linear-algebra

[autumn kraken]

nvm

I got what you're saying

thanks

<<M>> is the smallest subspace that contains <M>, and <M> contains instself, thus <M> is the smallest subspace that contains <M>, thus <<M>> = <M>

right?

[marble lance]

@autumn kraken yep

[autumn kraken]

I always think I need to write something down with math symbols

dunno why

<M ∩ N> = <M> ∩ <N>

This is false right?

I think I have a counter-example but not sure

[zinc timber]

yes false

M = {(1, 0)}, N={(2, 0)}

[autumn kraken]

hm I don't think I understand that example

isn't then <M>={(x,x) | x in R} and <N> too?

[dusky epoch]

no

<M> = <N> = {(x,0) | x in R}

M cap N = empty so <M cap N> = {0}

[autumn kraken]

oh

I see thanks!

[errant mist]

This is defined for $x \in \mathbb{R}^n, x \geq 0$. Im not sure I understand the equivalency described here. How exactly do we get $||x||_2 \leq 1$ here?

[stoic python]

Fredrikpiano

[marble lance]

Let y = x/norm(x)

[errant mist]

I believe it should relate the cauchy-schwarz inequality

[errant mist]

right, that works

we are allowed to make that replacement since norm(y) = 1 right?

[marble lance]

Yeah

Possibly you have x = 0

Then that doesn't work

But then norm(0) <= 1

So treat it as a separate case

[errant mist]

thanks)

[azure bolt]

Help Me please
i am a portuguese and didnt speak very well english i am sorry

[zinc timber]

what is the question?

find the true one?

do you know that $\det(MN) = \det(M)\det(N)$?

[azure bolt]

yes

[zinc timber]

use it to find which one is true

[azure bolt]

det
determinant of a matrix

do you understanding

[zinc timber]

yes I am

[keen carbon]

Hey guys, say I'm given a square matrix X but then I change only one of the columns so I have a new matrix X'. Is there a quick computation for the determinant of X' in terms of the determinant of X

Specifically, all the values in each matrix are uniformly distributed from 0 to 1, but then the new matrix X' will keep all the values the same except for one row which are new random values

[zinc timber]

I think there is something but not sure how helpful it will be

if say the det ≠0

[keen carbon]

It's alright, just something that can guide me in some direction would be helpful

yeah, the det is almost always not 0

[zinc timber]

then you can write the new col in terms of the linear comb of the col vectors

but finding the coefficient is not that fast so don't think it'll give you any speedup to begin with

[keen carbon]

so the new col in X' can be written as a linear combination of the col vectors of old matrix X

[zinc timber]

yes

[keen carbon]

ah I see

[zinc timber]

say your new col is v = au_1+bu_2+..

then the det will just be a*det(X)

[keen carbon]

yeah, but then I'd have to compute the coefficients in terms of the column vectors which may take some time you said

[zinc timber]

yes

if the coefficient aren't obvious, it'll be time consuming

[azure bolt]

help me please

I try to solve and give me 3 solutions...

[zinc timber]

which ones

[azure bolt]

the result it is correct?

[median ocean]

@zinc timber yes I’ve found the QR

[zinc timber]

yes?

[zinc timber]

So now you have $QRx = b \implies Q^TQRx = Q^Tx \implies Rx=Q^Tx$

[zinc timber]

can you solve now?

check the dimension, I've skipped through them

[azure bolt]

in the exercise only 0 , 1 or 2 solutions

[median ocean]

Do I use the Q matrix or R matrix

[zinc timber]

Q^T

[median ocean]

So I first do Q transpose QR*x

[zinc timber]

Q^tQRx = b

[median ocean]

Qtranspose x Q x R x X

[zinc timber]

yes

use the fact that Q is orthogonal (orthonormal? idk)

[median ocean]

@zinc timber do I multiply this by b

[median ocean]

That’s what I got it from

[azure bolt]

Ryuzaki#5323 please help me with this exercise

@zinc timber

[zinc timber]

??

[zinc timber]

where are you stuck?

[dark brook]

If v is orthogonal on every column in A, what subspace is v in?

[outer goblet]

translation:
let
where a, b, c, and d are real numbers, and let
show that p(A) is equal 2x2 0 matrix, such that
is

so obviosuly is i change lambda into A i get

det(A-A) = det(0) = 0

but why is 0 determinant allowed to be considered a 0 matrix

[dusky epoch]

you cannot change λ into A in det(λI - A)

[outer goblet]

but it legit tells me to

what else would p(A) mean

[ionic laurel]

Hi I have a question

[dusky epoch]

it means p(A) = α_2 A^2 + α_1 A + α_0 I

but plugging A into det(λI - A) makes no sense

A is a matrix and so is p(A)

p(A) is the zero matrix but not zero as a number

[stoic python]

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

[ionic laurel]

okay im pretty bad at inputting matrices using texit

[dark brook]

doing a 5x5 matrix?

[ionic laurel]

6x5

[dark brook]

Use #latex-testing if you want to 🙂

[ionic laurel]

oh

thank you

[signal mulch]

Can someone help with Part B? I know that I need to prove that M has n linearly independent eigenvectors but I'm not sure how to prove it

I'm done part A, and got 2 distinct values to be 0 and 1

[stoic python]

eswag

[lavish jewel]

to make sure, do you mean the span of the columns of the matrix, or the span of a set that contains the matrix?

[ionic laurel]

yea my bad i mean the columns of the matrix

im not good at using texit sorry

so the problem thats given to me is that those columns of the matrix span the subspace W of R^6

it says to find a basis for W but the columns are not linearly independent

And the hint is to treat W as the column space A

assuming A is a matrix made up of those 5 columns

is the basis just picking out the columns with pivots? @lavish jewel

[lavish jewel]

yep

[ionic laurel]

I don't understand what it means when it says treat W as the column space

oh okay

[dark brook]

C(A) 🙂

[lavish jewel]

the column space is the vector space spanned by the set of vectors that make up the matrix columns

[ionic laurel]

right

and the nonpivot columns can be written as linear combinations

of the pivot columns

[lavish jewel]

that is not the full story, but yes

[ionic laurel]

okay one more question

[signal mulch]

@lavish jewel Can you help me with this question?

[ionic laurel]

So if i make a new matrix out of those pivot columns into a matrix B

and do QR factorization

If I were to multiply Q * Q^T as well as the orthogonal projection of an orthonormal basis onto the perpendicular subspace, then would adding those two matrices result in the identity matrix?

[lavish jewel]

think of what happens to vectors y = Ax, and Ay = A^2x

[lavish jewel]

what's Q, and orthogonal complement to what?

[signal mulch]

Vector y? could u explain how u got that?

[ionic laurel]

Let's say if A is an mxn matrix with linearly independent columns using basis of W, A can be written as QR where Q is an mxn matrix whose columns form an orthonormal basis for ColA and R is an nxn upper triangular matrix

[lavish jewel]

any vector y that you get out of multiplying an arbitrary vector x with A

[ionic laurel]

And if we find Q from augmented matrix using the basis for W perp,

Would adding those two matrices equal to the identity matrix

[median ocean]

@zinc timber do I just multiply this by b

[lavish jewel]

yes

[ionic laurel]

Interesting

Thank you!

[lavish jewel]

you could see this by noting that QQ^T + PP^T (where P is a matrix whose columns form an orthonormal basis for the orthogonal complement of the span of the columns of Q) can be rewritten as a sum of rank 1 matrices, each of which is the outer product of one of the columns of Q or P with itself

so it's equivalent to augmenting Q and P into a matrix, let's say M, such that M = [Q P], which by construction has columns that form an orthonormal basis for the whole vector space, and writing MM^T

then we would have that M^TM = I, and since M is squared, MM^T = I as well

[median ocean]

@lavish jewel could you help me

[zinc timber]

I have different approach to b, given p^2-p = 0 means that f(x)=x^2-x is an annihilator of M. Now it means that the minimal poly must divide f(x) so possible forms of minimal poly are m(x) = x, x-1, x(x-1). either case, the minimal poly splits into linear factor and has only one factor of each, (forgot what it's technically called) so M is diagonalizable. It may not be solution you were looking for tho

[zinc timber]

I still have no idea what your question is

[wintry steppe]

i love this

best solution

[zinc timber]

[wintry steppe]

i really like arguments in linear algebra that work with the minimal/characteristic polynomials

very pleasing to me

[zinc timber]

yes same

[wintry steppe]

e.g. if T is an operator on a real vector space of finite dimension with T^2 = -I, then V has even dimension

[zinc timber]

I remember that one

[wintry steppe]

m(x) has to divide x^2 + 1, so it's just x^2 + 1, and since the minimal and characteristic polynomials have the same irreducible factors, the degree of the latter is 2n for some n

it's cute

there is an easier way to see this though

just define (a + bi)v = av + bTv and check that this extends the real structure to a complex one

fun stuff!

[zinc timber]

is det(M+eI) = det(M)+e tr(I) true here also?

[teal grotto]

this is going to distract me. musn't... think... about... why this is true

[zinc timber]

think you cn take Eii but idk a better approach

[teal grotto]

welp, ive been working on my cs project all day anyway

um

i was trying to just symbol bash it but i dont think that gets anywhere

[teal grotto]

this one is kinda crazier tho. every matrix can be well approximated by an invertible one

[zinc timber]

every matix is almost diagonalizable as well

[teal grotto]

[zinc timber]

in C tho

[teal grotto]

oh sheet true

[zinc timber]

[teal grotto]

so you were gonna show that PE_ii = E_iiP for all i

[zinc timber]

if A+B=AB prove AB=BA

can also use some help here

[teal grotto]

nah i dont believe this one dude

[zinc timber]

it's true

[teal grotto]

i

[zinc timber]

i don't either but

answer is true

so i must prove it

these guys do set some good problems

[teal grotto]

yea fr

[zinc timber]

there's more u wanna try?

I could use some help fr

[teal grotto]

sure

[teal grotto]

we can be stuck together

[zinc timber]

sent

[spring pasture]

Can be done by BA=A+B right

[zinc timber]

NO

well it must hold but how will you show it

[spring pasture]

By showing that

[zinc timber]

because A+B=AB doesn't necessarily mean B+A=BA

if you understand what I mean

[spring pasture]

A+B=AB
So AB-A-B=0
So (B-I)A-B=0

I guess I'm going right here

So (B-I)A-(B-I)-I=0

(B-I)(A-I)=I
(A-I)=(B-I)
I=(A-I)(B-I)

I=AB-A-B+I

Gimme a sec

[lavish jewel]

not sure what you did to get to (A-I) = (B-I)

[spring pasture]

I wanted to get BA

[lavish jewel]

the rest looks ok so far

[zinc timber]

so you showed B-I and A-I are inverses that I understand

[spring pasture]

Oh yeah I screwed there

[zinc timber]

just swap the orders lol

nice

think it's proven

[spring pasture]

(B-I)(A-I)=I is clear upto this

[lavish jewel]

yep

[spring pasture]

So yeah we swap it

[zinc timber]

(A-I)(B-I)=I next

and boom

[lavish jewel]

yep

well, (b-i)(a-i) = i is what you want

[zinc timber]

this was more ring theory than LA

[lavish jewel]

(a-i)(b-i) is what you started with

[spring pasture]

Yeah feels like that

[spring pasture]

I started here

[lavish jewel]

(b-i)(a-i) = i expands into ba = a + b, and you're done

[zinc timber]

yes

[spring pasture]

Yeah or (B-I)(A-I)=(A-I)(B-I)

[zinc timber]

so strange that it's actually true

[lavish jewel]

but you start with (a-i)(b-i) = i

just making sure that you get the order right

[spring pasture]

That's true but how do we come up with it

[zinc timber]

inverses

[lavish jewel]

the way you did it is fine, it's just that ryu said "then we just expand (a-i)(b-i)" but that's what you started with

[spring pasture]

I don't know how I can pull that out directly from knowing A+B=AB

[lavish jewel]

i'm saying you did it fine but you didn't do the last step right to end up with something that has BA instead of AB

you just multiplied out the same thing again

or well, ryu did

just be careful, you had it right

[zinc timber]

yes I figured the rest no worries

[spring pasture]

Yes ik that thing but I'm asking how would I think about (a-i)(b-i)=i intuition directly cuz I won't be thinking that I'll multiply (a-i) and (b-i) by knowing a+b=i
So I just wanted to know is there an easy way to figure out that we have to use such product

[lavish jewel]

you did some arithmetic to show that (A-I)(B-I) = I

if the matrices are square, this is enough to say they are inverses of each other

[spring pasture]

Yeah but is there a way I can think of it with intuition

Ok

[teal grotto]

i think for this one, we want to look at eigen spaces or something

something with eigenvalues

[zinc timber]

easy argument, P is diagonalizable, so for M to commute with P, it must have to be simultaneously diagonalizable. so asfhasekfa and M is scalar

[teal grotto]

damg how is P diagonalizable

[zinc timber]

^

[teal grotto]

oh that was for that problem

[lavish jewel]

it was for someone else's, but happy coincidence

[teal grotto]

o

[teal grotto]

lol is this one still up for grabs

[zinc timber]

yes

[teal grotto]

f(x) = det(A + xI) is a real polynomial in x right

[zinc timber]

yes

[teal grotto]

um

um

this is close i think

like

gotta be a maximal zero to this poly

[zinc timber]

think that should be tr(M)

not tr(I)

huh

[teal grotto]

wut

[zinc timber]

nah have to work it out

have to factor out x

[teal grotto]

why

wait

f(x) has at most n roots

so for small enough x, det(A + xI) should be non-zero

i think that does it

like theres only finitely many roots to this poly. you can just zoom in on a neighborhood of zero and miss all of them

[zinc timber]

lol

[teal grotto]

lol

is that the wrong way of thinking about it or

[zinc timber]

proof my imagination

[teal grotto]

lol

um. hmm.

if there are no zeros then were done, since A + xI is invertible for all x

if theres only one then we obviously done

otherwise, take the minimum distance between all the roots

call this r

fuck

[zinc timber]

[teal grotto]

nooooo its so close aah

if we can show this then we're good, cus we can get a sequence of A + x_n I converging to A

all invertible

agh

purple dots are the invertible ones that approximate some non-invertible matrix A

[spring pasture]

Is that mitochondria?

[teal grotto]

red squiggly is the det(A + xI) function

[zinc timber]

lol

[teal grotto]

why can i not explain this

i feel like a baby

use your words!

[teal grotto]

powerhouse of the cell

[spring pasture]

The only definition I remember

[rigid raft]

is this chat free to ask a question?

[zinc timber]

always is

[rigid raft]

is what ive done right to find the hermitian? is that what I should be doing to solve 1, and if it is what do i do from there? make it equal to T hat?

[worldly merlin]

Are there other vector spaces apart from {0} that only have a single basis?

[wintry steppe]

the field {0, 1} over itself?

[worldly merlin]

makes sense, thank you!

[worldly merlin]

I'm trying to determine in which cases there is only a single isomorphism from a vector space to its dual. Is it true that this is the case iff the vector space only has a single basis?

It holds for the above two examples and I know that when a vector space has multiple bases we can construct multiple such isomorphisms
but I'm not sure if there are spaces with a single basis for which we can still construct multiple such isomorphisms

[teal grotto]

try taking a basis of any vector space and making a new basis, but not by permuting the elements

[worldly merlin]

yes, then we can get a different isomorphism right?
but I'm trying to go in the other direction

[teal grotto]

no, this will show that if a vector space has exactly one non-empty basis, then it has to be isomorphic to Z/2Z

[worldly merlin]

alright

[zinc timber]

didn't understand the notation, ik bras and kets but what's ⟨..|+tj|...⟩

[vast iron]

Taking hermitian conjugate means you also reverse the order of multiplication, and also convert scalars into their complex conjugates:
$$(C \ket{A} \bra{B})^{\dagger} = C^{}\ket{B} \bra{A}$$
Applying that here yields
$$\sum t_j ^{} \ket{j+1} \bra{j} + t_j ^{*} \ket{j} \bra{j+1}$$

[stoic python]

Dovahkiin

[vast iron]

Where summation is over j on both terms, obviously.
Compare this with T, should be obvious under what condition they're both equal.
Probably don't use Dirac notation in math server lest you want to get murdered

[rigid raft]

so $t_j$ must be complex?

[stoic python]

nova

[vast iron]

t_j is a complex number. If you compare T and it's conjugate, you see that the condition for T to be hermitian is that t_j* = t_j, or that t_j is real

(or t_j is complex with zero imaginary part)

[rigid raft]

ohhhh okay, thank you

and for b do i just multiply by |i> ?

[vast iron]

Yeah and keep the Kronecker's delta in your mind.

[rigid raft]

okay thank you :))

[rigid raft]

wait so itd just be $t_j\ket{j+1}$?

[stoic python]

nova

[rigid raft]

because the first term goes to 0?

or am i extremely misunderstood

[signal mulch]

Hey! sry for the late reply

but any idea how I can solve it using rank and nullity theorem?

[vast iron]

Split the summation into two
$$\sum (t_j \ket{j} \bra{j+1} \ket{i}) + \sum (t_j \ket{j+1} \bra{j} \ket{i})$$
Which is equal to
$$\sum (t_j \ket{j} \delta_{i, j+1}) + \sum (t_j \ket{j+1} \delta_i,j)$$
The first summation is 0 except when $i-1=j$, and the second is 0 except when $i=j$, so
$$t_{i-1} \ket{i-1} + t_i \ket{i+1}$$

[stoic python]

Dovahkiin

[teal grotto]

@signal mulch
if k is an eigen value of M for non-zero x, then we have
Mx = kx --> M(Mx) = kMx = k^2x = M^2x = Mx = kx. so k(k-1)x = 0 thus k = 0 or k = 1.

it suffices to show that the two eigen spaces intersect trivially. if v_0 satisfies Mv_0 = 0 and v_1 satisfies Mv_1 = v_1, then if av_0 = -bv_1, we have 0 = -bv_1 so b = 0. but since v_0 is non-zero, then a = 0

the two eigen spaces form a direct sum of V, so M is diagonalizable (assuming that M has at least one eigen value)

[rigid raft]

thank you so much, I really appreciate it :))

[vast iron]

Np, have fun

[signal mulch]

after solving part a, I got that A has 2 distinct eigen values and those are 0 and 1

essentailly what Im trying to show using rank and nullity theorem is this:

[teal grotto]

M doesnt need to have two distinct eigen values. consider M = I

[signal mulch]

ohh i see

I have a quesiton

what do we know about matrix M? like how many pivot points does it have?

it would be n right?

[teal grotto]

it has rank(M) = n - null(M) pivots

[teal grotto]

this is rank nullity. there is nothing to show. you already know this

[zinc timber]

correction, eigen spaces intersect trivially doesn't mean it's diagonalizable

[crystal oracle]

How to show $(I+ST)^{-1}S=S(I+TS)^{-1}$?

[stoic python]

Phi

[zinc timber]

S(I+TS)=S+STS=(I+ST)S

[teal grotto]

eff bro. the dimensions gotta sum up to n

damn my bad homie

[zinc timber]

doesn't mean diagonalizable

[teal grotto]

it does. the sum dimensions of the eigen spaces is dim V iff it’s diagonalizable

[crystal oracle]

Ehh, and what then?

[zinc timber]

manipulate

[crystal oracle]

Oh, right, I can multiply by the inverses. Thanks

Waaait. I hoped for a proof that wouldn’t assume that if $(I+ST)^{-1}$ exists, then $(I+TS)^{-1}$ exists.

[stoic python]

Phi

[zinc timber]

if one is, the other is also invertible

try to prove?

[crystal oracle]

Lol this is the problem I was originally trying to solve.

[zinc timber]

oh lmao

[crystal oracle]

I was trying to prove that if one is invertible, the other one also is. I wanted to use Woodbury identityy, the proof of which used the identity I gave above.

[zinc timber]

well I don't want to spoil it but

u can try looking into sylverter's law of determinant

[crystal oracle]

Alright thanks. Although I hate deteterminant so perhaps I’ll try to prove it in some other way.

[zinc timber]

@crystal oracle can you prove that ST and TS have same eigen values?

if yes, and I+ST is invertible means -1 is not an eigen value of ST, therefore not an eigen value of TS as well, so I+ST and I+TS both are invertible

[crystal oracle]

This is clear. But so far I don’t see why they have the same eigenvalues. Thanks for help, I need to go now, will try to solve this later.

[zinc timber]

"left as an exercise"

[humble meteor]

I guess S and T are square matrices. Isn't it that $det(I+ST) = 0$ if and only if $det(ST) = -1$ ? As $det(ST) = det(TS)$ then likewise $det(I+TS) = 0$ only in that case. If the determinant is not zero and the result is a square matrix then it's invertible.

[stoic python]

𝓶𝓸𝓶𝓸𝓶𝓸

[crystal oracle]

Yeah, turns out it's trivial. If TSv=λv, define w=Sv, and then STw=λw.

[azure bolt]

to know which of the lines is correct

[zinc timber]

kind of,

[rigid raft]

can anyone point me in the right direction with this? im kinda stuck

[zinc timber]

not really

[zinc timber]

is (A+A*) hermitian?

[rigid raft]

i want to say no

[zinc timber]

am I using a wrong definition?

[humble meteor]

I thought determinants followed an addition rule. 😦

[zinc timber]

wish it was true

[rigid raft]

idk, id assume the hermitian conjugate to be A dagger not A*

[humble meteor]

But det(A - B) = det(A) - det(B), right?

[zinc timber]

then use A dagger

[zinc timber]

no lol

[rigid raft]

so you only have to add A and A dagger and then its hermitian?

[humble meteor]

I saw this https://math.stackexchange.com/questions/1943840/determinant-of-a-matrix-subtraction-addition

Is that not correct?

[zinc timber]

lol have you read the first answer?

that literally disproves it

[humble meteor]

Ah, true.

Don't laugh at beginners.

It makes you seem like an asshole.

[zinc timber]

wait what?

[humble meteor]

I didn't read the answer properly, so it's my fault. But don't openly laugh at me for making a mistake.

[zinc timber]

[zinc timber]

I can't really say for sure that it was the intended solution. not familiar with physics way

[rigid raft]

ah okay, thank you though

[pearl plaza]

Hermitian is similar to transpose, but for complex numbers.
If you can make matrices for transpose, this shouldn't be much of a problem for you.

[azure bolt]

the subset A is not vector subspace
the subset D is vector subspace

But I can't answer about subset B and C

please help me

[zinc timber]

for B take c=½ and look at p(x)/2

for C i believe they meant MM* not MMt

[azure bolt]

MMt means M * the Matrix Transposed

[zinc timber]

well you can show that 2M doesn't belong to C

[azure bolt]

how

?

[zinc timber]

just compute (2M)(2M)^T

[azure bolt]

so only D is vector subspace

?

correct?

[zinc timber]

hm

[azure bolt]

i dont understand?

hm

[zinc timber]

'yes'

[azure bolt]

iam a portuguese

iam sorry

I don't understand the abbreviations well

[zinc timber]

stop apologizing

[azure bolt]

thank you

[winged prairie]

is a linear operator from f: V to V the same thing as an endomorphism

?

[gray dust]

yes

[winged prairie]

ty

[winged prairie]

what does it mean for a field to have characteristic

like for example F_256 has characteristic two

what does this mean?

[dusky epoch]

the characteristic of a ring is the smallest number of copies of 1 that add up to zero if such a number exists, and zero otherwise.

if you're told a ring (or a field) has characteristic 2, it means that 1 + 1 = 0 in that field

[winged prairie]

ok ty

[fresh socket]

Does the kronecker product exist for tensors with rank bigger than 2?

I only ever see it defined for rank 2 tensors

Also is there a relation of the kronecker product to the outer product?

[azure bolt]

nulity matrix A

[zinc timber]

[azure bolt]

this is the last one

@zinc timber

[tawny grotto]

hello so III here is true but i need help with how to figure out the geometric multiplicity for the eigenvalues

i just know the algebraic multiplicity of each is 1

[zinc timber]

looks like an exam

which one you think is true?

[dusky epoch]

@tawny grotto is this an exam? Y/N

[tawny grotto]

no its not

its a practice exam

not the actual thing

[wintry steppe]

after a shear transformation of a square on a graph, how can i determine if the area is preserved?

i believe its an equation with its matrix?

[nocturne jewel]

what tells you how the area changes under a linear transformation?

[wintry steppe]

the determinant?

[nocturne jewel]

yes

area A becomes, under a linear transformation T, |det(T)|A

[wintry steppe]

if T is R^2 - R^2, what does that mean exactly

wait nevermind
so i take the determinant of the transformed figure and multiply that by the original matrix?

@nocturne jewel

[nocturne jewel]

No

You asked if a shear transformation (T) preserves area or not

ie you want an area A to, under T, remain with area A

so det(T)=+-1

since $A=\abs{\pm 1}A$

[stoic python]

Mosh

[wintry steppe]

ok i think i get it now

thank you!

[zinc timber]

not like one's gonna admit it even if it was

[wintry steppe]

what if i wanted to know if the perimeter is preserved?

[nocturne jewel]

you'd do a lot of vector norms

[wintry steppe]

hm

not totally sure i catch that

ok hold on this is my transformation

1 becomes 2

[lavish jewel]

you're presumably applying the transformation to the position vectors. you can measure the perimeter by making vectors that join pairs of points along the perimeter of the shape, and taking their norms to measure their lengths

mosh suggests you do this before and after transforming

[wintry steppe]

i seeee what you mean, im not familiar with vector norms so i had to search that up. so A B C and D are coordinates

[nocturne jewel]

norm aka magnitude aka length

[wintry steppe]

okok

and so i first find vector AB

and another is BC and CD and AD?

[nocturne jewel]

Sure, but you can just do distance formula

since you're in R^2

[wintry steppe]

right
distance between each set of points then i compare that to the values i get in the tranformation i guess

[nocturne jewel]

You find the perimeter of 1 (which is 4.. cause it's a square of side length 1) and find the perimeter of 2

which is something like 2+2sqrt(2)

[wintry steppe]

ok yeah that makes sense now
thanks mosh

[vestal yarrow]

If I have a system:

$\sum \vec{x_i^T} (\vec{w} \cdot \vec{x_i}) + \sum \vec{x_i^T} b = \sum \vec{x_i^T} y_i$

$\sum \vec{w} \cdot \vec{x_i} + \sum b = \sum y_i$

is there a way I can represent this using Az = c, where $z = \begin{bmatrix} w \ b \end{bmatrix}$

[stoic python]

THEBIGTHREE

[lavish jewel]

looks like X^T (Xw + b) = X^Ty

i would advice against using x = [w;b] since you already have x_i for something else

[still lodge]

what is the difference between inner product and dot product

i know inner product is a more "general" version

but how

[lavish jewel]

the dot product is an example of an inner product

an inner product is a map V x V -> F that's linear in the first (or both) argument (more generally sesquilinear) and positive definite

that means it takes 2 vectors and spits out a positive number that can be thought of as a length of sorts

[zinc timber]

I'm not sure what the problem is

[zinc timber]

oh

[lavish jewel]

for clarity, the columns of X are the x_i in the original formulation

actually

i have to ask to make sure

what is the dot representing there

elementwise product or dot product?

X^T (X^Tw + b) = X^Ty, on second inspection

[zinc timber]

only one option?

feels like multiple option to me tho

[vestal yarrow]

dot product

[lavish jewel]

X^T (X^Tw + b) = X^Ty, then

[still lodge]

so suppose we're working in M2(R)

i can define an inner product "however i want" ig

dot product is just an example of one in R^n right

[zinc timber]

not "however you want"

there are restrictions

[still lodge]

ah that's the linear part ig

[lavish jewel]

however you want as long as it follows the conditions i gave above

[still lodge]

and as long as it operates on two elements of the same vector space

[vestal yarrow]

yeah I guess I could write it like that, but I still need to solve for w and b

[zinc timber]

though every inner product on finite dims can be written in the form x^TAy where A>0

[still lodge]

what does it mean for the output to be in a field tho

i havent learned anything about groups, fields, rings, etc.

[zinc timber]

yes

R or C

can't say much abt finite fields

[lavish jewel]

then simply let A = [X^T, I] and now you have Ax = y

[still lodge]

ryu wdym

[tawny grotto]

srry i shud have sent the answer choices as well

[zinc timber]

seems bout right

[zinc timber]

what do you mean by 'wym'?

[lavish jewel]

"what do you mean"

[still lodge]

my b

but yeah i just wanna understand the rule on the output of the inner product if that makes sense

[vestal yarrow]

I think I phrased it wrong using Ax = b
in this case it would be A.[w, b] = c

[vestal yarrow]

and I dont think this can work since b is a scalar

[lavish jewel]

ah b is a scalar

[vestal yarrow]

which also probably means this can't work

[lavish jewel]

you're working with row vectors

that changes things, yes

this is kinda weird if this is really what you mean, btw

cuz you have a sum of b, but b is a scalar

the only way to get this is if you take a vector of 1s and multiply it by b

[vestal yarrow]

oh yeah i could also just write bn

[vestal yarrow]

why would I need to do that

[lavish jewel]

if there are different values of b, then it's a vector, not a scalar

[vestal yarrow]

no its a scalar

[lavish jewel]

then there is no b_n

are you sure there is a sum there?

[vestal yarrow]

sorry b*n

yes

originally it was this:
$\sum x_i^T(y_i - \vec{w}.\vec{x_i} + b) = 0$

$\sum (y_i - \vec{w}.\vec{x_i} + b) = 0$

i just broke it up to try and isolate the terms

[lavish jewel]

X^T w + [1]*b = y, then

is y_i scalar or vector valued

[vestal yarrow]

scalar

[lavish jewel]

what is the sum summing over, i?

[stoic python]

THEBIGTHREE

[vestal yarrow]

yeah

[lavish jewel]

ok

then yes, there's a vector of 1s

[vestal yarrow]

well the dot product of w and x_i should just be a scalar

so I don't think you need that

well ig removing the sum and representing the X and Y as matricies would work but it seems unncecesary(?)

[lavish jewel]

you have that \sum (y - X^T w + [1]b) = 0

this is the same as [1]^T (y - X^T w + [1]b) = 0

you can stack w and b together by augmenting X^T with a column of 1s

[vestal yarrow]

alright yeah that works

so then you can break that up:

[1]^T.(X^Tw) + [1]^T.[1]b = [1]^T.y

is that right?

[lavish jewel]

right, and you see that gives you the n*b you wanted

[vestal yarrow]

yeah

[lavish jewel]

hmm i think a few signs need to be double checked

the b should be negative

[vestal yarrow]

ah thats mb, the (w.x_i + b) should be in parenthesis up there

[lavish jewel]

ah

[vestal yarrow]

wait so is the system:
X^T.(X^Tw) + X^T.[1]b = X^T.y
and
[1]^T.(X^Tw) + [1]^T.[1]b = [1]^T.y

hm

[tribal hearth]

anyone know how to prove these i suck a proofs?

[nocturne jewel]

what have you tried?

[glad acorn]

this is basically the reconstruction of inner product by the polarization identity

for (a) and (b)

[teal grotto]

do you know what it means for two vectors to be orthogonal? do you know how inner products and norms are related?

[drifting scarab]

b) follows directly from a) if you substitute the right vectors

if you defined the norm with a scalar product then it's relatively simple

[sinful valve]

the book mentioned like setting j = 1

makes v_1 = 0 ofc however it mentioned changing the second part of the proof

im not too sure what it would change to though

[zinc timber]

resolved?

[zinc timber]

@sinful valve so what are you trying to prove?

[zinc timber]

this one?

you posted the proof before the lemma

[sinful valve]

yeah it was like a specific case they were on about or sumn

like if you selected j= 1

then you could still prove by altering it somehow

[zinc timber]

honestly I'm not following

the steps you are doing is as follows:

1. Check if v1 is 0 or not, if zero remove it, otherwise leave it,
2. check v2 is in the span{v1} or not, if yes, remove it, if no, leave it,
3. check for v3 ....
   .
   .
   .
   you'll eventually run out of v's because your set is finite

[sinful valve]

huh that's not what the proof is on bout

like they say the largest element of 1 to m

[zinc timber]

yeah you choose the first vector in you list that turns out to be a LC of the previous ones

[sinful valve]

oh shit yeah it's just till a_j not 0 then

ig

so the case they had in question was when the vector is the first 1

which makes the linear combo 0 ofc

how does b change for this then they said its obv

[gray dust]

$u=c_1v_1+\ldots+c_mv_m=c_2v_2+\ldots+c_mv_m$

[stoic python]

RokettoJanpu

[sinful valve]

why does that work though

[gray dust]

v1=0

[sinful valve]

oh yeah

so thats literally it ight calm

so yeah if j = 1 then the RHS of a which was 0 adds up ye

[winged prairie]

what does characteristic polynomial have to do with polynomials? i am struggling to see the connection

[dusky epoch]

it... is a polynomial in t lol

by the preceding remarks
you might want to show these if you want more explanation

[winged prairie]

i dont't understand this

[dusky epoch]

in particular, the determinant function is applicable to the ring M_n(F[t]) of matrices with polynomial entries, and it is obvious from the formula for determinants (7.3.2) that the determinant of such a matrix is itself a polynomial.

[winged prairie]

oh ok thanks

[lavish jewel]

what exactly are you struggling with

[winged prairie]

I didn't understand this "determinant function is applicable to the ring M_n(F[t])"

but now it makes sense

i just didn't read it propely lmao

[dusky epoch]

[lavish jewel]

fancy way of saying you can compute the determinant of square matrices whose entries are polynomials

[zinc timber]

proof of cayley hamilton (|xI-A|)(A) = |IA-A| = |A-A| = 0

[autumn kraken]

If M is a subspace of V, and f: V -> W is a linear map, how do I start checking if <f(M)> = f(<M>)?

<> being the linear hull

[gray dust]

unpack all relevant definitions

[winged prairie]

how do u know when to use det(A-lamba(I)) or det(lamba(I)-A) for eigenvalues ?

[zinc timber]

both give you same eigen values

[winged prairie]

so it doesn't matter which one you use?

[marble lance]

Yes it doesn't matter

[winged prairie]

ty

[pseudo maple]

ho should i prove this, i talked about assosiativity and then i reached to my answer but it does not seem right

[zinc timber]

can you write $0\in \mbb{F}$ as $\lambda-\lambda = 0 $?

[still lodge]

is it normal for the rational canonical form to feel kinda contrived

[zinc timber]

maybe at first

but if you know modules, it'll feel natural

[still lodge]

i do not know modules

[still lodge]

if i have the polynomial x^4 - 4x^3 + 6x^2 - 4x - 7, is this the right companion matrix

[humble meteor]

So, if $Q(X) = X^\top A X + B^\top X + c$ then $\vec{\nabla}Q(X) = 2 A X + B$

[stoic python]

𝓶𝓸𝓶𝓸𝓶𝓸

[humble meteor]

Where X is a vector, A and B are matrices, and c is a scalar.

Which looks a lot like the rules for polynomial differentiation.

But I don't know why $X^\top A X$ would be 'equivalent' to $ax^2$.

[stoic python]

𝓶𝓸𝓶𝓸𝓶𝓸

[humble meteor]

Is there a set of rules for this?

[zinc timber]

matrix cookbook

also you can try to derive them by hand by separating out

[dusky epoch]

is there a matrix equivalent for x^3? i can't think of one

[still lodge]

inf column of zeroes w a 1 in the 4th slot...?

[zinc timber]

think it's ok

[humble meteor]

What's that?

[zinc timber]

http://www2.imm.dtu.dk/pubdb/edoc/imm3274.pdf

[humble meteor]

Thanks! Is that the latest version?

[lavish trout]

why do I need to prove that T is a scalar multiple of the identity operator?

is it not sufficient to show that any v which is not a characteristic vector is a cyclic vector?

[marble lance]

I think you are misreading it. You are not using the second statement to prove the first. You are proving the first statement and then using that to prove the second.

[lavish trout]

No, I think I understand that. It's simple to show that T has a cyclic vector. I don't get why I would also have to show the scalar multiple part

the "either ... or ..."

is this more appropriate in #groups-rings-fields ?

[marble lance]

What do you mean why?

They are asking you to prove it

It's like an exercise

[lavish trout]

I don't understand how the second sentence is equivalent to the third

[marble lance]

Assume you have proved the second statement. Now, we want to prove that either T has a cyclic vector or T is a scalar multiple of the identity operator.

So suppose T has no cyclic vectors.

Then by the second statement, that means every nonzero vector is a characteristic vector

Use that to show T is a multiple of the identity operator

[lavish trout]

is it not saying that the second statement is equivalent to the third?

[ionic laurel]

Can anyone help me with whether my answer/work for this question is right?

[marble lance]

No?

[lavish trout]

then why the "hence..."

[marble lance]

A hence B, means that B follows from A

So they are using A to show B

[ionic laurel]

(From a past final)

Would you just do [5b1 + 3b2]_C so you would then break it up into

5[b1]_C + 3[b2]_C which is just 5[-1 4] + 3[5 -3] = [10 11] ?

[marble lance]

That works

[ionic laurel]

Are you talking to me

Lunasong

[marble lance]

Yes

[ionic laurel]

Oh nice

Thank you

[marble lance]

[lavish trout]

ok that is strange wording

so it's like two exercises

[marble lance]

Yes... It's like part a and part b

And you use part a to do part b

[lavish trout]

got it

thanks

[marble lance]

👍

[wispy pewter]

is there a methodical way to solve these with matrices?

[marble lance]

Not really with matrices. This is more about knowledge of trig

[wispy pewter]

$$ a + b\sin^{2}(x) = 2( 1 - 2\sin^{2}(x))$$

[stoic python]

EmilD

[wispy pewter]

$$a + (b+4)\sin^{2}(x) = 2$$

[stoic python]

EmilD

[wispy pewter]

b = -4 a =2 is that how you do that

[nocturne jewel]

yes

slightly roundabout way to do it, but yes

[wispy pewter]

what is the clean way to do it?

[tame mural]

I kind of understand why a dot product in R^n is a projection operation

But if I think of the axis as arbitrary

Why does it end up as $|x| |y| cos(\theta)$

[stoic python]

meow

[tame mural]

And not sin?

[humble meteor]

Does quadratic-algebra exist?

Nonlinear-algebra

[tame mural]

Most algebra is non-linear I believe

[humble meteor]

But "linear algebra" isn't just ax+c. It has non-linear things too.

[wintry steppe]

quadratic algebra
https://en.wikipedia.org/wiki/Quadratic_form

though these are usually studied in a linear algebra class, they come up in many places

[humble meteor]

Ah, interesting

[wispy pewter]

kinda stuck on setting up a matrix for this

I can see that 6ln(x) and 8ln(x) are scalar multiples but not sure which to toss or how to set up a matrix

I know that I can* introduce other equations with the derivatives

[humble meteor]

In fact, I did see $x^\top A x$ already today, in the context of quadratic implicit curves.

[stoic python]

𝓶𝓸𝓶𝓸𝓶𝓸

[humble meteor]

I didn't know where it came from exactly, so this is handy.

[wintry steppe]

you can get rid of either of them

you know they can't both be in a basis since they're linearly dependent

so that cuts your basis down to either two or three

after getting rid of one of them are the remaining vectors linearly independent, or do you need to remove more?

[wispy pewter]

so if I toss the first I have 5A+8Bln(x) + ln(6)Cx = 0

0 + 8B/x + ln(6)C = 0

how do I put that into a matrix?

[hot swallow]

quick question, similar matricies have the smae eigenvalue and eigenvectors right?

like do the eigen vectors hve to be exact same and the same amount of eigenvectors?

[tranquil steeple]

eigenvalues same, not same eigenvectors.

[hot swallow]

https://i.imgur.com/4TsNlBu.png

but these two have same eigen values but are not similar

[tranquil steeple]

correct

[hot swallow]

ahh i see

[tranquil steeple]

but similar matrices share eigenvalues. but obviously not eigenvectors

[hot swallow]

i understand now ty

[tranquil steeple]

yeah, so similar matrices implies that they share eigenvalues but sharing eigenvalues does not imply similarity

[wintry steppe]

no need to put it into a matrix. can you use this to show that A = B = C = 0, or do you need to do some further reduction of S?

[wispy pewter]

how are you making the connection that A = B = C = 0?

[wintry steppe]

that's what it would mean for {5, ln(x^8), ln(6^x)} to be linearly independent

[wispy pewter]

right

but how can you tell that from my equations

[wintry steppe]

are you asking how to show that?

[wispy pewter]

yeah I don't understand how

from my equations its clear that A = B = C = 0

[wintry steppe]

is it clear?

[wispy pewter]

I dont think so

[wintry steppe]

if 5A + 8B ln(x) + C(ln 6)x = 0 for all x in (0, 1], and you want to show that A = B = C = 0, then try playing around with the equation a bit

maybe you can put in some points of (0, 1] for x

or use some calculus

[wispy pewter]

you can plug in random points?

I thought you had to keep it as x

[wintry steppe]

5A + 8B ln(x) + C(ln 6)x = 0 is true for all x in (0, 1]

so you can plug in points

[wispy pewter]

I know that plugging in points was possible

I just thought we had to show that for all points

so plugging in any one point wouldnt be allowed

[wintry steppe]

if i let f, g, h, k be the functions on (0, 1] defined by
f(x) = 6 ln x,
g(x) = 5,
h(x) = ln(x^8),
k(x) = ln(6^x),
then S = {f, g, h, k}. you're looking at the subset {g, h, k} and want to show it's linearly independent. this means that if A, B, C are scalars with Ag + Bh + Ck = 0 (i.e. equal to the zero function no (0, 1]), then A = B = C = 0. and saying Ag + Bh + Ck = 0 is the same thing as saying that 5A + 8B ln x + ln(6)Cx = 0 for all x in (0, 1]

to be absolutely precise

[wispy pewter]

so ok

the only thinking to do here is to see that there is one we can toss

sometimes with vectors there might have been a non obvious linear combination

but you're saying thats not the case here

[nocturne jewel]

function spaces it's a bit more obvious when things are linearly independent

[sacred palm]

can someone help for d?

[nocturne jewel]

check if C is invertible

[sacred palm]

doesnt work

[nocturne jewel]

wdym doesnt work

[sacred palm]

determine is 0

[nocturne jewel]

Ok, so compute C(0,1,1)

then solve Cx = that vector

[sacred palm]

when u say compute (0,1,1)

what do you mean

a bit lost hehe

[nocturne jewel]

Compute C(0,1,1)

the product

left multiply the vector (0,1,1) by C

[sacred palm]

perfect

done

[nocturne jewel]

so then solve Cx=that vector

[sacred palm]

0 5 3
0 2 1
0 -1 1

= cx right?

[nocturne jewel]

???

no

C(0,1,1)=[8,3,0]

[sacred palm]

i think im misunderstanding the compute part (im from a french sector just switched to an english school

[nocturne jewel]

so solve Cx=[8,3,0]

which is turning it into an augmented matrix, then performing row reduction

[sacred palm]

i get it but how did u get the 8,3,0?

[nocturne jewel]

I did the multiplication of C and [0,1,1]

[sacred palm]

ah

i get it

i shouldve put it vertically and multiplied it to the vector

[nocturne jewel]

$\begin{bmatrix}1&5&3\0&2&1\3&-1&1\end{bmatrix}\begin{bmatrix}0\1\1\end{bmatrix}=\begin{bmatrix}0+5+3\0+2+1\0-1+1\end{bmatrix}$

[stoic python]

Mosh

[sacred palm]

Yup

I did that

[nocturne jewel]

so then solve $Cx=\begin{bmatrix}8\3\0\end{bmatrix}\to\begin{bmatrix}1&5&3&8\0&2&1&3\3&-1&1&0\end{bmatrix}$

[stoic python]

Mosh

[sacred palm]

got it

now I find x1,2,3,4

and name the free variables

[wintry steppe]

any1 knows the forumala to solve linear equation word problems?

[nocturne jewel]

yep

[wintry steppe]

tell

[sacred palm]

theres a lot of formulas

show ur problem lol

[nocturne jewel]

wasnt directed at you... however you just read the problem and form an equation/system and solve from there.

[wintry steppe]

lol 45 problems

[sacred palm]

what answer u got for that problem i sent you or you havent done it?

because I am looking at the answer sheet not sure why its like that

[nocturne jewel]

,w RREF{{1,5,3,8},{0,2,1,3},{3,-1,1,0}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Omegabet_#9771

[sacred palm]

got the same answer

[nocturne jewel]

yeah, so you have x_3 free

[sacred palm]

but not sure

why its like this

as the answer

[nocturne jewel]

x_3 is free, agree?

[sacred palm]

yea

theres no pivot

[nocturne jewel]

so x_1=1/2 - x_3/2 and x_2 = 3/2 - x_3/2

so x=[x1,x2,x3]=[0.5-0.5x_3,1.5-0.5x_3,x_3]=[0.5,1.5,0]+[-0.5,-0,5,1]x_3

[sacred palm]

i see

[nocturne jewel]

so the subspace is 0.5[1,3,0]+[-1,-1,2]t, t in R

the answer in the book likely made x_1 the free variable

[sacred palm]

okay perfect

thank you bro

really appreciate it

[wintry steppe]

help?

[nocturne jewel]

#prealg-and-algebra

[worldly bear]

looks pretty linear to me

[dusky epoch]

@worldly bear see pins

[worldly bear]

was just joking around lol

[dusky epoch]

was not clear you were joking

[zinc timber]

just calculate A and B first

[vast iron]

Or
A = M L M^-1, B = N L N^-1
=> L = M^-1 A M = N^-1 B N
=> A = M N^-1 B N M^-1
=> P = M N^-1

[wise pollen]

hello! so i was given this root / zero, and I was asked to write it in a polynomial equation and i'm not sure how to do it

[marble lance]

If a is a root of a polynomial, then x-a is a factor of the polynomial

Here you are given two roots, so that gives you two factors

[wise pollen]

okay 😄

what's after this?

i don't think my teacher got to discuss to us about imaginary roots so I'm a bit confused with this one

[lavish jewel]

it works exactly the same way as with real ones

[wise pollen]

wait so, I just ignore the imaginary roots and just solve it like a normal polynomial equation given the roots?

[dusky epoch]

you don't "ignore the imaginary roots"

[lavish jewel]

i don't think you were asked to solve any equation, you were asked to find a polynomial that satisfies an equation

[wise pollen]

the instruction says to convert the given roots to an equation. it basically like finding the roots but backwards, instead of finding the roots of the equation, you find the equation of the given roots!

[lavish jewel]

sure, so take the roots and use them to build factors of a poly

[wise pollen]

oh

Ohhhh

I get it now. thank you a lot!!

[marble lance]

[wise pollen]

I got the answer now. thank you from the 3 of you!

[brisk parrot]

can someone help pls?

[potent pewter]

Hey guys ! this was my first ever Notebook please go through the notebook and give me your responses. Thank you ! https://www.kaggle.com/supreeth888/linear-algebra-for-machine-learning-1

[lavish jewel]

maybe #math-pedagogy is a better fit for this?

it's not wrong here, but you could get more feedback if you also ask there

[potent pewter]

I don't have permission to send message there...

[lavish jewel]

read #get-advanced-access

[potent pewter]

Thanks mate !!

[wintry steppe]

Hi! I want to know if (P(e),∩) is a group, P(e) set of sets of e. I found P(e) as identity element ,[∀A,∈(⊂?)P(e) A∩P(e)=A] but does an inverse element exists or is this not a group?

[zinc timber]

#groups-rings-fields

[torpid socket]

Hello so for this question Let A∈M6×6(F) with A ^ 3−2I6 =0. Calculate detA.

Can someone tell me how to approach it

[dusky epoch]

is that bad copying?

did you mean $A \in M_{6 \times 6}(F)$ with $A^3 - 2I_6 = 0$?

[stoic python]

Kanga Gang Annihilator Ann

[dusky epoch]

@torpid socket

[gray dust]

@torpid socket pls dont multipost

[torpid socket]

Sorry

[torpid socket]

Yes!

[torpid socket]

May I ask how do you do that

$A$

[dusky epoch]

this is a mathematical markup language called LaTeX

[torpid socket]

Oh

[dusky epoch]

anyway

just to get this out of the way

what is F

[torpid socket]

Finite field Im guessing

You can ignore it

[dusky epoch]

...

so it doesn't say?

i mean ok, i guess we should assume F is a field that contains whatever numbers we need it to contain

okay so can you tell us what you know about determinants?

[torpid socket]

Mhm!

[dusky epoch]

some properties of the det function perhaps

[torpid socket]

What do you mean 😅

Are you asking me to define and stuff

To test my knowledge?

[dusky epoch]

no, i'm not asking you to write out the definition of the determinant.

i'm asking you to write out some properties of the determinant. or determinant laws, or whatever else you want to call them

[torpid socket]

Oh just trying to understand your question I apologize

[dusky epoch]

for example, how could one rewrite det(AB)?

[torpid socket]

det(A) det(B)

Right ?

These laws ?

[dusky epoch]

yes these laws

[torpid socket]

Oh yeah I know them!

[dusky epoch]

okay

so consider those laws and A^3 = 2I_6

maybe it might be good to know what det(2I_6) is

[torpid socket]

Is it not 2?

[dusky epoch]

no

[torpid socket]

Oh

Oh yeah I will check

Wait why not check 2I_6?

I know how to solve it thank you so much! @dusky epoch

Im sorry if my questions were stupid still trying to figure my way with determinat 😅

[zinc timber]

have you asked this question before? I feel like I have seen this one before

[torpid socket]

I asked it in another channel

Never before today tho

[ionic laurel]

Hi I have a question

Let's say given this span of vectors

with A being the matrix of those 5 vectors treat W as colA

however the rank of this matrix is 3

so to find a basis for W perp would i just take the columns with a pivot and transpose it (lets say matrix B) so that Bx = 0?

would that be the basis for W perp?

[lavish jewel]

why the transpose?

ah i see why the transpose lol

yes

i was mixing it up with the null space. you can do it as you say

[raven parrot]

What are some applications of Diagionalization?

I do understand the process of getting D from P^-1 A P

[golden ermine]

You can find powers of matrices since, if A = PDP^{-1}, then A^k = PD^k P^{-1} where D^k is the matrix where the diagonal entries are raised to the power of k

[wintry steppe]

can anyone help with this?

[vestal yarrow]

Does what I did here make sense?

[ionic laurel]

Hm

I’m just confused because

Would i transpose all 5 vector columns

Or would i just take the 3 columns with pivots and transpose that

Thats my concern

[tropic pebble]

I am wondering why c = [-1,-1,-2]. I thought x would be positive. Can anyone explain? TIA

[halcyon spindle]

You don’t really have to do much if you can see it has a negative z. There only one vector in the picture that satisfies it.

[tropic pebble]

@halcyon spindle I know. But, why x is -1

[halcyon spindle]

Tho it does look wrong

[tropic pebble]

?

[halcyon spindle]

Yeah that what I just noticed

[tropic pebble]

The graph is wrong?

[halcyon spindle]

Yeah

Or the option d

[tropic pebble]

Oh, I struggled with this question, and it turns out that the graph is wrong omg

so, which part is wrong?

[halcyon spindle]

You can’t be sure ask your teacher. My guess is it option d.

[vestal yarrow]

I think it just looks weird in a 2d representation, not sure how you could clearly show the -x component from that angle

[halcyon spindle]

Wait his right.

[tropic pebble]

So, why x is negative then

[vestal yarrow]

It’s negative in the graph, you just can’t really tell because of the angle it’s shown at

That’s what it seems like to me anyways

[tropic pebble]

So, the graph is wrong?

[vestal yarrow]

No the graph is right, but unclear

[halcyon spindle]

The graph right.

[tropic pebble]

I see. What if the question asks me the coordinates of c instead

or the sign of x, y, z

[vestal yarrow]

I would hope they wouldn’t from that angle. You could bring it up with your teacher ig

[tropic pebble]

Gotcha. Thank you guys so much!

[wintry steppe]

hi all. i am trying to prove something like the following: "let V be a vector space of matrices equipped with some norm, |.|. then there exists a C such that, for any A in V, A_{i, j}/|A| < C". is this a dumb idea?

[loud halo]

no, it's not dumb.

as long as V is finite dimensional, all norms on it are equivalent.

so you can lower bound by the L^infinity norm, that just takes the max absolute value entry of your matrix

[wintry steppe]

@wraith wren that is more or less what i am trying to prove (all norms are equivalent for matrix norms)

[loud halo]

hmm

it's true for any finite dimensional vector space

but i don't remember why

[wraith wren]

why are random people pinging me?

[loud halo]

i think they're trying to ping other kanga gang people and not recognizing there's many lol

[wintry steppe]

i have seen a proof for vector norms, but it relied on the particular properties of the 2 norm for vectors. oh yeah sorry my bad

[loud halo]

it's the same for matrix norms

matrix norms are just fancy vector norms

[wraith wren]

also, where's the question?

[loud halo]

[wintry steppe]

|.| is not necessarily induced by a vector norm

[zinc timber]

you can map a matrix to n² dim VS

[loud halo]

what do you mean

yeah

[dusky epoch]

all norms on findim spaces are equivalent and projection maps are continuous

[loud halo]

by vector norm i mean that any vector space of (finite dim, nxn) matrices is isomorphic to a vector space consisting of column vectors with n^2 entries

[zinc timber]

like finding an isomorphism to Rn² and use eqv norms

[loud halo]

not like operator norm induced by a vector norm

[wintry steppe]

ok, just wanted to call that out

[zinc timber]

very complicated way to define a norm but a norm nevertheless

[ionic laurel]

Hi I have a small quesiton

Gram schmidt can only work on a set of linearly independent vectors correct?