linear-algebra

[stoic python]

n/c

[wintry steppe]

This T is, in fact, nonlinear. But you can't just intuitively guess that

I agree the previous example has an easy way of looking at it, but that avoids the problem at hand I think

So why can't I convert to rectangular coordinates and get the result that way? I don't understand

[wintry steppe]

Let $V$ the vector space of all real functions continuous on $[a,b]$. If $f \in V$, $T: V\to V, g = T(f)$ defined by $$g(x) = \int_a^b f(t)\sin(x - t),dt$$ for $a \le x \le b$. Determine if $T$ is linear, and compute the rank and nullity of $T$.

[stoic python]

n/c

[wintry steppe]

T is linear and it's not difficult to show that, but how can I compute the nullity or rank of this? We need $g(x) = 0 = \int_a^b f(t)\sin(x-t) ,dt$

[stoic python]

n/c

[wintry steppe]

So we need $\sin(x-t) = 0$ or $f(t) = 0$

[stoic python]

n/c

[wintry steppe]

So f could be the zero function, or sin(x - t) = 0 when x - t is a multiple of pi....

I'm confused on how to continue this

[zinc timber]

are you sure you are asked to find the rank and nullity? because your V is infinite dimensional

also space of all functions is too vague, there are non integrable functions

[wintry steppe]

Non integrable continuous real functions?

[zinc timber]

oh you said continuous

so you have to solve the integral equation

[wintry steppe]

Yeah how?

[wintry steppe]

rank and nullity make sense for infinite dimensional too

[uncut quartz]

The B & W Leather Company wants to add handmade belts and wallets to its product line. Both belts and wallets require cutting and sewing. Each belt requires2 hours of cutting time and 6 hours of sewing time. Each wallet requires3 hours of cutting time and 3 hours of sewing time. The cutting machine is available for 13 hours and the sewing machine is available for 21 hours. The company makes $18 profit per belt and $12 profit per wallet.

what will the constraints be?

[zinc timber]

well there some methods you can try, first note that your kernel sin(x-t) is separable, so write it in terms of sin,cos

then blah blah blah

[uncut quartz]

if x = belts y = wallets

[wintry steppe]

@uncut quartz This is not linear algebra, can you please get another channel?

[uncut quartz]

ik but what channel would thi go in

[wintry steppe]

I don't know, try one of the questions channels

[dim epoch]

i mean technically it will be a system of equations

but the help channels suffice for that

[zinc timber]

it's fine

it's Linear Programming

[uncut quartz]

These are the constraints I got x < 8 y < 6 x + y < 13 x + y < 21

the underline is there in the <

idk how to add it

[nocturne jewel]

=<

[uncut quartz]

I got it

[zinc timber]

you should also have non-negativity constraints x,y >= 0

since it's 2 variable, you can pick any 2 equation of these and solve for x,y and see which point gives you the max (min)

[wintry steppe]

What about my problem?

[zinc timber]

.

[wintry steppe]

Oh okay

[zinc timber]

https://www.colorado.edu/amath/sites/default/files/attached-files/fredholm.pdf

[wintry steppe]

So $\sin(x-t) = \sin x \cos t - \sin t \cos x$

[stoic python]

n/c

[zinc timber]

yes

so $\int f(t)\sin(x-t) = \sin(x) \int f(t)\cos(t) - \cos(x) \int f(t)\sin(t)$ now assume $\int f(t)\cos(t) = c_1$ and $ \int f(t)\sin(t)=c_2$

just follow the article or google,

[wintry steppe]

So $x = \arctan \frac{c_2}{c_1}$ for nonzero $c_i$ (otherwise it's a trivial solution)?

[stoic python]

n/c

[wintry steppe]

I really don't understand the article

[wintry steppe]

so basically an exercise gives me a 4x4 Matrix and I have 3 linearly independent vectors that basically are basis if I am not mistaken. Then it asks me to find a 4th one that make up IR^4. How can I find the 4th one?

I hope I make sense

[keen sierra]

How do these 3 linearly independent vectors relate to the matrix? Or putting it another way, what does the matrix have to do with the question?

[wintry steppe]

The 4x4 Matrix consists of those 3 linearly independent vectors and one random one which is linearly dependent to those 3 independent ones

but I need to find a 4th one that is not linearly independent

[nocturne jewel]

so just scale one of your 3 vectors..

{v1,v2,v3,2v1} will have the 4th vector be dependent on the others

assuming {v1,v2,v3} form an independent set

[wintry steppe]

sorry mb , I meant to say a 4th one that is not linearly dependent

[wintry steppe]

implementing these in python for my package 🥲

[fringe fjord]

At least one of the four standard basis vectors will be linearly independent of the three you have. You can just check them in sequence.

[wintry steppe]

it's true that the operation "take a submatrix" distributes, right?

[fringe fjord]

Please ask your questions in the open channels.

[wintry steppe]

i.e. if $A = BC$ and $A_{(i_1, i_2, i_3)}$ represents the submatrix formed by taking the rows + columns $(i_1, i_2, i_3)$ of $A$, then $A_{(i_1, i_2, i_3)} = B_{(i_1, i_2, i_3)} C_{(i_1, i_2, i_3)}$

[stoic python]

capitalsigma

[wintry steppe]

can somebody help me with this?

are thes two formula's the same?

bd i tried to solve this and it gave me zero

i used the second formula

[wintry steppe]

How to check that given linear transformation has inverse ?

Got it

Can we find linear transformation whose matrix representation is given matrix . I mean is there any method do this ?

[gray dust]

for a real mxn matrix A define T:R^n->R^m by Tx=Ax

[wintry steppe]

Okay ty !

[wintry steppe]

Let $V$ the vector space of all real functions continuous on $[a,b]$. If $f \in V$, $T: V\to V, g = T(f)$ defined by $$g(x) = \int_a^b f(t)\sin(x - t),dt$$ for $a \le x \le b$. Determine if $T$ is linear, and compute the rank and nullity of $T$. T is linear and it's not difficult to show that, but how can I compute the nullity or rank of this? We need $g(x) = 0 = \int_a^b f(t)\sin(x-t) ,dt$ So we need $\sin(x-t) = 0$ or $f(t) = 0$

[stoic python]

n/c

[wintry steppe]

We can split it up into $\sin(x-t) = \sin x \cos t - \sin t \cos x$ and so $x = \arctan \frac{c_2}{c_1}$ but I don't know how to proceed from here

[stoic python]

n/c

[dusky epoch]

is V not infinite-dimensional tho

also what are c_1 and c_2

what are you even doing

[wintry steppe]

I'm expanding the integral and setting $\int_a^b f(t) \cos t = c_1$

[stoic python]

n/c

[wintry steppe]

and similarly for (...) * sin t = c_2

[dusky epoch]

what for

[wintry steppe]

I don't know

ryu sama said something yesterday

[dusky epoch]

if you don't know why you're doing something then why do it

[wintry steppe]

About the Friedberg Method or something

[dusky epoch]

still

[wintry steppe]

And to do this

But I didn't understand it

And I said I didn't understand it

[dusky epoch]

why are they having you compute rank and nullity in an inf-dim vector space

that's what i'm weirded out by

are you SURE thats what the exercise wants from you

[wintry steppe]

The exercise statement is verbatim

[dusky epoch]

i mean ok like... $g(x) = \sin(x) \int_a^b f(t)\cos(t) \dd{t} + \cos(x) \int_a^b f(t) (-\sin(t)) \dd{t}$

[stoic python]

Ann

[dusky epoch]

so the image of T is spanned by sin(x) and cos(x)

so one could say the rank is 2

that just makes it more obvious that the nullity is infinite however

[wintry steppe]

Why is it obvious?

[dusky epoch]

V is infinite-dimensional

[wintry steppe]

Ohhhh

Wait, so how do we prove the rank is 2?

[dusky epoch]

$g(x) = \sin(x) \int_a^b f(t)\cos(t) \dd{t} + \cos(x) \int_a^b f(t) (-\sin(t)) \dd{t}$

[stoic python]

Ann

[dusky epoch]

for every function f you have that Tf is a linear combination of sin(x) and cos(x)

thus im(T) ⊆ span{sin(x), cos(x)}

the reverse inclusion is annoying to show explicitly, and i would strongly advise against it unless you like prowling through page upon page of pointless algebra,

but it is in principle possible to find a function f that makes one of those integrals 1 and the other 0

[wintry steppe]

But f could also be something else

[dusky epoch]

wym

[wintry steppe]

You said we can find f to make one of those integrals 1 and the other 0

[dusky epoch]

there exists f such that Tf = sin(x)
and there exists f such that Tf = cos(x)

[wintry steppe]

But f could be something else as well

[dusky epoch]

ok f could be something else as well so what

my sub-goal here is to show sin(x) ∈ im(T) and cos(x) ∈ im(T)

this amounts to finding suitable inputs to T to produce sin(x) and cos(x) as outputs

[wintry steppe]

But that's not enough is it? We need every linear combination of sin(x) and cos(x) to be elements of im(T)

[dusky epoch]

once you have those, you can get any linear combination of sin(x) and cos(x) by linearity.

like

[wintry steppe]

ohh right

Because Im(T) is a vector space

[dusky epoch]

because T is linear

say you already have two functions $f_1$ and $f_2$ such that $Tf_1 = \sin(x)$ and $Tf_2 = \cos(x)$ then you have that $\alpha \sin(x) + \beta \cos(x) = T(\alpha f_1 + \beta f_2)$

[stoic python]

Ann

[wintry steppe]

Okay thanks I got it

Super good explanation I rate it 10/10

[dusky epoch]

i hope for both of our sakes that wasn't sarcastic.

[wintry steppe]

No it wasn't it clicked perfectly

The whole problem that is, I appreciate your help

[outer goblet]

can someone help me show that this is a linear transformation

i know its just aT(x)=T(ax), and, T(x+y)=T(x)+T(y)

but idk how to show that

[lavish jewel]

that's basically it, just do some substitutions

consider an (x1,x2,x3) and a (u1,u2,u3)

now try T(x1,x2,x3) and T(u1,u2,u3) and add them up

and then try T(x1+u1, x2+u2, x3+u3)

you already have an expression for these, so just apply your transformation T

[outer goblet]

Like this?

for scalar multiplicaiton?

its supposed to stay as minus i didnt mean to change it

[lavish jewel]

yeah

now, the preferred way is to write it in a way that shows you can reach one expression from the other

so you could write this as kT(a1,a2,a3) = k(a1-a2, a3) = (ka1 - ka2, ka3) = T(ka1, ka2, ka3)

and the procedure is much the same for addition

[outer goblet]

ah so just have an equals sign between the two expressions

ait thx thx

[lavish jewel]

in this case yes, assuming scalar multiplication works the usual way

just make sure to keep track of how scalar multiplication and vector addition are defined, since this might make it a bit trickier

but in this case, for usual vector addition and scalar mult, yes

[outer goblet]

ait thx

[brittle gyro]

Looks like I should find some sort of "Gram-Schmidt-ish" algorithm for finding such Q, but I am lost on the details... any ideas??

[frozen turtle]

This is my solution to the above question, is it sufficient? I was wondering if I should somehow show some algebraic operations to show their equivalent.

[wintry steppe]

If $b = \sum c_iv_i = \sum d_iv_i$ then $\sum c_iv_i = \sum d_iv_i$ or $\sum (c_i-d_i)v_i = 0$. Since $v_i$s are independent, this must imply that $c_i - d_i = 0$ for each $i$ or $c_i = d_i$

[stoic python]

n/c

[wintry steppe]

@frozen turtle

[frozen turtle]

Hmmm alright that works. But does that mean that my solution is incorrect per se? Or should i have also included what u said?

[wintry steppe]

Not sure I just think your solution is bulky

It just seems a bit handwavy as well

Mine is more concrete

[frozen turtle]

Hahah yea, i'm just v new to proofs so im not as concise as I'd like to be

[wintry steppe]

For example, you use the fact that there must exist a pivot in each column which then implies a unique solution, but this requires a proof

My proof doesn't use that, it only uses what we're given

[frozen turtle]

I see right

This helps, thank you!

[wintry steppe]

You're welcome

[brittle gyro]

<@&286206848099549185>

[frail timber]

Hello. Can anyone explain me what the image of a transformation is, and how to calculate it?

[zinc timber]

ok so one possible you can do it is by reversing the construction. given Q is orthogonal, we can multiply both side by Q and get $$QQ^TA-QDQ^T=QR \ \implies A-QDQ^T = QR$$

so A+QR = QDQ' which says it's symmetric

maybe work out from here

[zinc timber]

@brittle gyro ?

[zinc timber]

@sullen bay ok so you have 2b(a+b)=0, try find another equation by using their norm(v)=1

that was a weird typing experience ngl

[sullen bay]

I tried that

(thank you kindly for your help by the way)

Nothing really makes sense to me, I ended up with a bunch of algebra that didn't really got me anywhere 😅

[zinc timber]

like have u tried setting a²+3b²=1?

[sullen bay]

Yup

Lemme try again, maybe I missed something

[zinc timber]

then u have 2 eq and 2 unknowns, u should be able to solve

[sullen bay]

Bold assumption .. but I'll try 😄

Thank you kindly

[zinc timber]

[sullen bay]

Solved it thank you @zinc timber

I managed to deal with a 2eq & 2unknowns

Thank you for the help

[zinc timber]

[burnt hearth]

why are the "a1, a2, ... , a n" entries bold. isnt the x supposed to be the vectors and the "a" terms the weights (scalars)??

[gray dust]

no. the columns a1,..,an of A are vectors. x is a vector but its entries x1,..,xn are scalars

[burnt hearth]

got it thanks!

hi

[brittle gyro]

I agree, but Im not getting how this improves the situation though, still can't describe Q and/or R from A and D

[wicked rover]

help <@&286206848099549185>

[halcyon spindle]

This is not the right place for this. See #prealg-and-algebra.

[shell prairie]

hello. im guessing id be able to put differential equation questions here too?

or should i do that in calculus

[fringe fjord]

Differential equations go in #odes-and-pdes.

[slow scroll]

no

well, it depends what you mean by functional

its not a linear functional

are your functionals defined to be linear?

like, does a functional f need to satisfy f(cv + w) = cf(v) + f(w)

in that case, ye F(M) = det(M) is a good example

[subtle gust]

Is (AB)^n=(A^n)(B^n)?

[limber sierra]

not unless A and B commute

[subtle gust]

Ah

Meaning only if AB=BA right,

?*

So it's generally false

Interesting

Ty

[zinc timber]

[junior bane]

could i ask how do i prove the second theorem?

m not sure how to continue from here, cuz the cofactor of A has changed?

[cunning pier]

There doesn't exist a t such that A is not diagonalizable?
The geometric multiplicity of the Eigenraums is always 1.
The algebraic multiplicity of (-x+t)*(x-5)*(x+1) is always 1

Am I missing something?

[dusky epoch]

they're called eigenspaces in english, just for the record

but also have you considered that there are values of t for which the characteristic polynomial has a repeated root?

such as t=5

or t=-1

[cunning pier]

jup

thank you

@zinc timber

[zinc timber]

without context, what's jup?

[cunning pier]

jup [, that works, I'm dumb]

[zinc timber]

[spare widget]

If I have $C$ real skew-symmetric then I know that all of its eigenvalues are imaginary or zero. Let $C = B^HA + A^T\overline{B}$. Then it should follow that $C$ is zero correct? Since $C$ is real and thus the rhs must be real, and $(B^HA+A^T\overline{B})^T = A^T\overline{B} + B^HA$ implies that the rhs has only real eigenvalues.

[stoic python]

criver

[spare widget]

Or more simply $C^T = -C$ but $(B^HA+A^T\overline{B})^T = B^HA+A^T\overline{B}$

[stoic python]

criver

[zinc timber]

looks like it

[spare widget]

This only zero fulfills this (assuming C is real)

What happens if I set iC

Then the eigenvalues of iC are zero or real

But complex roots come in pairs

So it's indefinite

[zinc timber]

(though if the field is of char=2 then C may not be zero)

[spare widget]

I am assuming $\mathbb{C}$

[stoic python]

criver

[zinc timber]

yes

[spare widget]

So iC implies that the rhs must be fully imaginary

[zinc timber]

only matrix which is both symmetric and skew-symmetric is zero

[spare widget]

Can I say anything for the fully imaginary $D =B^HA+A^T\overline{B}, , D^T=D$

[stoic python]

criver

[spare widget]

If I identify $iC = D$

[stoic python]

criver

[spare widget]

Does this imply 0 only once again?

[zinc timber]

yes, for a short argument you can multiply both side of C=.. by i and conclude iC=D=0

thought there's a ^H in the exponent

[spare widget]

My question is why would I be able to conclude this

[zinc timber]

but D still D^T=D and D^T=-D so it's ok probably

[spare widget]

This is not the same as the initial problem

Ah it is

Yeah D turns into only imaginary eigenvalues

[zinc timber]

first by showing D is symmetric and then by assumption it's skew symmetric?

[spare widget]

yeah, I was missing the skew-symmetric part for D

Since in the former it wasn't

[zinc timber]

wasn't that the assumption?

[spare widget]

No the assumption is that iC = D

[zinc timber]

oh I see what you meant

[spare widget]

iC is indefinite in the above

Because C was skew-symmetric

[zinc timber]

but then the matrix turns into a complex one

[spare widget]

purely imaginary one*

[zinc timber]

skew symm has imag evs if it's real skew symm

it doesn't say anything about complex matrix tho

[spare widget]

If C was real skewsymm then multiplying by i

Would make it indefinite

Each eigenvalue gets multiplied by i

So I get real eigenvalues, but they were conjugate pairs, so I guess +-

Is that fine to this point?

The initial assumption was that C is real skewsymmetric, them iC must be fully imaginary and indefinite

[zinc timber]

yes, also we can have 0's

[spare widget]

Now requiring that $B^HA+A^T\overline{B} = D = iC$ does any being zero follow

[stoic python]

criver

[spare widget]

D is imaginary but D^T = D

I think it's simpler actually, if C was skew symmetric then

C^T = -C

[zinc timber]

yes

follows from the same argument tho

[spare widget]

Then (iC)^T= iC^T = -iC

So once again 0

[zinc timber]

$A = \m{0 & i \ -i & 0}$ then it's skew-symmetric but eigen values are +1,-1 which is not purely imaginary

[spare widget]

yes but C was assumed real

[zinc timber]

no I was talking for D

[spare widget]

Ah

I enforced that it must be purely imaginary since it kust equal a purely imaginary matrix

Finally if I have $C + iF = D = B^HA + A^T\overline{B}$, where $C$ skew-symmetric and $D$ either symmetric or skew-symmetric, does anything change

[stoic python]

criver

[spare widget]

It's clear that D is still symmetric, but now it doesn't have to be real or imaginary only

[zinc timber]

what's D? should be C+iD?

[spare widget]

D is a complex matrix now

Of the form on the right

[zinc timber]

oh sry didn't see it

[spare widget]

C is real skew-symmetrix, F is real skew-symmetric or symmetric

[zinc timber]

ok now $D^T = (C+iF)^T = C^T+iF^T = -C+iF^T$ do we know anything about F?

[zinc timber]

ok so either symm of skew

[spare widget]

let me see: D= R + iI = R^T + iI

[zinc timber]

if it's symm then we get $D^T=D$ meaning $C+iF = -C+iF$ equating real and complex parts we get $C=0$ so D is purely imaginary

[spare widget]

So I guess same thing

[spare widget]

I was doing this to verify that Ax cannot be the derivative of any energy of the form

$\langle Ax, Bx\rangle$

[zinc timber]

is the answer satisfying?

[stoic python]

criver

[zinc timber]

ok after this I have no idea

[spare widget]

Yes, considering I was trying to prove there exist no A and B resulting in a skew-symmetric derivative: Cx

The above is

$\frac{d}{dx}\langle Ax, Bx\rangle = (B^HA+A^T\overline{B})x$

[stoic python]

criver

[spare widget]

Which means that there is no energy of the above form (A,B linear) whose derivative $Cx$ results in a skew-symmetric matrix

[stoic python]

criver

[spare widget]

i.e. if I integrate it I get zero

I am assuming it acts like an odd function

I took x to be real though, I have no idea whether anything changes with a derivative wrt a complex x

[zinc timber]

ya that's what I was wondering as well, $\ip{Ax, Bx} = x^TB^HAx$

[zinc timber]

now the derivative follows, I see

[spare widget]

This all arose from trying to find an energy for

[zinc timber]

nice

[spare widget]

$\frac{d^m f}{dx^m}$ where $m$ is odd

[stoic python]

criver

[spare widget]

For even degrees they have the energy

[zinc timber]

probably something to do with the operator being self adjoint

[spare widget]

$\langle (-\Delta)^{m/2} f, (-\Delta)^{m/2} f\rangle$ resulting in differential operators $(-\Delta)^m f$

[stoic python]

criver

[zinc timber]

because for odd degrees ddx is not self adjoint

[spare widget]

Yes, odd degree derivatives are skew adjoint without extra boundary conditions

Which means there's no energy for those

Even degrees is the above

But if you set m odd, you get a pseudodifferential operator

And not d^m/dx^m

If I set boundary conditions that are not 0 at the boundary they show up in the adjoint though, so there the odd one is not exactly skew-adjoint.

[zinc timber]

(assuming proper space)

[spare widget]

I am an engineer

Most of our methods get regularised implicitly by just discretising. The above is not the case though since it results in a skew-symmetric matrix even in the discrete case, which turns singular with Dirichlet BCs. So what's done is that one takes unsymmetric derivative discretisations (upwind schemes). I am assuming that corresponds to one-sided derivatives in the continuous setting, but I am not sure.

Either way, thanks a lot for the discussion. 👍

[spare widget]

I figured out something else, let C and L be skew-symmetric matrices, then (LC)^T = C^TL^T = CL, that is, it is symmetric

To me this would imply that if I have another skew-adjoimt operator applied, then I would get something symmetric and thus derivable from an energy (or at least I think so). I'll have to think about what this means though. Some kind of integral of the derivative with a skew symmetric kernel.

[wintry steppe]

Let $U,V,W$ be vector spaces over the same field $F$. Assume $S$ and $T$ are in $\mathcal{L}(V,W)$, and let $c \in F$. For any function $R$ with values in $V$, we have $(S+T)R = SR + TR$

[stoic python]

n/c

[wintry steppe]

The wording is a bit weird, this means R is also a linear map right?

I suppose not

We can define $R: X \to V$

[lavish jewel]

you can come up with a counterexample

[stoic python]

n/c

[wintry steppe]

And then $[(S+T)]R(x) = (S+T)(v)$, where $V \ni R(x) = v$

[stoic python]

n/c

[wintry steppe]

And X need not be a linear space

[wintry steppe]

Likew hat?

[halcyon spindle]

come up with nonlinear function f whose value are in V. I am thinking of f : R -> V= R where f(x) = x^2.

[lavish jewel]

take something simple, like f(x) = (x^2, x^3 + 1)

and have S and T be some matrix of size 2x2

[wintry steppe]

How does that disprove it?

[lavish jewel]

f is nonlinear, but satisfies your conditions above

take R = f

[wintry steppe]

Oh you meant prove that R need not be linear?

[lavish jewel]

more like "disprove that R is linear"

[wintry steppe]

Sorry I thought you meant counter example to the theorem lol

What about the way I did it?

Since R(x) in V, say R(x) = v for v in V, then (S+T)(v) = (S+T)R(x)

And (S+T)(v) = Sv + Tv by definition

So Sv + Tv = SR(x) + TR(x)

[lavish jewel]

it's the same thing

[wintry steppe]

To prove the theorem?

[wintry steppe]

http://mathb.in/70442

Did I do these proofs correctly?

how can you check that some linear operator $L$ is linear if you have $L(uv)$ where $u,v\in V$? is there like a condition for this

[stoic python]

2M

[wintry steppe]

i thought it was $uL(v)+vL(u)$ but im not sure

[stoic python]

2M

[wintry steppe]

What's a linear operator?

@wintry steppe

[nocturne jewel]

uv is ill-defined

assuming both are vectors at least

[wintry steppe]

@nocturne jewel V doesn't have to be a vsp

[lavish jewel]

this V is not only a vector space. multiplication of vectors is extra structure

[wintry steppe]

Bumping this

[wintry steppe]

a map that obeys some properties

[wintry steppe]

Which properties?

[wintry steppe]

in my problem they are functions but i don think that makes any difference

[nocturne jewel]

yes that makes a difference, cause uv is actually defined

[wintry steppe]

Okay so that's the definition of a linear map, assuming V is a vector space

[lavish jewel]

it makes a big difference :x

[wintry steppe]

And we define uv to be the composition of u and v

[wintry steppe]

oh yea right, how does it look then if V is a function space

That is, let x in \dom V, and so (uv)(x) = u(v(x))

[wintry steppe]

yea mb forgot

[nocturne jewel]

uv is product not comp

[wintry steppe]

My text would define ST as the composition

[nocturne jewel]

Yeah, but u and v aren't the mappings

they're vectors

[wintry steppe]

ok ill state the question differently sorry: how can you check that some operator $L$ is linear if you have $L(fg)$ where $f,g\in V$? and $V$ is a function space is there like a condition for this

[nocturne jewel]

no condition

[wintry steppe]

I don't understand what you mean by linear operator is linear

You are saying it's linear already?

[nocturne jewel]

you need to check L(f+g)=L(f)+L(g) and L(cf)=cL(f)

[stoic python]

2M

[nocturne jewel]

L(fg) doesn't have any meaning when checking if L is linear

[wintry steppe]

Anyway

I'm gonna repost my proof

http://mathb.in/70442 can someone check it please?

[wintry steppe]

yea i just realized i read my text wrong

thx

[nocturne jewel]

This property does have a name though, but it has nothing to do with linearity

Called a Leibnitz rule

[wintry steppe]

leibniz rule no

yea

i read my text completely wrong lmao

sorry for that

[nocturne jewel]

But yeah, Leibnitz rule isn't linearity, linearity is additivity and homogeneity

[wintry steppe]

Hello. How do I find the values of a for which the inverse of A =

[nocturne jewel]

Use Fundamental Theorem of Invertible Matrices

[wintry steppe]

Okay, thank you. I am going to look that up

I am going to search for an example because I'm lost haha

[wintry steppe]

if you have x_0 + tv

that line would go through x_0 and be parallel to v right?

[spare widget]

Find when the determinant is nonzero

In your case det = -a

So for any a!= 0

[lavish jewel]

yes

[wintry steppe]

amazing tysm

[autumn lodge]

Hey, guys, I'm finding the eigenvectors for this matrix with eigenvalues 2 and 3, and I'll explain my question right after I send my work here:

Basically, I came to this as my solution - checking on Symbolab, they seem to have let t = 2 in order to come to integer solutions for the vector

[wintry steppe]

Thank you!

[autumn lodge]

Is that necessary to do? Is it just to keep the solution simpler or is there an overlying reason for it?

[spare widget]

It is not necessary to do

eigenvectors are unique up to rescaling and multiplicity

[autumn lodge]

okay, thank you!

[spare widget]

The definition is simply that $Av = \lambda v$

[stoic python]

criver

[spare widget]

So any non-zero rescaling works

You also get the $(A-\lambda I)v = 0$ from the above and the subsequent $|A-\lambda I|=0$ characteristic polynomial

[stoic python]

criver

[narrow nova]

so a vector space is a set of vectors. Could someone please assist with explaining what a basis is?

[wicked palm]

a basis set is a set of vectors that are linearly independent and span the entire space
Equivalently, you can write every other vector as a linear combination of all the basis vectors

[fringe fjord]

(add "in precisely one way" to the second line)

A fruitful way to look at it is that a basis gives you a coordinate system for the vector space. If the vector space is just R^n, you have coordinates for it already, of course -- but vector spaces can be weirder than that, and then it's useful to be able to write vectors as coordinates by choosing a basis.

And even if it is R^n, some problems become much simpler if you switch to a different coordinate system than the one R^n comes with out of the box.

[narrow nova]

thanks guys - that really helps actually to conceptualize

apologies if i am butchering this but I have to approach this like im a 5 year old.

Vector space is a set of all vectors. Basis is how we plot these vectors within the vector space

[wicked palm]

kind of

"plot" is a bit of an inaccurate word to use, I'd use "describe"

[narrow nova]

im rereading yours and Trops explanation. one moment please

it sounds like the basis is the plane on which the vectors lie?

[fringe fjord]

No, the basis is just a few vectors, not an entire plane.

[nocturne jewel]

A basis gives a way to describe the vectors in a space (think about vector form of a line that passes through the origin)

$\vec{x}=\vec{d}t$ is the equation of such line, and ${\vec{d}}$ forms a basis of the line

[stoic python]

Mosh

[fringe fjord]

For example we might say

I want to describe my vector space with a coordinate system with tree axes, such that I'll call this vector (1,0,0) and that vector (0,1,0) and that one over there (0,0,1), and everything else will have coordinates consistent with those choices".
Whether that actually gives you a description of the entire vector space depends on what "this" and "that" and "that one over there" vector actually are -- but IF it actually works, those three vectors are a "basis".

[narrow nova]

so it sounds like to me that the basis serves as a "template" for any vectors

[lavish jewel]

i don't know if "template", more like "pieces"

because the resulting vectors won't necessarily "look like" any of the vectors you started with

[narrow nova]

il brb in 10 min, watching 3blue1brown

so the basis "describes" the vector

[nocturne jewel]

a basis is an independent spanning set.

[narrow nova]

@nocturne jewel could you please describe an independent spanning set? I am unable to interpret. Im currently watching 3blue1brown linear algebra series

[nocturne jewel]

A set of vectors which are independent

and span the space

[narrow nova]

one step further please - "span the space"

[nocturne jewel]

span(B)=V

for basis B and space V

You can refer to your textbook for what span() is

[narrow nova]

that i will do - appreciate the help here fellas

do you have a preferred intro text on linear algebra?

@nocturne jewel the span of v and w (2 vectors) is the set of all of their linear combinations.

av + bw = linear combination
a and b are scalars which stretch or shrink the v and w vectors.

the span in this case is usually the entire 2D space.
"what are all the possible vectors we can get using vector addition and scalar multiplication?" = span

[nocturne jewel]

yes, span is the set of linear combinations

[narrow nova]

$$$ tysm

[nocturne jewel]

the span in this case is usually the entire 2D space.
Not always true, as you assumed Euclidean space

[narrow nova]

sometimes the vectors will lay on top of each other, in that case it would be a line. I mentioned usually because that is ususally the case (i think)

[nocturne jewel]

Yes, if the vectors are dependent, you don't have a basis

[narrow nova]

hecking heck its starting to make sense

[thin wing]

Likely a simple question (and me being stupid again); feel free to ping me.

[gray dust]

@thin wing u conflated set theoretic complement & subspace complement

[haughty berry]

What’s a subspace complement?

Like orthogonal complement?

Oh nvm

[gray dust]

this shouldve come up as a red flag anyway bc in nontrivial cases a set theoretic complement isnt a subspace hence it doesnt make sense to talk of its dimension

[thin wing]

Ah thank you!

[gray dust]

thats it

[fringe fjord]

It's a bit confusing of the text to write that complement as $\mathrm{ker}(\tau)^c$, as if it were a uniquely determined function of $\mathrm{ker}(\tau)$, though.

[stoic python]

Troposphere

[gray dust]

they do say ker(T)^c is a complement instead of the complement, but indeed its a bit misleading

[gray dust]

@thin wing see this too

[thin wing]

Yeah I was wondering about that earlier, since we know 0 isn't within the set complement of ker(τ), making it not a subspace. Thanks for that explanation!

[tired fossil]

hey guys, can someone explain this to me?

[thin wing]

Let me know if "linearly independent" is a new term.

[tired fossil]

I know what linearly independence is, but for my explanation I am confused as to what to write, should I just write out the linear combination

[thin wing]

Well the explanation is akin to explaining if 3 vectors can be linearly independent in a 2D space, and vice versa: can 2 vectors be linearly independent in 3D space.

Omg my Discord is freaking out, didn't mean to spam lol.. I kept pressing "retry" as it fails to send, but apparently that means "send 100 copies".

[thin wing]

Anyways, you follow?

[tired jungle]

How do I show this if I don't know if matrix multiplication commutes here?

If this were MEa it would be straightforward since Ea is guaranteed to b e in V, and L maps V to V but I'm not sure how to do this since M is in the middle

[gray dust]

@tired jungle does Ea denote a linear combo of e1,e2,e3 with the entries of a as coefficients?

[tired jungle]

yes

[gray dust]

from Lv=EMa it seems M should be a matrix representation of L

[tired jungle]

yea I got that but I'm not sure how it all works since shouldn't M be on the outside then

since v = Ea, L(v) would be MEa no?

if M is the matrix representation of L

actually I'm not even sure what the question is asking anymore

[gray dust]

V is an arbitrary space. Ea isnt necessarily in R^3 so MEa doesnt make sense as a product

so M should somehow represent L in the sense that Ma is the coordinates of Lv wrt E

[tired jungle]

wait why isn't Ea in R^3? I thought it would automatically be there since V is a 3-d vector space

oh wait nvm I get you

[gray dust]

if a=(r,s,t) then Ea=re1+se2+te3 is in V, which itself isnt necessarily R^3

[tired jungle]

ok yea I get you its an abstract vector space

So L(v) = EMa, how does this all work exactly? E is a basis (?), M the matrix representation of the linear map L, and a is a vector in R^3

if we evaluate it left to right, from E to M to A, i'm not even sure what comes out when EM is done

[gray dust]

yes E is a basis of V

a is a vector in R^3

M is a matrix

[thin wing]

M is the matrix representation, so if L:V→V, then M:V→V, so EM∈V

[gray dust]

@thin wing no

Ma is a vector in R^3

now recall i asked u this

does Ea denote a linear combo of e1,e2,e3 with the entries of a as coefficients?

u confirmed it

[tired jungle]

how do I know that a is in V?

since M is a mapping from V to V

[gray dust]

i just said a is in R^3

[thin wing]

I was wrong

[tired jungle]

ah yea I know a is in R^3 but I don't know if M can act on a if it isn't in V

since R^3 and V are different

[thin wing]

We want to preserve the basis, so M:ℝ³→ℝ³

[gray dust]

M is a matrix

[tired jungle]

oh

[gray dust]

not a map

[tired jungle]

yea lol

[gray dust]

specifically a matrix representation of L

but M isnt itself a map

[tired jungle]

just reposting so we can see it again

[gray dust]

a is a vector in R^3
M is a matrix
Ma is a vector in R^3
so EMa is the linear combo of e1,e2,e3 with the entries of Ma as its coefficients

use the hint that M should represent L to construct M

[tired jungle]

ok thanks will try

[gray dust]

more specifically this

so M should somehow represent L in the sense that Ma is the coordinates of Lv wrt E

[gray dust]

again wrong. M is a matrix, not a map, and the question requires lots of care to distinguish between the two

[thin wing]

It's because I'm implicitly talking about the "multiplication operator" between M and a, so ‧:M₃ ͓₃(ℝ)×ℝ³→ℝ³. I see that M itself is not the map. To be frank, I am thrown off by this question, so I'll take some time to ponder it as well.

Btw, sorry for interrupting. As you can see, it's also helping me learn heh..

[gray dust]

thats not the map either. again, M should represent L

[thin wing]

The binary operator is not a map? (M₃ ͓₃(ℝ) is to represent the set of all 3×3 matrices, not the same as M in the problem.)

[sleek sundial]

So I just have a question about this, MA outputs a vector in R^3. However, if this is not the same vector as the original vector a, how can it be guaranteed to land in V?

[gray dust]

EMa is in V

[gray dust]

the role of M here is to represent L, not a map R^3->R^3

[tired jungle]

hi this is part 2 of the question, what exactly is f() here?

this directly follows from the first

[tired jungle]

the first is found here

[thin wing]

I'm pretty sure that's a typo, it's meant to be F. If I'm wrong, I probably should move to Pluto.

[tired jungle]

yea you're probably right

[thin wing]

It is weird why now they're using parenthesis, so I am nervous.

[tired jungle]

Is this one of those symbol modification questions where I edit both sides of a matrix equation

seems like it since R is invertible

so E(Ma) = F(Nb)
E(Ma) = ER(Nb)

or am i going the wrong way

[thin wing]

Yeah that's what I did.

[stone wing]

how do u identify a polynomial

[nocturne jewel]

look at it and check if it meets the definition, however that's more apt for #prealg-and-algebra

[tired jungle]

how did you eliminate the b and a? since the question states for only R,R-1,M?

[thin wing]

Since you know v=Fb and v=Ea

[tired jungle]

was I not supposed to cancel the E's then

[thin wing]

Well what's b in terms of R⁻¹ and a?

You're almost done btw.

[thin wing]

(hint: Use the fact that F=ER)

[tired jungle]

oh lol

b = R^-1A

[thin wing]

Yep

[tired jungle]

💀

doesn't this lead to N being stuck in the middle then

$R^{-1}Ma = NR^{-1}a$

[thin wing]

wait why A, I thought that was a typo for a

[tired jungle]

oops

i meant a

[stoic python]

fake tree

[tired jungle]

but R^-1 is a matrix either way so I can't move it

without moving N

[thin wing]

So bring everything to one side

[tired jungle]

but wouldn't N still be stuck in the middle

i dont think I can consolidate terms

[thin wing]

In the end it wouldn't

(You can cancel something)

[tired jungle]

I don't think M or N are invertible though

[thin wing]

Additively, not multiplicatively

[tired jungle]

i have never seen that idk why

ok il try that

[thin wing]

There you go

[tired jungle]

but isn't R^-1 still stuck

it can't commute i dont think

[thin wing]

R⁻¹ has an inverse.

[tired jungle]

could you elaborate on that? So you want to move R^-1 to the other side?

RR^-1M = RNR^-1

M = RNR^-1

[thin wing]

How about the other way?

[tired jungle]

. wait

matrix multiplication is right to left?

[thin wing]

Matrix multiplication is associative, is it not?

[tired jungle]

yea

[thin wing]

So doesn't matter which order you multiply (hense no parenthesis)

[tired jungle]

Isnt that commutative

not associative

[thin wing]

Commutative means ab=ba

Associative means (ab)c=a(bc)

[tired jungle]

ok i get you now I forgot linear algebra thanks

[thin wing]

Yeah they're weird terms to learn the first time (elementary school doesn't count, even I forgot them from there).

[thin wing]

Btw, I think to be pedantic, a doesn't necessarily have an inverse so you can't really divide by it. So what really happens is since you know (R⁻¹M-NR⁻¹)a=0, and a isn't the zero vector (which is another case, which is trivial), then R⁻¹M-NR⁻¹ must be 0.

[tired jungle]

sorry, I get the associative part but how does it eliminate NR^-1 again?

(RN)R^-1 = R(NR^-1)

[thin wing]

Well if R⁻¹ has an inverse, then it has a left and right inverse.

[thin wing]

Nope, undo that move.

[tired jungle]

sure

R^-1M = NR^-1

[thin wing]

Yes, so what you did that took you to the wrong direction was you multiplied by a matrix R to the left, but you should multiply by a matrix to the right.

[tired jungle]

but how is that possible? using associativity I mean

I thought matrices were always multiplied to the left of an existing expression

[thin wing]

Not necessarily

As long as you do to one side you can do to the other then you're fine.

[tired jungle]

Oh ok thanks I guess that makes sense

since we can take R = R

and multiply that equivalent expressnion to both sides?

i mean R^-1M = NR^-1 to both sides?

[thin wing]

Note this is not the case for certain cases like with functions. i.e. take f∘x=x², then f∘x‧x doesn't make sense since (f∘x)‧x≠f∘(x‧x).

[tired jungle]

is that what you mean

so why is it applicable now is it only because it's invertible

[thin wing]

Because we're only doing matrix multiplication (which is "well behaved"), not function composition

[tired jungle]

oh

[thin wing]

So to be safe, always use parenthesis. In this case R⁻¹M=NR⁻¹⇒ (R⁻¹M)R=(NR⁻¹)R, but we know matrix multiplication is associative with itself, therefore (R⁻¹M)R=(NR⁻¹)R⇒R⁻¹MR=NR⁻¹R.

[tired jungle]

Sure matrices are hurting me today

[thin wing]

At least this problem's basically done.

[tired jungle]

yea I think that's the last of the matrices lol

[thin wing]

Cheers

[tired jungle]

ty for your help btw

[thin wing]

Np, just unfortunate I made a lot of big blunders earlier lol.

[tired jungle]

how do you get good at the abstract stuff

like I don't see the point of all this abstract space

like Hom(v,w) or something or like V* that maps like v to a real number

like sometimes I read the notes and it is so far removed from what I see on the hw like this slide

Will I ever need that stuff in a future class I mean, cause I low key just want to focus on the concrete stuff that I see in hw and leave the abstract stuff for a later class

[thin wing]

What's your major?

[tired jungle]

cs

[thin wing]

With practice. In my case, a good start was watching "Lectures on Geometrical Anatomy of Theoretical Physics" lectures by Frederick Schuller. It's a lot of abstract math, but I love the professor's way of teaching things so simply.

[thin wing]

Some documents may assume you know it (although for CS it's usually understood you may not have a strong background) since it's usually an important way to simplify/answer problems depending on your field.

[tired jungle]

pain

thanks though 💀 i will try to power through then

[tired jungle]

How do I show this without using a dimension argument?

I'm assuming that they don't want me to use one cause the sets orthonormal which probably has something to do with inner product idk

and cross product has not been introduced

<@&286206848099549185> any tips without a dimension argument

[thin wing]

Oh jeez, would I be correct to replace (e₁,e₂) with {e₁,e₂}, and [E] with span(E)?

[tired jungle]

yea 💀

notation is painful

[thin wing]

So then it's up to us to find a vector in ℝ³ that is linearly independent from {e₁,e₂}

[tired jungle]

isn't this like abstract though

no concrete stuff to work with

unless you mean like let e_1 = (a1,b1,c1) e_2 = (a2, b2, c2)

and try to find something

[thin wing]

I suppose, but this is standard for a linear algebra course.

[thin wing]

I mean, that could work.

Write the equation of dependence for those two, and use the conditions for them to be linearly independent and you should get a free variable, which can form the third basis (which is a nonzero vector not in the span of E).

[tired jungle]

by equation fo dependence, do you mean:
$\alpha_1 e_1 + \alpha_2 e_2$ = 0?

[stoic python]

fake tree

[thin wing]

Yea

But I may be jumping to the wrong thing, since they don't say e₁ and e₂ are linearly independent.

[tired jungle]

isnt orthonormal linearly independent

reposting from above

[thin wing]

I'm not sure why they included normal since the length shouldn't matter, but I'm not sure if "orthogonal" implies linearly independent was proven.

Anyways, yeah I think using the equation of dependence is okay, let me try it and see if it really works.

[tired jungle]

yea I think orthonormal implies independence since you can use inner product to prove it

[thin wing]

Yeah I'm pretty sure if you wrote an equation of dependence for e₁, e₂, and v you can find a v such that all the coefficients must be zero, meaning v is linearly independent from E, thus not in span(E).

[halcyon spindle]

Assume F is a field not of characteristic two, let $W_1$ be the subspace of $M_{nxn}(F)$ of all skew-symmetric matrices and let $W_2$ be the subspace of $M_{nxn}(F)$ of all symmetric matrices. I need to show $M_{nxn}(F) = W_1 \bigoplus W_2$ where $\bigoplus$ denotes the direct sum of $W_1$ and $W_2$.

[thin wing]

I meant I wasn't sure if they proved that for you yet, but anyways doesn't seem relevant. Or maybe I gave you a nuke, lol.

[stoic python]

Plegasus

[tired jungle]

just proved that 1 problem ago 💀

[thin wing]

Oh, gg

[zinc timber]

where are stuck exactly?

[halcyon spindle]

I am confused on how I should start this.

[zinc timber]

show their intersection is {0}

then any M_nxn can be written as a sum of symm part and skew-symm part

so W1+W2 = M

[tired jungle]

when you include v, do you mean that $a_1e_1+a_2e_2+a_3v=0$? Or do you mean re-arrangements of the first equation $a_1e_1+a_2e_2=0$ and try to wrangle something

[stoic python]

fake tree

[thin wing]

The first case, since you can find an arbitruary v such that a₁,a₂,a₃ are all zero, which implies L.I.

[thin wing]

Alternatively, for the second case what'd you do is a proof by contradiction: Assume to the contrary that v is an arbitruary vector in ℝ³, and that any v can be written as a linear combination of E: v=αe₁+βe₂. If you worked that out component-wise, you'll find that you have to restrict the components of v, which is a contradiction to the "any v can be written...".

[tired jungle]

sorry, could you elaborate on the first case? So you are trying to find an arbitrary v, from a1,a2,a3,e1,e2 such that the only way that the equation is 0 is if the constants are 0?

[thin wing]

Yes, so the equation of dependence is αe₁+βe₂+γv=0, if you prove that ∃v∈ℝ³: αe₁+βe₂+γv=0⇒α=0, β=0, and γ=0, then you proved e₁,e₂, and v are linearly independent, thus v∉span(E).

[tired jungle]

Ok I will try thatl, thanks

[wintry steppe]

Just to confirm, define $T(1,0) = (1,0,1)$ and $T(0,1) = (-1,0,1)$. Then the matrix with ordered bases (1,0) and (0,1) with cod bases (1,0,1), (-1,0,1),(0,1,0) will give the diagonal matrix, correct?

[stoic python]

n/c

[wintry steppe]

Need a sanity check

[zinc timber]

diagonal? you matrix isn't even square

though u can say a_ij = 0 if i \neq j

[wintry steppe]

Yeah that's how my text defines diagonal

[zinc timber]

then according to your text, it's diagonal

[wintry steppe]

it's the matrix

[1 0]

[0 1]

[0 0], right?

[zinc timber]

yes

[wintry steppe]

Thanks. Is there any software I can use to confirm this?

I know how to use matrices in wolframalpha but not how to specify bases

I meant in general, what if I have like a 6x10 matrix

Then it'd be really annoying to confirm it by hand

[zinc timber]

any software with mat mul support

ex-WA

you need to put the basis on your own

P^{-1}TP

is what you need to use

[wintry steppe]

The change of basis formula?

[zinc timber]

yes

[wintry steppe]

Okay thanks

[faint dune]

So this is the Gaussian-Approximation. We have an orrthonormalsystem psi.

On the fourier series we have this:

Isn't ak supposed to have the factor pi^(-1/2)

Same for bk.

[wintry steppe]

if you have a linear map $\phi: T\to D$ $T,D$ are vectorspaces with $v\mapsto D_v$ where $D_v$ is a differential operator, to prove this map is injective does it suffice to prove that $v = D_v$?

[stoic python]

2M

[wintry steppe]

for some value of D_v

[nocturne jewel]

$\phi(v)=\phi(u)\implies u=v$ is injectivity

[stoic python]

Mosh

[nocturne jewel]

@wintry steppe

alternatively you show the kernel is trivial

[wintry steppe]

hmm im follwing this proof and its supposes D_v = 0

and then it follows v = 0

then they say its injective

i was wondering why that also works

[nocturne jewel]

Yeah, Ker(T)={0} iff T is injective

[wintry steppe]

oh great

thanks

Let V have basis (1,x,x^2,x^3) and define T: V->V to be the map xp'(x) = T(p(x))

Let D be the differentiation map i.e. p(x) is mapped to p'(x) under D

I am trying to find the matrix for T^2D^2 - D^2T^2

I did this by computing D^2 first which gives us 0, 0, 2, 6x for each respective basis vector and then applying T^2 which then gives us 0,0,0,6x

Then I computed T^2 for each basis vector to get 0,x,4x^2,9x^3 and then applied D^2 to get 0,0,16,81x

So then $T^2D^2 - D^2T^2 = \begin{bmatrix} 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 6 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 & 16 & 0 \ 0 & 0 & 0 & 81 \ 0 & 0 & 0 & 0\ 0 & 0 & 0 & 0 \end{bmatrix}$

[stoic python]

n/c

[wintry steppe]

But this doesn't give the right result... I have the components in the right places, but it should be -8 and -48, not -16 and -75

Can someone find the mistake?

[dusky epoch]

hold on

what's the second derivative of 4x^2?

one would think it's 8, not 16.

@wintry steppe

[wintry steppe]

Thanks

I'm not sure how I missed that twice

[wintry steppe]

the function is $\phi: T_p\mathbb{R}^n\to\mathscr{D}_p\mathbb{R}^n$ such that $v\mapsto D_v$, how the hell does this prove surjectivity?

[stoic python]

2M

[wintry steppe]

carefully write down what it means for that function to be surjective

and you'll see that's exactly what the picture proves

yes thanks!

i got it now

[wintry steppe]

does this correspond to only this orthonormal basis or also for an arbitrary basis?

another basis (orthonormal or not) will correspond to a different set of derivations

you have an isomorphism of vector spaces. it takes any basis to a basis

so what would the set of derivations look like for a basis that is orthogonal and have a length of 2

?

maybe work it out

i think that'd be a good exercise

if you know what the standard basis is sent to under this isomorphism then certainly you can figure out what any basis is sent to

[wintry steppe]

care to send a hint?

i think this is something you should work out yourself

i will say that i don't have any particular result in mind

i have no idea how to begin

you have a map $\phi\colon T_p\bR^n \to \mathcal{D}p\bR^n$ given by $\phi(e_i) = \partial/\partial x^i|p$, which means $$\phi(v) = \phi(v^1,\dots,v^n) = \sum{i=1}^n v^i\left.\frac{\partial}{\partial x^i}\right|{p}.$$ this is a linear map. you are asking the following: "if $e_1',\dots,e_n'$ is another basis of $T_p\bR^n$, not necessarily the standard basis $e_1,\dots,e_n$, then what is $\phi(e_1'),\dots, \phi(e_n')?$

[stoic python]

TTerra

[wintry steppe]

maybe i don't understand the question

yes this is exactly my question

each e_i' can be expressed in terms of the standard basis

you get a change of basis matrix

you know how phi acts on the standard basis, so you should then be able to compute how phi acts on your new basis

this should all be review, it might be in tu's linear algebra appendix

[wintry steppe]

right, thanks!

[tired jungle]

What does it mean if a space is too rich

And rich enough

Also, I’m confused about abstract vector spaces and their dimension

Something like R^3 is obviously dim 3 but what about something like the space of continuous functions

If the functions have domain of R2 is that space also dim 2

[wintry steppe]

as far as i'm aware "rich" doesn't have a precise mathematical meaning here. can you give the context?

[wintry steppe]

infinite-dimensional

[tired jungle]

Sure it’s related to borel/lebesgue sets in R^n and general space

It was in probability class

But it was mentioned in passing and idk what that is supposed to mean

[wintry steppe]

hey i need help with my algebra 2 homework i am half way done i am just stuck on a few topic i just need some help understanding them

[sleek sundial]

This isn’t the algebra 2 channel, try #precalculus or something

[junior brook]

I'm really getting stuck on what exactly elemental matricies are

And it's making homework really difficult

I get the concept of elementary operations, but I'm not sure how to expand that the elemental matricies...

[junior brook]

Like this - I'm not 100% what it's asking

I can row-operate my way between A and B, but I don't think that is what is being asked here

[lavish jewel]

that's exactly it

matrices that represent those row operations

[junior brook]

Oh

Am I just stupid?

Could I get a (non-specific) example of the difference between standard notation and 3x3 format?

[lavish jewel]

what is standard notation

[junior brook]

That's the hangup here

3x3 makes sense to me, right? It's the standard notation that's tripping me up

[lavish jewel]

i would look back in your lecture notes, presumable in the lectures referenced there in the image

i don't recognize the term at a glance

[junior brook]

That's m ythough

Gotta be somewhere in there

[junior brook]

I somehow worked out how to do the standard first

Because it was just stating the row operations

[faint dune]

How can I interpretate this term?

We are dealing with gaussian-approximation. So we have the L2-Norm and a vectorspace over functions.

And psi is just a orthonormal basis of lets say the polynoms, and f any function C0[a,b], gN is the best approximation.

The basic idea I know is: We try to find a function g, that satisfies |f-g| < TOL. So we tolerate it if the error is smaller than that.

Maybe the square-error is easier to analyse. But whats the deal with TOL*(f,f)?

The prof said this is what we try to achieve, TOL is relative error.

[junior brook]

I need a nudge in the right direction, after someone helps Toge

Just like a general idea of how to approach the problem

[nocturne jewel]

what's the entire question?

[supple garden]

I need help computing the Jordan decomposition for this

So from my understanding it has algebraic mult. 3 and geometric mult 2 with eigenvectors (1,0,0) and (0,1,1)

So the Jordan canonical form is

3 1 0
0 3 0
0 0 3

but when I tried to find the change of basis matrix it doesn't work

Wouldn't it be
1 a 0
0 b 1
0 c 1

And then we do AW = WJ to find a b c?

but that gives 3a+1 = 3a

[dusk basalt]

Why is the cartesian product not a vector space?

[halcyon spindle]

It is a vector space.

[dusk basalt]

Yeahh I really thought so

But I am reading some book and it states. “Not all metric spaces are vector spaces. The cartesian product V x V can be turned in to a metric space ( another example which is not a vector space) bij defining some metric …..”

So it is implying that V x V is not a vector space which seems wrong to me, right?

[teal grotto]

just define addition coordinate wise so (a,b) + (c,d) = (a+c, v+d)
and c(u,v) = (cu, cv).

is V assumed to b a vector space?

[dusk basalt]

Yess

Yeah, I think it is certainly just a typo

Thx tho

[teal grotto]

weird. yea

[snow plinth]

ami just dumb or is there not infinte solutions for this problem

99% sure im doing something wrong

[halcyon spindle]

You are right if you row echelon matrix is correct(didn’t check), you just need to add no solution if k not equal to -4 and 4. Since you know there a unique solution of k^2 - 16 not equal 0, you can easily deduce when you have infinite solution.

[snow plinth]

ok but what about the k-24

dont i have to get all 0s for it to have infinite solutions

[halcyon spindle]

Only matters if k^2 - 16 = 0 which happens k = 4 or -4.

In which we have k - 24 is not zero, so the system has no solution.

[snow plinth]

holy shit it should be positive 10

not negative

@halcyon spindle thx for the help btw

[sleek sundial]

what is |$\phi$> in math notation lol

[stoic python]

Neko

[sleek sundial]

ok that didnt come out right

|v> is that just a matrix 💀 not familiar with notation for this physics class i'm in

[gray dust]

@sleek sundial vector. physics calls them kets

[fringe fjord]

https://en.wikipedia.org/wiki/Bra–ket_notation for details

[sleek sundial]

thanks

[fallen karma]

A linear operator T is continuous because |Tv-Tw| < norm(T)*|v -w| and the operator norm on T gives an upper bound on how a vector is scaled. Does this sound correct?

[fringe fjord]

If you have reason to know the operator norm is finite, yes.

[raven gull]

can somebody help me with a question in the voice chat?

[grave garden]

Hiii guys

I can't see why $\dim(U \times W) = \dim U + \dim W$

[stoic python]

Potato

[grave garden]

Could you explain me about that ?

I can't also find any article about this too

[zinc timber]

because it isn't

[grave garden]

Mostly they are group group theory not vector space

💀

[zinc timber]

haha

[grave garden]

The book I'm reading said so

[zinc timber]

because elements of $U\times V$ are of the form $(u, v)$

[zinc timber]

yes I know it's true

just x is not used in the context of vector spaces

[grave garden]

Ahhh I see you cross it out

[zinc timber]

what is your best candidate for the basis of UxV?

[grave garden]

$\dim U \times \dim W$ ?

[stoic python]

Potato

[zinc timber]

no

say ${u_1, u_2, \cdots, u_k}$ be a basis of $U$ and ${w_1, w_2, \cdots, w_l }$ is a basis of $W$

[zinc timber]

now think how should the basis of UxW will look like

consider any element $(\sum a_i u_i, \sum b_j w_j) \in U\times W$ and use linearity

[raven gull]

Can anyone help me with a question in voice chat?

I can't answer this question

[grave garden]

Here is my take on it

Let me think about what you tell me for awhile

[raven gull]

plz guys I really need help

[zinc timber]

that's not actually a basis because wj are being repeated

that's why it's written as U \oplus W not times

anyway your basis $(u, w) = (u, 0) + (0, w)$

[zinc timber]

you get the basis $(u_i, 0), (0, w_j)$

[raven gull]

Do you know linear algebra 2? @zinc timber

[zinc timber]

just put your question here

[raven gull]

the problem it is not in english

I hope you will understand

[zinc timber]

u can translate

[zinc timber]

what script is that

[raven gull]

lol

[grave garden]

Hebrew i think

[raven gull]

yes

[zinc timber]

lines and arcs

[raven gull]

in short you need to find all of the Jordan forms

[zinc timber]

this is what I understood so far, $A\in \mbb{C}^{5\times 5}$ with rank(A) = 3 and $A^4 = -A^2$

[grave garden]

what can $(u_i, w_j)$ be written into ?

[stoic python]

Potato

[zinc timber]

ok

[raven gull]

yes but there is another thing

wait a sec

[zinc timber]

can you translate the whole question

[raven gull]

the matrix is with real components

and that is all

[raven gull]

I mean you have this and the matrix is only with real components

[zinc timber]

is it given that A^4=-A^2 is the minimal polynomial or that's just an annihilator?

[raven gull]

ust an annihilator

j

[zinc timber]

assuming it's only an annihilator, $f(x) = x^4+x^2$, we have $m(x) | f(x)$ or $m(x) | x^2(x^2+1)$

[zinc timber]

you need to find all the possible minimal polynomials first

[grave garden]

( I could see why it is repeating now, thanks Ryu-sama )

[raven gull]

i think is has to be x^2

[zinc timber]

it doesn't have to x^2

[raven gull]

ok give another form

[zinc timber]

first find all the possible factors then we shall take into account the rank(A)=3

[zinc timber]

for ex m(x) = x

[raven gull]

but the rank is 3

so it is not possible

[zinc timber]

we are not considering that for now

[zinc timber]

yes, we'll come to it later

[raven gull]

ok

[haughty berry]

בהצלחה במבחן אחי

[brittle gyro]

Here are the steps of the Method of Conjugate Gradients. I need to prove that the all of the $d_k$ are $B$-conjugate, that is, $d_k^T B d_j=0$ whenever $k \neq j$. The "obvious" induction approach led me nowhere. It is easy to show $d_{k+1}^T B d_k=0$, but even the simplest case, like $d_2^T B d_0$ for instance, feels like a dead end...

[stoic python]

Sydd

[brittle gyro]

Here is f

<@&286206848099549185> let me know if there's another channel where this question would be more appropriate. Thanks!

[spare widget]

I am not sure whether there's a proof in Saad's book, but you can take a look there.

[wintry steppe]

מישהו העלה פתרון לאתר

[proper tundra]

Hello
I am in my first year of university. Today my maths teacher gave me a problem to solve and no one in the class understood.
It was an application problem. I knew that I had to proceed by double implication but I didn't know how to continue. Can anyone give me the solution to this problem?
Let F:E-->F E and F be finite sets
show that f is injective if and only if some A is a part of E, A=f**(-1)(f(A))
In the picture, you will find the clue in french given by our math teacher.Thanks you very much

[gray dust]

we dont give full solutions here. this is also more suited for #proofs-and-logic or #discrete-math

a hint is to unpack all relevant definitions (image, preimage, injective, etc)

[distant schooner]

Let F:E-->F E and F be finite sets
u mean F: E -> A?

oh wait

well F: E ->F still doesnt make sense right?

and this is relatively simple proof, think about what injective means in terms of invertibility of that function

@proper tundra ^

big hint: if $F : X \rightarrow Y$ is injective, it means that there is at most one $x \in X$ such that for a $y \in Y$, $f(x) = y$. that means that $F^{-1} : Y \rightarrow X$ right? if $A = F^{-1}(F(A))$, then $A = A$ because you are mapping from $X$ to $Y$ in $F$ and from $Y$ to $X$ in $F^{-1}$. can $F$ be injective if F^{-1}(F(A)) \not\in X?

[stoic python]

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

[hardy inlet]

im trying to think of B before A. That's not possible because that implies rank = nullity and by the ranknullity theorem,
dimV = dim range T + dim null T = 5, so rank = nullity 2.5, and we aren't dealing with fractals so thats not possible?

[paper fractal]

Hey guys, suppose youre given a matrix, how do you know what sequence of elementary row operations to do to get the row echelon form or reduced row echelon form

is there any guide or do i have to grow an intuition

[haughty berry]

There’s an algorithm, but it won’t always be the most efficient

You essentially reduce each column one at a time

[hardy inlet]

if doing it by hand; probably just try and filter by leading zeroes, then go column by column and zero all all but 1 row if possibile

Can someone give their opinion on if they think this is right? I emailed my teacher but im slightly unsure what it is to say the range and nullspace are "equal"

[haughty berry]

By equal they mean that they’re equal?
Anyways yes, I’m not sure why you used v_1… as coefficients in your vector (it’s not really an issue but it is confusing), but your proof is correct

[hardy inlet]

like just the subspaces formed by them are equal?

I guess since L(V,W) is R4 to R4, V = W

[haughty berry]

They are subspace themselves? Wdym “formed”?
(Also I’m assuming the range is the image and the null is the kernel?)

[hardy inlet]

yea

[haughty berry]

The image of T and the kernel of T are both subspaces

[hardy inlet]

i think the think that made me hesitant was that range is a subset of W and nullspace is a subset of V, but since V=W its okay

[haughty berry]

Oh I see

Yeah

[burnt hearth]

so for this answer in order to be consistent lambda has to = 1 and -2 so i include both solution sets right??

[gray dust]

@hardy inlet set equality, not much room to interpret anything else

[burnt hearth]

?

[hardy inlet]

is this 🧠

(is a fractal dimension possible for linear algebra in general tho?)

can someone please explain what it means to\
Find a $4 \times 4$ matrix $M$ so that the range of $M$ is spanned by $(1,0,1,0)$ and $(0,1,0,1)$.