linear-algebra

[faint lintel]

but I'm just getting this ugly summation inside a summation

[wintry steppe]

maybe sub J = N+D instead

[faint lintel]

yea so J^k = (N + D)^k

but don't I have to use the fact that ND = DN

can I just say that?

I'm just trying to see how JD = DJ implies that summation

[wintry steppe]

JD=DJ <=> ND+D² = DN+D²

[faint lintel]

wait what

[wintry steppe]

J = N+D

[faint lintel]

oh I see

[wintry steppe]

tho, you can see it more simply

D is a matrix of form \lambda I

those matrices commute with every other matrix

[faint lintel]

ye

that's how I proved DJ = JD

ty!

[wintry steppe]

If anyone wants to study for their LA finals, tag me and we can go into VC or sumn

[lament wasp]

i feel so lost on this question, my professor gave a hint saying use the point slope formula, but even with that im not really getting anywhere

with point slope formula, i have
y - b = m(x-a)
y - d = m(x-c)

from here, i dont know what to do. I tried rewriting the equations as
y - mx = b - ma
y - mx = d - mc
to maybe resemble something similar to a system of equations, but i dont think thats right

[autumn haven]

I need a tiny bit of help

[limber sierra]

can you write that determinant equation explicitly?

might be a good idea for a starting place.

[autumn haven]

the sum of the dimensions of all eigenspaces of an nxn matrix cannot exceed n

can someone help explain why

[lament wasp]

ok ill try that

[lament wasp]

i may be missing something but im not getting anywhere

xb - xd - ya + yc + ad - bc = 0

i tried factoring it but nothing linked to the first two equation

[limber sierra]

okay, now can you compute the slope?

(using the two points)

[olive grotto]

may i get help in one question

why is this wrong

[raw sand]

Multiplying A by a scalar will not make the determinant the product of that scalar and the determinant

the determinant will be scaled by a^n where a is the scalar the matrix is being multiplied by and n is the dimension of the matrix

so in this case it would be 2^(4)

[olive grotto]

ohhh

[raw sand]

det(2A) = det(A) * 2^4

[olive grotto]

thank you for explaining

[raw sand]

sure

[lament wasp]

okay ill try that and see where that takes me

[autumn haven]

ok one more question

$\begin{pmatrix}a & b \ -b & a\end{pmatrix}$

[stoic python]

Mr. Stewart

[olive grotto]

my brain hurts honestly

[autumn haven]

how do we determine the Q that makes this diagonalizable

well actually, if $$a, b$$ aren't distinct, i don't know if it even is diagonalizable

[stoic python]

Mr. Stewart

[olive grotto]

wtf the bot is talking

guys

i have a very simple question

how do you do this

[autumn haven]

can you row reduce?

[olive grotto]

im not sure tbh

sorry

i can attempt many times without an issue

but ofc i wanna understand the logic of doing this

idc about the mark

[pallid rampart]

Well there is a general method that works every single time, but in this case you can do a short cut

[olive grotto]

okay

im interested

[pallid rampart]

Ok so I will first talk about the short cut i guess

[olive grotto]

i appreciate it btw

[pallid rampart]

So the last equation tells you that x1 = -3-x4

[olive grotto]

yup

[pallid rampart]

now you can substitute this into the first equation and get $-3-x_4+x_2=-2$ or $x_2=1+x_4$

[stoic python]

Whoever

[pallid rampart]

Now substitute this expression for x2 into the second equation

you get x3 = 1-x_4

so now you have three equations: $x_1=-3-x_4$, $x_2=1+x_4$, and $x_3=1-x_4$

[stoic python]

Whoever

[pallid rampart]

So in this case x4 can be anything and then x1 x2 x3 are all determined based on x4

[olive grotto]

ohhh wow

give me a minute to absord

haha

[pallid rampart]

alright

[olive grotto]

ahh now i understand

after all theyre all simultaneous equations

[pallid rampart]

yes

[olive grotto]

thank you so much

[pallid rampart]

now just saying one thing

[olive grotto]

sure

[pallid rampart]

you see how the equation is written in the form v+sw where v and w are vectors

[olive grotto]

when u say vectors

is it like saying coefficients

[pallid rampart]

the column vectors

[olive grotto]

or u mean the variables

yeah i get it

i get the format

[pallid rampart]

the places where you can put answers

[olive grotto]

yes

[pallid rampart]

what what you were supposed to do was

[olive grotto]

write what the xs equate to

[pallid rampart]

find a particular solution to this equation

so like maybe x_4=1 and calculate x1 x2 x3 and put that in the first column vector

then you find the general solution if all four numbers on the right -2, 2, 1, -3 are 0's

[olive grotto]

i see

ill go experiment

again tysm really

[pallid rampart]

alright

[olive grotto]

btw

what does s mean

in the v+sw

[gray dust]

s is used as a free variable; here it means s is allowed to take on the value of any real number

[olive grotto]

thanks 🙂

[zealous junco]

is there a 1-1 correspondence btwn

[pseudo thicket]

We can tell easily that A can be diagonalized since it has distinct eigenvalues, however is the case for B, since B has a repeat of eigenvalue of 4?

[zealous junco]

all pos def matrices and inner products? in some finite dim vector space

[zealous junco]

you can check whether its diagonalizable by span of the second eigenspace

dimension*

so look at null space of A-4I, if its dimension is 2 then ur good

[mortal juniper]

@pseudo thicket wiz are u from ma1508e

[pseudo thicket]

what is that?

[pseudo thicket]

I'm not taught that yet, sorry!

[mortal juniper]

sorry i tot u were in the same math course as me

[pseudo thicket]

ntu?

[mortal juniper]

im nus

[pseudo thicket]

im ntu

[zealous junco]

eh idk how else to explain it then, sorry

[mortal juniper]

ah lol

but what is the difference between diagonalization and orthogonal diagonalization?

[zealous junco]

the idea is if you want diagonalizable then the eigenspace must have dimension n

in this case n = 3

[pseudo thicket]

im not there yet sorry

[zealous junco]

orthogonal diagonalization is when the matrix is normal

[pseudo thicket]

yup understood this, however from the picture we see that B has only 2 distinct eigen value, that means only 2 distinct eigen vector can be found, right?

[zealous junco]

there exist an orthonormal basis such that A = UDU^H

U contains the basis vectors

[mortal juniper]

just use symbolab it teaches u btr

[zealous junco]

no not necessarily

a repeated eigenvalue could produce an eigenspace that has larger than dimension 1

[brazen venture]

@mortal juniper what is symbolab 可以吃的吗？

[severe remnant]

Can anyone help me understand how to do part a

It's unlike any other basis question I've seen before and I'm not sure what the first step is

[mortal juniper]

looks complicated af

[severe remnant]

I thought you had to be given what x was and then use the basis to find what [x] should be

[mortal juniper]

yes

[pseudo thicket]

why did u think im from ur course tho

[pseudo thicket]

cant tell directly yeah?

[mortal juniper]

cos u sound very sg

can actually

[pseudo thicket]

ah haha

[pseudo thicket]

how though?

[mortal juniper]

We can tell easily that A can be diagonalized since it has distinct eigenvalues, however is the case for B, since B has a repeat of eigenvalue of 4?

yes

[lavish jewel]

find the eigenvectors and see if they're independent

[mortal juniper]

cos there are 3 distinct eigenvalues

and thus it can be diagonalized

B is not

cos it only has 2 distinct eigenvalues

[pseudo thicket]

I mean I am asking for matrix B

and we can't tell easily

compared to A

[lavish jewel]

time to do an evd in matlab then

[mortal juniper]

what do u mean easily^

[pseudo thicket]

oh wait your yes was to my question

[mortal juniper]

arent u supose to eigenvalue your matrix manually

[pseudo thicket]

like, we can just look at digonal unique entries boom we can easily tell it's diagonalizsanble

[pseudo thicket]

if it is diagonal matrix with unique diag entries we don't need to calculate eigenvalues

[mortal juniper]

oh i dont really like that method/assumption

but u can ask edd

[pseudo thicket]

it's a theorem that was just taught in my class haha

shortcut

[mortal juniper]

upper triangular matrix isit

[pseudo thicket]

upper and lower

both works

[lavish jewel]

any triangular

[pseudo thicket]

what course r u studying

[lavish jewel]

anyway, matlab says no

only 2 distinct eigenvectors

[mortal juniper]

ee, u?

[pseudo thicket]

oh insane course

cs

[mortal juniper]

lemme help u find that theorem

cs more rabs

[pseudo thicket]

ty dude, looks like it may or may not be diagonlizable

depending on the matrix

[lavish jewel]

pit depends on the number of lin indep eigenvectors

[lavish jewel]

you have no way other than finding them

[pseudo thicket]

thank you

[brazen venture]

can!

fibonacci sequence

[pseudo thicket]

i dont know why i am learning engineering math :^(

[mortal juniper]

cos ull learn that in coding ?

array matrix

[lavish jewel]

isn't comp sci a branch of ee?

[mortal juniper]

image compression uses coding and maths^

I think its called the spectral theorem

[lavish jewel]

also, the more theoretical part of computer science has some really fancy math

[pseudo thicket]

not really exposed to the practical application

also i dont see my future self applying calculus in cs

[lavish jewel]

what do you plan on doing with cs

[pseudo thicket]

software eng

[lavish jewel]

you didnt need cs then

[pseudo thicket]

ai and ml and things like that

[pseudo thicket]

need that in this country

[mortal juniper]

see if u can understand this @pseudo thicket

[lavish jewel]

software eng is super super not related to ai and ml

ai and ml is literally calculus, statistics, and linalg

[pseudo thicket]

depends which branch, no?

[lavish jewel]

no

[brazen venture]

no

r u y1 @pseudo thicket

[pseudo thicket]

yeap

nus can SU ntu can just gg

[mortal juniper]

i running low on SUs haha

thats y have to overload y1 to fully utilise the SUs

[brazen venture]

@lavish jewel are you my professor

u look sus

[lavish jewel]

anyway, in cs you'll learn how and why all of that stuff works. computational complexity, computability, the math behind ai and ml, etc

[pseudo thicket]

why here so many sg

[mortal juniper]

@brazen venture sus as in wat lol

[lavish jewel]

if you had studied informatics, you would only study how to use those things

[pseudo thicket]

yeap = cs = lowkey math course

[lavish jewel]

not how to make them yourself

and no, i'm not a professor

[brazen venture]

cos we are HELP vampires🧛

[pseudo thicket]

how to score straight As

._.

[mortal juniper]

lmao

[pseudo thicket]

do u have a lot of china profs in your uni?

[brazen venture]

depends on which course they are in?

[mortal juniper]

btw how can we find the limits of a matrix as the power of that matrix goes to infinity?

[lavish jewel]

diagonalize?

[pseudo thicket]

not in my school

[mortal juniper]

k nvm im asking a dumb qn

but can a matrix be a limit

lets say if they ask me find the ans for this question

and my ans is (0,0,0,0)

[zealous junco]

im pretty sure A can be normal too 🤔 for it to be orthogonally diagonalizable

[dusky epoch]

@mortal juniper you're asked to find the limit of a sequence of vectors

[zealous junco]

The spectral theorem states that a matrix is normal if and only if it is unitarily similar to a diagonal matrix, and therefore any matrix A satisfying the equation A*A = AA* is diagonalizable.

unless im not understanding what orthogonally diagonalizable means hm

[dusky epoch]

also uh. hmm

hold on

do we have the eigenvalues of A handy

3, -1, -1, -1

[mortal juniper]

so a limit is a value and it cant have vectors

[dusky epoch]

what?

the limit of a sequence of vectors, if it exists, is itself a vector.

[zealous junco]

yea i thought limits are only discussed when you have a normed space

[dusky epoch]

these things do make sense to talk about in real vector spaces.

we have a norm lol

anyway

A/6 has eigenvalues 1/2 and -1/6

[mortal juniper]

yes

[dusky epoch]

so yeah you are correct you'll have your sequence approach the zero vector

[zealous junco]

wait im confused whats the definition of convergence to vector? any norm right

[severe remnant]

Am I doing the second part correct?

I'm stuck at the last part for Cd

[mortal juniper]

@dusky epoch but the qn asks me to find the limit, so do i put 0 or (0,0,0,0)?

[dusky epoch]

(0,0,0,0)

[zealous junco]

ah i think i get it right

[dusky epoch]

i said "the zero vector"

[zealous junco]

what are like finite dimensional VS that is not a normed space, I thought you could always represent stuff in coordinates and that would produce a norm

unless the field is wacky?

[dusky epoch]

i mean

there's no canonical norm on the space of polynomials of degree <=k

you can have many different norms

for finite dimensional spaces all norms are equivalent

(in the topological sense)

[zealous junco]

oh, i thought the canonical one would be the absolute value lol

wait hm

right you need maybe something like $max_{x \in [a,b]}|f(x)|$?

[stoic python]

Anticipation

[zealous junco]

and ig that would no longer be space of polynomial but on an interval

[dusky epoch]

\max

[zinc copper]

how do we know that we can do the "in other words" part?

oh i think i might see why

yeah they establish that you can do this by induction basically, nvm my question

[zealous junco]

There is a 1-1 correspondence between all positive definite matrices and all possible inner product right? Given that you chose some basis

In other words, I would think if x,y in R^n and A is a positive definite matrix then (x,y)_A = (y,x)_A

and how about just semi-positive definite, I assume this also holds

its just that instead there is more zeros

[lavish jewel]

with positive semidefiniteness, it just changes the result from > 0 to >= 0

(when looking at the inner product of x with itself)

[zealous junco]

Is it true though that all inner products on C^n can be represented by positive definite matrices, given that you choose some arbitrary basis B? or I think that basis has to be orthonormal?

hm i believe if it was orthonormal basis then the PD matrix would just be identity

so i guess arbitrary basis

nvm this statement is true i found it

wait i got it

also how do you show d^(k) is indeed a descent direction

[lavish jewel]

you show that f(x - ad) < f(x)

the common approach is to use a 1st or 2nd order taylor expansion around x

[zealous junco]

ah thx

[dusk sage]

There's something in my notes about metric spaces which I think is a mistake

shouldn't this be the other way around?

so d(x,x) = 0

d(x,y) = 1

[stable kindle]

yeah

[dusk sage]

thanks

[coarse sandal]

Is it possible to find part C using part A and Part B

k is 3

[marble lance]

Assuming A inverse exists, just multiply that equation through by A inverse

@coarse sandal

Then the -4I term will have an A inverse that you can solve for

[coarse sandal]

oh okay

thank you

[mild current]

Can we use matrix multiplication on non-linear transformation

srry for a dumb question

my linear algebra course is mainly focused on computations

[dire thunder]

how would you define matrix of nonlinear map

[zealous junco]

you can for computing gradient

or i mean

jacobian and hessian in general

[jagged granite]

For a bilinear form isnt it true that if $(x,y)=0$ then $(y,x)=0$

[stoic python]

bert

[native rampart]

Only for symmetric bilinear forms

[jagged granite]

(x,y)=(x,y)+(y,0)+(-x,0)+(0,x)+(0,-y)=(y,x)

I am thinking something stupid

[spiral star]

adding 0 a bunch of times to (x,y) doesnt make it the same as (y,x)

consider a bilinear form on R² given by (x,y) = x1 * (y1 + y2) for x = (x1, x2) and y = (y1, y2)

then for the standard basis vectors e1 and e2 we get (e1, e2) = 1 but (e2, e1) = 0

[quartz compass]

break it up into individual steps, there's several illegal moves going on here

[mighty minnow]

can someone explain the last part in blue..?

[lavish jewel]

the transpose of a scalar is the same scalar, basically

[mighty minnow]

but they are all vectors,,

[lavish jewel]

Au dot v is a scalar

the dot product gives you scalars

also what you sent already has the explanation

Au dot v = v^T A u

it's in the first line with blue stuff highlighted

[mighty minnow]

so its like the same thing? as the last part basically?o-o

[lavish jewel]

v^T A u is a scalar, and its transpose is the same scalar

the whole first half of the page says the same thing in like 3 different ways

[mighty minnow]

okay, ill revise it

thanks!

[mortal juniper]

for this qn, what are we expected to show for part b given that we have solve part a for its A=PDP^-1

[lavish jewel]

look at what happens when you raise 1/6 A to an integer power

you'll get (1/6)^2 P D P^-1 P D P^-1 = 1/6^2 P D^2 P^-1 if you square it

i'm sure you can already see the pattern

[stone veldt]

Find the 20th term of the arithmetic sequence whose common difference is d=3 and whose first term is a1=5 .

[lavish jewel]

wrong channel

[mortal juniper]

so is it like this

where D=

[lavish jewel]

that looks ok

now multiply P D^n P^-1 to get the final matrix

[mortal juniper]

isnt the final matrix

[lavish jewel]

that's (1/6)^n D^n

you didn't multiply by P and P^-1

[quartz compass]

no you don't X out the P and P^-1 matrices

$A=PDP^{-1}$ means $A^2= (PDP^{-1})(PDP^{-1}) = PD^2P^{-1}$ notice how the inner $P^{-1}P=I$ cancels out here but the outer ones are still there? @mortal juniper

[stoic python]

Merosity

[mortal juniper]

this is my final matrix

and this is what i got when i multiply with a,b,c,d

[lavish jewel]

it's rank 1, as expected. that's all i can comment

[mortal juniper]

what does the a,b,c,d mean, is it even useful

[lavish jewel]

just some generic 4d vector

[mortal juniper]

so are they actually asking for the limit of A which is the 4*4 matrix of 1/4?

[quartz compass]

you can interpret it as meaning as you continually do the A transformation on any vector it eventually gives you something that looks like the average of all the entries of the original vector

well, (1/6)A I should say

[mortal juniper]

ok thanks!

[mortal juniper]

for this question

can i just neglect the imaginary part

or do i have to make a new variable such that it contains imaginary solution

[lavish jewel]

iirc you can absorb the imaginary part into the constant

[steel moon]

are all linear transformations diagonalizable?

nvm

how can you tell what has determinant 1

a e f look the moist similar to a

[lavish jewel]

the volume is preserved

what you noticed as "similar" in those letters is that the lengths were preserved

and the product of the lengths is the "volume"

notice B also looks similar to the original figures, but it wasn't only rotated, it was also enlarged

enlarging changes the volume

D and G had a length shrunk down to 0, so that has volume 0

i don't have a nice and easy argument for the shear in C

[steel moon]

for D and G, do they have eigenvaluie?

[lavish jewel]

probably sqrt(2) and 1 for one of the eigenvalues and 0 for the other

[steel moon]

oh ok

and im guessing b and f dont have eigenvalue because its orientation changing

[lavish jewel]

these things all have eigenvalues

just that some may be 0

[steel moon]

rotation by 90 degrees has eigenvalue?

i mean real eigenvalue

[lavish jewel]

ah, that's different

[steel moon]

i was asking if its true that B and F dont have real eigen values

thats what i guess at first glance im not sure if there are others

[lavish jewel]

you can create the corresponding transformations as matrices and check yourself

[steel moon]

does subset of a basis for R^2 mean something like pics A / B?

[winter acorn]

How would i go about balancing this equation? Havent done chemistry in a while but I feel like I should revive my linear algebra to complete it. Could I find the nullspace vectors and they would be my coefficients?

[steel moon]

is $A = PBP^{-1} the same as A = P^{-1}BP$ in the context of similar matrices

[stoic python]

add

[lavish jewel]

it isn't unless P is an involution

you can write one in the form of the other by doing a substitution for some other matrix V = P^-1 though

[manic oasis]

What's the simplest way to see this?

I guess one way to do it is to introduce a basis ${e_1,\dots,e_m}$ for $V$ seen as a \textbf{complex} vector space, and try to reason that ${e_1,\dots,e_m,Je_1,\dots,Je_m}$ is then a basis for $V$ as a \textbf{real} vector space. But this is done later in the chapter, which leads me to believe there is a simpler way to see it here. Let's assume that $V$ is finite-dimensional, I'm pretty sure this is an implicit assumption in this context.

[steel moon]

is this a matter of substituting each value?

[stoic python]

gustavn64

[steel moon]

is the linear transformation induced by a 2x2 elementary matrix mean the matrix is diagonalizable?

[tardy reef]

no idea where to start

[stable kindle]

w^3 = 1

[steel moon]

hi sorry to ask this again but i was wondering what it meant when for a subspace $V \subseteq R^2$, which sets are a subset of V

[stoic python]

add

[faint lintel]

Are you asking what it means for a subset to be a subspace of a vector space?

3 criteria

1: the subspace must contain the 0 element of the original space

2: it must be closed under addition of vectors meaning addition of any two vectors in the subspace results in a 3rd vector still in the subspace

[steel moon]

like which of those images are subsets of V?

[faint lintel]

3: it must be similarly closed under scalar multiplication

Subsets or subspaces?

[steel moon]

Which of the drawings in the image could be possible subsets of $V \subseteq \mathbb{R}^2$

[stoic python]

add

[steel moon]

where V is a subspace

[wary lily]

AFAIK, and I don't know much LA, all of A, B, C, F, G, and H are subspaces of R^2

@steel moon

[steel moon]

aren't all of them subsets of r^2

that wasn't the quesiton though

[wary lily]

sorry

[steel moon]

lol its ok

[wary lily]

I read your question from previous and restated it

give me a sec

they are subspaces of R^2

was what I meant

I'll edit the original above

[steel moon]

why not D?

[wary lily]

I'm not quite sure about it, it could be

let's call helpers on this

I'm curious

[wary lily]

<@&286206848099549185>

[rain echo]

is it a subspace of V

or a subset

so is the question "whihc of the following could possibly be a subset of a subspace of R^2"

[wary lily]

yes, that's what I understand

@rain echo

[rain echo]

well that would be what is written in a problem

unless V is a specific space given to you

[steel moon]

its not

oh well its fine

i had another question as well

[wary lily]

I didn't understand the answer to the problem

[steel moon]

can i say these two matrices are similar if their detertminants are the same?

[rain echo]

add, that is not polite

[wary lily]

which of those pictures are subsets of a subspace of R^2?

[rain echo]

i am in the middle of giving an explanation to 0.1

[steel moon]

it was my question lol

sorry

[rain echo]

shit

im sorry

i forgot

i thought you just came in

[wary lily]

yeah, but I tagged helpers

it's OK

[rain echo]

cause 0.1 was the one who pinged helpers

[wary lily]

explain it to me after you answered Add's recent question

[rain echo]

im sorry add

i will give you both answers

[steel moon]

its ok lol

[wary lily]

[rain echo]

first one: R^2 is a subspace of itself. so any subset of R^2 is also a subset of some subspace

you could even take a finite set of vectors

they dont have to form a subspace themselves

but if all it needs to be is a subset of a subspace

everything workd

works

2: no, two matrices are not necessaruly similar just because they have the same determinant

although, if they are similar, they will have the same determinant

[steel moon]

are those two matrices similar?

[wary lily]

seeing the determinant of a 2*2 matrix as the area of a whatever it is called, you can imagine two different shapes with the same area

[rain echo]

no

[wary lily]

what was the shape called for which the determinant of a 2*2 matrix is the area?

a parallelogram?

why this one isn't similar or why it isn't necessary similar when the determinants are equal?

[rain echo]

there are other properties preserved under similarity

[solid flower]

why is euclidean geometry annoying

[pallid rampart]

The two matrices are not similar

Every matrix is similar to a jordan matrix

Jordan matrices are made up of jordan blocks

And two matrices are similar iff the jordan matrices they are similar to have the same jordan blocks

The matrix B is made up of two jordan blocks with eigenvalues 2 while A is made out of 1 jordan block with eigenvalue 2, so they are not similar

Another way to see it is that $B=2I$ so any matrix similar to $B$ will be $M\inv BM\inv=2M\inv IM=2I=B$ and so $B$ is only similar to itself

[stoic python]

Whoever

[pallid rampart]

@wary lily @rain echo @steel moon

[rain echo]

yeah I just got busy and didn't come back

I didn't think I should explain jordan blocks or eigenvalues

[pallid rampart]

True

Ig the second explanation is just much easier

[wary lily]

thanks

[wintry steppe]

Could someone check my work?

[odd quest]

If I know the eigenvalues of a matrix, and I also know that the matrix is diagonalizable, can I get the characteristic polynomial from the eigenvalues without computing it the normal way?

[wintry steppe]

What is normal way? Determinant?

[odd quest]

Yea

[wintry steppe]

Yes, you can

[odd quest]

How can I be sure there isnt some constant factor outside, though

like a(x-lambda1)(x-lambda2) etc

[wintry steppe]

If it's diagonizable, simply look at the main diagonal

https://www.symbolab.com/solver/step-by-step/eigenvalues \begin{pmatrix}1%262%261\\ 6%26-1%260\\ -1%26-2%26-1\end{pmatrix}?or=sug

[odd quest]

but

so lets say the eigenvalues are 1,2, and 3, and its diagnoalizable, then it would be (x-1)(x-2)(x-3)

but how do i know its not -(x-1)(x-2)(x-3) or something

[wintry steppe]

IT will have only one solution for the polynomial given that you find the determinant correctly

[odd quest]

wdym

[wintry steppe]

https://matrixcalc.org/en/

[odd quest]

but the point is, im trying to find it without the determinant

so i know you can go chararacteristic polynomial -> eigenvalues

but im trying to figure out whether this is reversible

[wintry steppe]

https://www.youtube.com/watch?v=TQvxWaQnrqI

This will explain all

[nocturne jewel]

If you know it's diagonaliziable, iirc the diagonal entries will be the eigenvalues and the # of times they show up is the algebraic multiplicity

[odd quest]

right

so i at least know the polynomial without a possible constant

im trying to understand whether or not there could be a constant though

[nocturne jewel]

yes you know the polynomial up to a scaling factor

but the scaling factor doesnt change anything about the roots, and subsequently the spectrum

[odd quest]

right

[nocturne jewel]

so just say the scaling factor is 1

[odd quest]

i just wanted to verify there is indeed a scaling factor that is unknown when going from the eigenvalues

[nocturne jewel]

yes cause the only numbers you'd have to plug into the char poly are the eigenvalues

[odd quest]

great thanks that answers my question

[nocturne jewel]

but you already know $P(\lambda) = 0$ so they dont help find a

[stoic python]

moshill1

[odd quest]

right

tysm!

[zealous junco]

is there fast way to do multiplication if A is pos def and ur doing

$v^\top A v$?

[stoic python]

Anticipation

[zealous junco]

like non-naive way

cuz $v^\top v$ is O(n) and i would think potentially every inner product can be computed in O(n)..?

[stoic python]

Anticipation

[zealous junco]

but maybe that requires pre-computing the laws of the inner product

[reef prism]

did I do the maths correctly?

its least squares using a regularization term in the end (ridge regression, but that doesnt matter)

[lavish jewel]

i've never seen this written out as sums before, my brain melted

[unkempt bronze]

can someone help me with my optimization problem

[reef prism]

@lavish jewel oh you helped me before with some type of differential equation

[lavish jewel]

oh, really?

[reef prism]

yeah was like 5 months ago or something

[lavish jewel]

wow what

anyway, gimme a sec to write this out in matrix vector form and double check

[reef prism]

wait a second

the derivative

is it correct?

my next step is to write it in matrix form and i dont know what to do with the B_0 as it is a constant actually

[lavish jewel]

i will check that rn

[reef prism]

ahg no it wasnt you who helped me before i think but your name seems familiar

oups

[lavish jewel]

the overall form looks ok, but some things are not

the last part has a sum of 2 alpha * sum(beta_i)

but all the b_i where i =/= j are 0

so you don't need the sum, only 2 alpha beta_j

it's possible the left part also has a similar mistake

[reef prism]

what do you mean by b_i = 0, where i =/= j when the sum is from j = 1 to some number p?

[lavish jewel]

you're differentiating $\alpha(\beta_1^2 + \beta_2^2 + \cdots + \beta_P^2)$ w.r.t. $\beta_j$, yes?

[stoic python]

Edd

[lavish jewel]

so $\frac{\partial (\alpha(\beta_1^2 + \beta_2^2 + \cdots + \beta_P^2)#)}{\partial \beta_j} = 2 \alpha \beta_j$

[stoic python]

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

[lavish jewel]

the notation in the image is bad to begin with, because it may lead you to believe you differentiate w.r.t. EVERY b_j, but that would yield a gradient vector

so this is poor notation at work

do you want a partial derivative, or all of the partial derivatives?

[reef prism]

ai ai ai

k so i think its all Bj from 1 to p

my al is to minimize this least squares function

with regards to every b except b0, because b0 can be calculated seperately

so what would be the right notation? would the right notation be $\dfrac{\delta J(\beta_(1..p))}{\delta \beta_(1..p))}$

[stoic python]

Ronaldinho777

[reef prism]

this is my first time deriving anything in matrix form

[lavish jewel]

i would do it for a single beta_k at a time and put them in a vector

or directly in vector form, for the vector beta

[reef prism]

how would you incorperate the b0 in matrix form

[lavish jewel]

a vector of ones multiplied by b0, if it's just a scalar

[wintry steppe]

Sorry to interrupt, I have a quick question to clarify : If B is set of n-1 linearly independent vectors then clearly B is a basis of Lin(B). Then Lin(B) has dimension n-1. Is that correct?
My confusion arises from the fact that my textbook says Lin(B) has dimension at least n-1

[wintry steppe]

is Lin(B) the span of B

it didn't say B has at most n - 1 linearly independent vectors

it could have n

in which case the dimension is n

etc

@wintry steppe

oh i misread

well

[wintry steppe]

Yes

i guess the book just has shitty wording. you're right

either that or they actually meant to write "contains n - 1 linearly independent vectors"

[wintry steppe]

Thanks for the clarification @wintry steppe !

[wintry steppe]

Does anyone have an idea about proving this?

[wintry steppe]

PS: I would like to know how to prove that the trace is -a_n-1

[nocturne jewel]

do you want hints for both parts?

or just the trace part?

[wintry steppe]

I've done the determinant part, it's pretty straightforward with x=0

[nocturne jewel]

yes

[wintry steppe]

For the trace, I have the idea of expanding the determinant of (xI-A)

But that looks complicated

[nocturne jewel]

Trace of a matrix is equal to the sum of the eigenvalues

[wintry steppe]

I am interested in a proof from "first principles". While your statement is true, its proof relies on the statement I am trying to prove

[nocturne jewel]

I mean you just wanted to know how to prove it, so I gave you a hint

if you wanted a fully flushed out proof, probably best to specify that

[wintry steppe]

Apologies, should've done that 😄

[wintry steppe]

you could just write out the determinant of xI - A and stare at it to figure out what the linear term's gonna be

[edgy kelp]

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

This is true right?

It is linearly independen t

[wintry steppe]

(1 - t) + (t - t^2) = 1 - t^2, so no, this set is linearly dependent

@edgy kelp

[sage verge]

what is meant here by "consistent"

[gray dust]

@sage verge a solution to Ax=b exists

[sage verge]

so it just means there exists a solution x?

[gray dust]

that's what i mean

[sage verge]

alright thanks

sorry sometimes i just get confused and kinda need to restate it to make sure i dont have the wrong idea

[gray dust]

sure np

[tame mural]

Does anyone have a recommendation for intro to graph theory?

[dreamy iron]

If I have two basis vectors which are not equal. And I take one and subtract it from another, I am guaranteed to never et the zero vector.

Is this easy to prove?

[stable kindle]

if you have two non-equal vectors and you take one and subtract it from the other, then you never get the zero vector

[dreamy iron]

yeah i believe that

but whats the proof?

[stable kindle]

if you did get 0, then they have to be equal

contradiction

[dreamy iron]

it seems to follow immediately from definitions

lol....that feels too easy....

[stable kindle]

no, like

if you have two anythings and they're not the same, then the difference between them isn't 0

numbers, vectors, chairs, anything

this is exactly as easy as it deserves

[limber sierra]

suppose u - v = 0

add v to both sides

u = v

[dreamy iron]

okay, so it's the only thing i need to justify that if a linear map sends two basis vectors to the codomain, and theyre equal in the codomain under the transformation, then the linear map is sending those two basis vectors to the zero vector

TYTY

[wintry steppe]

im confused how to do this

[rotund jetty]

what did you try?

recall the definition of eigenvalue and eigenvector; an eigenvector is a vector $v$ so that $Av = \lambda v$ for some scalar $\lambda$, which is the eigenvalue associated with $v$. You'll just need to compute $Av$

[stoic python]

Nicholas

[mortal juniper]

Hi may i ask if i have this matrix that has the following eigenvectors

how do i proof that (1,i) is an eigenvector of the above matrix

i cant just flip the rows of the 2nd eigenvector right?

[gray dust]

@mortal juniper can you directly apply the def of eigenvector?

[mortal juniper]

no

[gray dust]

definition

by no do you mean you don't know the def?

[mortal juniper]

i know the definition but idk how u can just swap the rows of the eigen vector to proof that it is an eigenvector of A

[lavish jewel]

they are not swapped, they are multiplied by i

[gray dust]

can you compute A(1,i)?

[mortal juniper]

so i take i*(-i,1)?

and i get (1,i)?

[lavish jewel]

please listen to roketto

and test the definition

and yes

[mortal juniper]

yes

is this what u mean?

[gray dust]

yes. can you factor it as L(1,i) for some scalar L?

[mortal juniper]

L is 2

[gray dust]

are you sure that vector is 2(1,i)?

[mortal juniper]

R1= 2(1+i), R2= 2(-1+i)

[gray dust]

no, after factoring out L we must be left with (1,i)

again, we're finding L so that (2+2i,-2+2i)=L(1,i)

[mortal juniper]

sry im not so sure how to factor out a 22 matrix to a 21 matrix

i know that if i take this eigenvector

multiple to i

i will get (1,i)

[lavish jewel]

he is asking you to factor a scalar from a vector

there is some constant c such that c(1,i) = (2+2i, -2 + 2i)

if you show that such a constant exists, then (1,i) is an eigenvector of the matrix

by definition

hint: the constant is exactly that eigenvalue. try it out.

[mortal juniper]

but how do i find c

[lavish jewel]

by factoring out

your best hint there is that c*1 = 2 + 2i

so...

[mortal juniper]

2+2i

and the conj is 2-2i

[lavish jewel]

and what's 2 + 2i times i

[mortal juniper]

-2+2i

[lavish jewel]

so what conclusion have you reached?

[mortal juniper]

interesting

[gray dust]

a good way to find such an L is to focus on the entries of the vector separately

look at the 1st entry 2+2i. what can we factor out to get 1?

[mortal juniper]

2+2i

[gray dust]

look at the 2nd entry -2+2i. can we factor out 2+2i to get i?

[mortal juniper]

yes

[gray dust]

that means we can factor 2+2i from the vector to get (1,i)

[mortal juniper]

ok thanks a lot

[lavish jewel]

this should be a reminder to you that any scaled version of an eigenvector is also an eigenvector

[gray dust]

taking L=2+2i, we verified A(1,i)=L(1,i)

[lavish jewel]

the scaling was i w.r.t. the vectors you found previously

[gray dust]

ie (1,i) is an eigenvector of A (corresponding to an eigenvalue 2+2i)

[wintry steppe]

The sum of two subspaces is direct if their intersection consists of the zero vector only

Also, a minor correction in your argument : eigenvectors corresponding to distinct eigenvalues are linearly independent

if T is diagonalizable then you can find a basis of V such that the matrix representing the transformation with respect to that basis is diagonal. But isn't it the same as with normal diagonalisation of square matrices? The matrix [T]B is the same as the diagonal matrix you obtain with entries of the main diagonal corresponding to the eigenvalues. If you can't see this directly, you can use the formula for [T]B : the columns of this matrix are [T(u)]B for every u in the basis of B

But then if such matrix exists, that means you get just enough eigenvectors to span the whole space V

[sacred crescent]

If it was diagonalizable, then you could write any vector as a combination of eigenvectors where you can further group the eigenvectors into the distinct eigenspaces, where now if you assume that the vector is 0, then each eigenspace will have to be 0 because the eigenvectors are linear independent

[lavish jewel]

to be able to diagonalize a square mat, all you need is linearly independent eigenvectors

as an example, take the identity matrix of size 3x3

the eigenvalues aren't distinct

it is trivially diagonalizable as I I I^-1 though

[wintry steppe]

Enough linearly independent eigenvectors

Otherwise you get a JNF

[lavish jewel]

yeah, i meant all of them lin indep

[wintry steppe]

The key point here is that you should have as much lin independent eigenvectors as the size of the matrix

[dreamy iron]

So I’ve been trying to prove something all day that’s not true....

Given two distinct vectors in the domain, if the linear transformation sends both to the same vector, then they both get sent to the zero vector in the codomain..... which is not necessarily true

Like if you take R3 and send every point to R2, by a projection to the xy plane....

[lavish jewel]

that's indeed not true

[dreamy iron]

That’s linear, but an infinite number of distinct vectors in the domain get sent to non zero vectors.

[lavish jewel]

it's their difference that gets mapped to 0

[dreamy iron]

Huh....

Interesting.....

[lavish jewel]

do you know what the null space is?

[dreamy iron]

domain vectors that get sent to zero in the codomain.

[lavish jewel]

yep

so say Ax and Ay have the same image

Ax = Ay

then Ax - Ay = 0 (the zero vector)

by linearity, A(x-y) = 0

which means, by definition, that x-y is in the null space

so x and y share the same component in the complement of the null space

[dreamy iron]

oh...see i have that written out on my scratch paper. Ax-Ay = A(x-y). and i was trying to manipulate it somehow.

[arctic stone]

Is it true for diagonal matrices and for any function that f(M) would be just taking f(each diagonal component)?

like for example sth like this

is it actually true for any kind of function?

[limber sierra]

how are you defining f(M)?

if youre defining it as a matrix-valued function, then f(x) doesnt make sense in general for diagonal entries x

[arctic stone]

I was just thinking about generalizing things such as e^M

sqrt(M)

[limber sierra]

otherwise i dont think you can define f(M) in the general case

like if f(x) = x+1, whats f(M)?

[arctic stone]

oh I see the problem

[limber sierra]

you could, i suppose, define it as M + the identity matrix

but you can see that this might be hard to do with less well-behaved functions

if f(x) = |x|, whats f(M)? if f(x) = number of prime numbers at most x, whats f(M)?

that said, for "suitably well-behaved stuff" we often get properties like this

[arctic stone]

yeah it's not gonna work for all functions for sure

[limber sierra]

whats "really going on" is just that diagonal matrices multiply really well

so if your function boils down to multiplication

then yeah, its multiplication will be compatible with diagonalwise multiplication

matrix exponentiation does ultimately boil down to an (infinite sum of) matrix products, so that works

and the factorial works as well

[arctic stone]

yupp I know this too, I saw sin(M), ln(M) and other funny things so I thought that it could be generalized for more types of functions

now I understand that it's only those well behaved ones

[limber sierra]

yeah you can do some weird stuff for sure

[arctic stone]

with a power series expansion

[wintry steppe]

Thank you for your willingness to help last month, and your patience with me @wary lily , @lavish jewel , @nocturne jewel . Couldn’t have passed without your insights

[limber sierra]

typically we only bother to do this for things that behave well when generalized to matrices

and "diagonal matrices respect multiplication" is usually one of the criterias by which we judge that

so its kinda circular

[arctic stone]

cool stuff for sure

[dreamy iron]

What are the prototypical examples of infinite dimensional vector spaces?

1. F^∞
2. P(ℝ) set of all polynomials with real coefficients.
3. ℒ(V,W)

what else could I add that would be instructive?

[arctic stone]

I'd say that the set of all continuous functions on some interval would be good too

[dire thunder]

yep

mixed stuff up

set of integrable/differentiable functions

[merry imp]

Set of mxn matrices maybe

[wintry steppe]

ℒ(V,W) is only infinite dimensional if either V or W is infinite dimensional

[wintry steppe]

that's not infinite dimensional

[merry imp]

Oh I can't read

[wintry steppe]

In general, assuming axiom of choice, any infinite dimensional vector space is isomorphic to F^S where S is a set of infinite cardinality.

[arctic stone]

Do eigenvalues = 0 have any corresponding eigenvectors?

[dreamy iron]

Oh!! Very good catch. Tyty

[wintry steppe]

Do you know what it means for a number to be an eigenvalue?

[dreamy iron]

Axler waxes poetic about this. It kinda blew my mind.

[arctic stone]

Yeah I do, Mv = lv

l is the lambda eigenvalue

[wintry steppe]

not completely right

v has to be non-zero

[arctic stone]

OH true that's a necessary assumption

[wintry steppe]

a is eigenvalue of M iff there exists non-zero v such that Mv = av

What do you call such a v?

[arctic stone]

an eigenvector, right?

[wintry steppe]

yea

[arctic stone]

so an eigenvalue of 0 means that the span of those corresponding eigenvectors gets mapped to 0, right?

[wintry steppe]

M has an eigenvalue of 0 iff there exists a non-zero v such that Mv = 0

[arctic stone]

I think that's a yes then

[wintry steppe]

try to be sure then

[arctic stone]

yeah it all works out

ty for clearing things out

[wary lily]

I'm humbled 😅 . I don't really know much. Thank you

[dusk sage]

Would e1 here be the unit vector version of vector v?

furthermore what would be alpha here?

i'm really struggling to understand this slide

I don't get how Hv = v - u

how could v - u be a reflection in the plane with normal u?

[split oak]

here

[sharp idol]

What are you struggling to understand?

[split oak]

just it in general

[sharp idol]

Can you specify what you mean by matrix theory

If you're talking about grad school stuff, I'm not really able to help much

[split oak]

for example this: A( B + C ) = AB + AC

this is matrix algebra

[sharp idol]

What do you struggle to understand about it?

[split oak]

uhmmm its very hard to explain just it lel

1sec brb

[sharp idol]

Are you trying to prove these things?

[sharp idol]

And they're gone...

[dusky epoch]

@split oak asking to "explain the entirety of matrix algebra" is a very tall order

[split oak]

yes i agree with u
i could just hire a professor i might

[nocturne jewel]

"hire a professor"

[dusky epoch]

maybe you could ask more specific questions or issues that we could try to address

[split oak]

yes zoom call

[wintry steppe]

is the inverse of a transitive closure always equal itself? For a matrix

[stoic python]

squirtlespoof

[wintry steppe]

write (first matrix) = Q(second matrix)Q^(-1) and square the equation

these ones work though

let A be the first matrix and B be the second. B^2 = 0. if A = QBQ^(-1) then A^2 = QB^2Q^(-1) = 0. but A^2 \neq 0, so they can't be similar

@wintry steppe i thought the dude was giving an answer to my question for some reason, cause it was coincidentally both about matrices

more generally, if A is nilpotent and n is the smallest integer with A^n = 0, then the same goes for any matrix similar to A

so like

that's a good thing to look for

more generally even. two matrices being similar means they represent the same linear operator just on a different basis

which makes this property and many others immediate

hi guys

how can i use legendre polynomios to find a minimun of a function?

[olive tinsel]

can I always say that a transformation from one homomorphism to another is linear?

[cursive narwhal]

No

A homomorphism is a map

[olive tinsel]

Oh I mean

[cursive narwhal]

So it doesn't quite make sense to talk about transformations from one homomorphism to another

[olive tinsel]

the homomorphism as the vector space of the transformation

[cursive narwhal]

No

A vector space homomorphism is a linear map

They're two phrases used to describe the same mathematical object

[olive tinsel]

wdym I thought you can do a transformation like T: Hom(v,w) -> Hom(v,w)

[stable kindle]

can you give an example

[cursive narwhal]

Sure, you can

But you must realize that T is not an element of Hom(V,W)

If it is linear, then it is an element of Hom(Hom(V,W),Hom(V,W))

[olive tinsel]

so I'm asking if T is always linear, because linear transformations are matrices so I can always do (S+aU)v = Sv + aUv when S,U belong to Hom(V,W)

[cursive narwhal]

Of course not

T doesn't have to always be linear

T is just a function between vector spaces

In fact, you don't even need to consider Hom(V,W) as a vector space. It is just the set of all linear maps between V and W. You can choose to give it a vector space structure but this isn't necessary

So, you can always just talk about T: Hom(V,W) ---> Hom(V,W) as a map without ever bringing linearity or vector spaces or whatever extra structure you're thinking about into the picture at all

@olive tinsel

[olive tinsel]

isn't map the same thing as transformation

[cursive narwhal]

Sure

[olive tinsel]

though I think I can see why I was wrong now

thx anyways

[cursive narwhal]

For me at least, when I think about 'transformations', I usually assume that the mapping between two sets is also doing something to the operations on those sets (whatever they might be)

So, a transformation is a mapping with something else going on

But terminology may differ, obviously. Just use what you're most comfortable with and keep the definitions straight in your head

[wintry steppe]

Hi guys, I cant figure out why when applying the permutation I end up applying the inverse instead

Thank you in advance

[olive tinsel]

after 1 goes to 4 it stays at 4, because 4 doesn't appear again in the next tuples etc

this is probably #discrete-math tho

[wintry steppe]

so does it the first line read '4 is the new number 1' or '1 is the new 4'

i keep getting reversed results

i think '4 is the new 1', right?

[slim gyro]

i jus learned that vectors are linearly independent if the only linear combination of them that equals zero is the one where all the vectors are multiplied by zero/all their coefficients are zero

and it made me think about how a matrix A is invertible iff Ax=0 has only the trivial solution

could someone explain the connection between a matrix being invertible and it having linearly independent columns? like, why is that the case?

also if im misunderstanding anything feel free to point it out

[limber sierra]

see theorem 6.1 here https://inst.eecs.berkeley.edu/~ee16a/sp19/lecture/Note6.pdf

(note that it missed the requirement that A is square)

[slim gyro]

thanks!

[limber sierra]

basically the idea is

we can translate the expression Ax = 0 to a statement about the sums of columns of A

where the vector x is our scalar multiples

so saying "the only solution to Ax = 0 is trivial (i.e. all entries of x are 0)" is the same thing as saying "the only linear combination of the columns of A that equals 0 has all scalars 0"

the rest of the proof just does algebra using that fact

[slim gyro]

gotcha, i guess i was expecting the connection to be deeper but after reading it a few times and thinking ab it it just seems really obvious

thanks again

[limber sierra]

linear algebra is interesting in that a lot of the difficutly comes from parsing the definitions

if you really understand what the definitions are saying, a lot of things follow very quickly

without much work

[slim gyro]

nice, ill keep that in mind

[errant mist]

I'm having trouble understanding the last part of the highlighted section. Why do we take the square root of the eigenvalues and why can we write A as the product V sqrt "and logic symbol"?

[limber sierra]

thats just the construction they use

can you follow why AA^T = Sigma?

it requires the values to be set to the ones given

in order for the algebra to work out

also, just so you know: that ``and logic symbol" isn't actually a logical and $(\land)$, its a capital Lambda ($\Lambda$)

[stoic python]

Namington

[errant mist]

Thanks for the clarification with the symbols. For AA^-T we get a symmetric matrix, that I follow

[wintry steppe]

Question: Can someone please explain inner product spaces. I was a bit confused today in class

[limber sierra]

are you already comfortable with the notion of a vector space?

the idea is that we can "add on" more operations to vector spaces

and one type of operation that we often like to study is the "inner product"

which is a sort of generalization of the dot product

[faint lintel]

I need some help understanding a proof related to inner product spaces

Specifically where this one equality comes from

lemme get a screenshot

lmao I saw that message

anyways yea idk where this came from

it's using the induction hypothesis I suppose

[limber sierra]

from (1)

[faint lintel]

how tho

[native rampart]

If a=b+c then <a,x>=<b,x>+<c,x>

[faint lintel]

(1) is a vector and this bottom line is an inner product

Yes

but I'm strugging still to see the pieces

[limber sierra]

properties of the inner product

take drunk's and replace x with v_i

[faint lintel]

still struggling. I know the properties of the inner product but I can't see how it implies use of the induction hypothesis

[native rampart]

The induction hypothesis is used in the next step

[faint lintel]

wut

[native rampart]

You use induction to simplify that sum

Since <v_i,v_j> is non zero only when i=j

[faint lintel]

isn't the next step just <v_i, v_j> = 0 when i =/= j and v_i and v_j are orthogonal

[native rampart]

By induction

[faint lintel]

ok then I'm more lost >_>

[native rampart]

You assume {v_1,v_2...v_{k-1}} is an orthonormal set

[faint lintel]

It looks like they're using v_k = w_k - this summation

[native rampart]

And use that to show {v_1,v_2,...v_{k-1},v_k} is orthonormal

[faint lintel]

however isn't that what we're trying to prove

that v_k = w_k - that summation

[native rampart]

You are trying to construct an orthonormal basis

[faint lintel]

yes

oh wait that v_k = w_k - summation is part of our assumption

[native rampart]

So,You need to make sure at each step of adding a vector,the basis,the set remains orthonormal

Yes

[faint lintel]

We aren't proving that equality

ok then I get it

thanks

[faint lintel]

I got ||u_2|| = 1/sqrt(48)

so shouldn't x - 5/4 x^2 be multiplied by sqrt(48)

not 1/sqrt(48)

[native rampart]

Yea it should √48 times that

[faint lintel]

ok so the answer for the bottom one should be does not equal

cause the constant is wrong

cool so I gotta tell my teacher that it's wrong lol cause she has it that equals is the right answer

[errant mist]

@limber sierra could you further explain how AA^T is rewritten from yesterday (from the picture I sent) if you have time. (sorry for pinging!)

[lavish jewel]

i would argue that formulation is not the most intuitive

you could instead use the eigenvectors and eigenvalues of A (which is basically the same thing, but saves you the square roots) and build AA^T from that

[pseudo thicket]

what happens if we apply linear transformation repeatedly?

it depends on the type of linear transformation, right?

if it's just a scaling, it will keep extending, if it's a rotation, it just keeps rotating about x degree, right?

[lavish jewel]

you get closer and closer to a vector parallel to a scaled version of the singular vector corresponding to the largest singular value of the mat

[pseudo thicket]

could you explain in simpler term?

[dusky epoch]

how familiar are you with eigen-shit?

[pseudo thicket]

i have just started learning it

[lavish jewel]

if matrix prefer one direction, many matrix multiply prefer this direction too

[tame mural]

why use many operation when few do trick?

[wintry steppe]

I understand how to translate between different notations of permutations. But I fail to understand how to apply the permutations to the list

[native rampart]

1 gets mapped to 4

If permutation is (... a b ...) it means a gets mapped to b

@wintry steppe

So for example take (123)

That would map 1 to 2,2 to 3 and 3 to 1

[wintry steppe]

Thanks @native rampart. To clarify, for example, (...a b ...). A maps to B, right? OK So. Does that mean that in the new rearraged order, in the bth position I will find a, or in the ath position i will find b

in the example I have above, (1 4 8 6 5) Would I say, 1 moves to the 4th position or the 1st position is now occupied by 4

[native rampart]

whatever was in ath position will be in b th position now

So thing in 1st position moves to 4th position

[wintry steppe]

ok. let that sink in gime a minute

So then when I used that, I get 5 7 3 1 6 8 2 4

But I thought I should have gotten 4 7 3 8 1 5 2 6

[native rampart]

Yes

[wintry steppe]

I thought that because isnt the bottow row of the 2 row notation meant to be what it should look like when the permutations are applied. Am I wrong then?

[quiet adder]

this is correct, I think drunk meant that your initial thought was right

[wintry steppe]

I'd appreciate if someone could explain this to me via voice for 1 min. This is getting me too confused. Thanks for the effort though

[hallow kite]

is the conjugate transpose of a scalar product the equal to the scalar product of the conjugate transposes of the individual vectors?

[wintry steppe]

the scalar product of two (complex) vectors is a complex number so how do you define the transpose of a complex number?

[hallow kite]

i had to prove this identity (from an introductory lecture in quantum physics)

given this identity

[wintry steppe]

Hm. What is a|b?

[hallow kite]

this simply defines a scalar product in the vector space that were working with

but i tried it and it led to a pretty simple solution so i guess that idea was right

[wintry steppe]

Ok! In my courses the inner product is given by <a,b>

[hallow kite]

yeah quantum physics uses the Dirac notation which looks a little different

[wintry steppe]

workin on L2 (Hilbert), if i am given a dot product, how can i proof any subspace is orthogonal?

in this case, with a dot product i have to check if z^n is orthonormal

[native rampart]

You mean you want to show every vector space has a orthogonal basis?

[wintry steppe]

no

https://gyazo.com/bc2288d0dcc8c1cd87810e531b70f823

it sais

Proof z^n is orthonormal on this space

and that is the dot product

[native rampart]

You have to find a space such that z^n is orthogonal to that?

[wintry steppe]

and i am working on L2 Hilbert space

no

XD

again, i have to proof z^n is orthonormal

[native rampart]

Yes,just pick an arbitrary vector of the given vector space and dot with z^n

If that comes out to be zero,you get that vector is orthogonal to z^n

[wintry steppe]

what?

may i do the dot product of z^n * z^m ?

[native rampart]

Ok,The question is "show {z^n,n in Z} is an orthonormal set"

[wintry steppe]

yeah

yes, a set of vectors is ortonormal iff the dot product of z^i and z^j is the kronecker delta of i and j

[native rampart]

Yes

Show that will be 0 if n and m are not same

and 1 if n=m

[wintry steppe]

okey. One more thing, the integral has a T under it

it means the integral over that space, right?

does it affect the dot product somehow?

and also... whats the conjugate of z^m?

conjugate is a field homomorphism of complex numbers

so it preserves multiplication (and addition)

okey so it reminds the same?