linear-algebra

[nocturne jewel]

yeah, bases are independent vectors

[hollow finch]

you were given two of linearly independent columns of A. since A has rank 2 (because you were told it has a 2 dimensional column space) we know that all the other columns are just linear combinations of the ones you were given (meaning they span). since they're linearly independent and span, they form a basis.

[nocturne jewel]

Ok so the answer isnt whether b is in Col(A)?

[hollow finch]

you could also look at it as if we make the matrix C with the two columns you were given, the system Cx=b has only one solution since it doesnt have a null space (they are linearly independent)

its a two part thing

[nocturne jewel]

wait so is it inf or 1 solution?

[hollow finch]

we know there is at least one solutions because b is on Col(A)

[nocturne jewel]

You said infinite and only 1 confusion

[hollow finch]

we know there are infinitely many because Null(A) is not just the zero vector

wdym

[nocturne jewel]

You said an alternative way of thinking leads to only 1 soln

but with null(A) logic there are infinite

[hollow finch]

yeah if we only consider the two columns you were given

[nocturne jewel]

Ok so Ax=[5,7,-2]^T has inf solution?

[hollow finch]

we know there is at least one solutions because b is on Col(A)
we know there are infinitely many because Null(A) is not just the zero vector

[nocturne jewel]

ok so infinite

[fringe matrix]

what is the name of a Matrix such that

M^2 = M?

[slow scroll]

idempotent

[fringe matrix]

ty! 🙂

[slow scroll]

or... maybe a projection in this context

[fringe matrix]

lol that was exacly what i wanted

[slow scroll]

projection or idempotent?

[fringe matrix]

projection is a type of idempotent ^_^

i was looking at projection and wanted to give it another name

[slow scroll]

ah okay

[native rampart]

Well,All idempotents are projections

[fringe matrix]

in matrixes yea..

[slow scroll]

Well,All idempotents are projections
idk if i would call an idempotent matrix a "projection" outside the context of inner product spaces tho

[brittle juniper]

When I was being taught linear algebra for the first time, it went like

an endomorphism p on a vector space E is called
• a projection when E=Ker(p)⊕Im(p) and the restriction of p on Im(p) is the identity
• a projector when p²=p
(maybe it was the other way around)

and the important result was that projections are projectors and projectors are projections (and that's why I don't remember which one was which)

the prehilbert-specific stuff was introduced a lot later

a projection p on a prehilbert space E is said to be orthogonal when the direct sum E=Ker(p)⊕Im(p) is orthogonal

[thorny hemlock]

not really sure how this is done

rangeT = {Tv1 ... Tv_n}

[soft burrow]

maybe you could suppose that under some pair of bases, the matrix has exactly dim(range T)-1 non zero entries and derive a contradiction?

[thorny hemlock]

hm

[soft burrow]

in the best possible case you'd have some diagonal matrix with dim(range T)-1 non-zero entries and that should contradict the given value of dim(range T)

[thorny hemlock]

er

so

each Tv1 ... Tv_n corresponds to an entry in the matrix

so we get dimrangeT-1 = dimrangeT ?

but er

how does that contradiction work

[neat relic]

Hey guys, not sure if this belongs here but I've been asked to do a linear algebra online certification as proof of competence before taking on certain projects at a job..was wondering if do you guys have any in depth linear algebra certification you'd recommend pls? There's like so many it's tough to choose haha..

[wintry steppe]

are there any topics of linear algebra that you know you'll need in particular?

[neat relic]

ahh I don't know lol..it's to do with robotics and some machine vision so probably linear algebra topics to do with that..I think he wants proof of competence that I know linear algebra basically..

I guess personally I'm looking for like if it covers subspaces, row echelon, basis, dimension, nullspace, linear independence, eigenvalues eigenvectors, and the tougher stuff like orthonormal orthogonal gram schmidt etc..

[tame mural]

I never heard of a lienar algebra cert

that sounds so specific

[neat relic]

I guess kinda online course certification

that they want to show that I know linear algebra

[round coral]

@neat relic you mean a certificate course in linear algebra, am I right ?

you can look over edx or coursera courses on linear algebra, choose what is best for you

these give certificates of the universities they are being offered by, but are not credit eligible mostly, don't have any exams, just graded quizzes

i guess they may be useful for you

[hollow finch]

My LA class never touched on generalized eigenvectors so I'm hoping someone can help answer some questions I have about them.

1. How unique are they? I know eigenvectors in general are not unique because you can scale them, but from just messing around with GEs it seems adding any scalar multiple of the ordinary eigenvector(s) has no effect on whether or not it's a GE.
2. What would be the proper way to "introduce" them in a proof? Like if I have a general nondiagonalizable 2x2 matrix A with eigenvector v and generalized eigenvector w what is the formal way to state that?
   and if anyone has any good resources I could look into to understand them better that would be appreciated. The wikipedia page is a bit difficult to understand for me.

[limpid fiber]

why is f(x) = x + 4 non-linear

I've seen the proof, but intuitively it feels like a linear function

[wintry steppe]

it's "highschool linear"

that kinda function would be called affine, usually

linear function plus a constant

[round coral]

@hollow finch I think these questions will be easily resolved when you study some more about generalized eignevectors, I would recommend , you can read either Chapter 8 of Axler, he covers all about generalized eigenvectors, decomposition theorem etc. in great detail , chapter 7 of Friedberg Insel Spence also covers this in a good way , you can choose any one you like .

hope it helps

[hollow finch]

unfortunately i dont know when my next LA class will be
but thank you! i'll look into those

[wintry steppe]

your next la class is whenever you next open a linear algebra book

[limpid fiber]

Yeah it's highschool linear lol
thank you

[limpid fiber]

@wintry steppe if it were a line through the origin, that would be enough right?

Is that a consequence or a condition

[round coral]

if your given map, maps 0 to 0 , then it is not necessary that it's linear but if it is linear then for sure by linearity 0 will be mapped to 0

[limpid fiber]

I guess it's a consequence of scalar multiplication then

[stoic python]

slimvesus

slimvesus

[limpid fiber]

nice, thank you

[nocturne jewel]

bit bold to assume 0+0=0 \j

[wintry steppe]

linear functions........

[thorny hemlock]

if $v1,...,vm$ is a basis of V. we know there exists $T(a_1v_1 + .... + a_mv_m) = c_1v_1 + ... + c_mv_m$. Setting all scalars to zero we can get a specific vector so$ T(a_jv_j) = c_jw_j \rightarrow Tv_j = \frac{c_j}{a_j}v_j$

[stoic python]

Yes

[thorny hemlock]

does this answer the question?

[wintry steppe]

"there exists (equation)" what do you mean by this? what are these scalars? why are you setting these scalars to zero?

why are you writing your basis as v1, ..., vm when the space is 1-dimensional

[thorny hemlock]

lmao

idk

[wintry steppe]

you should start the problem by saying "let {v} be a basis for V"

then what can you say?

[thorny hemlock]

V = span(v) -> for all u in V, u = av for some a in F

[wintry steppe]

yes

(to be nitpicky, you mean for some a in F)

[thorny hemlock]

ok

we then make use of this ?

[wintry steppe]

i guess you can do it using that, yeah

probably not the easiest way to do it

[thorny hemlock]

oh

[wintry steppe]

but hey it could probably work out

[thorny hemlock]

how would you do it

[wintry steppe]

well you have a basis {v} of V

meaning any element of V is a scalar multiple of v

in particular, T(v) is

[thorny hemlock]

hm that works

[wintry steppe]

why?

[thorny hemlock]

basis' can span, T(v) is another something in V

av = anything in V

[wintry steppe]

you should write a full proof

[thorny hemlock]

i dont know how

[wintry steppe]

think about the problem some more then

remember you ultimately want to show that T equals scalar multiplication by some scalar (one you need to come up with!)

starting from a basis of V is how you find that scalar

i gave a big hint

[thorny hemlock]

ok

isnt it what i wrote tho

[wintry steppe]

at the start?

[thorny hemlock]

yah

[wintry steppe]

i don't understand it, i think it needs to be fleshed out a bit

[thorny hemlock]

jjust change the basis to just v1

oh ok

[blissful pagoda]

This matrix is singular because I found its determinant to be 0

Homogeneous systems all have the trivial solution

If it is singular, then it cannot have unique nontrivial solutions

Therefore it has no solutions or infinitely many solutions. The number of equations and unknowns are the same. So I say it has no solutions.

Turns out it has infinitely many solutions, how is that? (part b)

[wintry steppe]

the set of solutions to the system is the kernel of A. since A is singular, its kernel is nontrivial, thus infinite

[blissful pagoda]

Wouldn't that imply then that all singular matrices have nontrivial kernels and therefore all singular matrices have infinitely many solutions?

[wintry steppe]

all singular matrices have nontrivial kernels
this is true (and conversely for square matrices)

an nxn homogeneous linear system of equations Ax = 0 has a unique solution (zero) if and only if the matrix A is nonsingular, and on the other hand, has infinitely many solutions if and only if A is nonsingular. if the system is not homogeneous, then you can't say for certain whether or not there are solutions

in a nonhomogeneous nxn system Ax = b, you will have infinitely many solutions if you have one (b/c ker A is nontrivial)

[neat relic]

Yep that..and ok cool I'll look over the edx and coursera courses, and figure out the most suitable one..thanks..

[blissful pagoda]

Wait... so if I have a singular matrix (by finding determinant = 0), it will always have infinitely many solutions and never no solution?

(other than the trivial solution of course)

[acoustic path]

yep

[blissful pagoda]

Oh I did not know that. I was under the impression that if a matrix was nonsingular then it had one single unique solution. And if it was singular and if it had more variables than equations then it was infinite, else no solutions. So this is wrong?

[acoustic path]

no it can have no solutions

or infinite, it depends on the singular matrix

[blissful pagoda]

How can you tell

[4:06 PM] gotta go fast: Wait... so if I have a singular matrix (by finding determinant = 0), it will always have infinitely many solutions and never no solution?

[acoustic path]

ye i didnt read that first my bad

and wym how can you tell

just how a normal matrix can have zero, one or inf solutions, a singular matrix has zero or infinite, never one

[blissful pagoda]

I thought a nonsingular matrix can't have zero or infinite solutions 🤔

[pure tangle]

Can someone please help me understand why using linear combinations to find variables works? The proof I'm looking at says to think of it like a weighted average which kind of makes sense, but I'm having trouble verbalizing why if we have a weighted average, the point that the two equations intersect at will always be crossed by the equation of the weighted average

[smoky crystal]

@pure tangle can you elaborate on the find variable works part? maybe an example

[pure tangle]

@smoky crystal thanks for the response. Is it alright if i post a link to their explanation?

[smoky crystal]

im not sure but u can dm me

[pure tangle]

https://mathmaine.com/2013/12/26/linear-systems-why-does-linear-combination-work-graphically/#:~:text=To summarize%2C the linear combination,solution to the original system.

[smoky crystal]

@pure tangle this paragraph answers your question

what will be the weight average at the point (x1,y1) where it intersects?

well it will be exactly (x1,y1)

[pure tangle]

thanks so much for the response. so that's the paragraph i've been reading over and over and I'm still having trouble understanding

[smoky crystal]

think of it like this, let (x1,y1) , (x2,y2) be any points on the two lines respectively, if they intersect, then we must have (x1,y1)=(x2,y2) or x1=x2 y1=y2. Now if we take their weight average for the two lines, what will happen to the weighted value at the intersection?

[pure tangle]

does it not change?

[smoky crystal]

nope, it will not change

lets go the easy way and talk about average instead of weighted avg

that'll be easier to think of

[pure tangle]

thanks for being patient with me. I'm still having trouble understanding why that is

sounds good

[smoky crystal]

now, if you have x1=x2, y1=y2, say pick any points ,eg let x1=x2=4 and y1=y2=6

well the avg for x's will be (x1+x2)/2 correct?

[pure tangle]

yes

[smoky crystal]

and same goes with y's

[pure tangle]

yes

[smoky crystal]

well plug in the values we have for x's and y's,

so we have (4+4)/2 for x and (6+6)/2 for y

now our average x, y . wouldnt you agree that its the same as the old one?

[pure tangle]

oh so if they're equal they're not going change after averaging out our line equations

[smoky crystal]

yep

[pure tangle]

that makes perfect sense. Thank you so much for taking the time to help me!

[smoky crystal]

nope problem, and weight average is pretty much the same thing, you can try for yourself

the whole point of using linear combinations is that we can play around with the two equations, and somehow make a 3rd equation such that one of the variables dissapear, or in the context of line, that the line corresponding to it is independent of the values of either x or y

like the case we have here where the blue line is independent of y

[pure tangle]

that's great. Thank you so much for all the help and patience!

[smoky crystal]

however, just something that i personally dont like, is the way they call this the weighted avg, but yeah anyway gl

[pure tangle]

how would you better describe it?

[smoky crystal]

np, i just so happen to be bored after semester finished for me

well here it describes the weight avg for a particular case of us manipulating the equation, which is multiply both side by a costant before adding

but it doesnt have to be

are you a HS student?

i prob asked too much ,its okay, anyway gl lol

[wintry steppe]

linear maps..........

maps.....

topography

linear algebra = topology!

[primal cosmos]

@wintry steppe

[blissful pagoda]

I am so confused what eigenspace is. Can I have some clarification?

I have 1, -2, and 3 as my eigenvalues. My eigenvector corresponding to my eigenvalue of 1 is (1/3, -2/3, 1). My eigenvector corresponding to my eigenvalue of -2 is (-1/3, -1/3, 1). My eigenvector corresponding to my eigenvalue of 3 is (1, 0, 1).

Is my eigenspace {(1/3, -2/3, 1), (-1/3, -1/3, 1), (1, 0, 1)}?

[limber sierra]

well those arent your only eigenvectors

in the sense that any scalar multiple of an eigenvector is also an eigenvector, right? (unless that scalar is 0)

since they still satisfy Av = lambda v

your eigenspace is the space of all eigenvectors (as well as the zero vector) of a given eigenalue - so not only will it contain the three vectors you listed, but it will also consider any scalar multiples

if you're familiar with the notion of "span", then the eigenspace is the span of a given eigenvector

this is distinct from the set {(1/3, -2/3, 1), (-1/3, -1/3, 1), (1, 0, 1)} itself

since the span includes all linear combinations

[blissful pagoda]

Ok I see where ur coming from.

[limber sierra]

sorry my phrasing was misleading

edited

note that eigenspace is tied to a specific eigenvalue

we dont really talk about "the eigenspace"

since a given matrix/transformation may have multiple eigenspaces

we talk about "the eigenspace of the eigenvalue"

[blissful pagoda]

The thing I'm mainly confused about is this in my textbook. Isn't this implying that I have one eigenspace for each eigenvalue? The eigenspace being the eigenvalue's eigenvectors + zero vector

[limber sierra]

yes thats what im saying

[blissful pagoda]

I see

[limber sierra]

again my phrasing was poor

so if youre familiar with span

lets take the eigenvalue -2 as an example

the eigenspace of -2 would be span{(-1/3, -1/3, 1)}

or equivalently, the set {k(-1/3, -1/3, 1) | k is a scalar}

or equivalently, the set of eigenvectors associated to the eigenvalue -2, along with the 0 vector

theres a bunch of different ways to think of it

but they all result in the same object

and yes, each eigenvalue will have its own eigenspace

[blissful pagoda]

that makes sense. So my original thought of the eigenspace being all of the eigenvectors combined of all eigenvalues is just plain wrong. Got it, that's what I was mainly confused about lol

[limber sierra]

yeah

each eigenvalue has its own

(although some may be the same, if two eigenvalues have the same eigenvectors)

(if that's the case, then your eigenspace is multidimensional and it would be the span of multiple vectors)

(the same definition works though: the set of eigenvectors of a given eigenvalue)

[blissful pagoda]

it probably depends on instructor/prof but should I write the zero vector if I ever am asked to state the eigenspace?

[limber sierra]

yes.

or i mean

whatever set you write

should contain 0

so if you write something that doesnt contain 0, make sure to union it with {0}

[blissful pagoda]

gotcha

[limber sierra]

the definition they give - the set of eigenvectors together with {0} - is the best one

[although if you know what a "kernel" is there's another very handy definition]

[the eigenspace of a vector lambda is the kernel of (A - lambda I)]

[if you dont know what that means or if its weird to think about, dont worry about it]

[["kernel" and "null space" are the same thing]]

[blissful pagoda]

The kernel is defined in my textbook "the set of all vectors v in V that satisfy T(v) = 0 is the kernel of T and is denoted by ker(T)" but I don't know how it relates to eigenvalues and stuff

I do understand the kernel = nullspace thing though

[limber sierra]

well the idea is

if v is an eigenvector then it satisfies Av = lambda v

we can rearrange this to Av - lambda v = 0

and then factoring out the vector v

(A - lambda I)v = 0

where I is the identity matrix (of the appropriate size)

so now if we interpret (A - lambda I) as a linear transformation, call it T

this is the set of vectors v such that T(v) = 0

i.e. the kernel of T

[blissful pagoda]

Oh I see

[limber sierra]

where T corresponds to multiplication by (A - lambda I)

anyway, this isnt really necessary to worry about at first

but it might come up later on

(it's arguably a more "natural" perspective)

[blissful pagoda]

(lambda I - A) and (A - lambda I) are the same right (row equivalent in the end)

because my textbook has it as the former

[limber sierra]

oh yeah you end up with the same definitions

thats just what you get when you go from "Av = lambda v" to "lambda v - Av = 0" (instead of what i dod, going to "A - lambda v = 0")

the manipulations are the exact same

[blissful pagoda]

yeah okay gotcha

[karmic bronze]

How hard is this course

[thorny hemlock]

idk how to get (ST)^2 = 0

all those vectors that S maps are actually the zeroes in T so T(Sv) = 0 for any vector in V

[native rampart]

If TS is 0 ,STST is 0

[thorny hemlock]

i see

so the first T gives a vector and then S gives another vector, the third T makes it zero yes?

[native rampart]

Yes

[thorny hemlock]

i see

[gray dust]

maps V->V associate under composition so we can simplify STST whatever way, STST=S(TS)T=S0T=0

[thorny hemlock]

i was not aware

thanks

[past meteor]

For the question:
Show that any set of five points from the plane R 2 lie on a common conic section, that is, they all satisfy some equation of the form ax2 + by2 + cxy + dx + ey + f = 0 where some of a, . . . , f are nonzero.

The solution said:
*On plugging in the five pairs (x, y) we get a system with the five equations and six unknowns a, . . . , f. Because there are more unknowns than equations, if no inconsistency exists among the equations then there are infinitely many solutions (at least one variable will end up free).

But no inconsistency can exist because a = 0, . . . , f = 0 is a solution (we are only using this zero solution to show that the system is consistent— the prior paragraph shows that there are nonzero solutions).*

I'm confused about the last line, why does using the zero solution show the system is consistent? The question says "where some of a, ..., f are nonzero" but how does this translate to a = 0, ..., f =0 being solutions?

[marble lance]

The a to f are the variables in your system of equations

Now if there are fewer variables than equations, then a system of equations has either no solutions(if there is an inconsistency) or infinitely many distinct solutions

Because there is one solution (the solution where a to f are all 0), that eliminates the possibility of there being no solutions

So then there must be infinitely many solutions (and in other solutions, some of the variables are nonzero)

Unknowns probably better word than variables

@past meteor

[past meteor]

I'm still a littlee confused, so if the system has either no solutions or infinitely many, I don't get how there's still the one solution when all a to f are 0

[marble lance]

Do you not think making all the unknowns 0 is a solution?

[past meteor]

how does that make it solution

[marble lance]

Take an equation, make them all 0, and the equation is true

[past meteor]

ahhhhh, that makes sensee

right, yeah the rest makes sense, thank you so much!!

[marble lance]

Np

[thorny hemlock]

shouldnt dim(nullT) = 2 ?

[gray dust]

@thorny hemlock how

@round coral let him answer

[round coral]

alright

[thorny hemlock]

the basis of the null space is of length 2

?

[gray dust]

no. we're given info on U, not ker(T). dim(U)=2

@marble lance not yet

[marble lance]

I don't think I want to give anything away.

You could do a proof by contradiction and assume there is a T where U = null T, but this is NOT what the solution is doing.

[gray dust]

it's the multiple helpers thing

[thorny hemlock]

but isnt the question saying that the kernal equals (x1 x2 x3 x4 x5) where x1=3x2 and x3=x4=x5 ?

[marble lance]

Yeah, yeah, but I still owe you for last time and I thought that was worth pointing out. Go ahead.

[thorny hemlock]

Vectors of that form get sent to zero

[gray dust]

the exercise is showing that for any T in L(F^5,F^2), ker(T)**!=**U

[thorny hemlock]

i see

[gray dust]

keyword !=. reread 'does not exist' in the text

[thorny hemlock]

What if we wanted to do it by assuming that there is a linear map from F5 to F2 such that the null space is that

then we would get that dim(RangeT) would be 3

then dimRangeT > dimF2

which is contradiction

[gray dust]

that works

back to the book, the problem is reading too fast

[thorny hemlock]

So in the solution, they did a kinda direct proof?

[gray dust]

if that's what you call any proof not using contradiction

[thorny hemlock]

im bad at proofs

[wintry steppe]

sad

[thorny hemlock]

:(

[gray dust]

it 'directly' aims to show any linear map F^5->F^2 doesn't have U as its kernel

[old flame]

One question, how does $<Tv,v>=\overline{<Tv,v>}$, makes $<Tv,v> \in \mathbb{R}$

[thorny hemlock]

which is more commonly used? null or kernal ?

[stoic python]

Otoro

[gray dust]

@thorny hemlock axler uses nullspace. i use kernel bc it used in more contexts in abstract algebra

[native rampart]

If a=conjugate(a) a is real

That is a property of complex numbers

[old flame]

yeah ?

[native rampart]

(Take a=x+iy,x,y in R
conj(a) will be x-iy, x+iy=x-iy implies y=0
)

[old flame]

ah I see what u mean there

so since inner product gives out a scalar, the scalar is equal to its conjugate

got it thanks

[wintry steppe]

psst

\langle \rangle

$$\langle x, y \rangle$$

[stoic python]

MoonWolves-T-

[old flame]

Could someone explain how does the third line transform to the fourth ?

[wintry steppe]

the definition of the adjoint

$\langle T^* Tv, v \rangle = \langle Tv, Tv \rangle$ by the definition of $T^*$, and that's just $|Tv|^2$. similarly for the other side

[stoic python]

MoonWolves-T-

[wintry steppe]

@old flame

[old flame]

ahhhh

so the vector in the definition is instead $Tv$ and $T^{*}v$ in the other case then

[stoic python]

Otoro

[wintry steppe]

i guess so

[old flame]

I mean by definition I just know that $\langle Tv,w \rangle = \langle v, T^{}w \rangle$. So from that we would have $\langle TT^{}v,v \rangle = \langle T^{}v, T^{}v \rangle$, So I guess its because the conjugates are equal, so the last one could swap places right ?

[stoic python]

Otoro

[wintry steppe]

i'm not sure what this has to do with conjugates

[old flame]

The one im talking about is $\langle T^{*}Tv,v \rangle

because from the definition the adjoint is in the right

[wintry steppe]

ah i see where you're coming from

right

ya you got it

[old flame]

so self adjoints have this property $\langle v, T^{}w \rangle = \langle T^{}w,v \rangle$ ?

[stoic python]

Otoro

[wintry steppe]

nno i don't think so, try T = identity, so that that fails if <v, w> is not a real number

maybe i misunderstood what you meant by "because the conjugates are equal, so the last one could swap places"

let me write something out

[old flame]

sure thing

cause I just don't know why the places could be swapped

[stoic python]

MoonWolves-T-

[wintry steppe]

sorry i forgot how to tex there

the second equality sign is the definition of the adjoint

where it's on the right argument of the inner product

then you can undo the conjugate because swapping the arguments in <Tv, Tv> changes nothing

and then you just get the norm squared of Tv

[old flame]

oh

[wintry steppe]

does that make sense?

[old flame]

yes, so instead the swapping occurs during Tv,Tv, so it doesnt matter as its the same anyways

[wintry steppe]

mmhm

[old flame]

so I guess whenever I see T^* up front, should try to think of conjugating it to use the definition

[wintry steppe]

probably a good thing to keep in mind, yeah

[old flame]

alright, thanksssss

[thorny hemlock]

i dont get the solution for the second part

they just chose a specific transformation so all the u's get sent to zero and say nullT = U ?

[wintry steppe]

you should write out the proof that U = null T

in excruciating detail

[thorny hemlock]

theres too many problems

If i wrote full proofs for each one

it will take me hours

[wintry steppe]

well then at least convince yourself that null T = U by writing it down

[thorny hemlock]

i did

[wintry steppe]

so what part of the solution is confusing you

[thorny hemlock]

i think ive understood now

for the second part,

We just need to show that there exists a Linear transformation such that nullT=U under assumptions given right

[wintry steppe]

yes

[thorny hemlock]

yeah i think ive got it

[thorny hemlock]

$Forward: \ $T is injective \Rightarrow dimV \leq dimW \ $define $T(v_j) = w_j $ (This is injective and linear)\ Define S so $Sw_i = v_i$ for $i = {1...n}$ $Sw_i = 0$ for $i = {n+1....m}$ w1...wm is basis of W then $STv_j = v_j$

[stoic python]

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

[thorny hemlock]

does something like this work ? the solution does it a different way

[wintry steppe]

what are the v_j's? w_j's?

[thorny hemlock]

basis

v1..vn basis of V w1...wm basis of W, j is just a specific something 1<= j<= n

@wintry steppe is this ... somwhat ok?

[native rampart]

Yea, That's fine

[thorny hemlock]

[slate fox]

Let V be an infinite dimensional vector space over a field F. Suppose that T : V → V
is a linear transformation such that Ker(T) is finite dimensional. Prove that Im(T) is infinite dimensional.

I was able to do this by finding a finite basis for V

but i remember seeing a clever solution using the first iso theorem that i cant recall

anyone know?

[thorny hemlock]

:0

[steady fiber]

a finite basis for an infinite dimensional vector space?

[slate fox]

yea contradiction

[native rampart]

That's Pretty direct,ig

[slate fox]

if i assume both ker and Im are finite dim

[steady fiber]

oh I see what you mean

[native rampart]

im(phi) iso to V/(ker phi)

[thorny hemlock]

[slate fox]

yea it used that

somehow

i dont remember how tho

and i cant find it

[native rampart]

Do you know dim(A/B)=dim(A)-dim(B)?

[slate fox]

yes

[native rampart]

dim(im(phi)) would be dim(V) -dim(ker phi)

Since dim(V) is infinite ,dim(im(phi)) is also infinite

[slate fox]

doesnt this only work if B is a subspace tho

[native rampart]

ker phi is a subspace of V

[thorny hemlock]

did the normal chatting channels get deleted?

[native rampart]

No,You have the studying role

So you can't see them

[thorny hemlock]

oh lmao

[slate fox]

hm

[brisk fractal]

apologies for the weird notation (it's just the linear maps as matrices in their respective spaces)

[spice storm]

looks like linear done right book

[brisk fractal]

I'm a bit confused in the end result, is B^T = A or is B = A?

[brisk fractal]

this is basic algebra by knapp

[spice storm]

what does that cover?

[brisk fractal]

it's an abstract algebra textbook for the most part but he spends like 4 chapters covering LA in an abstract setting to build motivation

[spice storm]

oh cool

[brisk fractal]

I was reviewing his sections on dual spaces and quotient vector spaces so I would have a better understanding of inner direct products of groups

[spice storm]

sounds like grad school is making you busy during winter break

[brisk fractal]

I'm putting myself through this actually, I've felt kind of bad for my woeful lack of algebra knowledge

[spice storm]

oh nice. Is it so far worth it

[brisk fractal]

Knapp is seemingly very well written

I haven't done any abstract algebra beyond linear algebra and although it's aimed at graduates it's a fairly pleasant read

I'm actually enjoying myself despite pretty much only doing analysis/topology/geometry

[wintry steppe]

contragredient

this notation is really confusing

contragredient makes me feel like i'm gonna be eating food soon, too close to "ingredient"

but all i'm fed is shitty indices

shouldn't the end result be A^T = B (matrix transpose) anyways

i agree with you slim

not in the statement of the proposition, unless i've failed to interpret that notation from the context

some weird way of writing the matrix that represents a linear map

what else?

lol

@brisk fractal typo in the statement probably, end result should be B^T = A

i checked against another book (friedberg) to be sure

bacono can utell us what the notation means

pls

i don't like this

[stoic python]

slimvesus

[wintry steppe]

bacono depriving us rn

$[T]_\beta^\gamma$ means the matrix of $T \colon V \to W$ with respect to the bases $\beta, \gamma$.

[stoic python]

TTerra

[wintry steppe]

friedberg

fries burger

yes

i don't think friedberg contains much more than axler

😔

roman talks about infinite dimensional space stuff

neat

[brisk fractal]

hello I am back on my computer

yeah so I think it is a typo tterra

the reason I ask is because in the proof it claims that A, B are just the matrices given in the statement, but then it proves that A^T = B instead of A=B

[wintry steppe]

[brisk fractal]

[wintry steppe]

[brisk fractal]

don't worry about it

[wintry steppe]

.

just reinforces what you said

"read a linear algebra book"

[steady fiber]

there was a typo in my textbook yesterday, I spent half an hour trying to prove something impossible

typos do make me mald

if they're critical

I only found out in some comment of a math stackexchange post

that it was a typo

fixed in later versions

@wintry steppe I'm on the cringe part of the Sylow section

[wintry steppe]

have fun!

[pallid rampart]

😩

[wintry steppe]

i like how all of these require you to find the prime factorization

[steady fiber]

I just have a prime factoring website open

[thorny hemlock]

idk how to do this :/

any hints?

<@&286206848099549185>

[stoic python]

slimvesus

[thorny hemlock]

-1 as in inverse?

[stoic python]

slimvesus

slimvesus

[thorny hemlock]

i havnt learnt all of this yet :0

maybe theres an easier way of doing it

no

doing the axler book

[stoic python]

slimvesus

[thorny hemlock]

oh yeah

lol

lmao

ok

T-1 takes a vectors from W to U ?

confusion

er

T-1(nullS) = u \in U where u is in nullS ?

ok

i see, then use rank nulitiy thing

yeah got it

[sweet vine]

what the hell 'positive definite' mean? @ me

[sweet vine]

ok 🤔

[limber sierra]

if your matrix is hermitian that means

all its eigenvalues are positive

[sweet vine]

alright thanks Namington

[old flame]

How does he find the diagonal matrix with respect to the eigenvectors from the orthonormal basis ? what is the calculation procedure ?

[native rampart]

Just find the eigenvalues

He found the diagonal matrix and used that to find the eigenbasis

[old flame]

so use the basis to find the eigenvalues ? then find the corresponding eigenvectors right ?

[native rampart]

He didn't use the basis at all

He probably used the char eqn

[old flame]

oh but that is taught in chapter 8

[native rampart]

Anyway,The point is that there will always be a diagonal matrix,similar to a normal matrix

[old flame]

I mean I was trying to verify and that was stated, but not sure how lol

yeah well I get his point, which its literally the complex spectral theorem up next

[old flame]

one to one correspndance ?

[native rampart]

Yes

Well,upto rearrangement of basis

[old flame]

I see, thanks

[old flame]

I am just wondering why is the decomposition in terms of null spaces ? This I still don't really fully understood, even though I have seen this a couple chapters back

[slow scroll]

well null(...) are just the eigenspaces

[stoic python]

Otoro

[native rampart]

null(T-xI) will be the subspace of eigenvectors with eigenvalue x

[slate fox]

Let V = W = R^3 and T : V → W be the linear transformation T(x, y, z) =
(x + 2y + 3z, y + 2z, 2y + 4z). Find a basis BV for V and a basis BW for W such that the matrix representing T with respect to the bases BV , BW is a diagonal matrix

so i need to find bases {v1,v2,v3} and {u1,u2,u3} such that T(v1)=au1, T(v2)=bu2 and T(v3)=cu3

is this just guess and check from here?

[slate fox]

it doesnt look very linear

[thorny hemlock]

Ah

[slate fox]

<@&286206848099549185> any hint?

[thorny hemlock]

Any hints on how we should define S?

T1 and T2 are linear maps from U to W

[slate fox]

how do I show that a cyclic operator T such that T^2=T cant be invertible

[native rampart]

If T is not I,T is not invertible

Because det(T)=0

[slate fox]

det(T)?

u mean of the matrix representing T?

[native rampart]

Well,You can take it that way

[slate fox]

how do we know the det is 0 tho

[native rampart]

If T^2=T,The minimal polynomial of T is either x or x-1 or x(x-1)

In first case,T=0 ,in second case T=I and in the final case,T is diagonal in some basis,with zero as an eigenvalue

[slate fox]

we havent covered eigen values or vectors yet

not sure what a minimal polynomial is

[native rampart]

Minimal polynomial of T is a polynomial p,such that p(T)=0(i.e, p(T)v=0 for all v)

[slate fox]

yea havent met yet that

[native rampart]

What do you mean by cyclic operator?

T on a Tcyclic subspace?

[slate fox]

T is cyclic if there exists a v such that the set {v,T(v), T^2(v)...} spans V

[native rampart]

You can conclude,V is atmost 2 dimensional in your case

Because {v,Tv,T^2v...}={v,Tv}

Since T^2=T

And then consider matrix of T in basis {v,Tv}(if 2 dimensional)

If V is one dimensional,T=I,and T is invertible

[slate fox]

shouldnt it be at least 2 dimensional?

nvm its not

ok yea if it has dimension 2 then it cant be invertible

but if it has dimension 1 doesnt that mean the question is wrong?

cuz the identity operator is invertible right

right? @native rampart

[crimson pelican]

if U,V and W be subspaceses or R^n . Is U∩(V+W)=(U∩V)+(U∩W) true ? If it is true how could I proof that ?

<@&286206848099549185>

[native rampart]

Yes

[slate fox]

ok thank you

[thorny hemlock]

Doesn’t look very true

[crimson pelican]

why how can I conclude that

[thorny hemlock]

Finding an example that fails

[wintry steppe]

it is very false, not hard to come up with a counterexample in R^2

[crimson pelican]

what if I defined a element of V and b element of W then define x= a+b then if x element of U∩(V+W) then x also element of U and x element of (V+W) . But (U∩V) does not contain x and (U∩W) does not contain x.

@wintry steppe @wintry steppe @thorny hemlock

[steady fiber]

find a case when the intersection of U with V+W is small, but the intersection of U with one of V or W is large

[crimson pelican]

included two vectors which have two row and one column

this vectors must hold 3 condition

okay what are the types

lines through the origin

lets choose x+y=0 and x+2y=0

[wintry steppe]

This is a physics question that has to do with vectors, so I thought this may be the right channel. If not, please lmk.

My question is why can't I multiply the magnitudes of F and ∆d together to get work? Why do I have to use the dot product?

[crimson pelican]

@wintry steppe we are trying solve the quesiton and you send quuesiton bro c'mon

[steady fiber]

#old-network physics server might be more useful

[wintry steppe]

It's a conceptual question, so I don't think the answer should matter

[steady fiber]

they could give you a better conceptual understanding of force and distance and work

[wintry steppe]

Gotcha - thanks!

[crimson pelican]

ı dont understand

[stoic python]

pancakehammer

[crimson pelican]

(0,0)

[marble lance]

And the sum?

[crimson pelican]

sum is still (0,0) but now we have not false statement

[marble lance]

No

[crimson pelican]

yes

[stoic python]

slimvesus

[crimson pelican]

i think its cant be zero

no sorry it included zero

[wintry steppe]

[crimson pelican]

left hand side can be equal ${0}$

its vector addition ı dont know

its line through the origin nvm ı cant find it

[wintry steppe]

[crimson pelican]

yes

span{u,w}

yes

so?

nope ı cant figure out thx for helping

[wintry steppe]

v and w are linearly independent, so their span is 2 dimensional. what are the 2 dimensional subspaces of R^2

[crimson pelican]

lines through the origin

@wintry steppe

[wintry steppe]

[crimson pelican]

then its just (0,0)

ı m going to crazt

y

[crimson pelican]

it is empty seyt

@wintry steppe

[hoary osprey]

thats not a subspace

[crimson pelican]

and what is

this

[wintry steppe]

the only 2 dimensional subspace of R^2 is R^2 itself.

so the span of v and w is R^2.

[crimson pelican]

and left hand side is R^2 but right hand side is 0

[wintry steppe]

you mean of U∩(V+W)=(U∩V)+(U∩W)?

[crimson pelican]

yeap

[wintry steppe]

almost

[crimson pelican]

U are included R2

[wintry steppe]

that's an intersection

not a union

[crimson pelican]

yes but (V+W) included R2

[wintry steppe]

?

[crimson pelican]

so the span of v and w is R^2.

Let's write W= span{w} and V=span{v}

[wintry steppe]

V + W is R^2, but U is a subset of R^2, so the intersection is just U

one side is U, one side is empty

[crimson pelican]

ohh okey you are right

ım sorry

thanks

[wintry steppe]

it is true

[fallen karma]

if a finite dimensional vector space V with dim(V)=n has a linear map T with n distinct eigenvalues, where v_1,v_2,...,v_n are the corresponding eigenvectors, then is any invariant subspace under T just the span of some collection of those eigenvectors?

just yes or no will help me

[fallen karma]

nm think I figured it out

[wintry steppe]

was it yes or no

i am interested

[fallen karma]

haha idk yet but I suspect yes. I'm trying something with quotient operators

[wintry steppe]

i'm tempted to say yes

cause if you diagonalize T then it's clear that the invariant subspaces (identifying V with R^n) are just the axes, and then something about changing bases and how that interacts with subspaces idk puts away crack pipe

[fallen karma]

that's an interesting take

[wintry steppe]

lmao

forgot how to LA there for a second LOL

[fallen karma]

it'd be nice if it was true so it is true right??😆

what was wrong with what you said?

[wintry steppe]

i wanna check it before i un-strikethrough it just to make sure i'm not saying complete bs

cause also the entire space is invariant and so is {0} so i have to be a bit more careful with that claim

[fallen karma]

ok valid

but it seems worth exploring

[wintry steppe]

maybe "the only non-trivial proper T-invariant subspaces are the axes" ?

just thinking about how like, a diagonal matrix is just scaling along each axis, it makes sense

[quartz compass]

I figure just represent anything in the eigenbasis, it will automatically be trapped in the span

[fallen karma]

pesky {0}

i shoulda specified when I posed the question "and {0}"

[wintry steppe]

skimming over my LA book to see if there're any nice results about invariant subspaces that could be used here

alright so if you restrict T to any invariant subspace, you get an operator whose characteristic polynomial must divide that of T. now, by assumption, the char poly of T splits, so the char poly of T restricted to any invariant subspace splits

i.e. T restricted is also diagonalizable with some subset of the original eigenvalues

(still doesn't really answer the original q tho)

[fallen karma]

that's cool I didn't know that

[wintry steppe]

try proving it, you take a basis of the invariant subspace and extend it

[quartz compass]

what's wrong with what I said?

I think you're overcomplicating it

[wintry steppe]

overcomplicating is fun

[fallen karma]

i tried representing in the eigenbasis earlier but it got messy. so i went looking for another way

[quartz compass]

eh?

[wintry steppe]

Let $W \subset V$ be a $T$-invariant subspace of $V$. Since $T$ is diagonalizable, $T|_W$ is also diagonalizable in $\mathrm{End}(W)$. That is, $W$ is spanned by some of the eigenvectors of $T$.

[stoic python]

TTerra

[wintry steppe]

@fallen karma some relevant facts/exercises

(from section 5.4 of the linear algebra book by friedberg, insel, spence)

[quartz compass]

I was thinking of something a bit simpler

suppose you take any invariant subspace, then all its vectors can be represented by a linear combination of some set of eigenvectors. Then let's show it actually contains the entire span of these eigenvectors

[limpid fiber]

Sorry to interupt, why is it true that if $Ax = 0$ only has the trivial solution then A cannot have a free variable?

[stoic python]

pancakehammer

[quartz compass]

so take some u in the invariant subspace, it has coefficients, $$u = \sum_{i=1}^n a_i \vec e_i$$ now we can get the vectors in our invariant subspace $Au$, $A^2u$,..., $A^{n-1}u$ to get $A^k u = \sum_{i=1}^n a_i \lambda_i^k \vec e_i$ which gets us a system of linear equations

[stoic python]

Merosity

[quartz compass]

it's a nice vandermonde matrix that we can invert which has det !=0 since the eigenvalues are distinct, so we can solve for any eigenvector as a linear combination of the A^k u

and by being a subspace we have all scalar multiples

[fallen karma]

oh thanks

[wintry steppe]

you should explore mero's idea

i think it is good

[fallen karma]

yeah bet, I didn't think about applying the operator more than once

[wintry steppe]

just wait until you get to cyclic subspace nonsense

[fallen karma]

thanks for your help guys

and gals and nb pals

[wintry steppe]

How could I forget this place exists?

[half ice]

Lots of alcohol, like me

But I always come back

😫

[tame mural]

Does the set with only the 0 vector have a name?

[quartz compass]

Dunno, maybe 'the trivial vector subspace' or something like that maybe

[tame mural]

mmm

thx

[smoky crystal]

its just the zero space

[tame mural]

if a transformation preserves the dot product, does that also mean it always preserves angles?

[blissful vault]

Correct or not? How is my writing?

[quartz compass]

well, do you know any formulas that relate the dot product to the angle? @tame mural

[native rampart]

It's correct

[tame mural]

Sure, it's the dot product of two vectors divided by the product of their norms

[blissful vault]

Is there anything I can improve in my writing?

[quartz compass]

yeah pretty much, and the norms, can those be written in terms of dot products?

[bold garden]

hello, i had a question about the following proof:

I understand the general point of the proof, but I was confused because I thought the stuff in the blue brackets might be too informal

I also don't understand what the finite-ness of S has to do with anything - why couldn't this work for a potentially infinite S (besides the fact that the theorem would be false)?

[marble lance]

I mean, you could end up choosing all the vectors of S if they are all linearly independent

But that's it, then you can't choose any more

[bold garden]

sure, but what I'm saying is why the statement "continue, if possible, choosing vectors such that $${u_1, u_2, ..., u_k}$$ is linearly independent" is the right way to do it

[stoic python]

kirafa

[bold garden]

shouldn't there be some kind of induction involved?

[marble lance]

I don't know what you mean. It's basically "in a finite number of steps, you can either choose vectors so that you cannot choose another so that's its linearly independent or you have chosen them all."

There's nothing wrong with saying you can do something a finite number of times

[bold garden]

oh, i wasn't aware of that, in that case i guess it's fine then

thank you

[thorny hemlock]

Tv = a1w1 + .... + amwm where a's/in F

S1(v) = a1 .... Sm(v) = am

thats correct right?

[wild fern]

there is $$T_1, T_2 \in L(V,W)$$, prove inversible transforms $ R \in L(V,V)$ and $S \in L(W,W)$ exists such that $T_1 = ST_2R$ iff $ dim(N(T_1)) = dim(N(T_2))$

[stoic python]

Amirali

[native rampart]

Yes

[wild fern]

any help will be appreciated

[round coral]

@wild fern you know change of basis, it follows from that

except that what is N in dim(N(T_1)) = dim(N(T_2)) , you didn't define that

[wild fern]

N is nullspace

[round coral]

well seen that notation for first time, either use null or ker , I don't think N is the official notation for nullspce

[wild fern]

its in linear algebra by carl d. mayer

[round coral]

ok, I am really ignorant, but that Q follows directly from change of basis of a linear transformation

hope it helps

[wild fern]

thank you ill try

[round coral]

https://math.stackexchange.com/questions/228516/matrix-of-a-quadratic-form @potent elm

[soft burrow]

that's the inverse process, the quadratic form of a (e.g. real) matrix $A\in \mathcal M_{n\times n}(\bR)$ is $q: \bR^n\times \bR^n \to \bR$ given by
$$
q(x,y)=x^{\mathsf T}Ay.
$$

[stoic python]

derivada.schwarziana

[soft burrow]

all it does is associate a degree two polynomial to the matrix. It appears naturally in mathematics, e.g. inner products can be defined by quadratic forms. One proves that quadratic forms are esseentially the same as symmetric square matrices (in fact each quadratic form has a unique such matrix)

[native rampart]

That's the bilinear form

The corresponding quadratic form is q(x,x)

Yes

[soft burrow]

ah that's right I keep confusing them

[native rampart]

Gram-Schmidt will not always give you an eigenbasis

Just Take a non eigenbasis(Start your construction with a non eigenvector) and apply gram schmidt,you wouldn't get a eigenbasis

I don't think that's true in general

[stoic python]

pancakehammer

[wintry steppe]

that's not null 2 minus 3

that's null of the vector (2, -3)

[limpid fiber]

yes, sorry about that typo

[wintry steppe]

or better, the null space of the 1 by 2 matrix (2 -3)

had a brain fart for a sec but that's what it should be

[limpid fiber]

lol yes

but isn't this just a vector not a matrix?

[wintry steppe]

the elements of V are vectors

what does Nul(2 -3) mean to you

that may help clarify things

[stoic python]

pancakehammer

[limpid fiber]

I am confuse because $[\begin{bmatrix}
2 \
-3 \
\end{bmatrix}]$ is in $R^2$ and $[2 -3]$ is in $R^1$ but it feels like they liberally switch between them?

[round coral]

what's so confusing about it ?

[stoic python]

pancakehammer

[wintry steppe]

how is that second one in R^1

wat

the distinction between column vectors and row vectors / tuples is usually blurred

just something you have to get used to

[round coral]

do you know how the vectors in R^2 and R look

[limpid fiber]

yeah, I guess they'd both be lines?

that's good to know though I appreciate that @wintry steppe

[round coral]

this is an important point #linear-algebra message

[limpid fiber]

but the former line is in R^2 and the latter R^1?

[round coral]

see it clearly

[limpid fiber]

okay I will look now, thank you

[wintry steppe]

addendum: for the sake of multiplying vectors by matrices, they are column vectors. however, they will sometimes be written as row vectors to save space, which is why sometimes you see authors write things like [2 -3]^T inline

[limpid fiber]

Does the same thing hold for a 2x2 matrix, I know the row 'picture' and vector/columns 'picutre' are the same , but I'm not exactly sure why yet

or is that different from what y'all are talking about?

[round coral]

these are equivalent

actually the 3) is the one , from which 2) comes out as I studied in the beginning, but you can do from 2) to 3) as well

doesn't make a difference

[limpid fiber]

nice, I hope this is related to what you were saying

yes what you say about 2 and 3 makes sense I think

I still cannot connect 1 and 3 or 1 and 2 though, I know the proof that they must have the same solution set which I know is true but still feels unenlightening

is that something that will become clearer as I progress in Linear algebra?

[round coral]

1 -> 2 are what you first study in any book, how to write a system of linear equations in matrix form

as 2 and 3 are equi 1 and 3 are equi , you just need to look at it

[limpid fiber]

yeah you're right I can see 1 and 3 must be the same by definition now that I look

but in my mind 1 is the equation for two planes, wheres 3 is a linear combination of vectors

[round coral]

although I do not like it, you can read Strang's Chapter 1 , he has discussed about geometry of linear equations well

@limpid fiber

[prisma tide]

anyone done quaternions here?

[soft burrow]

(possibly in #groups-rings-fields depending on what your question's about)

[prisma tide]

ok

@soft burrow no acess

nvm just got it

[hollow finch]

is there a special name for the matrix $\pm\begin{bmatrix}0&1\-1&0\end{bmatrix}$?

[stoic python]

nix

[hollow finch]

i know its the matrix representation of the complex number plus minus i but does it have a name?

[soft burrow]

definitely not a unique standard name

some authors use $\mathbf{1}=\begin{pmatrix} 1&0\0&1 \end{pmatrix}$ and $\mathbf{i}=\begin{pmatrix} 0&1\-1&0 \end{pmatrix}$

[stoic python]

derivada.schwarziana

[hollow finch]

i see. thank you 🙂

[winter harbor]

Given a real vector space $V$ and being $\mathscr{A}{n}(V)$ the space of all alternating n-forms $\varphi : V \times \hdots \times V \rightarrow \mathbb{R}$. We can define $A^{#} : \mathscr{A}{n}(V) \rightarrow \mathscr{A}{n}(V)$ as $(A^{#}\varphi)(v_1,\hdots,v_n) = \varphi(Av{1}, \hdots, Av_{n})$
\
\
The important thing to know is that since $\dim \mathscr{A}{n} = 1$, then the LHS is just equal to $\varphi$ times a constant, which we define as the determinant of the linear operator A.
\
\
So $\det A$ satisfies by definition:
\
\
$\det A \cdot \varphi(v{1}, \hdots, v_{n}) = \varphi(Av_{1},\hdots,Av_{n})$

[stoic python]

MisterSystem

[winter harbor]

I just wonder if this A hash operator has a name

or is just a trick we use only once to define the determinant of a operator

I also wonder what's the intuition behind this definition

if it has some geometrical insight if it's R^n maybe

oh, V is finite and has dimension n

I forgot to say that

But how can we prove using this definition that $\det A = \int_{ [0,1]^{n}} 1 dx$

[stoic python]

MisterSystem

[winter harbor]

(Damn I'm horrible at typesetting)

[wintry steppe]

pullback by A

[wintry steppe]

it's pretty important, i'd say it's got a lot more applications than just this

e.g. in slightly more generality, if you have a linear A : V -> W, then you can define the pullback A^* : {k-tensors on W} -> {k-tensors on V} in an entirely analogous way

particularly useful in smooth manifold theory

also i think you mean to write $$\det A = \int_{A([0, 1]^n)} 1$$

[stoic python]

TTerra

[wintry steppe]

in that case, you can use the change of variables theorem for integrals

[thorny hemlock]

[wintry steppe]

\begin{align*}
\int_{A([0,1]^n)} 1 , dx^1 \wedge \cdots \wedge dx^n &= \int_{[0, 1]^n} A^(dx^1 \wedge \cdots \wedge dx^n) \
&= \int_{[0,1]^n}\det(A),dx^1\wedge\cdots\wedge dx^n \
&= \det(A) \int_{[0,1]^n} 1 , dx^1 \wedge \cdots \wedge dx^n \
&= \det(A)
\end{align}

[thorny hemlock]

[wintry steppe]

can you wait?

[thorny hemlock]

yep sorry

[stoic python]

TTerra

[wintry steppe]

@winter harbor

eh tbh you don't need to do the pullback and differential forms nonsense to prove that, that just follows pretty easily from the classic change of variables theorem

but i wrote it like that to try and emphasize your definition of det A, i guess

overkill is fun

[thorny hemlock]

integral :o

[winter harbor]

oh yeah, you are right, I meant A([0,1]^{N})

[primal cosmos]

is that a lebsegue integral?

[winter harbor]

thanks

[wintry steppe]

np

[winter harbor]

riemman integral in R^n

[wintry steppe]

differential forms are fun

[primal cosmos]

?

[primal cosmos]

are these lebsegue integrals?

[wintry steppe]

no

[thorny hemlock]

no

[winter harbor]

nah

[primal cosmos]

oh

[thorny hemlock]

rein

[wintry steppe]

plain old riemann integrals

[wintry steppe]

[winter harbor]

this is just the riemman integral in R^n as I have said before

[thorny hemlock]

lmao

[primal cosmos]

why don't they have an upper bound?

[limber sierra]

theyre multidimensional

[wintry steppe]

because you're integrating over a region in R^n

[winter harbor]

yup

[primal cosmos]

woah

[winter harbor]

I mean

[thorny hemlock]

oke, for my question, 1/x^3 , 1/2x^2 , x , 1 for p3 and x^2 , x, 1 is correct ?

[primal cosmos]

@thorny hemlock what was your question?

[primal cosmos]

oh woah

[winter harbor]

when you have an integrable function $\varphi{phi} : [a,b] \rightarrow \mathbb{R}$, and you want integrate this function over an its domain, you could just write it as $\int_{[a,b]} \varphi(x) dx$

[stoic python]

MisterSystem

[winter harbor]

it means that you are interating from a to b

or more generally

integrating this function over the region [a,b]

[thorny hemlock]

integrals are scary

[acoustic path]

integrals 😠

[winter harbor]

it's just that we are used historically to write it as $\int\limits_{a}^{b} \varphi(x) dx$

[stoic python]

MisterSystem

[winter harbor]

but in general

[wintry steppe]

Define $F \colon Vect_k \to Vect_k$ by $F(V) = {\text{$k$-tensors on $V$}}$ for objects $V$ and $F(T) = T^* \colon F(W)\to F(V)$ for $T\in Mor(V,W)$ by $T^*\omega(v_1,\dots,v_k)=\omega(Tv_1,\dots,Tv_k)$. Then $F$, the pullback, is a contravariant functor

[stoic python]

TTerra

[wintry steppe]

if only texit displayed emotes

[primal cosmos]

integrals are not always so bad to set up, but the scary thing is the integration

[winter harbor]

when you are dealing with integral operators over more complicated regions in a way more complicated space as R^n (say differentiable manifolds) we just write the region we are integrating on below the integral symbol

for example

stokes theorem for a general Manifold M may be written like this

$\int_{\partial \Omega} \omega = \int_{\Omega} d \omega$

[wintry steppe]

\integral isn't a thing, use \int

[winter harbor]

I forgot it lol

[wintry steppe]

you got it right on one side

[stoic python]

MisterSystem

[steady fiber]

why are we doing stonks theorem in lin alg

[wintry steppe]

why not?

[winter harbor]

I mean

[limber sierra]

manifolds are just vector spaces on drugs

so

[winter harbor]

it has to do with linear algebra

[wintry steppe]

namington not wrong

tbh

[steady fiber]

I'm convinced

[wintry steppe]

manifolds are just the union of the basepoints of their tangent spaces

so

basically linalg

[steady fiber]

everything is basically lin alg

or actually, alg

[wintry steppe]

algebra...

[steady fiber]

algebra is the only real math dont @ me

[wintry steppe]

the term "semidirect product" came up when i was reading IRM today and it felt nice to know what it meant

maybe algebra isn't so bad

[steady fiber]

lol, I love semidirect products

[thorny hemlock]

whats IRM

[wintry steppe]

intro to riemannian manifolds

textbook

[thorny hemlock]

nice

[limber sierra]

i still dont really know what a semidirect product is

[wintry steppe]

it's a weird way of writing down a group action by automorphisms

[limber sierra]

okay actual confession

i always forget which way to write

the fish symbol

for semidirect product

i get that its like "the same as" the triangle for normal subgroups

but

still

it just looks wrong

[winter harbor]

little omega is just a differential form, which is a kind of linear-form for general vector spaces, and d little omega is just the exterior derivative which is a linear operator.

Also

[steady fiber]

isn't it a way of writing a "product" of groups by a homomorphism

and it happens to have some relationships to automorphism groups at times

[winter harbor]

uhh I said before ''the manifold M'' but I wrote Omega in the latex

and I didn't even say it's an orientable smooth manifold with boundary

[steady fiber]

if you let the homorphism be from K -> Aut(H) for H (weird x) K

[winter harbor]

what a shame

[wintry steppe]

If $\varphi\colon K \times H \to H$ is a group action of $K$ on $H$, written $\phi(k, h) = k \cdot h$, satisfying $k \cdot (h_1h_2)=(k\cdot h_1)(k\cdot h_2)$, then define $\tilde{\varphi}\colon K \to Aut(H)$ by $\tilde{\varphi}(k)(h) = k \cdot h$. Then we have a semidirect product $H \rtimes_{\tilde{\varphi}}K$, and conversely.

[stoic python]

TTerra

[wintry steppe]

woh gives a fuck about semidirectproduct

just study everything as group action

[steady fiber]

no

[wintry steppe]

dumb fish symobl

[steady fiber]

group actions are too abstract

let's unabstract a bit

[winter harbor]

Also TTerra

[wintry steppe]

let's abstract to the group action ur mom had last night

[winter harbor]

I have studied about Pullbacks and Pushforward in the context of manifolds

but I just don't get the concept at all

I guess that's why I forgot the definition

what's the intuition or the motivation?

[wintry steppe]

honestly off the top of my head, i don't know what the intuition or motivation is, other than "it's a really useful thing that comes up everywhere"

i'd have to think

[winter harbor]

it just happens that it's useful when studying 1-differential forms of a smooth manifold

and now it happens to be important to define the notion of the determinant of a linear operator

[acoustic path]

merry christmas my linear algebra people

[thorny hemlock]

its the 26th here, :0

merry christmas

[winter harbor]

@wintry steppe why is it that when you are integrating over A([0,1]^n) it's the same as integrating the pullback of the wedge product of those differential forms over [0,1]^n ?

[wintry steppe]

change of variables theorem

https://mathworld.wolfram.com/ChangeofVariablesTheorem.html

[winter harbor]

for some reason i really can't see it

lemme read the statement again

[wintry steppe]

you don't actually need the differential forms stuff to do it, there's a simpler proof that avoids it (despite being basically the same thing)

[winter harbor]

wait aaaaah

i didn't know the change of variables theorem for general manifolds

[wintry steppe]

you can do it with just this guy

also

looking at this and the page i linked, i was a bit sloppy with the signs

you have to be careful to account for the case that det A < 0

[winter harbor]

yeah you are right

[wintry steppe]

orientations 🙄

[winter harbor]

I really meant |det A|

but i didn't know the change of variables theorem had something to do with pullbacks

[wintry steppe]

well the scalar you get by doing pullback by A will still be det A (without absolute values out front), just when you start integrating you'll pick up a negative sign if det A < 0, whence the absolute value

[winter harbor]

that's some motivation for me to study it

I really didn't get the point until now

[wintry steppe]

i am pretty sure the change of variables theorem on manifolds is just the pictured one in disguise

[winter harbor]

https://www.youtube.com/watch?v=H9-bAbRFG8Y&t=3170s time to rewatch this again

so yeah I know some of the motivations now

but I still don't get the intuition behind the concept

maybe math stack has something concerning this

[wintry steppe]

maybe you can mess around with the 1-dimensional case of pullbacks, in which case it's just composition by the operator

[hollow finch]

if $A=QS$ where $S$ is symmetric and $Q$ is skew symmetric, will it always be the case that $\operatorname{tr}(A)=0$?

[stoic python]

nix

[round coral]

yes

[hollow finch]

why? how could i prove that?

[round coral]

yes

wait a min typing

[hollow finch]

just saying tho thats freakin awesome btw

[round coral]

tr(AB) = tr(AB)* = tr( B* A*) = tr(-BA) = -tr(BA) = -tr(AB) => tr(AB) =0 here I used * for transpose in order to keep it neat

tr(AB)* = tr( B* A*) this can be proved

[wintry steppe]

nice

[native rampart]

Or just writing and expanding the matrix product works

Not as nice, But this approach is kinda obvious

[round coral]

yeah, that is a good way too by writing the matrix product

[hollow finch]

so tr(QS)=tr(S*Q*)=tr(-SQ)=-tr(SQ)
because tr(SQ)=tr(QS), we get tr(QS)=-tr(QS) so tr(QS)=0

[hollow finch]

i dont quite get what the transpose of a trace is though. i thought that was a scalar quantity

[wintry steppe]

the matrix inside is being transposed

not the trace itself

[hollow finch]

oh so tr((AB)*)

?

[wintry steppe]

mmmhm

[hollow finch]

got it

damn that sexy

[native rampart]

It's also nice,that transpose is the only functional on matrices such that f(AB)=f(BA)