linear-algebra

[static sparrow]

so,..

[zinc timber]

what are the possible eigen values, the odd dim matters

[static sparrow]

hm...

Maybe it's because I use google translate that makes it difficult for you to understand, really sorry for the inconvenience

[zinc timber]

the possible eigen values @static sparrow

[static sparrow]

I think there exists a matrix X such that X^2 = -I . But I don't know how to prove it

I only know to prove that there is a matrix X when X^2 = I , but proving that doesn't help much.

[zinc timber]

spoiler: ||there isn't||

[static sparrow]

really?

I also thought that there was a matrix X

[dusky epoch]

do you know what the word "eigenvalue" means

[static sparrow]

The above post does not require searching for "eigenvalue" as you say.

There is only proof that there exists a matrix X satisfying the condition or not

I don't know if you understand what I mean.

huhu

[wintry steppe]

unhelpful answer: ||if c > 0, x^2 + c is an irreducible polynomial annihilating the matrix X, so the minimal polynomial of X if x^2 + c. but the characteristic and minimal polynomials share irreducible factors, so the former of X is a power of x^2 + c, hence of even degree. that is, X must be of even size.||

i only write this because i think it's a cute argument

the bystanders, if they ever encounter almost complex structures, will benefit from this

[static sparrow]

Oh... thank you

Your ideas enlighten me many things. ❤️

😘

[wintry steppe]

because A - λI sends a non-zero vector to zero if and only if it has determinant 0

(A - λI)x = 0 for some non-zero x <-> A - λI is not injective <-> det(A - λI) = 0

to be precise

[wintry steppe]

it's pretty easy to prove. a square matrix is not invertible <-> its columns are linearly dependent <-> the determinant evaluated at the columns (so just that of the matrix) is zero

[empty ibex]

need help with this

[tropic nest]

Hello! Do any of you have any more accessible material for calculating the n-row determinant

[winged prairie]

does anybody know what this means

[zinc timber]

looks like an anti-symmetric n-linear form

[teal grotto]

these are helpful in giving an explicit formula for the determinant and the dimension of the space of alternating k forms over an n dimensional vector space. one of my fav proofs tbh

[halcyon spindle]

Hoffman and kunze introduced characteristic of a field and then said don’t worry too much about it. Finding the least n such that adding addtive identity n times is equal to the zero identity. Why consider this for a field besides it being the definition?

[lavish jewel]

it can become interesting in modulo arithmetic

for example when working in Z/2Z you get that 1 + 1 = 0, and this makes everything very annoying

[halcyon spindle]

I see, I just keep this in the back of my mind until I encounter them again.

[empty ibex]

^^

[teal grotto]

once they start talking about affine spaces, their are some weird counter examples

[green sorrel]

@everyone can someone help me

[zinc timber]

linear algebra? everyone??

[halcyon spindle]

#prealg-and-algebra

[heavy crown]

how do I show that a matrix is not normal?
Only by showing that AA* != A*A or there's another way?

[zinc timber]

depends

[heavy crown]

I mean are there other techniques to show that a matrix isn't normal?

oh I found it, if I show that A isn't diagonalizable by a unitary matrix over C => A isn't normal

thankyou

[wintry steppe]

eigenvectors corresponding to distinct eigenvalues are orthogonal. that's one necessary condition for a normal operator

and probably a bit easier to check than showing non-diagonalizability

[harsh lake]

hello, I am trying to code something and there's this equation a + 11b + 121c + 1331d = 0.106378. how do I find a, b, c and d?

[lavish jewel]

if that is all you have, there are infinitely many solutions

do you want all of them or any one?

[heavy crown]

you're right thank you!

[harsh lake]

I guess that any one will work out

[lavish jewel]

then you can pick a b and c arbitrarily and solve for d

[harsh lake]

well I also know that a + b + c + d = 0.099833

and

[lavish jewel]

ok that is completely different, then

a system of equations is completely different from a single equation

do you have 4 equations?

[harsh lake]

i've got 3

[lavish jewel]

all right

[lavish jewel]

then i would suggest to do gaussian elimination

[harsh lake]

thanks, i'll try it

[lavish jewel]

you'll have a free parameter, so beware

[harsh lake]

I think I've made a mistake in last one it should be 753571d

[heavy crown]

I'm having something weird here hmm
can I say that A = {{i, i}, {i, i}} is not unitary matrix because column1=column2 so the columns don't form a basis (and specfiically orthonormal basis)?

[lavish jewel]

sure

[heavy crown]

okay thank you cleared my thoughts)

[harsh lake]

@lavish jewel

Is it right?

[lavish jewel]

idk, if you input the system correctly, probably

[noble swan]

Hey, I'm having trouble getting through these conceptual problems

How is it possible that dim(Nul(A)) = 0 if m > n?

din(Nul(A)) is the number of free columns

They say it's 0, but if m > n, there has to be free columns, right?

[lavish jewel]

this looks like an evaluation

is it?

[noble swan]

What does that mean?

Like is it a test? No, this is actually review for my test

Title says study guide

[lavish jewel]

aight

well, if m > n, that does not necessarily mean there are linearly dependent columns

as a simple example, consider this chunk of the identity matrix: $\begin{bmatrix} 1 & 0 \ 0 & 1 \ 0 & 0 \end{bmatrix}$

[stoic python]

Edd

[noble swan]

Ohhh, shoot

Forgot it's row x column

[lavish jewel]

so your claim is in general false

(it can be true in some cases tho)

[noble swan]

I keep getting mixed up because coordinates are (x, y)

Agh

Okay, that helps. I'm seeing that these questions link to each other, but I guess I'm not able to visualize it well enough

How do I go about answering a?

I know for part b, A is linearly independent because dim(Nul(A)) = 0, now I know it doesn't span R^m, because m > n meaning there's a free row

[lavish jewel]

to be completely honest, they don't give you enough info cuz they don't specify the field

but i assume you worked mostly in R^n

in that case, it would be R^n -> R^m

oops

[noble swan]

Ah, gotcha

[lavish jewel]

the notation is T: domain -> codomain

where the codomain need not equal the image of T

as you say, your matrix is rank deficient and won't span R^m, only a subspace of it

but R^m is still the codomain. the specific subspace is the image

[noble swan]

I see

So I know Rank(A) = n, but how do I figure out the last 3 parts?

[lavish jewel]

well

[noble swan]

Wait, nvm, f is infinite since it's LI

[lavish jewel]

when does a system have no solutions, one sol, or infinitely many?

[noble swan]

I don't get d & e, though

[lavish jewel]

f is not infinite

[noble swan]

Aww, crap

Oh wait

Only the trivial solution?

No solution is when you have a free row that doesn't equal 0

One solution is when you have a single pivot equal to whatever on the right side

And infinitely many is when you have a free row equal to 0

[lavish jewel]

they're telling you straight up that b is in the column space of A, so there is at least one solution

what's the dimension of the null space?

[noble swan]

0

[lavish jewel]

so there is only 1 solution

[noble swan]

Ohhh, I see

So then e would be no solution?

[lavish jewel]

right

[noble swan]

Okay, so how come f isn't infinite?

m > n, so you have a free row

[lavish jewel]

mhm

[noble swan]

And since you're setting it equal to a 0 vector, you have a free row equal to 0

[lavish jewel]

setting what equal to a 0 vector

[noble swan]

The free row

[lavish jewel]

why?

[noble swan]

Because Ax = 0, right?

[lavish jewel]

right, so we have Ax = 0

and there is only 1 sol to that

[noble swan]

Only one? How come?

[lavish jewel]

they told you

the dim of the null space is 0

linear transformations satisfy that T(0) = 0

so we know the trivial solution works

[noble swan]

So we can just ignore the free row equal to 0?

[lavish jewel]

what free row equal to 0

[noble swan]

Because m > n, right?

[lavish jewel]

sure, and what about it

[noble swan]

So you have something like this, right?

[lavish jewel]

mhm

and what about it?

[noble swan]

Does the last row now imply infinite solutions?

[lavish jewel]

no

[noble swan]

Huh, okay

[lavish jewel]

the previous 4 rows tell you that your 4 variables are exactly 0

the last row tells you nothing new

in fact, if you use the last row as you say, and let, for example, x = 5, you get that 5 = 0

this is called an overdetermined system of equations, where the extra rows tell you nothing new

[noble swan]

Okay, I see

It's the pivot columns that tell you the solutions

[lavish jewel]

you are thinking about the problem wrong from the moment you default back to rows and columns anyway. they gave you all the information you needed by describing the subspaces related to the matrix

in this context, having infinitely many solutions comes rather from having a null space with dimension greater than 0

this means there are nonzero vectors such that Ax = 0

then if you have that Ay = b, and that Ax = 0, by linearity, A(y+cx) = b as well for any c

this gives you infinitely many solutions

ofc you can always relate it back to rows, pivots, columns, etc, but there was no need to in this case because they already told you no such vector x exists

and so Ay = b has a unique solution

[noble swan]

I see

[lavish jewel]

b = 0 included

[noble swan]

Okay, one last question

I don't know how I start this at all

[lavish jewel]

simply apply the definition of eigenvector

if x is an eigenvector of A with eigenvalue lambda, then Ax = lambda x

now let's consider A-5I instead

we would like to find some mu such that (A-5I) x = mu x

but x can be distributed in that product. we know that x is an eigenvector of A, and all vectors are eigenvectors of the identity matrix rather trivially

[zinc timber]

wait what?

[lavish jewel]

so that (A-5I) x = Ax - 5Ix = lambda x - 5 x = (lambda - 5) x

so x is an eigenvector of (A - 5I) with eigenvalue mu = lambda - 5

[noble swan]

I followed until you replaced A with lambda

Ax - 5Ix = lambda x - 5 x
This part here

[zinc timber]

"show that x is ... and eigen vector of A-5I"?

[noble swan]

Yeah, professor must have typo'ed the d in there

[lavish jewel]

prolly a typo

[noble swan]

Show that x is an eigenvector of A-5I

[lavish jewel]

so what part are you stuck on

[noble swan]

How are you able to change the A to lambda?

[lavish jewel]

...

it's the definition of a x being an eigenvector of A?

[dusky epoch]

sounds like somebody has a big misunderstanding of the definition of an eigenvector.

[lavish jewel]

with eigenvalue lambda

[dusky epoch]

oh nevermind

just read the log

[noble swan]

OH

I got it lmao

[lavish jewel]

[noble swan]

Okay, I think I got it now

Thank you so much!

[heavy crown]

help?

[lavish jewel]

• here is complex conjugate transpose, yeah?

[heavy crown]

yes

[lavish jewel]

this means that the elements of the main diagonal don't move when being transposed, but only get conjugated

and this means that a_ii* = - a_ii

we remove the subindices for simplicity

so that a* = -a

and a is in general complex here, so we substitute (x + iy)* = -(x + iy)

do you know what is needed for two complex numbers to be equal?

[heavy crown]

right, they stay at same position because they're part of the main diagonal and only minus adds when conjugated

[heavy crown]

I don't know sorry

[lavish jewel]

the real parts are equal, and the imaginary parts are equal

that means (x + iy)* = -(x + iy) can be interpreted as 2 separate equations

x = -x, and -iy = -iy

[heavy crown]

aha I understand you so x = -x, y=-y

[lavish jewel]

careful with the y

there's a complex conjugate

[heavy crown]

oo right yea so (x + iy)* = (x -iy) and then -yi = -yi

[lavish jewel]

right

so that pretty much means y can be anything, it's trivially true

[heavy crown]

is it allowed to just pick (x+iy) to replace A like this?

or it was only to explain me

[lavish jewel]

it's perfectly valid, and i think needed in this case

because we have to look at the other part still

x = -x

[heavy crown]

aha I see what you mean

[lavish jewel]

by the way, you could've used the polar form of complex numbers if you prefer that, too

but yeah

[heavy crown]

no thanks haha

[lavish jewel]

you see that, if x is nonzero, then we can divide both sides of the equation by x

and we get that 1 = -1

which... is obviously not true

this means we made a mistake by saying x is nonzero

[heavy crown]

so x is zero

[lavish jewel]

so we arrived at the following: a_ii = a = x + iy

where y can be anything, and x = 0

[heavy crown]

well done

[lavish jewel]

so that a_ii = a = iy

and now we end it by letting r = y

so that a_ii = ri, as desired

[heavy crown]

thanks that helped me a lot )

[zinc timber]

,av 289503597581041665

[stoic python]

username#1754's Avatar

Click here to view the image.

[heavy crown]

hahah well said

[noble swan]

Can someone explain how I got this one wrong, please?

[lavish jewel]

i'm guessing you have been working with column vectors in class

[noble swan]

Yep

[lavish jewel]

you can see by associating the products that your matrix first rotates, and then reflects

instead of first reflecting, then rotating

Reflection(rotation * x)

[noble swan]

Sorry, I don't see it

[lavish jewel]

your matrix is reflection * rotation, yes?

[noble swan]

Yeah

[lavish jewel]

well

let's multiply this by x

reflection * rotation * x

what are you doing to x first?

[noble swan]

Well, you said I want to rotate first

[lavish jewel]

no i didn't

i said that's what you're doing

[noble swan]

Oh

LOL

So how do I reflect first?

I thought I did that

[lavish jewel]

rotation * reflection * x

you swap them, as they told you with that cryptic double ended arrow

order matters when you multiply matrices

stuff to the right happens first

[noble swan]

So I do the rotation around the origin first

[lavish jewel]

that is what you are doing

[noble swan]

So was I supposed to end up with a 2x1 matrix?

[lavish jewel]

what the fuck

please sit down and read what i wrote carefully

[noble swan]

I'M SORRY I DON'T GET IT

[lavish jewel]

dude

you wrote AB

you needed to do BA

stuff on the right happens first

A(B(x))

B(A(x))

also both B and A are 2x2, so no matter what order you put them in, you get a 2x2 matrix as a result

[noble swan]

Shit

[lavish jewel]

you wrote AB

you needed BA

[noble swan]

I'm braindead

So the correct answer just has the sign of the two sin's swapped

[lavish jewel]

in this case yes, but if the reflection had been along a different line, things could be vastly different

[noble swan]

I don't know why that took me so long

[lavish jewel]

better think of it as you took the transformations in the wrong order

[noble swan]

Can you explain why that is?

Wording had it reflect first

[lavish jewel]

no, i can't because i just did like 5 times

someone else will have to

[noble swan]

Sorry :(

I got it now, I think. I'm super sorry again

[still lodge]

BA is not necessarily equal to AB when it comes to multiplying matrices

[ionic laurel]

Think of it as compositions of functions

Fog is f(g(x))

Gof is g(f(x))

[forest quiver]

Help

[mystic frigate]

when diagonalizing a matrix (A=PDP^-1), why does matrix P have to have integer values in an nxn matrix

cause when I put fractions, it says that its wrong

[wintry steppe]

it doesn't. that sounds like a problem with whatever software you're using

[mystic frigate]

oh ok

[heavy crown]

if I need to find diagonalizable by a unitary matrix over A, and I found that A's eigenvectors corresponding to distinct eigenvalues, and I see they aren't orthogonal.
I can conclude that A isn't diagonalizable by a unitary matrix?

I mean found this and I see that <(i,1) , (-i,1)> = 2 != 0

[wintry steppe]

.

[heavy crown]

yes you told me earlier about it I'm sorry but I didn't know if this is how I'm supposed to do it

like in terms of the steps

we didn't do any example about it in class and the teacher said like only that if matrix is normal then <u,v> =0 but he didn't say the other way around is true also, that if <u,v>!=0 then A isn't normal but it makes sense to me

thank you terra :]

[wintry steppe]

what you did works

[heavy crown]

thank you I understand now, also I got it wrong the < >
it is <(i,1) , (-i,1)> = 0 indeed

[wintry steppe]

if $A = U^DU$ for $D$ diagonal, then $$AA^ = U^(DD^)U = U^*(D^*D)U = A^*A$$, so $A$ is normal. so eigenvectors corresponding to distinct eigenvalues are orthogonal.

[stoic python]

TTerra

[wintry steppe]

second equality because all diagonal matrices commute

[heavy crown]

thanks for pointing out, so it works both ways. Now I just used it in other exercise like you said. Appreciate your help

[cedar dawn]

Hello, i have a question I hope you can help me answer and understand:

[tidal pumice]

can anyone help explain #3 pls 🙂

<@&286206848099549185>

[zinc timber]

find the equation of the subspace orthogonal to the intersection of the 2 given planes

[tidal pumice]

how can i do that?

[zinc timber]

find the intersection of the planes

[tidal pumice]

https://mathinsight.org/intersecting_planes_examples

by doing this?

[thorn cypress]

is it sufficient to say by inv thm it must have a nonunique (ie 0 or infinite number of) solution?

[zinc timber]

that's not a complete answer

[thorn cypress]

must I note that A is not invertible?

nvm, i have got it

[raven badger]

I have a question in my notebook:

If a set of vectors are orthonormal, are they always linearly independent?

My answer is yes; Because they will be orthogonal to eachother, i.e not in eachother's span.

But is there any way to prove this?

[limber sierra]

the answer is yes because theyre nonzero orthogonal

[dusky epoch]

you could take an arbitrary linear combination of them and take its dot product with itself

[limber sierra]

this hints that the proof will rely on that fact in some way

[dusky epoch]

$v = \sum c_k x_k \implies \ang{v,v} = \sum c_k^2$

[stoic python]

Kanga Gang Annihilator Ann

[limber sierra]

and yeah, just as ann said

[dusky epoch]

where {x_k} is your orthonormal set

if you are talking about a complex vector space replace c_k^2 with |c_k|^2

[dawn rain]

Hi as suggested there, can anyone help me with this?
#help-6 message

[zinc timber]

you forgot to mention that A is symmetric

[tranquil steeple]

a typo (missing a 2)?

[zinc timber]

what is your problem exactly?

that should be v2

[lavish jewel]

yeah judging by how it was written, seems like the rhs should've been v1^T v2

[honest grove]

Why is that we only have to do row operations and can't do colomn operation in reducing matrix into echelon form? What happens if we did column operation?

[teal grotto]

column operations don’t preserve the kernel

they preserve the image

and you can do them to accomplish some different things, they can column reduce a matrix, but the reduced form you get tells you different information

i believe they’re also used in some row reducing optimization methods

[honest grove]

What exactly are you telling? I didn't understand..

[winged prairie]

do u know any good resources for learning this stuff, cuz im a bit lost

[outer goblet]

if i know that a 2x2 matrix has 2 different eigenvalues. Is there a possiblity of that matrix to not have 2 linearly independant vectors?

so basicailly, can two different eigenvalues give me linaerly dependant eigenvectors

[zinc timber]

given 2 eigen values are different, the eigen vectors will also be linearly independent

[worldly bear]

yes the eigenvectors from distinct eigenvales are always linearly independent

[drifting ravine]

Why is only c. a subspace of R^3

[lavish jewel]

try testing the definition of subspace

as a hint, scalar multiplication will trip it up

[drifting ravine]

thank you, that makes sense. Just missunderstood the task

[wintry steppe]

with the given formula how can i find the basic eigenvectors of A corresponding to the lambda value?

this is so confusing to me lol

[lavish jewel]

you can RREF

you'll get at least 1 free parameter if 2 is an eigenvalue of A

[wintry steppe]

so do i subtract the lambda value from the diagonal digits and then RREF?

or RREF A first

[lavish jewel]

subtract first

[wintry steppe]

ok

[lavish jewel]

solve this

[wintry steppe]

but if I RREF there will just be leading 1's with the rest zeroes... what do i do with that

[lavish jewel]

nope

do it and see

[wintry steppe]

okok im doing it rn

[lavish jewel]

and well, even if it were the case, that'd give you an answer

[wintry steppe]

does this look right

[lavish jewel]

idk if right, but certainly possible

we can test at the end though

do you know what to do from there?

[wintry steppe]

not really lol

[lavish jewel]

for one, you forgot to augment the matrix

[wintry steppe]

as in write a column of zeroes on the right?

[lavish jewel]

mhm

[wintry steppe]

ahh i see

so now im here

not sure how exaclty to use the formula now. so i subract A from 2I?

[lavish jewel]

what formula

you did RREF (A - 2I), yes?

[wintry steppe]

oh right i got confused

yes i did that

[lavish jewel]

mhm

[wintry steppe]

i think?

lmfao

[lavish jewel]

and now you found a vector v = x4[0,2,-1,1]

this vector v satisfies that (A-2I)v = 0

[wintry steppe]

where did the last 1 come from?

[lavish jewel]

so you're done

well, what values did you find for x1,x2,x3, and x4?

[wintry steppe]

wow that was so straightforward. i overthought it

i understand now thank you soo so much!!

[lavish jewel]

i'm not convinced 😛 but ok

[west saddle]

My linear algebra class also does differentials. right now we have just learned the first shifting theorem. I have to do a problem that involves sin^2(x). Can I find the laplace transfor or do I need to change it to cos with the half identity

[wintry steppe]

linear algebra class

[dusky epoch]

the vector (1,0,0) is reflected into (1/3, y, z) for y and z yet unknown

the vector (1,0,0) - (1/3,y,z) is parallel to the normal of pi

and hence you have that (2/3, -y, -z) and (1, -1, 0) are orthogonal

you also have (1/3)^2 + y^2 + z^2 = 1

@wintry steppe

[ionic laurel]

Hi I have a small question

Why is this statement true?

Wouldn't the conjugate only be an eigenvector

[dusky epoch]

w = (1+i)v

dunno about common use

all i did was use my knowledge about reflections

namely that the vector connecting a point and its reflection is perpendicular to the plane of reflection

that's it

[muted loom]

If someone could pop into help-14 it would be greatly appreciated! It is a linear algebra question.

[ionic laurel]

can you explain this by any chance?

im confused

Like why is this an eigenvector if you don't mind

[dusky epoch]

if v is an eigenvector of A then kv is also an eigenvector of A for any nonzero scalar k

[ionic laurel]

right

so you mean to say the scalar k is 1+i

[dusky epoch]

yes the scalar is 1+i in your case

[ionic laurel]

fuck

thats such an easy question and i got it wrong

thank you nonetheless

[winged prairie]

yo does anybody know a good textbook

for understanding determinans and multilienar forms

[random axle]

idk any textbooks, but the MIT playlist by gilbert strang seems very good. i will make extensive use of it when i start my university linear algebra course next year

[winged prairie]

ye but i need something more proof based

[wintry steppe]

determinants and multilinear forms
open any differentiable manifolds book and jump to the multilinear algebra chapter :^)

[teal grotto]

i do not. google would be my best resource

[sinful valve]

wait how is this proving it exactly?

i dont follow

0 on lhs is scalar, rhs is vector

like its additive inverse meaning subtracting 0v so lhs becomes 0 vector and rhs becomes 0v + 0v?

0 = 20v?

[halcyon spindle]

you agree 0v = 0v + 0v? Which comes from 0 = 0 + 0.

[sinful valve]

yeah acc tbf

like adding 0v to 0 gets (0+0)v

like 0 on rhs adding the additive inverse of 0v is just (0 + 0)v right thats all they saying

yeah 0 = 0 + 0 def

[halcyon spindle]

Ok

[sinful valve]

just they on about adding the additive inverse

to both sides

[halcyon spindle]

Yeah, they are using vector space axiom to show 0v = 0. They assert that 0v = (0+0)v which is true

Since 0 + 0 = 0.

They distribute it and add the inverse of 0v to both side. So another two axiom in use.

That shows 0v = 0.

[sinful valve]

oh yeah so they negate one of the 0v

i thought they were they talking about the equation above the proof

confused as

makes sense though cheers

[halcyon spindle]

This looks like Axler.

[sinful valve]

yeah it is

[tribal moss]

I was wondering if someone could help me prove this, by giving a hint or something 🙂

[sinful valve]

$\alpha$ has some inverse $\alpha^{-1}\alpha = 1$ and $\overrightarrow{a} = 1 \cdot \overrightarrow{a}$

[stoic python]

shriller44

[sinful valve]

given $\alpha \neq 0$ ofc

[stoic python]

shriller44

[sinful valve]

you can use that to show that the a vector gets the 0 vector

[stoic python]

plegasus

[halcyon spindle]

#❓how-to-get-help or one of the calculus channels

[sinful valve]

i swear you cant technically get help if its an exam?

against this servers like rules or sumn i think

[halcyon spindle]

Assume one is nonzero the show the other one is zero then do the same thing but reverse

[sinful valve]

this specific proof stuff is kinda hard to formulate sometimes i presume it gets easier on practice

[tired fossil]

Hey guys, anybody know how i find teh determinant of the characteristic polynomial of this matrix

[ionic laurel]

you can do cofactor expansion

[tired fossil]

sorry i mean this matric

matrix

[ionic laurel]

yes do cofactor expansion

[tired fossil]

Of the original matrix of the Ca(x)?

[zinc timber]

what is 'determinant of characteristics polynomial'?

[halcyon spindle]

I am trying to show every subfield of the set of complex number C is a field. I am using contradiction by assuming there exist a rational number q $\neq$ 0 such that q $\notin$ F thus we have q is not a complex number. We have the set Q is a subfield of R and R is a subfield of C, this implies q $\in$ R thus we have q $\in$ C. This contradicts our assumption that q was not a complex number therefore we have every rational number in F.

[zinc timber]

"subfield"

[stoic python]

plegasus

[zinc timber]

it's already on the name

[halcyon spindle]

I know it obvious but the book says show it.

[wintry steppe]

Guys I need help with this

I dunno how to multiply infinite matracies

[still turret]

why is this wrong

[wintry steppe]

I think im close but I forgot to transpose this first matrix...?

what in the world

I literally want to cry

[sinful valve]

is a direct sum just saying like add the elements of individual subspaces to obtain the new subspace

like a long ass way of saying adding up each vector/list of the contributing subspaces

[zinc timber]

(MtM)_ij = m_ki*m_kj

[wintry steppe]

Does this look good

I think this is m_ki*m_kj

[zinc timber]

no

[wintry steppe]

rip

I mean looks like m_ki*m_kj to me just transposed

[zinc timber]

that's a summation notation over k

[scarlet willow]

Hi guys, I was wonderring if someone can point me in the right direction with this. Given the observations in this table I need to use least squares to find the polar curve that best fits this data. I need to use things like design matrix's etc.

[halcyon spindle]

Yes but realize each element in the direct sum can only be written as a unique sum of element in U_j.

that is for any k $\in$ $U_1 + ... + U_m$ where $U_j$ are subspaces of V if we have $k = x_1 + ... + x_m$ and $k = w_1 + ... + w_m$ where $x_j, w_j \in U_j$ we must have $x_j - w_j = 0$ for all j = 1, 2, ..., m.

[stoic python]

plegasus

[halcyon spindle]

thus x_j = w_j for all j.

[halcyon spindle]

you can also show that it is a subspace of V by verifying it has the 0 identity and addition, scalar multiplication is closed.

[sinful valve]

yeah tbf i should have read on it cleared the idea of unique representation tbf

[sinful valve]

thats exact proof yeah

[ionic laurel]

i think they meant det(A-(lambda)I_n) = 0

[versed yew]

guys

what does this mean

the equation of a plane

and The Equation

is the equation = ax + by +cz + d =0?

[halcyon spindle]

For a) do you know how to find the normal vector?

[versed yew]

yes

[halcyon spindle]

the normal vector n is perpendicular to the plane so for any point x in the plane we have vector x-p is tangent to the plane so what do you think?

[versed yew]

oh my question is like what does it mean "the equation"

is it like the general solution ?

[halcyon spindle]

Yeah.

[versed yew]

ohhh im too dumb to see it mb

[solemn lotus]

What does $P_v(u)$ usually denote?

[stoic python]

CoolShot

[solemn lotus]

Ctx: inner products

[wintry steppe]

orthogonal projection of u onto v?

idk

[scarlet willow]

Hi guys, I was wonderring if someone can point me in the right direction with this. Given the observations in this table I need to use least squares to find the polar curve that best fits this data. I need to use things like design matrix's etc.

[zinc timber]

what is your function exactly?

r=c θ+d?

[scarlet willow]

it mentions in the "hints" section of the assignment that ;
"The graph we are fitting has the formula : r=beta+e(rcos(theta))"

[zinc timber]

don't think linear algebra is the way to go

tho you can convert the coordinates back to Cartesian and solve the least square and then to polar

[scarlet willow]

I believe thats what Im supposed to do

[zinc timber]

[scarlet willow]

I think I need to possibly plug in the r and theta values im given into the said equation, and then make a design matrix

although this is mind boggling atm

I plugged the points into desmos, and got it to match up with a parabola, but this isnt necessarily a good way to prove my maths

[zinc timber]

what's U?

[solemn lotus]

Is there any relation between $\dim V$ and $\dim V^\perp$?

[stoic python]

CoolShot

[zinc timber]

V as a subspace of U?

otherwise dim V \perp =0

but as a subspace of U, dimU=dimV+dim V\perp

@teal grotto

I've seen all

[lavish jewel]

as ryu says, they're orthogonal complements

[teal grotto]

lol i was like, what is U? and then it made sense

[zinc timber]

shape of U

[tribal moss]

@halcyon spindle @sinful valve thank you both for the answers to my question

[winged prairie]

why is the dimension n^k, surely it should be k*n

[dusky epoch]

is n the dimension of X?

kn would be the dimension of the space of linear maps from X^k to K.

not k-linear.

[winged prairie]

yes

[winged prairie]

so what is n^k the dimesnion of

[dusky epoch]

the space of k-linear maps on X^k

a 3-linear map on X^3 satisfies f(a+a', b, c) = f(a,b,c) + f(a',b,c) while a linear map does not

[winged prairie]

ok makes sense

so what equation was applied

to get n^k

[winged prairie]

nvm i got it

thanks

[azure bolt]

hello

can someone help me in this exercise please

Without calculating the determinants show
Imagem

[outer goblet]

For an orthogonal matrix, why is Ax=xA^t

[zinc timber]

can you write out bc = (abc)/a, ac=(abc)/b, ab=(abc)/c?

[zinc timber]

xA^t is not even defined

[outer goblet]

what

A^t is transpose of a

[zinc timber]

x = Nx1, A = NxN

matrix multiplication is not compatible

[outer goblet]

im just trying understand this

[zinc timber]

that's not (xA^t)

$x\cdot A^TAx$ is very different than what you have written

[outer goblet]

how do they go from Ax dot Ax to x dot A^t A x

yeah mb

but then question still stands why can they do that lol

[zinc timber]

that's the property of inner product and adjoint

[outer goblet]

adjoin?

inverse of transpose?

i forgot

[zinc timber]

$\ip{Ax}{y} = \ip{x}{A^Ty}$

[outer goblet]

so its just a theorem

[zinc timber]

for reals

well, that's the definition of transpose ( adjoint*) but alright

[outer goblet]

aith

[zinc timber]

oh let's step back then

[azure bolt]

the teacher said to perform the operations on the columns

[zinc timber]

can we write $\norm{Ax}^2 = Ax\cdot Ax = (Ax)^T Ax = x^T (A^TA) x$?

[outer goblet]

yes?

or

idk

[zinc timber]

why the '?'

[outer goblet]

i guess only if Ax=(ax)^t

[zinc timber]

well no, for real spaces a.b = b^Ta

[outer goblet]

ait so its just a theorem i forgot

[zinc timber]

that's true

[outer goblet]

ait ait

anyone got a collection of all the theorems for linear algebra?

so i can memorize them lol

[zinc timber]

that's not a theorem, well kind of is ig, Reize theorem you can say

[outer goblet]

ait ait

[zinc timber]

you don't know it?

[outer goblet]

so much to memorize

[zinc timber]

a, b column vectors btw

[outer goblet]

im just forgetting everything the more i study for the exam

[outer goblet]

your take is good enought for me :)

[zinc timber]

unless you know why they are true, I won't really call them helpful

[outer goblet]

wish there was a collection of all the theorems in teh book in a tight list

and definitions

[zinc timber]

[azure bolt]

ryu,
the chat is very fast.
I'm Portuguese and I don't speak English very well.

the teacher said to perform the operations on the columns

i am sorry

[zinc timber]

only on the columns??

also why are you apologizing

[azure bolt]

because for my bad english

did you understand the exercise?

i have to justifiably show the solutions

[zinc timber]

$$\mqty|1 & a & bc \ 1 & b & ac \ 1 & c & ab|\to\mqty|1 & a & (abc)/a \ 1 & b & (abc)/b \ 1 & c & (abc)/c|\to abc\mqty|1 & a & 1/a \ 1 & b & 1/b \ 1 & c & 1/c|$$

[zinc timber]

this is first step

[azure bolt]

ok

[zinc timber]

try to multiply a,b,c with each row strategically

[azure bolt]

I can't solve it help me please

[zinc timber]

$$\to \mqty| a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 |$$ now rearrange

[azure bolt]

Thank you very much

[azure bolt]

can you help me in one more exercise? of the subsets, which are vector subspaces?

[halcyon spindle]

A is clearly not a subspace since it is not closed under scalar multiplication. For example take a scalar c =-1.

[azure bolt]

ok

[whole solstice]

I think D is a sub space

It’s closed under addition and scalar multiplication

[teal grotto]

its the kernel of a linear map

[whole solstice]

I don’t know what that means

Am I wrong

[teal grotto]

no, its just a different way of seeing it

[whole solstice]

I know very little linear algebra

Sorry

[teal grotto]

all good

[versed parrot]

wait what is D? what does the "e" mean?

[teal grotto]

tho, if interested, it is the set of all points (x,y,z) such that the matrix
1 1 0
0 1 1
sends those points to the origin in R^2

[teal grotto]

and

[azure bolt]

#linear-algebra message

can you help me in one more exercise? of the subsets, which are vector subspaces?

[versed parrot]

you mean $\mqty[1 & 1 & 0 \ 0 & 1 & 1] \mqty[x \ y \ z] = \mqty[0 \ 0 ]$ ?

[dusky epoch]

\bmqty

[teal grotto]

ye

[dusky epoch]

also [0; 0] not [0; 0; 0]

[stoic python]

Ninja [Euler notation gang]

[azure bolt]

no

[winged prairie]

can something be linear but not multinear and vice versa?

[marble lance]

Yes, in fact, they are usually not the same:
If linear,
f(a+b, c+d) = f(a,c) + f(b,d).

If 2-linear,
f(a+b,c+d) = f(a,c) + f(b,c) + f(b,d) + f(a,d)

[winged prairie]

ok thanks!

[viscid lagoon]

Let $R$ be a ring and $p=\sum_{i=0}^{k} a_{i} x^{i}$ and $q=\sum_{j=0}^{\ell} b_{j} x^{j}$ polynomials such that $a_{i}, b_{j} \in R$ for all $0 \leq i \leq k$ and $0 \leq j \leq \ell$. Define the product of $p$ and $q$ as the polynomial
$$p q:=\sum_{r=0}^{k+\ell}\left(\sum_{i+j=r} a_{i} b_{j}\right) x^{r}
$$
Let $R=\mathbb{Z}{6}$ (quotient ring), $p=a x^{2}+b x+c$ and also $q=d x^{2}+e x+f$ such that $a, b, c, d, e, f \in \mathbb{Z}{6} \backslash{0}$. Determine all the possible polynomials $p$ and $q$ such that the product $pq$ is the zero polynomial $\sum_{i=0}^{4} 0 x^{i}$. Is there any way to solve this elegantly instead of breaking it down into separate cases?

[stoic python]

lewis

[versed parrot]

what if f(x,y,z) is 2-linear

what does that mean?

[lavish jewel]

that it's linear in 2 of its arguments?

[zinc timber]

[versed parrot]

but a function being linear doesn't mean it's only linear in one of its arguments right?

[lavish jewel]

it could be linear only in one, why not

consider f: R^2 -> R defined as (x,y) |-> xy^2

without going so far, the scalar product of vectors in C^n behaves this way

linear in one of the two args

when there are multiple args, you have to specify

[versed parrot]

but then f(a+x,b+y) != f(a,b) + f(x,y)?

[lavish jewel]

right, because it is only linear in one argument

[versed parrot]

but this def seems different from what luna said

[lavish jewel]

what did luna say

[versed parrot]

this

[lavish jewel]

i added an extra parameter that is neither linear nor multilinear

it does not conflict with what luna says

[versed parrot]

it doesn't?

[lavish jewel]

it doesn't because i didn't say it was linear in the second arg

[versed parrot]

ohh you can't just say linear or 2 - linear

you have to specify what arguments its n-linear for

[lavish jewel]

mhm

[versed parrot]

so f(a+b, c+d) = f(a,c) + f(b,d) means it's linear in both arguments?

[lavish jewel]

yes

[versed parrot]

wait.. is it f(a+b, c+d) = f(a,c) + f(b,d). or f(a+b, c+d) = f(a,b) + f(c,d).

[lavish jewel]

i actually don't get lunas 2 linear def

multilinear means that if you consider one param and keep al others fixed, the one under consideration is linear

and that this can be done for many params

[versed parrot]

what would you say if f(a+b,c) = f(a,c) + f(b,c) and f(a,b+c) = f(a,b) + f(a,c)

[lavish jewel]

then i would say it's bilinear

[versed parrot]

oh so bilinear doesn't mean 2- linear?

[lavish jewel]

idk what 2-linear is

i'm talking about multilinearity

https://en.wikipedia.org/wiki/Multilinear_map

[versed parrot]

so bilinear is linear in each, but linear would be linear in both? like at the same time

[lavish jewel]

arguably if it's bilinear it's linear in both lol

like the dot product in R^n

oh

maybe that's what luna meant in their example

cuz then expanding out the product gives you 4 terms

but that need not be the case, cuz you could just add vectors

addition of 2 vectors is bilinear

[winged prairie]

yo guys i have a question. What does it mean that the dimension of alternating forms is at most one and at least 1?

like i know it implies that the dimension is 1

but i don't understand what it means that the dimension is least 1

[marble lance]

I meant bilinear:
f(a+b, c+d) = f(a+b, c) + f(a+b, d) [by linearity in second argument]
= f(a,c) + f(b,c) + f(a,d) + f(b,d) [by linearity in first argument]

So the idea was to show how you get extra terms there if it's bilinear as opposed to linear @versed parrot

So bilinear and linear maps on VxW are in general not the same

[lavish jewel]

for whatever reason i couldn't see that immediately until i worked out the scalar product example lol

[marble lance]

👍

[rigid sorrel]

How would you represent the standard matrix of a linear transformation of polynomials?

say you had T(c+bx+ax^2)=(3c+a)+(2b+3a)x+ax^2

[lavish jewel]

depends what you mean by standard and which linear transformation

first pick a basis and go from there

the choice {1,x,x^2} makes it straightforward

[rigid sorrel]

Okay say if I choose {1,x,x^2} as my basis, where would i go from there?

[lavish jewel]

what's the coordinate vector of (c + bx + ax^2) in this basis?

[rigid sorrel]

wouldn't it jus be (c,b,a)?

[lavish jewel]

sure thing

and now we make a matrix that does what we want it to

first, you can see that nothing happens to the a component

so the matrix's bottom row is [0,0,1]

oops, i meant bottom row

the linear term in the output of the transformation is (2b + 3a), so the second row would be [0,2,3]

and the constant term is 3c + a, so that'd be a row of [3,0,1]

[rigid sorrel]

I'm a little ocnfused though. For the second row there's a 3 next to a, so wouldn't it be 3 2 0? not o 0 2 3

[lavish jewel]

well, you chose the coords as (c,b,a) instead of (a,b,c), so

[rigid sorrel]

i see, but if I chose (a,b,c) it would be First Row: 1 0 3 Second Row: 3 2 0 Bottom Row: 1 0 0 ?

[lavish jewel]

sounds about right

[rigid sorrel]

okay thank you!

[ripe birch]

Is there any intuitive or meaningful reason why the inner product of 2 complex vectors demands that we take the complex conjugate of the 1st vector? e.g.: (A,B) = a*b where "a" and "b" are just the i'th index of the resultant matrix. sorry i'd write it better but im not familiar with the bot commands

is it so that (A,A) >= 0? kind of like, we needed to define it as a magnitude or something?

[lavish jewel]

this is needed for positive definiteness, as you wrote

or in a related fashion, because we'd like inner products to be associated to a norm

[dim lotus]

does the positive definiteness assert <A,A> greater than or equal to zero?

yea the norm has to be nonzero since we can't talking about the notion of negative length

[ripe birch]

is there any rigorous way of coming up with this or is it just "we want this operation to mean this, therefore it must be defined as X"

[lavish jewel]

both

we want to have lengths associated to vector spaces over different fields, and to construct them, we need norms that have certain properties

[solemn lotus]

if $U$ is a subspace of $V$, and you have $B = (u_1, u_2, \dots, u_n)$ as an orthonormal basis of $U$, is it possible to get an orthonormal basis for $U^\perp$ wrt $B$?

[stoic python]

CoolShot

[lavish jewel]

wdym by wrt B?

[solemn lotus]

hmm actually nvm that part

is it possible to find just a general orthonormal basis of U^\perp?

[lavish jewel]

sure

start with an orthonormal basis for U and extend it to an orthonormal basis for V

all the extra vectors are a basis for U perp

[solemn lotus]

o

right that
makes sense

ok yea ig i just need to prove that you can extend the basis to be orthonormal and also that this is the basis of U^\perp
ayt gotcha ty

[lavish jewel]

should be straightforward using the definition of a subspace and orthonormal basis, along with orthogonal complement 😛

probably easier to do it in the other direction, starting from the ambient space

[solemn lotus]

so you mean by taking an orthonormal basis of V?

[lavish jewel]

mhm

[gray dust]

@lavish jewel @solemn lotus u cant guarantee it contains a basis of U

[lavish jewel]

can you elaborate a little? i don't think i see it yet, i'm not sure what i missed

[gray dust]

not every basis of V contains a basis of a given subspace

[lavish jewel]

aha, right

then it's better to start from U and extend it to V?

[gray dust]

yes. gram schmidt on the extension gives an orthonormal basis of U^perp

[versed parrot]

a complex number is actually itself a matrix and when you take it's tranpose it's the same as taking the conjugate

$z = a+ ib = \mqty[a & -b \ b & a ]$

[stoic python]

Ninja [Euler notation gang]

[versed parrot]

when you tranpose a comlex number you take it's conjugate

so any matrix with a complex number is actually a block matrix

(or at least can be viewed as one)

[restive vector]

don’t u just fucjin rotate it

i’ve learned that transpose doesn’t mean whag i thought it meant

[nocturne jewel]

Transpose of a matrix is swap rows and columns.

Conjugate transpose is transpose and make the entries their conjugates

[restive vector]

so@pretty much rotation

a b
c d

becomes

[nocturne jewel]

not at all

[restive vector]

a c
b d

right

no

that’s wrong

wait

[nocturne jewel]

yes, but that isnt rotation

[restive vector]

that’s right

[restive vector]

it is to my heart

[nocturne jewel]

then your heart is wrong.

if you require associating a transformation with ^T, then it'd be reflection along the main diagonal at best

[restive vector]

my heart is correct because it says it is

i know it’s not a rotation HOWEVEE

it look kind of like one so

[nocturne jewel]

It doesnt

[restive vector]

i win

[restive vector]

i respectfully disagree

[nocturne jewel]

and I respectfully think you're trolling.

[restive vector]

i swear i’m literally not

it look like rotation time

to me

and i know i’m wrong but in my mind i am right

[wintry steppe]

[median ocean]

can someon help me with b

do i just multiply the top matrix by part b matrix/

[wintry steppe]

presumably you have to use the factorization you found in part a

[median ocean]

i have a Q equation and R equation

[ripe birch]

@versed parrot sorry I was driving. How can a complex number be written as a matrix? I'm not really following that bit. Doesn't seem intuitive

[wintry steppe]

algebraically, $$\bC \to M_{2\times 2}(\bR), \qquad x + iy \mapsto \begin{pmatrix} x & -y \ y & x \end{pmatrix}$$ is an isomorphism. geometrically, this says that the complex number $x + iy$ is the same as the linear transformation $\bR^2 \to \bR^2$ which scales by a factor of $|x + iy|$ and rotates by $\mathrm{arg}(x+iy)$

[stoic python]

TTerra

[ripe birch]

That's a bit dense. I'm not too hip with all the math lingo (I'm a physics guy originally) so it's a bit hard to follow. I feel like I'm so close to getting it but something isn't clicking

[wintry steppe]

it's just saying complex numbers are the same thing as how they scale and rotate when you multiply by them

scaling and rotating is a linear transformation, so it can be written as a matrix

so a complex number can be represented as a matrix

let $z$ be any complex number. we can write it in polar form as $$z = re^{i\theta}, \qquad r \geq 0, \theta\in\bR.$$ if we have another complex number $z' = r'e^{i\theta'}$, then $$zz' = (rr')e^{i(\theta+\theta')},$$ so $z$ acted on $z'$ by scaling by a factor of $r$ and rotating counterclockwise by an angle of $\theta$. written in terms of matrices, this is $$r\begin{pmatrix}\cos\theta &-\sin\theta \ \sin\theta & \cos\theta\end{pmatrix}.$$ since $$z = re^{i\theta} = r(\cos\theta + i \sin\theta) = r\cos\theta + i r\sin\theta = x + iy$$ we just get the matrix $$\begin{pmatrix} x & -y \ y & x \end{pmatrix}$$

[stoic python]

TTerra

[solemn lotus]

hey so im trying to complete the proof i was doing earlier, though im struggling a bit in one part

$U$ is a subspace of $V$ and $(u_1, \dots, u_n)$ is an orthonormal basis of $U$. $V$ then has an orthonormal basis $(u_1, \dots, u_n, v_1, v_2, \dots, v_r)$. the basis of $U^\perp$ is claimed to be $(v_1, v_2, \dots, v_r)$, and i want to prove this

[wintry steppe]

U^\perp?

[solemn lotus]

oops yeah sorry

[stoic python]

CoolShot

[solemn lotus]

so since (v1, v2, ..., vr) is orthogonal, it is also linearly independent

so the only thing that needs to be shown is that (v1, ..., vr) spans U perp

ive been trying this for some time but im not able to make any progress

[wintry steppe]

any vector in U perp can be written as a linear combination of u_1, ..., u_n, v_1, ..., v_r. take some inner products and see what you get

[solemn lotus]

hmm ok ill try that

[median ocean]

Hmmmm

[solemn lotus]

if we take $u\in U^\perp$, $u = a_1u_1 + \cdots + a_nu_n + b_1v_1 + \cdots + b_rv_r$, then $<u, v_j> = b_1 + b_2 + \cdots + b_r$

[stoic python]

CoolShot

[solemn lotus]

where 1 ≤ j ≤ r
i have a feeling that this should mean that (v_1, ..., v_r) is a basis of U^\perp but im not exactly sure how to phrase that

[wintry steppe]

not sure about that computation

a more direct hint is that for any $v \in V$, $$v = \sum_{i=1}^n \langle v, u_i \rangle u_i + \sum_{j=1}^r \langle v, v_j\rangle v_j.$$ what happens if $v \in U^\perp$?

[stoic python]

TTerra

[solemn lotus]

hmm ig $\sum_{i=1}^n \langle v, u_i \rangle u_i = 0$?

[stoic python]

CoolShot

[wintry steppe]

yeah, but why?

[solemn lotus]

since $\langle v, u\rangle = 0$ for all $u \in U$, and $u_i \in U$

[stoic python]

CoolShot

[solemn lotus]

(when v in U perp)

is this right?

[wintry steppe]

yes

i know what you mean

[solemn lotus]

so we have $v = \sum_{j=1}^r \langle v, v_j \rangle v_j$, which implies $span({v_1,v_2,\dots,v_r}) = v$, so its a basis

[stoic python]

CoolShot

[solemn lotus]

what is that alignment lol

[ripe birch]

@wintry steppe fabulous explanation on the rotation matrix explanation. That makes perfect sense. Thank you!

[wintry steppe]

[tribal moss]

I proved one side that if the scalar is non zero then the vector must be zero, but was wondering if this was correct or if there was a more standard or beautiful way to prove this

[stable kindle]

,rotate

[stable kindle]

if alpha != 0, you can divide both sides by alpha and get a = 0
if alpha = 0, clearly a can be anything

[tribal moss]

Yes this one I have proved thank you, but I'm kinda struggling with the one where you have to prove that the scalar is zero

[stable kindle]

?

[tribal moss]

I mistyped

So you are talking about if alpha is different from zero, but I want the one where the vector is different from the zero vector and then prove the scalar is zero, you see what I mean?

[stable kindle]

if alpha * a = 0, then:
if alpha != 0, we can divide both sides by alpha to get a = 0, but this is a contradiction
so alpha = 0

[tribal moss]

Oooh so you use proof by contradiction I see, thank you!

[gray dust]

thats not the logic goes in the division of cases

the cases are alpha=0, alpha!=0. if alpha=0 then im done. if alpha!=0 i can divide by it to get a=0

[median ocean]

part b please

[sinful valve]

how do you verify it has no solutions ?

like i have calculated one a_1 and a_2 and of course it does not hold for the 5 in this case

is that enough to say its got no solutions

[nocturne jewel]

do you get 5=5?

[sinful valve]

uh not the one i did i may have fucked one sec

which equations did u combine to get that result?

mine gives 5= 19 with the , -4 and 5 a_1 a_2

thats what i want though as there are no solutions

actually tbf the bottom two lines should intersect as shown on desmos but the last one does not

last one as being 17 = ...

yeah i should just stop trying to do stuff in my head and rely on solvers tbf

[nocturne jewel]

if you get bullshit, then there's no solution

[sinful valve]

nah the bottom two were fine

the a_1 and a_2 they gave did not solve the top one though

so its got no solutions

well 17 = not 17 i didnt solve it

but i saw on graph

[sinful valve]

what does it mean one representation like

like each vector can only be composed of a single combination of the other vectors with some scalar?

like i assume they mean like setting that vector to 1 and others to 0 is the only representation

[agile bronze]

Only one representation means if there is a vector in the span of a lin. ind. set that can be written as two different linear combinations of the basis, then the original v_1, ..., v_m can't be linearly independent

The easiest way to see it is if some vector x is written two different ways, subtracting the two different linear combinations is 0, but a nonzero linear combination of the coefficients gives 0 which immediately implies linear dependence

[sinful valve]

lin. ind ?

o

[agile bronze]

linearly independent is lin. ind.

[sinful valve]

not sure what you mean in the second message exactly

like you have a representation of two different linear combinations which equals 0

this is not linear independence then?

like the subtraction of them i mean*

oh you said linear dependence thought u meant independence

[agile bronze]

It applies to higher dimensions the same, if x = a1v1 + a2v2 = b1v1 + b2v2, then x - x = 0 = (b1 - a1)v1 + (b2 - a2)v2 where a_i and b_i are not equal

Which implies linear dependence

[sinful valve]

yeah but for the linear independence unique representation

wouldnt that just be what i said where the coefficient of the vector you want is 1?

and the others are 0 that works right

just they mentioned one rep of a lin. comb of v_1, ..., v_m

but i presume you were showing the method to show there are more than one ways of representing

as like a method of proof for linear independence if it fails those

actually tbf i think the single representation is generic ofc like my idea was wrong just thinking

[zinc timber]

how is this false?

[noble swan]

I need to prove this theorem, starting with b first. How do I do this?

[versed parrot]

it makes more sense if you write it in polar form

$z = a+ ib = \mqty[a & -b \ b & a ] =\mqty[r \cos(\theta) & - r\sin(\theta) \ r\sin(\theta)&r\cos(\theta) ] =r\mqty[\cos(\theta) & - \sin(\theta) \ \sin(\theta)&\cos(\theta)]$

[stoic python]

Ninja [Euler notation gang]

[zinc timber]

$(Ux)\cdot (Uy) = x\cdot (U^TUy)$

[zinc timber]

otherwise you can use |Ux|= |x| and polarization identity

[median ocean]

part b help please

[zinc timber]

have you found the QR?

[zinc timber]

@lavish jewel apparently $\m{.5 & 2 \ 0 & .5}\mqty[0 \ 1] = \mqty[2\0.5]$ proves it wrong

[zinc timber]

fuck

[lavish jewel]

what's this

lemme see

[zinc timber]

the |Av| < |v|

eigen values are .5

my brain hurts now, for real

[lavish jewel]

the operator norm is a lie

oh!

hmm wait

[zinc timber]

[lavish jewel]

ah right

the operator norm is in terms of the singular values

and A need not be symmetric, so svd \neq evd in general

so yeah, seems fair

[zinc timber]

yeah I knew the singular value one, I used to think the matrix norm is the largest abs eigen value

and the singular val formulation is extra work

[lavish jewel]

nah it's |Av| <= \sigma_1 |v|

so yeah

it checks out

[zinc timber]

see told you the questions are counter intuitive and correct as well

[lavish jewel]

yep

[zinc timber]

yes svd shows the truth,

[lavish jewel]

you saw some honorable tripped over it as well 😛

[zinc timber]

eigen values doesn't

[lavish jewel]

my condolences

[zinc timber]

how am I supposed to clear it then lol

[lavish jewel]

[zinc timber]

[marble lance]

Every time a question gets deleted, a fairy dies :(

[outer goblet]

were trying to understand change of basis

first they write that vector [v]_B has to be multiplied on the left by
a c
b d

but then they conclude that the matrix has to be multiplied on teh left by

a b

c d

why do they transpose the matrix? is this a typo

[marble lance]

Typo

[outer goblet]

ait

[cedar crescent]

hi, i was looking for a demonstration for this. I had no idea how to proof it even though i tried to form the final result.

thanks in advanced

For more information: I'm trying to find the matrix A which represents the least squares regression line y = a0 + a1.x

$e_i = y_i - f(x_i)$ (which stands for the error in the approximation of $y_i$ by $f(x_i)$

[stoic python]

usr/bin/kannaaa3

[lavish jewel]

what one can do is write this as the minimization problem $$\min_X \Vert AX - Y \Vert_2^2 = \min_X (AX-Y)^\text{T} (AX - Y)$$

[stoic python]

Edd

[lavish jewel]

you can find stationary points by taking the gradient and setting it to 0. i won't show it, but this problem is convex (the hessiam is symmetric positive semidefinite), meaning that if we find one local minimizer, it is global (but not necessarily unique)

so if we take the gradient of this and set it equal to zero, we get that $$ \nabla_X (AX-Y)^\text{T} (AX - Y) = 2A^\text{T}AX - 2 A^\text{T} Y = 0 \implies A^\text{T}AX = A^\text{T} Y$$

[stoic python]

Edd

[lavish jewel]

then when A^T A is invertible, the result follows immediately

as an additional note, when A^T A is invertible, the problem is strictly convex and the minimizer is unique

[fallow gazelle]

Hello everyone. I hope y'all are having a great day.

I'm currently calculating the altitudes of a triangle, and i have a problem.
I have to find the equation of altitudes, and i know how to do that but...
In one altitude, i'm getting this result:

Anyone knows how do i continue, considering that i should not get decimal numbers as a result

[zinc timber]

#prealg-and-algebra or #geometry-and-trigonometry

[fallow gazelle]

okay, thank u

[autumn kraken]

Let $V$ be a vector space and $M \subseteq V$, $\langle \langle M \rangle \rangle = \langle M \rangle$

[stoic python]

madmike

[autumn kraken]

Can someone help me with how to start proving this?

[marble lance]

Define <M>

[autumn kraken]

smallest vector space that contains M?

[marble lance]

Okay

Then what is <<M>>

[autumn kraken]

smallest vector space that contains <M>

[marble lance]

And <M> is a vector space, and it contains itself

And that means <<M>> = <M>

[autumn kraken]

oh rly, so <M> \in <M>?

[marble lance]

What?

[autumn kraken]

<M> is an element in <M>?

[marble lance]

No

contains here means is a superset of

<M> is a subset of <M>

[autumn kraken]

ahh okay

$\langle M \rangle \subseteq \langle M \rangle \Rightarrow \langle \langle M \rangle \rangle = \langle M \rangle$

[stoic python]

madmike

[autumn kraken]

Is that what you're saying?