linear-algebra

[paper egret]

let x1 and x2 be some vector in X, and that y1 and y2 be some vector in Y

x1 + x2 = u for now, and y1 + y2 = w for now

when they say vector addition

does that mean

u must exist in X, and w must exist in Y?

[slow scroll]

well yea, since they are subspaces

[paper egret]

ok

fuck

how do i show addition holds in the intersection of X and Y

[slow scroll]

lol yea you went a bit off the rails when you let x1,x2 in X and y1, y2 in Y. You're on the right track, but if we just take arbitrary elements from each of those spaces, then we can't say anything about what happens in their intersection. Any ideas?

[paper egret]

i wanna say let's take an element from X intersect Y

call that u and w

but then i don't know what to go from there

wait

can we say that

[slow scroll]

yes. That is what we are trying to show: that for u,w in XnY, u+w is in XnY.

[paper egret]

yea i dont know what to do from there

[slow scroll]

use definitions: what does it mean for u,w to be in XnY?

[paper egret]

if u and w are in X and Y, that means u exists both in X and Y, and w exists both in X and Y

hmm

yep that's all i gd

got

[slow scroll]

Ill write it a slightly more organized way: u,w in X and u,w in Y. We want to prove that u+w is in XnY...

[paper egret]

u, w in X implies u + w is in X
u, w in Y implies u + w is in Y
thus u + w exists in X intersect Y

wait

that's it?

nANI

[slow scroll]

eyyy

[paper egret]

bro

wtf

how tf did i do that

[slow scroll]

pro skill

[paper egret]

wait this is all just definitions

lots and lots and lots of definitions

[slow scroll]

yea. ur not done yet. do scalar multiplication now

[paper egret]

aight

we gotta show that

wait

what do we have to show

for scalar multiplication

if u is in X intersect Y, u have to show c*u is in X intersect Y?

[slow scroll]

yup

[paper egret]

ok, u is in X intersect Y

means

u is in X, and u is in Y

implies cu is in X, and cu is in Y

that means cu must exist in X intersect Y

[slow scroll]

yep thats it. If you're writing things out for homework or to explain to an audience, I would be sure to include details like since X is subspace, its closed under scalar multiplication so u in X implies cu in X.

[paper egret]

damn vector spaces are weird

it's just a set of vectors

[slow scroll]

ikr algebra be like that.

[paper egret]

have u taken abstract alg

[slow scroll]

nah, ive tried to do a little group theory on my own, but I've barely had time since school started. I want to take it next year tho!

[paper egret]

any math classes u doin atm?

[slow scroll]

i take diff eq and a proofs class

[paper egret]

👀

[north sierra]

so when you say R^n --> R^m, the n is the number of entries in x and the m is the number of entries in matrix A right?

in like Ax=b

[dusky epoch]

if you mean a linear transformation from R^n to R^m, then a matrix representing such a transformation is m by n

it's the number of rows in it

[half ice]

f : R^n → R^m
Represents a function with a domain R^n and a codomain R^m.

This function takes real vectors of length n, and outputs real vectors of length m.

Often, such a function is a multiplication by an m×n matrix.

@north sierra

[north sierra]

okay thanks

[leaden fiber]

I don't understand nor know how do this, ree

what is closed form

[half ice]

Like, find a matrix A^n

There will be n in the matrix

[leaden fiber]

there will be n in the matrix?

[half ice]

For example, find me a matrix that you can just plug n = 3, you get A³

[dusky epoch]

they want you to write an expression for A^n in terms of lambda and n

[leaden fiber]

ah

what does the hint mean though? Do I have to find B?

[dusky epoch]

i mean it should be somewhat obvious what B is

[half ice]

What's λI?

What could you add to it to get A?

[leaden fiber]

ohh right I'm a giant dummy

[dusky epoch]

you may find it useful to note that B is nilpotent

[half ice]

Ahh clever, I didn't think of it that way

[leaden fiber]

what is nilpotent?

[half ice]

There is some power of B that is 0

[dusky epoch]

specifically, in this case, it is the fourth power

[leaden fiber]

ohhh yeah!

[dusky epoch]

and λI commutes with everything, which is another convenient point

[leaden fiber]

I can't equate A^n=(lambda*I)^n+(B)^n right

[dusky epoch]

no

you can't

freshman's dream

[half ice]

You have the binomial theorem for that

[leaden fiber]

what does commuting mean :o

crai

[dusky epoch]

two matrices A and B are said to commute if AB=BA

[leaden fiber]

ohh

it commutes with everything because it's basically the identity matrix times a scalar right

and identity matrix times anything gives that thing, no matter the order?

[dusky epoch]

yes

[leaden fiber]

lemme do some thinking, gonna ask more questions in a bit

ohhhhhhHHHH

got it!

tysm

[north sierra]

when is a transformation (or mapping) of T not linear?

is it ever not linear?

[dusky epoch]

there are plenty of nonlinear functions out there

for example, f(x,y) = (x^2 + 4y, -7e^x + sin(xy^11) + cos(y-55))

[north sierra]

true lol

so if it has a degree of 1 itll be linear right

even if it goes in a lot of dimensions

[dusky epoch]

degree?

[north sierra]

like the exponent

[dusky epoch]

there exist maps that aren't polynomial in the input coordinates, yknow?

not all functions are polynomial

[wintry steppe]

this needs to be true for f for it to be linear.

[dusky epoch]

yes that's the definition of linearity.

[north sierra]

how can i tell if its linear by just looking at

is there something easy i can look for

[dusky epoch]

you can tell if it's linear by checking it against the definition of linearity

don't try to come up with shortcuts

[wintry steppe]

sometimes its very visible, sometimes you just need to check if the above is true

[dusky epoch]

those are a complete waste of time

just

check it

against

the fucking

definition

[north sierra]

okay

sounds good lol

[dusky epoch]

it's really not that hard at all

and you'll get an intuition for which maps are linear and which ones are not

[north sierra]

aright

thank you

[wintry steppe]

@north sierra for example, let's say we have a function f(x)=x^2. we need to have f(x+y) = f(x) + f(y) for every x,y. if we put x=y=1 we have f(1+1)=(1+1)^2=4 which is NOT f(1) + f(1) (because f(1) + f(1) = 2)

[north sierra]

true

yeah

[dusky epoch]

yeah and the conclusion from that is that f(x) = x^2 is NOT linear.

[north sierra]

i dont understand this

why does e1 have 2 entries

isn't it only supposed to have 1

[dusky epoch]

why would it only have one

e_1 is just the first of n vectors in the standard basis for R^n, where in your problem n=2

[north sierra]

but how can you do multiplication on (5,-7,2) (1,0)

isn't x supposed to equal the number of columns in A

[dusky epoch]

why would you be multiplying [5, -7, 2]^T by [1, 0]^T

where in the problem does it say you're supposed to do that

[north sierra]

idk im just tryna wrap my head around this

earlier in the book they had this

[remote fable]

Is the following characterization of linear adjoint operator correct? The adjoint of T is an unique map T* s.t : For all vectors v and w, we have : <Tv, w> = <v, T*w> ?

[dusky epoch]

that's... the definition

[remote fable]

ok thanks. Since the definition in the textbook is a bit verbal, I just want to make sure I get it right

[remote fable]

@north sierra if the map has the form ax+by+cz+... in each component, then it is linear (but if it is of the form ax+by+c, i.e free constant without variable then it is not linear). Indeed there is a theorem for that, but I can't remember how exactly it is stated.

[cobalt tartan]

For this question, part c, where the hell do I even start?

[quartz compass]

I see, it partially follows from part b

hint: what do you know about the number of real roots an odd degree polynomial has?

[cobalt tartan]

At least one real root?

[quartz compass]

exactly

so where does the polynomial come from

[cobalt tartan]

Oh

If P is odd-dimensional

Err

The polynomial is the characetristic polynomial?

[quartz compass]

yeah

[cobalt tartan]

Wait would'nt this be true regardless of if it's odd or even?

[quartz compass]

no cause an even degree polynomial might not have a real root

think, x^2 +1 just hypothetically

[cobalt tartan]

Hrm

Hrm

[quartz compass]

regular 2x2 rotation matrix only has i and -i as eigenvalues

[cobalt tartan]

Oh

[quartz compass]

well, eigenvalues are roots of the characteristic equation

[cobalt tartan]

Ah

[quartz compass]

didn't want to make my edit unclear there

[cobalt tartan]

So now there's actually a theorem in the book that they did'nt want you to use for part b that explicitly states, "All real eigenvalues of P are 1 or -"

[quartz compass]

yeah, but I think it's safe to assume b is true while using that result to work c

[cobalt tartan]

Yea

So something like "By part b, we see that any real eigenvalue of P must be +/- 1, and that because n is odd, we must have an odd-degree charactersitic polynomial which will in turn have at least one real eigenvalue which must be either 1 or -1"?

[quartz compass]

yup

[cobalt tartan]

A friend scared me

Because he made it sound much harder lol

Wait I have a qusetion with regards to part a, can I not just show that if n = 2, then I can go and plug in pi/2 and that both P matrices are orthogonal??

[quartz compass]

idk I didn't read part a one sec

[cobalt tartan]

SHore

And that you could do the same for part d

[quartz compass]

idk what constitutes "show" here

[cobalt tartan]

Yea

Nor do I lol

[quartz compass]

like I could draw it out by geometry

take (1,0) and rotate it by some angle, find where that ends up, same thing for (0,1)

[cobalt tartan]

I could just

Go and show that if theta = pi/2

Then P is orthogonal

My original method

Was to show it with cases

[quartz compass]

I think that wouldn't show it, that seems like a special case

[cobalt tartan]

But that is really long

Like this but there's like 8 cases

And then for part d that'd be even more cases lol

[quartz compass]

ahh that's kind of painful

maybe there's a way to cheat with calculuss

I don't know if it would end up better or not though, but the idea I have in mind is take a vector, transform it by P, the length doesn't change

try to reverse engineer that down to a single parameter

probably not too dissimilar from what they're doing here

[cobalt tartan]

idk

That is my method

like

I tohught taht I could cheat by going "Similarily" but idk if that'd be fine LOL

[quartz compass]

hmm

it might be if you then diagonalize it

and then raise it to a power

then powers of the eigenvalues would end up corresponding to the angles

[cobalt tartan]

uhh

How would you diagonilazie it...?

[quartz compass]

so like, I guess you could take the special case when theta = pi/2

then your matrix is something like

$P=\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$

[stoic python]

Merosity:

[quartz compass]

or at least you could show this is one possible orthogonal matrix, doesn't matter what rotation this actually is

but once you diagonalize it, then you can raise it to powers

[cobalt tartan]

pi/4 is also

Orthogonal

Err wait I think that Imeant that if theta = pi/4

Then it's always orthogonal

[quartz compass]

all rotations, reflections, and permutations are orthogonal

it's really not enough to do it just for one example, gotta show all are possible

alright I think that way is not really too exciting, so maybe another way is

[cobalt tartan]

Wait

Wait

[quartz compass]

start with P, show it works, then add on an extra bit and show that extra bit has to be 0 maybe, that might work

[cobalt tartan]

hrm

What if

A matrix is orthogonal

If (P^T)(P) = I

Right?

[quartz compass]

yep

[cobalt tartan]

WHat if I showed that (P^T)(P) = I

Regardless of what theta is

Would taht be "Showing"

[quartz compass]

yeah you can do that and that will partially work

the problem is then proving there isn't some third possible form that could also work

[cobalt tartan]

Why only partially

Oh right

THis requires that P is always one of those fgorms

oh then I guess that there is the diagonilazibility thing

Wait

Hrm

[quartz compass]

I don't know if this diagonalization thing is the best approach since it actually won't get around this either I think

[cobalt tartan]

hrm

Yea

what we want to show is that every single orthogonal matrix in R2 can be expressed in that form

Yea?

[quartz compass]

yeah

[cobalt tartan]

Hrmmm

Idk

Cases is the only way that I could find

But then there'd be 8 cases which is painful lol

[quartz compass]

yeah, I think there's no way around it, you're on your own, I wish you good luck and that you get it over with quick haha

[cobalt tartan]

Rip

Any ideas for part d then?+

[quartz compass]

looks like they're choosing an axis of rotation then rotating around in that plane that keeps the first vector component in this basis fixed

[cobalt tartan]

Wait

IT just shsays "Show that there is an orthonormal basis"

So on this one

I can just plug in a value of theta that makes it true?

[cobalt tartan]

Hrm

A hint that I was given is to go and use the unit circle

I still got nohting 🤔

[narrow haven]

For vectors...perpendicular is same as orthogonal right?

[cobalt tartan]

Yes

Oh wait

@quartz compass the answer is the unit circle

Oh

It's because a unit vector

The rows and columns are an orthonormal basis

So you have some vector (cos theta sin theta) as one column

Then there's only 2 possibilities for the other column

[narrow haven]

Anyone wanna help me with some vector stuff?

[wintry steppe]

[narrow haven]

Well

Its not for a specific problem

Just want some help getting less confused when it comes to vector geometry

Like what methods to use in which situation

Like i know that to find an equation of a plane passing through points you do cross product of P1P2 and P1P3 and if it's a line then it's the coeffecients in front of t

But when you start talking about parallel and orthogonal i get confused

Andi havent done Linear in a while

(I'm doing Cal3 now)

Also is it always cross product? Which situations do you need to use dot product and magnitude?

@wintry steppe

[quartz delta]

Hey, I can try to help.

You're mainly asking about using techniques.

For any technique, I can sort of explain "why" it works.

The "cross product" is useful in 3D space.

It will find you a third vector orthogonal to the input vectors.

With length equal to the "area formed by the input vectors"

A dot product takes two vectors and returns a scalar.

It's useful for measuring the "magnitude of things".

You can measure a vector's magnitude in its own direction which is its total length.

You can also measure a vector's magnitude in different directions.

Which will be the length of the vector along some dimension.

@narrow haven .

That's just vectors.

[undone pier]

What are the advantages and disadvantages of using matrices to rotate objects? or in programming?

[quartz delta]

Matrices have lots of advantages.

One is that fast algorithms exist for matrix-math.

And fast hardware.

Matrices are very general.

They can describe rotations in high-level systems.

For instance, if you have a sequence of hinges, you can use matrices to describe rotations on them.

It's a very general questoin.

Like a pneumatic hand has lots of angles that depend on other angles.

You can describe all the angles as a high dimensional vector with associated rotations. @undone pier

[steady fiber]

disadvantages, you can get gimbal lock

which kinda sucks

[quartz delta]

What's that?

[steady fiber]

and so you use quaternions to describe the rotations instead

you lose a degree of freedom of rotation

when the rotation matrix takes on certain values

[quartz delta]

Hm?

[steady fiber]

wikipedia has a good example of it

[quartz delta]

Link?

[steady fiber]

https://en.wikipedia.org/wiki/Gimbal_lock#In_three_dimensions

look at "loss of degree of freedom with Euler angles"

[quartz delta]

Appreciate it.

[steady fiber]

you can see that instead of having 2 degrees of freedom when keeping one variable constant

you instead are only left with 1 degree of freedom

this always happens with whatever matrix you'd like to choose

if you can know that your object wont rotate in certain ways, you can use specific rotation matrices to side step the problem

[quartz delta]

Pretty cool fact.

[steady fiber]

but quaternions just handle every case properly

and do not suffer from problems like this

which is why they're used for rotations basically everywhere

[quartz delta]

So it was worth it for Hamilton to go insane over them.

[steady fiber]

it indeed was

quaternions are amazing

[undone pier]

Thank you everyone

[wintry steppe]

for this

do i just subtract the left matrix from right then take inverse?

[quartz compass]

if you're solving for A, sounds like the way to go

[wintry steppe]

ty

for this one

now that i have the inverse

what do i do

also does anyone know a method so i can check if my solution for the inverse is correct

i got it oh just multiply both sides by the inverse

[half ice]

@wintry steppe
You have a system Ax = b
You can solve for x here. x = A'b
' meaning inverse

[wintry steppe]

i see thanks

[wintry steppe]

https://gyazo.com/544ff92f12c243da211b3e9d946b7cfc

can someone help me with this

[livid crow]

how would i get the average of a multitude of vectors

(x,y,z)

Have objects in a plane want to get a normalized average position vector of all the objects

I added the vectors up and divided by the number of them but for some reason i am thinking this is not correcft

[dusky epoch]

why would that not be correct

[livid crow]

i dunno

[rancid sonnet]

Is the d/dx operator in matrix form the same as the matrix form of d/dx operator for functions in Hilbert space? Assuming that both have the same dimension

[wintry steppe]

https://gyazo.com/8c56795a88ea297aebe1d02f79017bad

can someone help me w this

[pallid swallow]

crossposted in #help-1

please continue discussions there

[urban bough]

so, the inner product of a matrix looks like Tr(X(transpose)Y)

and if it's like that, it's always a scalar, right?

[pallid swallow]

hmm yeah it's a scalar

[urban bough]

on what?

are you saying that it can also be a matrix?

[pallid swallow]

trace is a scalar

[urban bough]

yes correct

so the trace of two matrices X Y multiplied together like X^T*Y

will also be a scalar, correct?

[pallid swallow]

trace is a scalar, yeah

[urban bough]

okay

so then here's my question

How can the R and S subspaces be with “respect” to something that’s only a scalar?

Or am I misreading this problem?

[pallid swallow]

ah, it's not a scalar

it's an inner product

take it as a function of pairs of matrices

@urban bough

So basically, what the question is asking you to prove is that the inner product of a matrix in R and a matrix in S is 0

and that R is the maximal set of matrices with that property

(which is obvious because any matrix A can be written as a matrix in R + a matrix in S)

[urban bough]

Wait, but if the inner product involves a trace, how can it be a scalar?

[pallid swallow]

the inner product evaluates to be a scalar

trace of a matrix is a scalar

[urban bough]

Ohhh

Yea

What do you mean that R is the “maximal product”, in this case?

[pallid swallow]

Did I say "maximal product"?

R is the maximal set of matrices
?

Yeah, it means that for any other set of matrices A that satisfies inner product their matrices and a matrix in S = 0, if R is contained in A, R=A.

@urban bough ?

[urban bough]

Yea

Is this proof sufficient?

[pallid swallow]

not yet

For any other set of matrices A that satisfies inner product their matrices and a matrix in S = 0, if R is contained in A, R=A.

You need to show that

alternatively, you can note that the dimension of R and S...

@urban bough

[urban bough]

Why’s that?

[pallid swallow]

because checking orthogonality is not enough

span (0, 0, 1) is orthogonal to span (0, 1, 0), (standard inner product) but span (0, 0, 1) is not the orthogonal complement of span (0, 1, 0)

[urban bough]

Right right

How does showing the whole R=A thing prove orthogonality though?

[uneven bloom]

This is easy by a dimension counting argument. Write a matrix in A as R_1+S_1. Since Tr((R_1+S_1)S)=0, we get Tr(S_1S)=0. Selecting S=S_1 gives a contradiction unless S is the 0 matrix.

@pallid swallow is this what you intended?

[pallid swallow]

I didn't intend that

you need to show the complement bit

@urban bough

[uneven bloom]

Showing the direct sum result isn’t too bad

What did you intend

[pallid swallow]

the dimension counting argument

I don't see any mention of it

[uneven bloom]

The dimension counting argument is to prove the direct sum result

[pallid swallow]

yeah, that too

we need to prove the direct sum result

but the proof given doesn't seem to have that

[uneven bloom]

True

But the proof isn’t hard

[pallid swallow]

yeah, it isn't hard

[uneven bloom]

@pallid swallow Here’s the other direction. P has dimension n(n-1)/2 and S has dimension n(n+1)/2 generated by matrices with two nonzero (equal or opposite) entries in positions a_(ij) and a_(ji) (except when i=j, but 1 obviously spans R^1) Since each entry a_(ij) appears in one basis matrix of P,S, it’s enough to show that
[1 1], [1,-1] span R^2, which is trivial. Write a matrix in A as P_1+S_1. Since Tr((P_1+S_1)^TS)=0, we get Tr(S_1S)=0. Selecting S=S_1 gives a contradiction unless S is the 0 matrix. Hence A=P.

[pallid swallow]

we need to differentiate R and $\bbR$

[stoic python]

Element118:

[uneven bloom]

I let P be the space of skew symmetric matrices in R^(n*n)

[pallid swallow]

other direction?

[uneven bloom]

He already did the other direction

You have cancelling terms in the trace (ab and -ab)

[pallid swallow]

I mean, what's your "other direction" in

Here’s the other direction

[urban bough]

Can I just show that because the sums of the dimensions of 2 matrixes in nxn which are skew and symmetric, respectively

[uneven bloom]

Oh I’m proving the orthogonal set to S is contained within P

[urban bough]

Add up to account for all of the dimensions of R(nxn)

[pallid swallow]

oh I see

[uneven bloom]

Since we already know P works

[pallid swallow]

yeah

both works

[urban bough]

Oh epic!!!

And I came up with that one in my won

Own

Thanks guys

And then a matrix in R(nxn) has dimensions n^2 yes?

[uneven bloom]

The space of matrices has dim n^2

You should be able to find the basis

[urban bough]

Yea

[charred stirrup]

is R ^ 2 a plane?

and consequently does that make R ^ 3 an infinite "stack" of planes?

[half ice]

Kinda yeah

I don't see why not

It's a real line of planes

[charred stirrup]

sorry thanks I was having trouble picturing it

how do you do this without row reduction :/

should I use this theorem?

[half ice]

R³ is best seen as 3D space, but it's also like taking R² and making R copies of it

[charred stirrup]

The theorem doesnt say anything on independent, but, can I make an assertion that if it is not dependent, then it MUST be independent?

like arguing that independence is a binary state for a set of vectors

[half ice]

You'll want to look at an earlier theorem. Independence is when c1v1 + c2v2 + c3v3 = 0 has no solutions other than the trivial c1 = c2 = c3 = 0. Dependence is the lack of Independence

[charred stirrup]

RIP i tried to be sneaky lool

didnt know how to make a matrix with c1 c2 and c3

but thank u

[urban bough]

Is a pseudo inverse A+ equivalent to a transverse AT

[west mulch]

If you have a matrix equation like AX=B, and both det (B) and det (A) = 0
How to deduce weather the equation has an infinite amount of solutions or no solutions at all?

[clever cedar]

is it normal to do exams without a calculator

i was caught off guard when i had to find inverse of a matrix without a calculator

[brittle juniper]

Well yeah

you're expected to be able to do some calculations by hand

solving a linear system of equations, calculating a determinant, finding eigenvectors, that kind of stuff

[undone garnet]

A is 2011x2012

B is 2012x2011

prove that det(BA)=0

any idea

[halcyon garnet]

interpret A,B as linear maps. Then A : K^2012 -> K^2011 can't have a trivial kernel, so

what about the kernel of BA

[dusky epoch]

BA is 2012x20112

it doesn't have a det

it's not square

[undone garnet]

sorry

typo

😄

[dusky epoch]

ok so

rank(BA) <= max(rank(B), rank(A)) <= 2011 < 2012

[undone garnet]

😮

damn

thank you

:p

[whole parcel]

Hello! I need to prove the linear independence of those functions

I know that I need to use the lim as x tends toward 0 but that only gives me that a0=0

[halcyon garnet]

multiply a common denominator. Then you have a polynomial which can be defined on the whole real line, now plug in -k1,...,-kn

[whole parcel]

That denominator would be a sum?

[halcyon garnet]

,rotate

[halcyon garnet]

ok we also need the fact that a polynomial defined on $(0,\infty)$ can be uniquely extended to $\mathbb{R}$

[stoic python]

leoli1:

[paper egret]

reviewing for lin alg, question, given some set of vectors, how does one show that the set of vectors span, aka show that every vector can be obtained

[sonic osprey]

This question doesn't quite make sense

[paper egret]

wait sry ik i asked it weird

how do u prove linear independence and spanning

i know the definition of it

[sonic osprey]

The second still doesn't quite make sense

[paper egret]

ah shit

idk

wait hol on]

uhh

[brittle juniper]

[paper egret]

Let S be a set of vectors in vector space V. How do I show that S is linearly independent,

and how do I show every vector in V, can be expressed as a combination of the vectors in S

[sonic osprey]

Okay that's better

[paper egret]

squints eyes

u torture me

[sonic osprey]

Sorry

Uh for the first one

[paper egret]

if i wanted to show otherwise, all i have to do is come up with a counter example, but how do i properly show something is true

in this case

[sonic osprey]

So you have $v_1, v_2, \dots, v_n \in V$ and you're trying to solve the equation $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0$ with $a_1, \dots, a_n \in \mathbb{R}$

[stoic python]

Zopherus:

[paper egret]

for linear independence, the trivial linear combination is the only linear combination that is equal o 0

like how do i show that that's the ONLY combination

uniqueness?

[sonic osprey]

You show that for any solution

You must have that a_1 = a_2 = ... = a_n = 0

[paper egret]

how would i do that? uniqueness?

[sonic osprey]

Take the equation $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0$ and show that you must have that

[stoic python]

Zopherus:

[paper egret]

but how do i show the "only" part

[sonic osprey]

If you take $a_1, \dots, a_n \in \mathbb{R}$ and show that if $a_1 v_1 + \cdots a_n v_n = 0$, then $a_1 = \cdots = a_n =0$

[stoic python]

Zopherus:

[paper egret]

but u could run into the possibility where some a_i is not equal to 0

[sonic osprey]

But then it can't be a solution to $a_1 v_1 + \cdots a_n v_n = 0$

[stoic python]

Zopherus:

[paper egret]

linearly independence says that a1 = a2 ... = an = 0 is the only solution, how do i show that THAT is the only solution and that there is no other soln

that's the problem im having atm

[dusky epoch]

but u could run into the possibility where some a_i is not equal to 0
you want to show that that DOESN'T happen in order to establish linear independence.

[paper egret]

how would i show that it doesn't happen

proof wise

[dusky epoch]

depends on what your vectors are

[paper egret]

uhh hlemme pull an exampel

[dusky epoch]

i mean idk you could do contradiction if you want?

but then again

it's honestly like

easier to start with sum a_i v_i = 0

[paper egret]

like it seems obvious sometimes, but im trying to find a way to formalize it properly

[dusky epoch]

and show that each coefficient is forced to be zero

directly

more or less

what's your example

[paper egret]

S = {[1,3], [2,7]}

i know it's incredibly obvious

[sonic osprey]

The point is that

If you show that if $a_1 v_1 + \cdots a_n v_n = 0$, then $a_1 = \cdots = a_n =0$

[stoic python]

Zopherus:

[sonic osprey]

Then you can't run into the situation where one of the a_i is not 0

Because if it's not, then it can't be a solution to $a_1 v_1 + \cdots a_n v_n = 0$

[stoic python]

Zopherus:

[dusky epoch]

@paper egret ok so like

for convenience we might wanna call our coefficients here a and b since there's only two

so you want to show that if a[1,3] + b[2,7] = 0

then a=b=0

[paper egret]

but normally in proofs, that's not enough, is it?

[dusky epoch]

why would it not be enough

[paper egret]

i thoguht you had to do something like uniqueness, where you believe there is a second set of solutions, and then solve it again to find that the first = the second soln

idk, it didnt feel enough for some reason

[dusky epoch]

it's lit the defn of linear independence

like

it's obvious that the trivial solution is indeed a solution

[sonic osprey]

The point is basically showing that every solution has to be the zero solution

[dusky epoch]

^

[sonic osprey]

So there's no real probem with uniqueness here, because there's only one possible solution

[paper egret]

OH

bruh

im stupid

its linear

b r u h moment

so in short

either there is a set of solutions or there isnt

[sonic osprey]

It being linear isn't a huge deal here

The zero solution is always a solution to that equation

[paper egret]

yea i gotchu now

thanks for putting up with my bs lmao

and ima assume for spanning, it's the same idea, except you show that Ax = b, that there is a soln for any b

[sonic osprey]

Not Ax = b

[paper egret]

ah right lemme be a bit careful

damn my terminology is ass

every vector an be expressed as a linear combination

[sonic osprey]

Yeah

[paper egret]

it either is, or it either isnt

ok i realized i made it much harder than it needed to be

thanks for the help

[dusky epoch]

yeah this kind of thing is like

ridic easy to overthink

[whole parcel]

Thanks @halcyon garnet

[paper egret]

i did kinda forget it, but anyone wanna explain the intuition behind why A homogeneous system of m equations in n unknowns with n > m has a nontrivial solution

[paper egret]

mmmmm i'd say #prealg-and-algebra

[surreal horizon]

Gotcha

[paper egret]

all good

[lone quail]

How do i find the equation of the plane which passes through a line r {x=3, y=t,z=t} and is parallel to: {x=2, y+2x=-1}?
https://cdn.discordapp.com/attachments/359052604149465088/632295929998737429/unknown.png
line r is formed when h=-1

[ruby sigil]

Can someone explain the vandermode matrix to me, and its relationship to the discrete fourier transform?

[steady fiber]

vandermode matrix is just a matrix that looks like this

for some constant alpha

if you choose alpha to be the Nth root of unity, then it becomes a DFT matrix for an N-point DFT

if you multiply that matrix by your N-point data, you get the N-point DFT

[reef iris]

I was wondering why my reasoning is wrong here. Shouldn't the shape's points expand rather than shrink?

I followed a similar math tutorial (second option is shown as the correct answer)

[wintry steppe]

For anyone here thats proved the cross product formula for themselves, did you do it this way http://prntscr.com/pii40f ?

e_ is a unit vector

epsilon is levi-civita

[quartz compass]

@reef iris are you familiar with when you have a function and you want to translate the graph of f(x) to the right by 3, so you do f(x-3) ?

you are committing the same kind of error here if you were to try to do it by f(x+3) to mean to the right by 3

[reef iris]

@quartz compass whoops sorry late response. I think the math's correct, but I just realized the program was affecting the actual "canvas" holding the shape, not the shape itself. That's why it halved instead of doubling (like zooming out on a camera)

My mistake on not mentioning the software.

[quartz compass]

yeah, that's what I mean by my comment

seen this kind of question before from people working substance designer or houdini that kind of thing

[reef iris]

oh gotcha, ya all the years of slacking in math class catching up to me

😐

[quartz compass]

haha it's fine, at least you're doing math 😛

[uncut forge]

I'm having some trouble grasping Unitary transformations

Is the matrix of a unitary transformation always unitary independent of basis?

[dusky epoch]

no, if the basis isn't orthonormal then the matrix of your transformation in it will not be unitary

but if the basis IS orthonormal, then yes

[uncut forge]

Hmm I see

Ah right

Because [U*]=[U]^H only for orthonormal basis

So we can't just use any basis on U*U=I

But if I show that given a matrix U, (U^H)U does not equal I, (standard basis) then I can conclude that the linear transformation induced by U is not unitary right?

[dusky epoch]

what's ^H

transjugate?

[uncut forge]

Hermitian

[dusky epoch]

ok so conjugate transpose

aka transjugate

[uncut forge]

Ah yeah

[dusky epoch]

if U is the matrix of your thing in the standard basis then yes it's not unitary

[uncut forge]

Ok that's great

[lone quail]

How do i find the equation of the plane which passes through a line r {x=3, y=t,z=t} and is parallel to: {x=2, y+2x=-1}?
https://cdn.discordapp.com/attachments/359052604149465088/632295929998737429/unknown.png
line r is formed when h=-1

(Thanks in advance and sorry for repost)

[serene axle]

the first step of the solution involved assuming V a vector space has 2 elements x1,x2 in it

why is this ok to assume?

im confused

[quartz compass]

x1, x2 aren't elements of the vector space, they're components of your vector which is itself a single element of the vector space

just like (1,-1) is a vector in R^2

(x1, x2) is a vector in R^2

[frail ember]

@serene axle it suffices to check if the first and third vectors span R^2 there

if they're linearly independent then they span R^2

[wintry steppe]

Anyone can help me understand this

Nevermind i understand it now

[clear spoke]

This seems like a stupid question but

$A \cdot X = 0$

[stoic python]

Autistic Hoodie:

[clear spoke]

Where A and X are 0

The only possible solution for X that I see is a null matrice

Is this a trick question or am I missing something?

[shrewd kernel]

I'm confused, you said where A and X are zero, and then mentioned solving for X. what are we going for here

[clear spoke]

SOrry

Where A and X are matrices

[shrewd kernel]

hmm, i suck at writing matrices in latex, but

[clear spoke]

Maybe they meant

$A \cdot x = 0$

[stoic python]

Autistic Hoodie:

[clear spoke]

where b is 0

[shrewd kernel]

typically X is a vector

[clear spoke]

yes

Um

Maybe they are looking for the solution of the system of equations where each equation is equal to 0

[shrewd kernel]

there are matrices like this with more than one solution for zero.

1 1
1 1```

[clear spoke]

Oh really?

[spring urchin]

Probably because if they were matrices then youd have different ways of making the product null

And itd be a weird question

So I'm pretty sure they mean x

As a vector

[clear spoke]

Yeah...

[shrewd kernel]

so that matrix multiplied by the vector [1,-1]

and also [0,0]

and every multiple of [1,-1]

[clear spoke]

Hmm

[shrewd kernel]

comes out to [0,0] if i haven't messed up

[waxen bay]

?

[shrewd kernel]

have you tried multiplying by 22

[clear spoke]

Maybe they didn't mean vector

[waxen bay]

No

[shrewd kernel]

well, vectors are just 1 by x matrices

or x by 1 depending on preference

[clear spoke]

1     -1     1    1    -2
0    -2    -1    2    3
0    0    0    0    0
0    0    0    0    0

[shrewd kernel]

this has many solutions for the zero vector

[clear spoke]

What's the trick to find one?

[waxen bay]

How do you work it out

[clear spoke]

I can see that the top one will be 0 if multiplies by one

[shrewd kernel]

parameterize your solution

first of all

[clear spoke]

Hmm

[shrewd kernel]

as in, finish gaussian elimination

[waxen bay]

@shrewd kernel how do you work it out

[clear spoke]

It's 136/100

and just shorten it

[waxen bay]

Okay

[clear spoke]

136/1000

[shrewd kernel]

i think this is the wrong channel for you anyways

linear algebra is actually a field with matrices and vector theory and application

[clear spoke]

Oh the dots mea it's repeating

nvm

[shrewd kernel]

still try multiplying by 22 for part of it, and then show that the part that carries over to another part makes it 3

not sure if that's what your professor wants or if that's rigorous enough

[clear spoke]

Hmm, the solution is supposed to have 3 parameters

[shrewd kernel]

ok, if you have a parameterization of the gaussian elimination then if you plug in any reals for the parameterization you will get a vector that when multiplied by the matrix yields zero

the set of all vectors that do this is called either the kernel of the matrix or the null space of the matrix

[clear spoke]

The last time I did it was last year but I remember it being something like this

[shrewd kernel]

that looks like a vector decomposition of something

[clear spoke]

It's the solution for the matrice

I think I found it

$\begin{pmatrix}a\ b\ c\ d\ e\end{pmatrix}=\begin{pmatrix}0\ :0\ :-\frac{1}{3}\ :4\ :1\end{pmatrix}$

[stoic python]

Autistic Hoodie:

[clear spoke]

Let's test it...

Ok... doesn't seem to check out

[vital torrent]

I need help with this question, I am not sure why this is the correct answer. https://i.imgur.com/1GtReCq.png

[clear spoke]

Okay

Do you know what linear dependency is?

[vital torrent]

ya

i get that the last row would need to be 0's for this to be independent. But both c and d would make it row reduce that way.

[clear spoke]

No

They are linearly depended if one can be written as a multiplication of another

[vital torrent]

and if it rrefs to have a pivot in each column right?

[clear spoke]

$\begin{pmatrix}a\ b\ c\end{pmatrix}=x\begin{pmatrix}a\ b\ c\end{pmatrix}, x \in R$

[stoic python]

Autistic Hoodie:

[clear spoke]

Currently you have

$\begin{pmatrix}-5\ -6\ 5\end{pmatrix},\begin{pmatrix}20\ 24\ h\end{pmatrix}$

[stoic python]

Autistic Hoodie:

[vital torrent]

k

[clear spoke]

$\begin{pmatrix}-5\ :-6\ :5\end{pmatrix},-4\begin{pmatrix}-5\ -:6\ :\frac{h}{-4}\end{pmatrix}$

[stoic python]

Autistic Hoodie:

[vital torrent]

where are you getting a negative 4 from?

[clear spoke]

$-4 \cdot (-5) = 20 $

[stoic python]

Autistic Hoodie:

[vital torrent]

oh ok ya i see what you are doing

[clear spoke]

Which is the first row of the first column

Now

For which number will the vectors be lineary dependent?

[vital torrent]

-20

[clear spoke]

Correct

They can be either lineary dependent or lineary independent

If they are lineary dependent for h = -20

They will be lineary independent for?

[vital torrent]

h /= -20

[clear spoke]

Yes

For everyone other number besides h = -20

[vital torrent]

i get how to do it when i ref to get:

$\begin{pmatrix}-5\ 0\ 0\end{pmatrix},\begin{pmatrix}0\ 44\ h+20\end{pmatrix}$

[stoic python]

Rm7gaming:

[vital torrent]

i just dont get how you did it?

[clear spoke]

Do you not see the correlation between these two?

The second one is just a multiple of -4 of the first one

[vital torrent]

ya i see that now

[clear spoke]

$5\cdot(-4) = -20$

[stoic python]

Autistic Hoodie:

[vital torrent]

oh ok so -20 makes them dependent... i get it

so they are dependent when once can be scaled to equal the other right?

[clear spoke]

Yes!

[vital torrent]

and they are independent when there is just the trivial solution right?

[clear spoke]

I don't understand what you mean with trivial

If they are dependent only for -20 they will be independent for every other number besides -20

[vital torrent]

ya im just talking in general

with any matrix

[clear spoke]

Yes

Pretty much

Are you in high school?

[vital torrent]

no college

[clear spoke]

Oh

[vital torrent]

why lol?

[clear spoke]

I though linear algebra was in high school in the us

nvm

I am too just having linear algebra in college

This is my linear dependency theorem

[vital torrent]

so how do you do this one without reducing it?

https://i.imgur.com/b4XuWMO.png

[clear spoke]

Hmm

Only the last two seem to be linearly dependent

[vital torrent]

ya i think you just have to ref it and go from there... thats what i have been doing for these

here is the question with the answers/correct answer: https://i.imgur.com/49FUYpr.png

[clear spoke]

That makes no sense to me

They are linearly INDEPENDENT for all h

[vital torrent]

nah because if you plug in any number for h and then rref it, the matrix has free variables

right?

[clear spoke]

Hmm

They indeed are

[vital torrent]

i just dont really get how you figure out the solution for that one

[clear spoke]

I'm looking something up

[vital torrent]

k

[vital torrent]

ohhhhhhh!!!!

i remember now

its because there are more columns then rows so there has to allways be a free variable

it isnt even about h really

right?

[clear spoke]

I am not sure

This is unfamiliar to me but I'd love to know

They put the vectors in a matrice and try to find a solution

If a solutions exists where they are all equal to 0 then they are independent

[vital torrent]

do you understand:

onto

and

one-to-one?

[half ice]

@clear spoke
Cut half of it off. Linearly independent if the only solution to that is c1 = c2 = ... = 0

[marsh cedar]

that's not a good definition

it's a good sufficient and necessary condition for proofs, but it's not quite what it should be thought of as meaning

when a set of vector is linearly dependent, that means you can linearly obtain one from the others, and so there's a sense in which one is unnecessary for spanning the span, because whatever its coefficient is in the sum, it can be replaced with other vectors in the set (which is the same as changing the other vectors' coefficients accordingly)

more formally, $v_1, ..., v_n$ is linearly dependent if you can present any one (say $v_1$) as a linear combination of the rest as such:
$v_1 = {\lambda}_2v_2 + ... + {\lambda}_nv_n$ which has solutions iff ${\lambda}_1v_1 + ... + {\lambda}_nv_n = 0$ has solutions that aren't all 0.

[stoic python]

Intel:

[marsh cedar]

More intuition for this is that a set of vectors being linearly dependent means that some vectors are redundant, which is expressed both in the ability to remove a vector without changing the span, but also in the representation of each vector in the span as a linear combination of a set of vectors being unique being equivalent to the set being linearly independent

Suppose a set of vectors $v_1, ..., v_n$ is linearly independent, and the vector $v$ can be presented as their linear combination in two ways:
$a_1v_1 + ... + a_nv_n = v$
$b_1v_1 + ... + b_nv_n = v$
Subtract the equations to obtain:
$(a_1-b_1)v_1 + ... + (a_n-b_n)v_n = 0$
Since the set is linearly independent, this implies $a_j - b_j = 0, \forall j \in {1, ..., n}$, and so the representation of any vector as a linear combination of a linearly dependent set is unique.

[stoic python]

Intel:

[solar orbit]

So I am really having trouble with understanding vector spaces. More specifically proving that something is a vector space. I get the concept that you need to prove all 10 axioms but once I am given an example with non standard multiplication or addition it is right over my head. Does anyone know any good resources with examples of how to prove these types of problems. I have looked through the appropriate sections of my textbook and online but haven't really had any luck. I apologize in advance if I am not making much sense and might not be using proper terminology.

[sonic osprey]

Your terminology makes sense

Do you have an example we could work through or something

[solar orbit]

Ya give me one second I am working on an assignment and am just trying to see if I can find a similar example in my textbook

Suppose scaler multiplication is defined as x = x+y and addition is defined as x^k and is defined over R^n (-infinty,0)

[steady fiber]

that makes no sense

scalar multiplication of what

x = x+y?

what is the scalar being multiplied

what is the vector

[sonic osprey]

Also, what exactly do you mean by R^n (-infinty,0)?

[steady fiber]

and what is k

there's far too many things lacking

to make sense of that

[quartz compass]

scalar multiplication is defined as an addition and addition is defined as an exponentiation, seems logical

[steady fiber]

adding a scalar and a vector total sense

I, too, regularly add scalars and vectors

[solar orbit]

See that's the part that got me I am working on an assignment and was trying to change what I am given so we aren't working on the same problem. I cant find anything similar in my text

[quartz compass]

that's a good sign

[steady fiber]

can you like take a picture of the question

because if that's the question, then it's literally undoable

there's not enough information to do anything with that lol

what's x?
what's y?
what's k?
what and where is the scalar being multiplied in the definition of scalar multiplication?
what and where are the vectors being added and vector addition?

[solar orbit]

The part that I am reallly struggling with is what it means by it being defined over the interval of (0,infinity)

[half ice]

Literally just the positive reals

This set includes 1, π, 0.7
But does not include 0, -7, -π

On your paper you jumped to a counter example by assuming there's negative numbers in the set. That won't work, sadly

[solar orbit]

So the set means that both x and y have to be positive?

[half ice]

All of your vectors are positive real numbers

Same with your scalars, as it would look

[solar orbit]

Okay that makes more sense

[half ice]

Vector addition is real multiplication
Scalar multiplication is exponentiation

To put it vaugely

[solar orbit]

Oh okay I am about to have dinner and then take another stab at it thanks @half ice

[half ice]

Good luck! Let me know if you need any help with it

[rocky hill]

I'm rusty on LA, how many solutions will we have for a singular matrix $A$ in the system $Ax=b,;b\neq 0$

[stoic python]

kickpuncher:

[rocky hill]

since it's singular we can say we'll have infinite solutions

but there's also the possibility of having no solutions

I'm not sure what $b\neq 0$ signifies

[stoic python]

kickpuncher:

[slow scroll]

@rocky hill youre right that there is a possibility of infinitely many solutions or no solutions. It depends on whether b is an element of the column space of A.

[rocky hill]

so b neq 0 is just a red herring? lol

[slow scroll]

idk. there is always a solution for b=0, so it would kinda trivialize the problem i guess

[rocky hill]

ahhh, duh

[sonic osprey]

A singular matrix implies that there is always a non-trivial solution when b = 0

[slow scroll]

yea ^

[silent dune]

How would I go about 5b

[steady fiber]

show that they're linearly independent

n linearly independent vectors in R^n span R^n

[silent dune]

am I doing the same thing I did for a but instead of setting it equal to 0 set it equal to a variable like x

[steady fiber]

you can also do it by representing each "dimension" with a basis vector or something, so you get:
(a+b+c)i + (b+c)j + (a+c)k = di + ej + fk

where d, e and f can be any arbitrary constants, and i,j,k represent the three rows

and then prove that you can select a,b,c to make any d,e,f possible

[frail ember]

@steady fiber that might be a fact he needs to prove.

the n LI vectors spanning R^n thing

[steady fiber]

oh true

well ya, you can prove that

[frail ember]

I love that fact

it's so handy

and like k > n vectors not being linearly independent in n-dim space

stuff like that

[steady fiber]

that too

there's a lot of simple theorems about linear independence

that are just so damn useful

I just assumed that was a given because of that

[half ice]

@silent dune
What will end up happening is this system:
Ax = b
Where A is the matrix created by u, v, w as columns.

Every 3D vector can be represented if, given a b, you can always find an x. That happens when the system is consistent, and that happens when A's determinant is non-zero.

[solar orbit]

@Kaynex is vector multiplication commutative or does this not prove that this not a vector space?

[half ice]

Real number multiplication is commutative

x ⊕ y = xy = yx = y ⊕ x

So vector addition is commutative, and that axiom is good

[solar orbit]

Awesome thanks

So for the next axiom would I be right in assuming x ⊕ y ⊕ z =xyz so I can use the associative property and prove that axiom?

[half ice]

x ⊕ y ⊕ z is meaningless until you prove associativity

What order am I supposed to do the ⊕ in?

You just want to prove
(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z)

That happens by simplifying both sides, and showing they simplify to the same thing. You'll also need that real multiplication is associative

[solar orbit]

Right okay, doesn't axiom 4 the additive identity prove it is not a VS since x ⊕ 0 = x(0) =0?

[half ice]

What is 0?

There's no 0 here

[solar orbit]

Oh right since it's not included in the set

[half ice]

So in the axioms we aren't talking about the literal "0", but an element that we call the "additive identity"

It's an element that satisfies this:
x ⊕ e = x

There is an additive identity in this vector space

[solar orbit]

It would be 1 wouldn't it? x ⊕ 1 = (x)(1) = x.

[half ice]

It would be 1!

1 is this vector space's zero vector

And that proof is all you need

[uneven bloom]

You need distributivity as well

[half ice]

Oh we need a ton of things. We're nowhere near done

Luckily they all take like one line to do

[uneven bloom]

Well the (a+b)v=av+bv is slightly annoying

[solar orbit]

I am working through a few more axioms right now is it cool if I show you a picture of my proof when I am done?

[half ice]

Sure if you'd like

[uneven bloom]

Because (a+b)v=v^(a+b) and av+bv=v^a+v^b=v^(a+b) but one uses the standard plus sign and the other uses the new plus sign.

[half ice]

Same idea as associative, simplify both sides and show they simplify to the same thing

[uneven bloom]

But associativity is completely trivial.

[solar orbit]

I know I definitely need to play around with 9 a bit

[half ice]

Change the order in which you wrote 3

5 can't be right. What is 0?

[solar orbit]

Right I cant use 0 it's not in the set

[half ice]

I don't remember the order of all of the axioms, lol. I'm not sure what you're trying to prove with 6 or 7 or 8

[solar orbit]

This is what I am working g with

[half ice]

Ok, you want to prove that kx is a vector. That is, you want to prove that x^k is a real number

Positive real, I should say

Well, you can't exactly prove that, but can probably state that

x^k ∈ R^+, therefore kx ∈ V

[solar orbit]

Yes wait since 0 is not in the set doesn't that mean there is no zero vector since you cant make zero from two positive numbers

[half ice]

Remember, your zero vector is not 0

It's 1 in this case

[remote fable]

I have trouble understanding the third equality in this proof

How can we argue that a square root for T*T, with T being a general linear operator, always exist?

[half ice]

What's the book, if I may ask?

[remote fable]

Linear Algebra Done Right by Axler

Oh, it is because T* T is always positive definite ( because <T*Tv,v>= ||T v|| square, means >=0 , and by the definition in the book it is positive definite). And according to a previous theorem in the book, a square root for positive operator always exists and is unique

We talk about the row space as the orthogonal complement of the null space. But in a vector space without an inner product, the orthogonal complement is not defined. Does that mean the row space doesn't have any meaning other than "span of matrix rows" then?

never mind it was just my late night musings

[thorn tangle]

Linear algebra is kicking my but ,any good videos to watch?

[uncut forge]

I'm trying to follow a proof that a finite-dimensional vector space can be written as a direct sum of the generalized eigenspaces

They take an eigenvalue $$\lambda$$

[stoic python]

AoiKunie:

[uncut forge]

Then they define $V_1$=Ker$(T-\lambda I)^n$

[stoic python]

AoiKunie:

[uncut forge]

$V_2$=Ran$(T-\lambda I)^n$

[stoic python]

AoiKunie:

[uncut forge]

Another theorem has shown that the finite-dimensional vector space V is a direct sum of $V_1 and V_2$

[stoic python]

AoiKunie:

[uncut forge]

But then they say, and this is what I have trouble understanding

If lambda is the only eigenvalue then $V_2 = $ {0} (set of the zero vector)

[stoic python]

AoiKunie:

[uncut forge]

Why is that true?

(n=dim(V))

[uncut forge]

<@&286206848099549185>

[remote fable]

How is the parameter n defined?

As I see n is a free parameter here

[uncut forge]

n was dim(v)

I realized this belonged in the questions channel so it has been answered there now

[remote fable]

nice

[rocky hill]

Can I get a TL;DR on induced matrix norms? I'm getting lost in the weeds of notation-heavy explanations and am not grasping the concept.

[dusky epoch]

they're really operator norms in disguise

like say you have a linear map T between two normed vector spaces X and Y

each with their own norm

and the thing is

if X and Y are finite dimensional

the factor by which T stretches a given vector - that is, the ratio of the norm of Tx in Y to that of x in X - is bounded

this can be proved by many means but regardless this is a fact

and the least upper bound of all these factors is the operator norm of T induced by the norms on X and Y

now replace X and Y with R^n and R^m, each equipped with its own norm - and replace operators from X to Y by m*n matrices

@rocky hill

[rocky hill]

mmk I'm like... 35% there lol

"now replace X and Y with R^n and R^m, each equipped with its own norm"

I'm not totally understanding this: "each with its own norm"

I took linear only last semester, and we didn't even talk about norms 😠 so I really don't have a feel for it, just a vague understanding that it has to do with ways to measure distances between vectors...??

[dusky epoch]

hrgh

wait ok so like

aight

a norm is basically a function that assigns lengths to vectors

it has to satisfy three properties to be called a norm

1. it must sign zero length to the zero vector, and positive length to any other.

2. it must be homogenous: ||c*v|| = |c| * ||v|| for all scalars c and vectors v

3. it must satisfy the triangle inequality: ||v+w|| <= ||v|| + ||w||

[rocky hill]

ok yeah I'm with you

[dusky epoch]

if a norm satisfies these definitions, a metric can be recovered from it in the natural way: the distance between v and w is just ||v-w||

so the norm you're prob most familiar with is the euclidean norm

sometimes called the 2-norm

$|v|_2 = \left(\sum_i |v_i|^2\right)^{1/2}$

[stoic python]

Ann:

[rocky hill]

mhmm

[dusky epoch]

this is the "length" of a vector from R^n in the everyday sense of the word

but other norms can be considered

for example

the euclidean norm is just one of a whole family of norms known as p-norms or Hölder norms

$|v|_p = \left(\sum_i |v_i|^p\right)^{1/p}$

[stoic python]

Ann:

[dusky epoch]

with $p=1$, you have what is called the \textit{taxicab norm}: $$|v|_1 = \sum_i |v_i|$$

[stoic python]

Ann:

[dusky epoch]

there's also a norm that arises when you let p approach infinity

and is accordingly called the infinity norm

$|v|_{\infty} = \max_i |v_i|$

[stoic python]

Ann:

[dusky epoch]

these are of course just well known examples

[graceful osprey]

Hey just a quick question. If I am given a matrix that I am determining if it is a linear combination of three other matrices, is there another way to figure that out other than making a system of equations of the three matrices equal to each entry in the matrix at question?

[dusky epoch]

basically none

[graceful osprey]

Ooof