess i was thinking of orthogonal backwards, since we assume Tv,v isn't 0 then they aren't orthogonal but that doesn't matter

so to be clear <x,y> > 0 implies x and y are nonzero?

i think its true but im trying to think of any edge cases

[wintry steppe]

what would happen if one of them was zero?

[wintry steppe]

same logic

[spare widget]

inner product is bilinear

[hardy inlet]

if any component is 0 the whole thing is zero

[wintry steppe]

exactly!

[hardy inlet]

but the converse isn't true

[wintry steppe]

right, orthogonal vectors are a thing

you're using only the correct implication in your proof

[hardy inlet]

which is what i was thinking earlier, but that doesn't have an impact here

i think the thinking <Tv,v> = 0 means Tv = 0, v = 0, or Tv is orthogonal to v was throwing my brain for a loop

[wintry steppe]

<Tv, v> = 0 only says that Tv is orthogonal to v and nothing more

(in general)

[hardy inlet]

true, since the 0 cases are covered by 0 is orth to everything

this is just defining some weird subscript notation?

[wintry steppe]

it's just notation they're using for a bilinear function defined in terms of T

[hardy inlet]

what is a bilinear

[wintry steppe]

...

[hardy inlet]

2 args?

[wintry steppe]

do you remember what an inner product is?

they're examples of bilinear functions

[hardy inlet]

yeah it maps 2 vector elements to 1 scalar

but we never called them that

[wintry steppe]

god i hate axler

so much

a bilinear function on a vector space $V$ is a map $f\colon V \times V \to F$ (where $F$ denotes the underlying field, in this case $\bR$ or $\bC$) such that $$f(av+bw,u) = af(v,u) + bf(w, u)$$ and $$f(v, aw + bu) = af(v, w) + bf(v, u)$$ (i.e., is linear in each argument when the other is fixed)

[stoic python]

TTerra

[wintry steppe]

you should recognize inner products as being an example

[hardy inlet]

yeah so i know all that just not the same 😅

[wintry steppe]

well, complex inner products satisfy a modified version of this

where scalars being pulled out of one of the arguments get conjugated

but i hope you get the idea

[hardy inlet]

right

[hardy inlet]

so we wanna show these, but im thrown off by the _T

[wintry steppe]

go verify all the things in axler's definition of an inner product

not what i wrote, since (as i remarked and as you acknowledged) that's a little imprecise in the complex case

[hardy inlet]

Ye so these

and this would be the forward direction? so the properties are assumed ye?

[wintry steppe]

that's typically how "if and only if" proofs start, yes

[spare widget]

You would need that T is hermitian for the conjugate symmetry part I think

does the definition of positive in the above also include symmetric?

[hardy inlet]

positive includes hermation

aka SeLf aDjOiNT

[wintry steppe]

homogeneity

[hardy inlet]

wdym by symmetric

we have

[spare widget]

You have to prove those

the last one is conj symm

[hardy inlet]

An operator $T \in \L{V}$ is called \boldit{positive} if $T$ is self-adjoint and $$\inner{Tv,v} \geq 0$$ for all $v \in V$

[stoic python]

MattDog_222

[spare widget]

I have seen people call T positive when only <Tv,v> > 0 is satisfied

not often, but it happens

[hardy inlet]

well (0, \infty) is a subset of [0, \infty)

[spare widget]

It's like how positive definite usually means symmetric positive definite

[hardy inlet]

but only in the reverse direction correct? Since forward we assume all this is true 0_0

[spare widget]

But you can have non-symmetric operators for which <Tv, v> > 0

[spare widget]

They say if and only if

So both directions

[hardy inlet]

yeah but showing all the stuff like conjugate symmetry is in the reverse direction

[spare widget]

So T positive invertible, <,> inner prod -> <,>_T is an inner prod

And vice versa, <,> inner prod, <,>_T inner prod -> T positive invertible

[hardy inlet]

idk if my arrows are flip flopped, not that it affects the end result

[spare widget]

Suppose <,>_T is an inner product. We will prove that T is an invertible positive op wrt <,>. ...

[hardy inlet]

can an inner product be negative

[spare widget]

it can in physics, not in your book

Ah

You mean wrt diff vectors

Sure

[hardy inlet]

i just meant in general so yeah that. but i need <Tv,v> >= 0 for positivity

[spare widget]

which part are you proving?

T positive invertible?

Then <Rv, Rv> >= 0

From positivity of the original inner prod

[hardy inlet]

i think i have the positive part? this one feels trivial

[spare widget]

ok so the opposite dir

[hardy inlet]

but I didn't do the invertible

[spare widget]

From definiteness you should get invertibility I think

[hardy inlet]

unless thats what u meant about using positivity

[spare widget]

Since <Tv, v> = 0 would imply v = 0

[hardy inlet]

Tv and v can't be orth?

[spare widget]

you assumed <v,v>_T is an inner prod

This means that <v,v>_T = 0 <=> v = 0

so <Tv, v> = 0 => v = 0

[hardy inlet]

ah right sec

like this?

[wintry steppe]

where's the "self-adjoint" part of "positive" in this?

[hardy inlet]

oops 👀

[spare widget]

use the symmetry property

[hardy inlet]

this?

[wintry steppe]

what else

[hardy inlet]

idk i have Tu v = Tu v

[wintry steppe]

you need to use conjugate symmetry of <., .>_T, not of the original inner product

i would suggest writing out what conjugate symmetry of <., .>_T means and exploring that

[hardy inlet]

$\inner{u,v}_T = \overline{\inner{v,u}}_T$?

[stoic python]

MattDog_222

[wintry steppe]

that's right

you assumed this

so now see what this says about T

[hardy inlet]

idk what the RHS equals in terms of the inner product

[wintry steppe]

why

[spare widget]

You do know

[hardy inlet]

in terms of this

[spare widget]

<a,b>_T = <Ta, b>

[wintry steppe]

ok so write that out

[spare widget]

now set a = v, b = u

[hardy inlet]

$\overline{\inner{v,u}}_T = \overline{\inner{Tv,u}}$?

[stoic python]

MattDog_222

[spare widget]

yes

[hardy inlet]

weird... ok i'll see where this gets me

[spare widget]

note that <Tu,v> = <u, T^* v>

And note that the original inner prod also has the symmetry prop

[wintry steppe]

let them do it

[spare widget]

you should also write out what you're trying to prove

i.e. what self-adjoint means

[hardy inlet]

idk if this is a thing, gtg eat tho brb

[hardy inlet]

can i pull a T* into the subscript or is that not allowed

[hardy inlet]

where are you stuck

[wintry steppe]

please do not

[hardy inlet]

ooh ooh i think i got something:

$$\inner{Tu,v} = \inner{u,v}_T = \overline{\inner{v,u}}_T = \overline{\inner{Tv,u}} = \inner{u,Tv}$$

[stoic python]

MattDog_222

[hardy inlet]

would that work?

[wintry steppe]

there you go

[hardy inlet]

watching tiktok for an hour and coming back "fresh" is OP ngl

[wintry steppe]

you're two steps away from showing that T is self adjoint

[hardy inlet]

wait is that not enough for <Tu,v> = <u,Tv>?

[wintry steppe]

what?

[wintry steppe]

then it's fine

i just wanted to make sure you knew how to go from <Tu, v> = <u, Tv> to T = T*

[hardy inlet]

yea

is the forward good? Now i just gotta show the linearity and stuff the other way?

[wintry steppe]

this looks ok

[hardy inlet]

idk if this is what i'm trying to show, or if we can use the equality at the top

[wintry steppe]

this is correct

[grim cliff]

Let A be a matrix and J(A) be it's Jordan block representation. Do the blocks of J(A) correspond to the eigenspaces of A?

[hardy inlet]

so I can go from <Tu,v> to <u,v>_T and use InnerProduct operations on <Tu,v> before the conversion?

this feels even shadier

[wintry steppe]

yeah i'm not terribly convinced of this

[wintry steppe]

because <., .> is an inner product, yes

[hardy inlet]

yeah im not too conviced either

I tried reducing it to <image> but a 90° rotation easily violates this, so i need to bring in positivity

also the dagger was

but that used the other hypothesis

so cant do that

[wintry steppe]

here's a mega hint: ||to use positivity to show definiteness of <., .>_T, use the fact that T has a square root. also, you showed earlier that an invertible operator with a square root necessarily has an invertible square root; that should come in||

[hardy inlet]

like this?

oops i'lll clarify R invertible bc T

[wintry steppe]

other than the typo in the first line, this is exactly correct

[hardy inlet]

what typo

[wintry steppe]

you can find it

[hardy inlet]

like the "positive" square root?

[wintry steppe]

[hardy inlet]

cause axler says T positive has a self-adjoint positive square root

[wintry steppe]

that part's correct

i'm not gonna proofread your work

[hardy inlet]

oh lol

i see

R²,v

thanks

[hardy inlet]

assuming no typos, is this the right idea?

[wintry steppe]

this is fine

[wintry steppe]

mate you posted 7 questions at once

be more specific

[hardy inlet]

for c → d can I just claim gram schmidt? This questions weird cause its basically what we already know

[hardy inlet]

Av = λv = 6v. try to multiply Av and find abc such that the A(3,2,1)^T = 6(3,2,1)^T

solve this smaller system

first, what should
3a + c = __ ?
11b-a+2c = __ ?
-2b-2a+6c = _?

I know theres gotta be a cheaky change of basis argument for this...

Av = λv, and λ=6 and v = (3,2,1) then what should the result be as numbers, e.g. (5,10,12)

[uncut dust]

i just blew my own mind rn

and u want to get

im going to solve it rn as a system of linear equations

[hardy inlet]

Av = λv

applying the transformation matrix A to vector v scales v by λ

yes technically but thats a slightly more convoluted multiplication

an then you're solving
3a + c - 18 = 0
11b - a + 2c - 12 = 0
-2b - 2a + 6c -6 = 0

(A-λI)v = 0 is one way yes. do the constant terms in that system mean anything? -18, -12, -6? what was v

[uncut dust]

v is 3 2 1

errr

[hardy inlet]

3 2 1 and 18 12 6 is there anything special there

[uncut dust]

yes

18 12 6 is a stretch of 3 2 1

[hardy inlet]

by

[uncut dust]

6

[hardy inlet]

yes. λ = 6. the definition of eigenvector/eigenvalue

[uncut dust]

OHOHOOHOHHH

OHH

oH

ill send u my work hold up so i solve for A

OHH

i see

i just didnt know how to solve for A

it looked really complicated

[hardy inlet]

you're solving for a, b, c

you have a system of 3 equations with 3 unknowns

[uncut dust]

yeah

a b c = A

[hardy inlet]

this

how do you solve that using linear algebra

[uncut dust]

u dont set it equal to 0 I think

[hardy inlet]

hint: a matrix

[uncut dust]

Because a b c will keep equalling 0 and thats why i got nowhere

[uncut dust]

i see

with a 3x3 matrix

There is 9 ps

P1-P9

[hardy inlet]

ok, on a smaller scale, 2 equations, 2 unknows.
2x+3y -5 = 0
3x + 4y + 7 = 0

what size matrix do you use and what do you put where

[uncut dust]

A 2x2 matrix

put -5 on the other side

7 on the other side turns to negative 7

[hardy inlet]

ok, so if u used a 2x2, what matrix what would the entries be

[uncut dust]

-5 turns to postive

[uncut dust]

It would be

Lemme do the paper work rq

[hardy inlet]

u can just type it as rows

a b
c d

not sure how ur gonna cram all these into a 2x2 tho

[uncut dust]

i get this

2x+3y=5

3x+4y=-7

now we need to find a way

To eliminate x or y to find the lather

so

Common denominators r:

4x3 is 12

So

but an easier way is to eliminate x 2(3)=6

so:

[hardy inlet]

im not asking for pure algebraic solving

as a matrix

what entries

[uncut dust]

i got y=29 then x=-41

[uncut dust]

uhhhh

I think

A1= 2x A2=3y B1=3x B2=4Y

Right?

[uncut dust]

OHH

yea

over here

[hardy inlet]

lol nice those are actually the solutions, didn't know it has integers

does this maybe ring a bell

<@&268886789983436800>

[uncut dust]

stop

<@&268886789983436800>

@delicate lichen

plz

[hardy inlet]

perhaps if i draw a line

[uncut dust]

im staying up all night

[delicate lichen]

got em

[uncut dust]

thankls

[uncut dust]

dude

OHHH

[real dawn]

yay!

[uncut dust]

i remember now

my prof gave us a website for row reducing

but i forgot what the method was called

[hardy inlet]

so u can use a calculator?

[uncut dust]

yeah

any online ressource

24 hours

desmos and symbolab allowed and wolfram alpha

idk how to use wolfram

[hardy inlet]

if u tell me the correct matrix like
a ...
b...
c...
i'll give u the output

and then how to wolfram it

[hardy inlet]

so like
2 3 5
3 4 -7
for this

whats the one for ur hw going to look like

hint: 3 0 1 __

[uncut dust]

18 12 6?

[uncut dust]

i managed to get what u got

i used row reduction

[uncut dust]

12

[uncut dust]

18 12 12?

[hardy inlet]

idk which is it

its a 3x4 matrix

i'll give u that much

[uncut dust]

AHHH

i see

[hardy inlet]

3a + c - 18 = 0

• a + 11b+ 2c - 12 = 0
• 2a -2b + 6c -6 = 0

so start like u did in the 2 equation one that u got 29 and 41

[hardy inlet]

this was RREF { {2,3,5}, {3,4,-7} } for example

on wolfram alpha

i'll be back in 15

i'll give u that they're all single digit positive values

for the a,b,c

[uncut dust]

how did u get -18, -12 and-6?

[hardy inlet]

the result of multiplying this

and then the = 0 components are because you're saying that they all equal zero by (A-λI)v = 0

[cedar dew]

if a set of vectors generate a vector space, do they have to be linearly independent?

[uncut dust]

OHHH

i get it now

Also

[hardy inlet]

No. (1,0), (0,1) and (1,1) can generate R2 and are linearly dependent

[cedar dew]

oh hmmm

if they form a basis, then they have to be linearly independent

[hardy inlet]

Which was

[cedar dew]

i got the two mixed up it seems

thank you!

[hardy inlet]

Basis yes

Basis by definition is a linearly independent spanning/"generating" set

Linear Independence just means they arent combinations of one another. (1,0,0) and (1,1,0) are LI but dont span R3

I think thats correct i shut off my pc tho

Looks familiar

Whatever
3 0 1 18
-1 11 2 12
-2 -2 6 6
Gives

Are u supposed to reuse the a b and c?

[cedar dew]

yeah got it

another thing, the set of generators of a vector space is always a proper subset of that vector space, right?

[hardy inlet]

I guess? I meam Vector space V is infinite vectors

[cedar dew]

oh right

wait a vector space cannot have finite elements?

fuck im dumb sorry

got it thanks for the help

[hardy inlet]

No it can

[cedar dew]

wait what

[hardy inlet]

But u said always

[cedar dew]

a vector space can have finite elements?

oh

V = {O}

i guess

[hardy inlet]

U can have a vector space over Z/2Z where 0 and 1 are the only elements and if itz 2D then (0,0) (0,1) (1,0) (1,1) are the 4 elements afaik

[cedar dew]

huh...

idk what that means

[hardy inlet]

But yes 0 is also a vector space with finite elements

[cedar dew]

ok nice

[wintry steppe]

no, a vector space always generates itself

there's no one unique set of generators, so you can't say "the" set of generators

[cedar dew]

OH right right

[cedar dew]

i meant any arbitrary set of generators

thanks that clears things up

[wintry steppe]

@hardy inlet should see this too

[grave garden]

Guys, how to find kernel and Im of this ?

I have this but why its image is 2 also ?

[quartz compass]

what's the dimension of your vector space

[grave garden]

I have just this

[quartz compass]

right, so what's the dimension of the vector space M_2(R)?

maybe try figuring out what a set of basis vectors might be

[peak marsh]

@sharp idol done more work on the reflections

just need to explicitly write it out but

the basic idea is finding the angle the direction vector of the reflection 'axis' makes with the positive x axis

rotating the entire plane by said angle

constructing the reflected function using the rotated functions' difference in y coordinates

and then undoing the rotation

[pseudo maple]

what notation is this?

[dusky epoch]

this is just another variable

in this case it probably stands for a matrix

[pseudo maple]

ohhh ok thanks

[dusky epoch]

can you show the full context?

[pseudo maple]

yeah sure

[dusky epoch]

okay yeah

[pseudo maple]

Great, thanks alot!

[wraith monolith]

If W in V, both of them are vector spaces and dimW=1, dimV=2. Then is it possible that there exists two linearly independent vectors such that either one of them are not in W

Possibly true

[spare widget]

Let V = span{(1,0)}, and W = span{(0,1), (-1/2,1/2)}

the two vectors for W are linearly independent and neither is in V

V corresponds to a single line through the origin, W is a 2D plane containing that line

it's clear that you can pick two non-colinear vectors on that plane that do not coincide with the line V

[pseudo maple]

would the circled part be a characteristic equation?

[gray dust]

@pseudo maple thats called the characteristic polynomial

setting it =0 gives the characteristic equation

[pseudo maple]

ohhh ok sweet thank you

[wintry steppe]

Anyone able to explain the answer of a?

I'm thinking of p = identity minus [1,1...][1,1,1]T

Which is 0 on the diagonal

And abunch of 1s

But I think this question and Its answer has some principle I'm missing

[gray dust]

@wintry steppe u is any unit vector, not a specific one

[wintry steppe]

Yes, i was just doing this to make it concrete

[gray dust]

use what it means for 1 to be an eigenvalue

take any corresponding eigenvector x

Px=1x

then use the definition of P

[wintry steppe]

Man, definitions of p?

I've had 20 different definitions

[gray dust]

its given here

P=I-uu^t

[wintry steppe]

Ahh identity minus uu

Uut is just a projection right

[gray dust]

yes onto u

[languid sphinx]

does anyone have any tips if I keep messing up in linear algebra

like I think I got rank as the number of LI row/columns in a matrix in my head fine (and other basic facts)

But I don't think I get diagonalisation/symmetric properties down

Should I go back to a textbook/reference book

[stuck tendon]

how do you show uniqueness of the adjoint?

[tacit pelican]

why does it have be A^T?

why can't i just row reduce A

since if it has a pivot in every column, its obviously going to have a pivot in every row

since hte matrix is square

like i dont see how that will change the result

since det A = det A^T

and the entries along the main diagonal don't change after matrix transpose

[oblique prairie]

i don’t think the book tells you that yet

[tacit pelican]

yeah i saw

i’m proving it rn in fact

it was in the page right after it

lol

[sick sandal]

assume its not and you'll reach a contradiction
||say f* is the adjoint of f and suppose g* is as well
<f(x), y>=<x, f* (y)>=<x, g* (y)> for all x, y --> f *(y)=g *(y) for all y f * = g * ||

[stuck tendon]

gotcha, thank you

[wintry steppe]

Is there an easier way

To find the last eigenvalues

Or is there an easy way

[zinc timber]

note that sum of each row/col is same (a+2)

try the vector [1 1 1]^T

@zinc timber

Next, note that subtracting a-1 from each diagonal makes the matrix singular, so a-1 is another one

[wintry steppe]

Yes a plus two

So 2 is one eigenvalues

[zinc timber]

No

a+2, not 2

[wintry steppe]

Okay.

[zinc timber]

but don't listen to me

[wintry steppe]

What do you mean by sing6lar?

[zinc timber]

try multiplying (1,1,1) see you'll get (a+2)(1,1,1) making it an eigen hector

vector*

[zinc timber]

0 det

also A-(a-1)I_3 has nullity 2

so Eigen values are a+2, a-1 and a-1

[wintry steppe]

Sorry trying to understand this

I understand a +2

As the row sum

[zinc timber]

you can also try to do it this way

A-xI then you'll get
$\m{a-x & 1 & 1 \ 1 & a-x & 1 \ 1 & 1 & a-x }
$
so after you do R1=R1+R2+R3 you'll get
$$\m{a+2-x & a+2-x & a+2 - x\
\vdots & \vdots & \vdots }$$

[hollow thorn]

Hello

[zinc timber]

Now you can factor out a+2-x from first row and that'll give you
$$(a+2-x)\m{1 & 1 & 1\ 1 & a-x & 1 \ 1 & 1 & a-x}$$

[wintry steppe]

Ryu sama

Thank you

[zinc timber]

from this
R2=R2-R1 and R3=R3-R1 will give you
$$\m{1 & 0 & 0 \ 1 & a-1-x & 0 \ 1 & 0 & a-1-x}$$

[wintry steppe]

I understand what you mean by singuglar now too

[zinc timber]

now this gives char poly $(x-a-2)(x-a+1)^2=0$

[zinc timber]

that was a big brain way of calculating the evs,

this is step by step

you'll get used to those as you practice

did you understand the row operations?

[wintry steppe]

I appreciate the help man

[zinc timber]

np

[zinc timber]

Lol I tagged myself there

[wintry steppe]

Luckily i was paying attention

[hardy inlet]

still lost on approaching this

[hardy inlet]

im not sure if theres a way to use similar matrices/change of basis for it

T1 and T2 are similar, but S*T2S isn't necessarily similar 🤔

[stuck tendon]

I think you can choose a basis of eigenvectors for T_1, such that T_1 is diagonal, and then find a unitary matrix that diagonalises T_2

[hardy inlet]

wdym by unitary matrix

[stuck tendon]

UU* = I

By the spectral theorem for normal operators, you can find such a U

[hardy inlet]

but i thought that wouldn't hold if it was real

[stuck tendon]

Why?

[hardy inlet]

it says for Real Spectral T is self-adjoint/hermatian

normal isn't necessarily SA

[stuck tendon]

https://www.math.ucdavis.edu/~anne/WQ2007/mat67-Ll-Spectral_Theorem.pdf Thm 5. Normal operators are diagonalisable

[hardy inlet]

yeah, complex

[stuck tendon]

Oh, that's LA Done right. What exercise is it?

[hardy inlet]

this isn't restricted to complex

7c11

[proven torrent]

Given the following hyperplane, x_1 + 3x_2 + 4 = 0, what is the normal vector?

[stuck tendon]

Well fine. Don't use the spectral theorem. But you have 3 eigenvalues, and since the operator is normal, the eigenspaces corresponding to those are orthogonal. So you can find an ONB of eigenvectors

[hardy inlet]

how do u know they're orth without the CST

[stuck tendon]

The operator is normal

[hardy inlet]

oh i see

pg 214 on axler

The eigenvectors for distinct evs are orth

pog

so we have an orthonormal basis by dimV=3

[zinc timber]

@hardy inlet still need help?

[hardy inlet]

with this one? Yeah. 134631 got me to the ONB part

i also added T2 having an ONB to the above

[stuck tendon]

Don't know where your V comes from. The ONB is for F^3.
Let {e1, e2, e3} be an ONB for F^3 from eigenvectors of T_1, and {f1, f2, f3] be an ONB for F^3 composed of eigenvectors of T_2, correspoding to eigenvalues of 2,5,7 for each.
Then define Sei = fi for i = 1,2,3.
Check S is unitary.
After having checked that, we can write S^-1 = S*.
So S*Sei = S*fi = ei.
Prove S has the required property.

[hardy inlet]

oh V is F3 right

how are we allowed to just define Sei = fi?

when u say unitary do u mean axlers "isometry"

or lenth 1

[stuck tendon]

We are meant to prove existence of the map. So we have to give the map

[hardy inlet]

ok so we just throw it out there, then show the conditions hold for that

namely isometry and S*T_2S=T_1

S should be an isometry from that ^

[hardy inlet]

not sure if I went down the right road for showing all v, i think its right but it feels a bit drawn out

[brazen sinew]

Hey guys

Anyone know how to get started on this?

[wintry steppe]

well

start by assuming S is a basis

and then show the thing in (i) holds

[brazen sinew]

Ok i started with that then i took an arbitrary vector in V and claimed that it can be written as a linear combination of the vectors in our basis S

Im assuming that the next thing to do is to define a finitely supported function which is equal to the coefficients in that linear combination?

[wintry steppe]

yes

[brazen sinew]

That makes sense

Is uniqueness implied or do i have to show it

[hardy inlet]

i think im in need of T* but dunno how to get it

[hardy inlet]

i got T* but cant get a clean sqrt T*T

[modest cliff]

Does anyone know why matrices have to hold only linear functions? Why can't they hold non-linear functions like sin(x)?

Like... I don't see why they can't

[hard drum]

Well what does it mean to you for a matrix to 'hold' a linear function?

[hardy inlet]

any idea on how to get the sqrt of this to equal that top result

[modest cliff]

Because
[[2, 6], [9, 3]] [x,y] = [2x+6y, 9x+3y]

Those are linear functions

[hard drum]

Sure. So the thing is that any function you define in this way is automatically linear by the way matrix multiplication works

Because M(x+y) = Mx + My and so forth.

If you want a proof of that, you can just look componentwise

[modest cliff]

Oh I see

I guess I can't call it a 'matrix' then

But is there a name for the sort of thing I'm describing?

[hard drum]

I'm not really sure what you intend to describe

Linear functions, or functions defined through matrices like that etc

[modest cliff]

Like:

[f(x,y), g(x,y)] [2, 6] = [f(2, 6), g(2, 6)]

And let's say f(x,y) = x * sin(y), and g(x) = x * y

Then [f(2, 6), g(2, 6)] = [2 * sin(6), 2 * 6]

[hard drum]

Well that's not really anything special, that's just a function R^2 -> R^2 written in a nonstandard way

Like i'd just write that as a function h: R^2 -> R^2 where h(x,y) = (x sin(y), xy)

[modest cliff]

Ohh

I see

So we can just write functions with multiple outputs

I didn't know that

Thx

[meager harness]

how to do this

[hardy inlet]

as a start, Av = 3v, so lambda=3 for eigenvector (1,2,2)

[native breach]

I'm kind of stuck

[wintry steppe]

where?

after the last step, you should be able to conclude what you want. (if the rest of the proof is correct, that is; i didn't read it)

[little frigate]

Is anyone familiar with the distinction between column and row approach?... I can barely find anything about it online and this terminology only has been used thrice in this server

[wintry steppe]

given that a lot of introductory linear algebra classes focus a lot on row and column reduction, the distinction is probably in that

[little frigate]

column reduction huh

yeah I think that's the word I'm looking for, thanks

[native breach]

@wintry steppe how do I argue that it's mu * i_N though

idgi

like, YES I'm SURE what I have there is enough

but I don't know how to say it

like r/n it's like "ok so the answer to 2+2 must be even, so 4 is a cndidate"

but it' not THERE afaict

[grand knoll]

Is there anyone who is proficient in subspaces and stuff here

I can't seem to understand the concept of it clearly

For these two questions, I am assuming (b) is a subspace while (c) is not a subspace

Is this correct? The only reason I assumed no for (c) is because (0,0,0) won't exist in R3

[cedar dew]

"assumed"

but that reason is correct yes, the scalar product and sum conditions are also not satisfied

[grand knoll]

more like deduced

what about for (b)?

[cedar dew]

does it satisfy the sum condition?

[grand knoll]

scalar is satisfied there right

[cedar dew]

for example if i give (0, a, b) and (0, c, d)

[grand knoll]

yeah it satisfies sum condition,

[cedar dew]

so it is a subspace

[grand knoll]

even the scalar is satisfied cuz x1 is always 0?

[cedar dew]

yeah

[grand knoll]

okay this is an easy topic, i was overthinking it

alright thanks a lot mate

[cedar dew]

np :)

[hollow silo]

    float a = r.direction.normal().length2();
    float b = 2.0f * vec3::dot(r.direction.normal(), oc);
    float c = oc.length2() - radius * radius;

    float d = b * b - 4 * a * c;

    if (d < 0) return false;

    float sq_d = sqrt(d);
    float root = 0.0f;

    // find the nearest root that lies in the acceptable range
    float r_a = (-b - sq_d) / (2.0f * a);
    float r_b = (-b + sq_d) / (2.0f * a);

    if (vec3(r.ray_at(r_a) - r.position).length2() < vec3(r.ray_at(r_b) - r.position).length2()) root = r_a;
    else root = r_b;

can someone tell me why this does not return the closest point of impact for some reason

I'm trying to calculate the closest point of intesection a sphere and a line

a = n.n
b = 2 * n.(o - c)
c = (o - c).(o - c) - r*r

o : ray origin
c : center of the sphere
r : radius of the sphere

[spare widget]

the above is correct, it's likely a bug in your code or you are actually getting the closest

Also note that for a ray t<0 intersections are invalid

where r(t) = o + t * d

||o+td-c||^2 = r^2 -> 
t^2 ||d||^2 - 2t <c-o, d> + ||c-o||^2 - r^2 = 0 -> 
A = ||d||^2, B = <c-o, d>, C = ||c-o||^2 - r^2
D = B^2 - AC
D < 0 -> return INF
t1 = (B - sqrt(D)) / A
t2 = (B + sqrt(D)) / A
t1 = t1 < 0 ? INF : t1
t2 = t2 < 0 ? INF : t2 
return min(t1,t2)

Also if d is normalized, then A=1 and you can omit A from everywhere in the above pseudo-code

Then the returned value is the distance from o to the intersection

Note that I return infinity when there is no intersection

[languid wigeon]

simple just copy the 1st and 2nd column.

[spare widget]

nvm I realized they mean sth else

[languid wigeon]

coz it's written that $A$ is a matrix for $T$ with respect to $B$.

[stoic python]

vin100

[spare widget]

The matrix in the canonical basis is: [A]_E = V^{-1} * [A]_B * V

[languid wigeon]

you don't need that

[spare widget]

V^{-1} [T(v)]_B

From what I wrote above

[languid wigeon]

think abt this: suppose that $B := {v_i}_{i=1}^n$. $$[e_i]_B = ?$$

[stoic python]

vin100

[spare widget]

Or wait, is it the reverse

I think it's the reverse

[languid wigeon]

now you get $[T(v_i)]_B$ from part (a) with $i \in {1,2}$. What would you do to get rid of the squared brackets $[ \cdot ]_B$?

[spare widget]

V [u]_E = [u]_B

[stoic python]

vin100

John doe

[languid wigeon]

oops sorry i wanted to say $[v_i]_B$.

[stoic python]

vin100

[languid wigeon]

$$e_i = [v_i]_B$$

[stoic python]

vin100

[languid wigeon]

column vector with $i$-th entry $1$, zero elsewhere

[stoic python]

vin100

[spare widget]

Then [[A]_B [u]_B]_E = V^{-1} [A]_B V [u]_E = [A]_E [u]_E = [T(u)]_E

[languid wigeon]

say $e_1 = (1,0)^T$, $e_2=(0,1)^T$. depends on the dimension of the vector space

[stoic python]

vin100

[spare widget]

I think I am reversing something, since you mentioned it's the two columns, but then [A]_E = V [A]_B V^{-1}

at the same time I was getting [A]_E = V^{-1} [A]_B V

am I messing something up?

we have [u]_E = V [v]_B

[languid wigeon]

wait what is your $V$?

[stoic python]

vin100

[languid wigeon]

$V = B^{-1}?$

[stoic python]

vin100

[spare widget]

nvm, I was messing up this

[languid wigeon]

that's why i dun remember stuf

[spare widget]

It is [u]_E = V [u]_B

V is made of the two v columns

But they are wrt E

That's what I messed up

You are correct

[languid wigeon]

i abuse notations. $$B:={v_k}_k$$ is a basis (i.e. a set of basic vectors)

[spare widget]

[A]_E = V [A]_B V^{-1}

[stoic python]

vin100

[languid wigeon]

but i also write $$B = [v_1 v_2 \cdots v_k]$$ as a matrix

[stoic python]

vin100

[spare widget]

Then it's clear that [T(V)]_B = V^{-1} [A]_E V = [A]_B

[languid wigeon]

to minimize the number of variables used

you're typing on phone?

[spare widget]

T(V) = [A]_E v1 = V [A]_B V^{-1} V = V [A]_B

[languid wigeon]

if u r on computer, you may try wrapping your math with a pair of dollar signs

[spare widget]

@chrome radish that's what you need

[languid wigeon]

$T(V) = [A]_E v1 = V [A]_B V^{-1} V = V [A]_B$

[stoic python]

vin100

[spare widget]

Phone, it's a pain

$B = \begin{bmatrix} \vec{v}1 & \vec{v}2 \end{bmatrix} \implies \ [u]{E} = B [u]{B} \implies \ [A]{E} [u]{E} = B [A]{B} B^{-1} [u]{B} \implies [A]{E} = B [A]{B} B^{-1} \ T(u) = [A]_{E} u$

Wat

doesn't work for some reason

I am not sure, I think it's the []_X

It breaks the latex bot here for some reason

[stoic python]

criver

$B = \begin{bmatrix} \vec{v}_1 & \vec{v}_2 \end{bmatrix} \implies \\ [u]_{E} = B [u]_{B} \implies \\ [A]_{E} [u]_{E} = B [A]_{B} B^{-1} [u]_{B} \implies [A]_{E} = B [A]_{B} B^{-1} \\ T(u) = [A]_{E} u$
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   u
l.57 ... & \vec{v}_2 \end{bmatrix} \implies \\ [u]
                                                  _{E} = B [u]_{B} \implies ...
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

[spare widget]

Yeah, it hates the [...]_B I think

Broken latex

Either way the point was that [A]_E = B [A]_B B^{-1} where B has columns v1, v2

T(u) = [A]_E

Thus T(B) = [A]_E B = B [A]_B

This is for b)

The formula for T(u) is T(u) = [A]_E u = B [A]_B B^{-1} u

So you will need to compute B^{-1} for c)

[languid wigeon]

nope, it's abt the latex bot's treatment with markdown i think

see this

,, B = \begin{bmatrix} \vec{v}1 & \vec{v}2 \end{bmatrix} \implies \ [u]{E} = B [u]{B} \implies \ [A]{E} [u]{E} = B [A]{B} B^{-1} [u]{B} \implies [A]{E} = B [A]{B} B^{-1} \ T(u) = [A]_{E} u

[stoic python]

vin100