linear-algebra

[dusky epoch]

are you sure you got that down correctly?

[cold topaz]

oh

[dusky epoch]

$AA^{-1} = I$ is the definition of $A^{-1}$, if anything.

[stoic python]

Ann:

[cold topaz]

AA^T = I

[dusky epoch]

AA^T = I

and not A^TA = I?

[cold topaz]

my book has both way

[dusky epoch]

ok great well

A^TA = I will be easier to verify with what you've calculated so far

[cold topaz]

so my second question is, to prove a matrix is ortho, the matrix multiplication that leads to I is enough, or i have to shows that row and comlumns are orthonomr?

[dusky epoch]

if you like doing redundant work and want to force yourself to do redundant work then yes you do

if you're a sane person then no you don't

[tall moon]

(it's because right inverse is also left inverse: https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i)

[cold topaz]

that's the answer to which part of my question? @dusky epoch

[tall moon]

If you check rows are orthogonal, you are done.

[dusky epoch]

that's the answer to your entire question.

[tall moon]

same for columns

[cold topaz]

during the quiz, I dont have the entire time in the world. I want to show complete answer, and save the maximum time. Thanks for mocking though.

[quartz compass]

@cold topaz something that might help to see why that is sufficient is to consider this

if A is a matrix of column vectors then A^T is a matrix of those same vectors but as row vectors. then the product A^T*A is every possible combination of dot products of the column vectors with themselves

the dot products on the diagonal of the resulting matrix is every vector dotted with itself, so just 1s on the diagonal

0 everywhere else, because those are orthogonal

so the result is an identity matrix

[elder robin]

Is the column space just the span of the column vectors of a matrix ?

[tall moon]

yes

[elder robin]

Or am I oversimplifying. Trying to understand this lecture I'm watching lol

Is it just used to find the basis?

[tall moon]

well, I guess you can use it that way, if the column space is the entire space, then it's a basis

[elder robin]

ah

thanks

[gray dust]

if the column space is the entire space, then it's a basis

@elder robin a set of vectors S is a basis for a vector space V if span(S)=V and S is linearly independent

[elder robin]

right, I think I'm just about to get into that stuff in my next problem

but iirc you basically remove vectors from S that are linearly dependent? Maybe I'm using the wrong vocabulary here.

but we try to get S to only have vectors that make it linearly independent

[gray dust]

if span(S)=V but S is LD, S is not a basis for V but there exists a proper subset of S, call it W, where span(W)=V and W is LI

[elder robin]

Oh yeah, I think I'm describing the process for finding a basis

is it possible to have multiple basis'

Oh yeah it should be

because you could just multiple the vectors by constants

[gray dust]

sure

[cold topaz]

since P^-1<x, y> = <x', y'>,
so P<x', y'> = <x, y>?

[tall moon]

what's P here?

[cold topaz]

rectangular xy-coordinate system counterclockwise through an angle

[slow scroll]

for any function that has an inverse, its true that f(x) = y implies f^-1(y) = x

[red shard]

thats the definition of an inverse?

you can even formulate it for arbitrary functions, f(x)=y implies x\in f^-1(y) when f^-1 is not inverse function but preimage function

inverse function is just the special case of preimage function only take sets of size 1 as values

[slow scroll]

r u responding to me? I was answering sm's question

[autumn kraken]

I'm supposed to determine if U_4 is a subspace and closed under addition.

I started with $(f + g)(x) = f(x) + g(x) = f(-x) + g(-x)$

[stoic python]

madmike:

[autumn kraken]

but I'm a bit stuck here D: I can't properly add them because I don't know which functions they are

so I don't know how to come up with $(f + g)(x) = (f + g)(-x)$

[stoic python]

madmike:

[autumn kraken]

any tipps?

wait

[wintry steppe]

what does Abb(R, R) mean

[autumn kraken]

is this as simple as $\(f + g)(x) = f(x) + g(x) = f(-x) + g(-x)\(f + g)(-x) = f(-x) + g(-x) = (f + g)(x)$ ?

[stoic python]

madmike:

[autumn kraken]

I think Abb(R, R) means all functions that go from R to R

Abbildung = function in German, sorry

[wintry steppe]

why would it generally be true that for any function (f+g)(x) = f(x) + g(x) ?

or am I dumb

[autumn kraken]

I don't know, someone here said yesterday it's the same thing but written different, could be wrong though

[wintry steppe]

yeah I guess it's different than f(x+y) = f(x) + f(y)

[autumn kraken]

yeah it is

[wintry steppe]

yeah it sounds about right that you can add functions

hard to think of a case where it wouldn't be true?

[autumn kraken]

I basically need to determine that if $f(x) = f(-x)$ and $g(x) = g(-x)$ that also $(f + g)(x) = (f + g)(-x)$ I think

[stoic python]

madmike:

[autumn kraken]

not sure if what I wrote above is right

[wintry steppe]

a subspace also has to preserve the original zero

[autumn kraken]

yeah

[pale coyote]

$(f+g)(x)$ is defined to have value $f(x) + g(x)$

[stoic python]

moonside:

[pale coyote]

$(f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x)$

[stoic python]

moonside:

[pale coyote]

since $f, g \in U_4$

[stoic python]

moonside:

[autumn kraken]

ok perfect

thanks

[pale coyote]

you also need to check closure under scalar multiplication

[autumn kraken]

yeah I'm trying it right now

$f(x) = f(-x)$, since $f \in U_4.\f(x) * \lambda = f(-x) * \lambda$

[stoic python]

madmike:

[autumn kraken]

is this enough to be able to say $f(x) * \lambda \in U_4$?

[stoic python]

madmike:

[autumn kraken]

Also for the neutral element, is it really fine to just check if U_4 is non-empty?

Or do I need to determine the exact zero-element

[pale coyote]

a little more precisely

if $c$ is a scalar, and $f \in U_4$ then $(cf)$ is the function for which $(cf)(x) = c\cdot f(x)$

[dusky epoch]

i mean it is enough to check U_4 for nonemptiness

[stoic python]

moonside:

[dusky epoch]

but establishing that 0 ∈ U_4 is not that hard

0 here of course refers to the zero function on R

[pale coyote]

does the $0$ function satisfy the required property?

[stoic python]

moonside:

[dusky epoch]

ofc it does

[pale coyote]

i know, it was a question directed at @autumn kraken

[autumn kraken]

🤔

oh you answered to both questions

yeah I see

so what I wrote about scalar multiplication is correct because $f(x) * \lambda = (f * \lambda)(x)$?

[stoic python]

madmike:

[autumn kraken]

and $(f * \lambda)(x) \in U_4$

[stoic python]

madmike:

[autumn kraken]

about the neutral element, I don't really understand why checking if it's nonempty is the same as checking if it has the neutral element

[pale coyote]

it isnt

[torn silo]

http://www.math.kit.edu/iag3/lehre/la2info2015s/media/linearealgebraskript2012.pdf

@autumn kraken page 87

[autumn kraken]

danke 🙂

so the neutral element of a subspace is the neutral element of the scalar multiplcation?

[torn silo]

no you need the neutral element from the addition

and you need to be able to do scalar multiplication

[autumn kraken]

ahhh

[torn silo]

see if you don't have a zero you can't a have a subgroup with addition

so that's why if you don't have a zero you can stop looking because it's not a vector space

it tends to be the easiest check, it's why it's most of the time the first step

[autumn kraken]

what is the zero-function of $\real$?

[stoic python]

madmike:

[autumn kraken]

lol wrong symbol

[torn silo]

The zero function is the function that points everything to 0

$\phi(\mbb{R}) = 0$

[stoic python]

deekaan:

[autumn kraken]

I see, so just $f(x) = 0$?

[stoic python]

madmike:

[autumn kraken]

sorry I dont know the notation that you used

[torn silo]

yes for every x in R

[autumn kraken]

thank you

[torn silo]

no worries

[fossil quest]

So there is this System of equation of the Field F3 and it is a subspace of (F3)^4

And i have to find the Basis of it and i have no clue what i got to do :( can anyone pls help

[torn silo]

do you know gaussian elimination?

[fossil quest]

Yes

[torn silo]

good that's what you have to do

start by writing down a matrix

[fossil quest]

Ok

[torn silo]

$\begin{pmatrix} 1 & 1 & 2 & 1 \ 2 & 0 & 1 & 1 \ 2 & 1 & 1 & 0\end{pmatrix}$

[stoic python]

deekaan:

[torn silo]

should look something like this

[fossil quest]

Right

[dusky epoch]

deekaan im gonna warn you straight up that taehyun has specified this is happening over F3

[torn silo]

ya

thats next

so because we're on F3 we have 2 + 1 = 0

right?

[fossil quest]

Right

[torn silo]

good now you use that to clear the first entry from the second and third row

[fossil quest]

Huh i dont get that part

[torn silo]

you add the first row to the second row

[fossil quest]

Ok

[torn silo]

and the first row to the third row

remember like ann said its F3

[fossil quest]

Oh right got it now

Ok so i get
1 1 2 1
0 1 0 2
0 0 0 0

[torn silo]

with just those two operation?

[fossil quest]

I did it nice more

Once*

[torn silo]

ahh okay

good

now add the second row to the first twice

[fossil quest]

1 1 2 1
2 0 1 1 ?

[torn silo]

no like you did in the first three steps

you take 0 1 0 2 and add it to the first row twice

[fossil quest]

Ok

1 0 2 2
0 1 0 2

Right?

[torn silo]

$\begin{pmatrix} 1 & 0 & 2 & 2 \ 0 & 1 & 0 & 2 \ 0 & 0 & 0 & 0\end{pmatrix}$

[stoic python]

deekaan:

[torn silo]

yes

okay now add another row of zeros at the bottom

[fossil quest]

Not sure why.. but ok

[torn silo]

$\begin{pmatrix} 1 & 0 & 2 & 2 \ 0 & 1 & 0 & 2 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{pmatrix}$

[stoic python]

deekaan:

[quartz compass]

that's too many rows

[torn silo]

wait you don't actually need the kernel

oops sorry

thought this was a kernel question

[fossil quest]

I dont even know what a kernel is

[quartz compass]

no this is right, you're finding how the coefficients of a generic vector are dependent

you just got it into reduced row echelon form

[fossil quest]

I need the basis or generating system whatever its called :/

[quartz compass]

which just means you were doing elimination on your equations really

[torn silo]

ya you can just take the two rows you have

sorry if you had needed the kernel you could have used this form to easily get it

[fossil quest]

Ok

And that's the basis? That's it??

[torn silo]

ya you could check it

basis means you can use those two vectors to generate the previous three

[fossil quest]

Dang ok thanks

[torn silo]

sorry about the wasted steps

[fossil quest]

Not a problem at all👍

[elder robin]

If I'm trying to find the basis for R^2, I can just choose any two vectors that are linearly independent right?

[elder robin]

two vectors with two rows, 1 col

[tall moon]

@elder robin, yeah.

something like (1,0), (0,1) suffices.

But I'm sure you can come up with more exciting ones.

[elder robin]

lol, thats exactly what I put

[eternal finch]

@hearty cliff df_i / dx = (df_i / dx_1, . . ., df_i / dx_n), where x = (x_1, . . ., x_n).

[wintry steppe]

What does it mean if the equation of a line is such:

r = (1, 1, 0) + t(1, -1, 2)

Is it the vector (1, 1, 0) added to another vector that's multiplied by t?

but then what's t ?

[hallow cliff]

yes

you basically just have a line that is the span of (1,-1,2) and then is translated by (1,1,0)

[wintry steppe]

I see

Hmm, by span do you mean that it's basically all the points that vector (1, -1, 2) could reach ?

[hallow cliff]

yes

[wintry steppe]

Cool, thank you

[hearty cliff]

So what dimension would this partial derivative be?

In the pic above?

[eternal finch]

Well, it isn't the same for f_1, f_2, and f_3.

You have to look at the dimension of the input.

[hearty cliff]

Oh so i have to find the dimension of each function? Or to generalize it

[wintry steppe]

Given the lines (1+t, 1-t, 2t) and (2-s, s, 2) how can I find the equation of the plane they create ?

[sonic osprey]

because P_2 is a 3 dimensional vector space

That's kinda right

[pale coyote]

@real plaza Thats actually what a colum vector is. Vectors are abstract objects, and can take any form not just columns of numbers (here they are polynomials). However, once I decide on a choice of ordered basis for my vector space, I can express every vector as a unique column vector of numbers: For example if ${v_1, \ldots, v_n}$ is my basis, then by the definition of basis any other vector $v$ can be written as as a unique linear combination $v = c_1v_1 + \ldots + c_nv_n$ for some $c_i \in \mathbb{R}$. We can therefore identify vectors uniquely by their coefficients $c_i$ in reference to this chosen basis and write $v = \begin{bmatrix} c_1 \ c_2 \ \vdots \ c_n \end{bmatrix}$.

[stoic python]

moonside:

[pale coyote]

Similarly, if I have a linear map between two vector spaces $T: V \to W$, once I choose ordered bases for $V$ and $W$ I can represent $T$ as a matrix, where the columns of $T$ correspond to the result of applying $T$ to the basis vectors in $V$ written in terms of the basis vectors in $W$.

[stoic python]

moonside:

[limber sierra]

[this is assuming you can choose a basis - while bases always exist, they cant necessarily be constructed. that's outside of the scope of this problem/discussion, but it's worth mentioning in case you ever work with functional analysis or what-have-you]

[eternal finch]

@hearty cliff Yes, I think you have to find the dimension of the derivative of each function.

[sick dragon]

can someone explain this please

[limber sierra]

A is the matrix representation of the transformation T

every linear transformation has a unique matrix representation, and vice versa

this is just stating that kernel and column space coincide

and similarly range and null space do

[eternal finch]

I'm confused.

Why would the kernel of T be equal to the column space of A?

Wouldn't the range of T be equal to the column space of A?

[limber sierra]

...oh yeeah

lmao

they've mixed stuff up

idk why i mentally just assumed the image was correct

[eternal finch]

Yeah, I did too at first.

But then I read it again and my brain was like ???

[sick dragon]

Wouldn't the range of T be equal to the column space of A?
why is that

nm kinda makes sense

i dont get the secodn part though, th nullspace of A

[eternal finch]

It's also incorrect.

[sick dragon]

so nullspace is just ax=0?

[eternal finch]

Null space and kernel are synonyms, btw.

[sick dragon]

cinamons 🙂

[eternal finch]

The null space is the set of all vectors v such that Av = 0.

[sick dragon]

ok so nullspace is like the unit vector

[eternal finch]

No.

[sick dragon]

unit matrix...whatever it is called

[eternal finch]

The null space is the set of all solutions to Ax = 0.

Where is the unit coming from?

[sick dragon]

so it can be anything=0?

7=0?

[eternal finch]

Ok, how about an example?

A =
     1     0
     0     0

What are the solutions to Ax = 0?

[sick dragon]

x1

x_1=0

on top

x_2 can be anything i guess

[eternal finch]

Ok.

That's right.

So, any vector (0, x_2), where x_2 is any number, is a solution to Ax = 0.

So, the set of all vectors of that form is the set of solutions to Ax = 0.

That's your null space of A.

[sick dragon]

what if you end up w/ somethign that odesnt equal 0

[eternal finch]

Ax not equal 0 means x is not in your null space of A.

Yes, that's right.

Well, you'd only need to show that A is similar to B.

A = P^-1 B P

B being similar to A follows immediately.

So, you're trying to prove similarity by diagonalization?

[dusky epoch]

If a question asks whether or not matrices A and B are similar, that's just asking whether or not they represent the same linear transformation, right?
not quite

first thing to do is check their eigenvalues

if they don't match, the matrices cannot be similar

[eternal finch]

represent diagonalized matrices?
I'm not sure what you mean by "represent diagonalized matrices".

[quartz compass]

similar meaning a similarity relation, represent meaning representative of the class in the partition

or maybe I'm being too generous

[dusky epoch]

,w eigenvalues [[-1,6],[-2,6]]

[stoic python]

Display results online and refine query

Requested by Ann#0413

[quartz compass]

don't compute the eigenvalues

[dusky epoch]

,w eigenvalues [[1,2],[-1,4]]

[stoic python]

Display results online and refine query

Requested by Ann#0413

[dusky epoch]

why not lmao

[eternal finch]

similar meaning a similarity relation, represent meaning representative of the class in the partition
I see.

[quartz compass]

the determinant is equal to the product and the trace is equal to the sum of eigenvalues

[limber sierra]

don't let BIG EIGENVALUE get you down

[quartz compass]

so it just suffices to check those instead

[dusky epoch]

it's a 2 by 2 matrix

i'm not allergic to quadratics unlike you

the eigenspace matrix? what

[quartz compass]

space of the eigen

[dusky epoch]

mero pls

[quartz compass]

sorry lol

[dusky epoch]

you don't rly care about that... you care about the eigenvalues themselves

the eigenvectors are secondary

[eternal finch]

Yes, but all you care about is if they're similar.

[dusky epoch]

bruh

there's no such thing as "having to" do anything

as in an unconditional "You Have To Do This And I Am Forcing You To"

[eternal finch]

It's more work than you need to do.

If A and B have the same eigenvalues, then they'll be similar to any diagonal matrix with the eigenvalues on the diagonal and thus similar to each other.

[dusky epoch]

you're insisting on taking a detour

[eternal finch]

Yeah.

[dusky epoch]

the same linear transformation wrt 2 different bases, perhaps

or rather two matrices are similar if there EXISTS such a transformation and two such bases that your two matrices are the matrices of the transformation in those bases

[rapid prism]

hi, so i'm taking a computational LA course next semester

and i felt that before taking the course, i should learn LA first

i've seen two books most commonly recommended (Strang vs Axler) ?

i guess they've been recommended so much that I assume both are very good and it just gets down to style of the books i guess

if anyone has used LA for like stats/data science etc, i'm just wondering which ones you found most helpful

[quartz compass]

I think Strang is more applied, personally I don't like Axler cause determinants are too powerful

[torn hornet]

i liked strang when i did physics personally. axler is more for pure math

(plus you can supplement with his lecture series for strang which is nice)

[rapid prism]

thanks for the feedback!

i'll probably start out with Strang then( i'm familiar with his Calculus Book) then look into Axler later

[pale coyote]

I like Axler personally, but I'm a pure mathematician, but I think Strang might be more applicable for ML stuff

[torn hornet]

suck2015 also look into gilbert strangs lectures on mitocw @rapid prism . it supplements the book very well

[rapid prism]

ok thanks!

[pale coyote]

yeah his full course is on there

[rapid prism]

axler seems more proof based(atleast from the short description i read)

[torn hornet]

yeah axler is meant to be for pure math(although i still would recommend hoffman-kunze over it if thats the route you are going, i dont like what axler does with determinants)

[pale coyote]

hoffman and kunze is great too

[elder robin]

Is it possible for {[1;0;0],[0;1;0]| to be a basis for this? Or is it only the 1st and third column of the original matrix?

the way it was explained to me is first and third column have pivots, so first and third col. of A is a basis

but i don't see why the first and third column of rref couldnt be a basis

[pale coyote]

Basis for the nullspace?

[dreamy iron]

yeah axler is meant to be for pure math(although i still would recommend hoffman-kunze over it if thats the route you are going, i dont like what axler does with determinants)
@torn hornet

I’m going through Axler right now. Just finishing up the p-set of chapter 1. Dets and Trace are not until the very end. What don’t you like about how he treats Dets and Trace? (I’m looking for some critiques of LADR, but MSE only has fe.w.)

[elder robin]

Oh i left out part of the problem

the column space

[pale coyote]

A basis for the column space will be given by the columns in the original matrix corresponding to the pivot Columns in rref

[elder robin]

Ok

[torn hornet]

well its really the not using it part lol. i mean tbh it is a fine book, i just dont personally like some of a things (for example it defines char poly as the poly with roots as eigenvalues, instead of with det). But again my not liking it stems from me having learnt it completely differently

its still a good book

so if ur doing it and liking it, keep doing it

its a matter of taste i think

[dreamy iron]

I’m a physics dude, and this level of proofyness is really a touch out of my comfort zone, but I’m getting the hang of it, truly.

Also, i read an MSE answer saying LADR BEFORE Hoffman/Kunze.

So that’s the route I took.

[torn hornet]

oh if your a physics guy and want to do proof stuff, i reccomend doing some intro to proof book into those

[dreamy iron]

I took a formal math based intro to proof course at uni.

[torn hornet]

oh i see so u just need practice which sounds like you are getting

[limber sierra]

yeah axler doesnt use determinants because determinants are kinda weird

which is true

and he argues that you can do most of linear algebra without em

which is also true

[pale coyote]

I do think determinants obscure what's really going on

[limber sierra]

but theyre still really useful

[dreamy iron]

I’m having to pull a lot of rusty set theory tricks from dusty notes. but its a fun grind.

[limber sierra]

idk conceptually determinants are definitely strange but the "motivation" isnt too bad

[rapid prism]

yeah i also wanted to do proof stuff but my knowledge of proofs is kinda trash( i've really only done induction proofs)

[limber sierra]

ok well

sometimes the motivation for determinants is

actual dogshit

[torn hornet]

yeah fair enough

[limber sierra]

like anyone that tries to justify determinants by somethingsomething area of a unit polygon somethingsomething

fuck that

the geometric intuition is useless

[torn hornet]

lol, for me determinant is just a test of invertibility

its suppose to be the

[limber sierra]

just say "matrix multiplication is hard, but scalar multiplication is easy. determinants take certain parts of information about matrix multiplication and translate it to scalar multiplication"

[torn hornet]

volume of a parellilipied (??? spelling)

[limber sierra]

this is a vague motivation, but its the motivation nonetheless

and it leads naturally into common results on determinants, like

"how would we talk about invertibility with determinants? well, what's the only real number that doesn't have a multiplicative inverse? 0. so noninvertible matrices have 0 determinant"

sorry, went on a tangent there

[dreamy iron]

no, keep going plox. this is very instructive.

Det related tangents are greatly appreciated.

[torn hornet]

i mean yeah conceptually dets can get a bit messy

but they are very powerful atleast computationally

but this is also coming from the side of me that does physics lol

[dreamy iron]

All I know about dets is it eats matrices and spits out a “number”. And i feel that a lot of info is lost in that kinda of mapping, but I’m a noob.

[rapid prism]

wait

is a scalar triple product just a determinant

[pale coyote]

Determinants are useful yes but also obscure just why Det = 0 is the same as being noninvertible

[quartz compass]

yes

cross product is a determinant operator waiting for a 3rd vector to be acted on by dotting

[limber sierra]

scalars satisfy $\det(AB) = \det(A)\det(B)$ (when multiplying $AB$ makes sense) and $\det(I) = 1$ (where $I$ is the identity matrix)

[stoic python]

Namington:

[west spade]

How are determinants useful past the fact that they're multiplicative?

[limber sierra]

this is literally, formally, all there is to them

[west spade]

oh

just got answered lol

[pale coyote]

Often people just remember some awful formula without developing any geometric intuition

[limber sierra]

but its natural to see why this would cause them to encode some "multiplication information"

since det(AB) is the same as det(A) det(B)

[dreamy iron]

scalars satisfy $\det(AB) = \det(A)\det(B)$ (when multiplying $AB$ makes sense) and $\det(I) = 1$ (where $I$ is the identity matrix)
@limber sierra

Strang on MITOCW made a big deal about this, and didn’t prove it. He did prove other algebraic properties of the det though?

[stoic python]

ninnymonger:

[quartz compass]

One way to think of it is as the volume of the image of the unit cube

[limber sierra]

this hence gives us facts like $\det(A^{-1}) = (\det(A))^{-1}$

[stoic python]

Namington:

[west spade]

Does a notion of size exist for transformations without a known basis?

[torn hornet]

i mean its prolly not a proof anyone would give in a lecture

[west spade]

or do you literally need the matrix to calculate it

[torn hornet]

its uninteresting and takes up time

its better to just read the proof somewhere

[limber sierra]

ok so

theres 2 facets to the proof that this characterizes the determinant

[pale coyote]

I think the eigenvector/value viewpoint is of more importance for matrices in general though and is perhaps a better viewpoint to take from the start

[limber sierra]

i mean it depends on how you define "determinant" in the first place

but generally it involves:

• proving that your construction satisfies the properties we want the determinant to satisfy
• proving that no other construction does

[quartz compass]

seems useful to have the determinant in the first place when deriving the characteristic equation for eigenvalues though @pale coyote

[limber sierra]

[unless that construction is equivalent]

[torn hornet]

mero you can define char eqn as poly with roots= eigen

[pale coyote]

Not needed

[quartz compass]

"can" but "should"?

[torn hornet]

but im p sure u need det to actually compute these, but for theory you dont

[quartz compass]

not in my book

[limber sierra]

the correct way is to define $\det(I + AB) = \det(I + BA)$

[stoic python]

Namington:

[limber sierra]

actually wait

i dont even know if det is unique

if you define it like that

lmao

it might not be

[quartz compass]

no point in waffling around avoiding determinants

[limber sierra]

let me think

[torn hornet]

i mean you can define det= product of eigenvalues

[limber sierra]

unique nonconstant function such that A = 0 or B = 0 implies det(I + AB) = 1

i should say

but even then its not unique

so rip

[pale coyote]

I'm not saying avoid completely, it just obscures a lot of whats happening and aren't necessary

[torn hornet]

i think for non-computation stuff, i use the det=prod of eigenvalues thing more than using an actual matrix to det thing

[limber sierra]

it might be unique among linear functions though

[quartz compass]

give an example of something that it obscures happening

[limber sierra]

well yeah, det = prod of eigenvalues encodes multiplicative information

in the same way

[quartz compass]

I don't know what you're referring to

[pale coyote]

I just think eigenvalues/vectors can be argued to be more fundamental quantities associated to a matrix as opposed to some number you compute with some complicated formula. I'm not anti-determinant but I at least can see an argument for why centering elementary linear algebra around them could be confusing

That being said I do believe the determinant free approach seems it requires more mathematical maturity, so am not sure if it'd be more effective for nonmathematicians or not

[half ice]

Oh god the Axler conversation

[pale coyote]

Itshappenin

[half ice]

I think this is a rare case where "best construction" ≠ "most helpful construction".

Yes, eigenvalues is the least arbitrary way to first define the determinant.

But knowing that there's a multiplicative function that tells you so much about a matrix really helps with "understanding them", especially when matrix invertibility can be hard to pin down. I don't see why an author would want to pass that up.

On the other hand I remember being an engineer student and wondering what a determinant was even supposed to represent. Haha, those were the days. I guess "be careful" is what I'm saying? Idunno I didn't write a book

[pale coyote]

I've never taught out of axler so I have no idea how good the approach is for beginners, I may just like it now after having seen all the material

I think the main issue is perhaps that determinants are taught poorly, not something inherently wrong with them. I like approaching them with students doing lots of exercises involving using determinants rather than having to actually compute them

[limber sierra]

i think i computed a total of like 8 determinants in my intro linear algebra course

and most of those were to find the char poly

[torn hornet]

i did it for physics so mine had much more

[pale coyote]

Some courses have you do a lot of determinant computations

[half ice]

Engineering was all determinant computations lol. I would often have to set up a matrix just to get the determinant of it

[torn hornet]

for physics we had to find a lot of eigenvalues

which equaled computing determinents

(this was for a QM class iirc)

[half ice]

Even though I don't really do that anymore, I still think det(A) ≠ 0 ⇔ invertible is one of math's best theorems

[pale coyote]

It's nice

[limber sierra]

the best theorem of linear algebra is that the different definitions of dimension coincide

unless you could first iso as a "theorem of linear algebra"

[pale coyote]

Tho one direction isnt as nice as the other

Nah it's gershgorins circle theorem

[half ice]

I mean yes, knowing a matrix is invertible doesn't make me want to exclaim that the determinant is 0 lol

[limber sierra]

honestly its still kinda wild to me that a set is spanning only if it is at least as large as the maximal lin ind set

like i guess it's just a generic statement that

linear combinations are a powerful tool for expressing things

[west spade]

I recently had to break down and look up a proof for the fact that all bases have the same size. I'd proven like ten different important intuitive things that all followed from that fact but I just could not manage to prove the thing itself, however obvious it seemed. God, how ugly that proof is for such a simple and important fact

[half ice]

Oh yeah that's a huge one too

Linear algebra is just really good

[limber sierra]

honestly its legitimately weird to me that like

in popular culture, linear algebra is a "course about matrices"

[torn hornet]

yeah true never thought about that too much

[limber sierra]

when matrices are like the least interesting part of LA

theoretically

[torn hornet]

but yeah that would be wierd if i wasnt familiar with it

[limber sierra]

and most statements about matrices are trivial

or secretly just a statement about vectors in disguise

[half ice]

It's a course about the algebra of linear transformations imo. But that's also matricies

[torn hornet]

personally i like cayley-hamilton as my favourite LA theorem

[limber sierra]

bleh

[torn hornet]

bc it applies quite nicely to algebra stuff (showing algebraic numbers form a field for example)

[limber sierra]

your opinion is wrong

[torn hornet]

your opinion is wronger

[half ice]

Yes okay good justification

[limber sierra]

if yure gonna say something like that, at least say like

spectral theorem

[half ice]

Also works on modules right?

[torn hornet]

smh

[limber sierra]

has to be over a comm ring

[half ice]

Ahh yes comm

[torn hornet]

i havent gotten to apply spectral really, so i wouldnt call it my fav

its cool and all

[limber sierra]

the spectral theorem literally spawns entire fields

[torn hornet]

yes

[limber sierra]

you cant say that about any other theorems other than like

[torn hornet]

but i havent

studied

[limber sierra]

all bases are the same

[torn hornet]

any of them

[half ice]

So time to embarrass myself wtf is the spectral theorem

[torn hornet]

(ok i used a bit of spectral in physics but w/e)

statements about eigenvalues of certain types of matrices is tl;dr

[limber sierra]

hermitian implies unit equiv to diagonal matrix

more generally

[half ice]

Because I know that Spec(Z) is a thing

[torn hornet]

different spec

[limber sierra]

generalize this to hilbert space

[half ice]

It's a spherical thing

[limber sierra]

also yeah different meaning of "spectral"

how do you know what spec(Z) is

but not what hilbert spaces are

[half ice]

I know what Hilbert spaces are lol

[limber sierra]

those are less generalized versions

basically, on hilbert spaces, compactness implies orthonormal basis of eigenvectors

[half ice]

That's not how I learned that though. I learned what Spec() is as a topology on modules

[torn hornet]

yeah im familiar with the less versions actually havent seen too much generalization into hilbert spaces

[half ice]

Which is probably less powerful

[torn hornet]

its called the zariski topology, its nice but i havent seen many applications of it yet rip

(atiyah macdonald keeps making me prove things about spec without showing much if any applications of it rip)

[limber sierra]

all of math is secretly the study of hilbert spaces

hmm

now i wanna go through atiyah macdonald and just see how many times

"the below theorem is a generalization of [x theorem from linear algebra]"

occurs

[half ice]

I laugh that "what is Spec(Z) lol" is a chapter 1 question as if there's any chance many could answer

[limber sierra]

it must be like half the textbook

[torn hornet]

hmmm the module chapter prolly had a bit of that

[wintry steppe]

hello

so my prof told me that this is false while i think this holds true.... but i think my prof is wrong about this... can someone tell me why this may be false?

[pale coyote]

Think of different matrix sizes

[wintry steppe]

but @pale coyote wont u, in most cases, have a free variable?

$\begin{pmatrix}
1 & 2 & 0\
0 & 0 & 1\
0 & 0 & 0
\end{pmatrix}$

Rip latex

[pale coyote]

It's not about most cases, this is a question about always

[stoic python]

christina:

[wintry steppe]

but the wording is has, not only has

[pale coyote]

What's that Christina

[wintry steppe]

It's a matrix with bottom row 0 and no solution

[pale coyote]

Ah it's augmented

[wintry steppe]

yes, i understand that it could have no solutions

but what im trying to prove is that has is not a good word to state that its all the time (always)

[pale coyote]

No, that's the correct wording

[wintry steppe]

Well implicitly you should put a "for all matrices such that" in front of the statement

because has just means contain, but is not disregarding the fact that it could potential have other solution

its vague and its not certain like only has

[wind yacht]

Are we talking about the first or second has

[wintry steppe]

the system has infinitely many solutions

[pale coyote]

Linear systems only have no solns, exactly one solution, or infinitely many solutions

The problem is asking if the rref has a row of zeros then this implies the last case

This isn't true

In fact, knowing you have a row of zeros tells you nothing

[wintry steppe]

It tells you the system can't have exactly 1 solution

[limber sierra]

[note that "infinitely many" only holds if your vector space has infinitely many elements]

[pale coyote]

Yes it can

[wintry steppe]

Oh right it can

Yeah xD

[pale coyote]

:)

[wintry steppe]

@limber sierra can you elaborate?

[pale coyote]

🙄

[limber sierra]

are you working over arbitrary vector spaces

or just over R^n or C^n or whatever

[wintry steppe]

R^n

[limber sierra]

then ignore me

[wintry steppe]

what im confused is how the wording implies that the system only has infinitely many solutions

because i thought it holds true cuz indeed, the system could have infinitely many solutions

[limber sierra]

it could but it doesnt have to

the question is phrased as an implication

if A, then B

but it's possible for A to be true and B to be false

so it's false.

[wintry steppe]

so what do you mean by if A, then B

sorry for the trouble

[limber sierra]

like if i said

"If someone owns a pet, then they must own a dog"

that'd be a lie

since it's possible for someone to own a pet but not own a dog

for example, someone might own a cat

and no dogs

hence it's false

[wintry steppe]

but this is not really the case

[pale coyote]

Yes it is

[wintry steppe]

but i slowly see what u are saying

this statement is saying that it has to have infinitely many

but thats not the case

[pale coyote]

If x > 0 then x = 3

[wintry steppe]

can u confirm my answer, i just need one more small push

what the sentence is saying is that If A, then B holds
but B is not always neccessarily tru, therefore false

[pale coyote]

Yes

[wintry steppe]

ok, i understand now

it is indeed my mistake lol, thank you for helping @pale coyote @wintry steppe @limber sierra

[pale coyote]

Np

[nimble raft]

If I can't invert a matrix, does it mean it's dependant?

Or not nececelery

[dusky epoch]

"nececelery"

[nimble raft]

ye

forgot how to spell

[dusky epoch]

did you mean necessarily

anyway

no, because it doesn't make any sense for a matrix to be (linearly) dependent.

a matrix can be singular.

which is the same as saying its columns are linearly dependent.

(ditto for its rows.)

[nimble raft]

Oh, right.

So are all non-invertible matrices singular?

(Doesn't sound true imo tho..)

[dusky epoch]

"non-invertible" and "singular" are literally synonyms.

[nimble raft]

Huh really?

So if I got a non-invertible matrix, I should be able to find at least 2 columns

Which are dependent.

Well not 2

But I should be able to find columns (or rows) that are dependent?

[dusky epoch]

your entire set of columns is linearly dependent.

you can find at least one nontrivial linear combination of them which sums to zero.

whether or not it'll involve two columns or more than two columns really depends on the matrix.

[nimble raft]

Right.. well time to see if I can find this "nontrivial linear combination"

[dusky epoch]

what is your matrix

[nimble raft]

Im looking at:

$$\begin{pmatrix}3&3&2\ ::6&3&8\ ::9&3&14\end{pmatrix}$$

[stoic python]

Soap:

[nimble raft]

ok thats a big image lol

wait maybe its invertible...

Though the determinant is 0, so I don't think so.

[torn silo]

how do you get a det 0?

[dusky epoch]

one moment

[nimble raft]

I got the matrix of minors, and -,+'d them up

[eternal finch]

The determinant is indeed 0. So, yeah, it's noninvertible.

[dusky epoch]

yeah the det is 0

[nimble raft]

But I can't see where nontrivial linear combination is.

[dusky epoch]

lemme see

[torn silo]

subtract the first row twice from the second and three times from the third

the next step should be clear

[nimble raft]

R2 - 2*R1
R3 - 3*R1?

[torn silo]

ya

[dusky epoch]

-3*(col 1) + 2*(col 2) + 1*(col 3) = 0

[nimble raft]

oh dam

How dyu figure that out?

I was plannin on using Gauss jordy

[dusky epoch]

tried guessing a solution of the system with the first two rows of this matrix as the coeff matrix

[nimble raft]

Right, that werks. Thangs ya'll

[peak nova]

how did it turn from ( 4 ) into ( 0.58 )

[cursive narwhal]

@thin hazel ? What? Where have you gotten till in your working?

[eager burrow]

Seems like exactly what you want, right? Then every element in V admits a decomposition like you want, and the sum is even direct.

Yeah, that's not super difficult; use that T^2 = id and just play around with the object a bit

Start with "Assume v - T(v) is in V_+, then..." and just see where this goes

(you might need that your field does not have characteristic 2 :^) )

[cursive narwhal]

Consider $V_+ \cap V_{-}$. Let $v \in V_+ \cap V_{-}$. Then, $v \in V_+$ and $v \in V_{-}$. So:

$T(v) = v = -v \implies v = 0$

This is where you have to assume that your underlying field doesn't have characteristic 2. Otherwise, the implication above doesn't work.

Now, you just need to prove that $V$ is the subspace sum of those two sets. Clearly, the subspace sum is a subspace of $V$. So, let $v \in V$. Then, we claim that $\frac{v+T(v)}{2} \in V_+$. To prove this, notice that:

$T(\frac{v+T(v)}{2}) = \frac{T(v)+T(T(v))}{2} = \frac{T(v) + v}{2}$

A similar argument can be made for $\frac{v-T(v)}{2} \in V_{-}$. With that, you've proven that the subspace sum is equal to the vector space. This proves the desired result.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@thin hazel

By the way, that first part just proves that the intersection of the two spaces is {0}. That's part of the definition of the direct sum that I typically use

If you're using another definition, then just change the proof to fit that instead

You're welcome.

[thorn prairie]

Hello im doing linear algebra for my 1st year in computer science. Is there anywere where i can find all the theory in one place? Like for matrixes , determinants and valuables.

[limber sierra]

a textbook

[thorn prairie]

?

[torn silo]

your professor probably suggested one

or has a script

[thorn prairie]

He has some notes but its mixed with a lot of examples and exercises and isnt really to the point

[torn silo]

well you could look for better scripts

or make your own

[viscid kernel]

Anyone what the different between columnvectors and rowvectors ?

[dusky epoch]

col vectors are cols, row vectors are rows

col vectors are acted on by matrices from the left, row vectors from the right

[viscid kernel]

Yeah but, If you have a 2x2 matrix, I always thought thr columns represented vectors cuz 3b1b showed it like that

[dusky epoch]

it's CONVENTION that we represent vectors as col vectors, because then linear maps can be represented by matrices acting on your vectors from the left, which resembles function notation

so if a linear map T is represented by a matrix A, it's T(x) = Ax
if we used row vectors everywhere it'd be T(x) = xA, which is less convenient

[viscid kernel]

Hmm, okay

Still confused xd.

[dusky epoch]

it's just convention, don't overthink it

[viscid kernel]

I dont understand why you switched Ax to xA

[dusky epoch]

if we used row vectors everywhere

i didn't switch shit

i was telling you that the representation of linear maps as matrices would work somewhat differently

[viscid kernel]

Thanks for your help, It helped me, but I think I just need to think more.

[jovial sigil]

There is a saying " one can't be taught matrix , one must see for himself what it is"

[limber sierra]

?

[subtle walrus]

jeez, it's a saying namington

[storm python]

What does it mean?

[molten pumice]

it's from a movie....

[storm python]

Do we need to physically apply a linear transformation to a person for them to see for themselve

Yeh but not everyone watches that particular movie

[molten pumice]

but its use here seems pretty clearly as a meme to me

[storm python]

How is it a saying if it's a quote from a movie?

[molten pumice]

meme as in a pop culture reference joke

[storm python]

Anyway, I just wanted to know the meaning of that quote, not its origin

[molten pumice]

in context, it means that one must observe/experience "the matrix" to understand it, not be told about it by someone else

here it could mean that the best way to gain understanding of these things is through practice using them, rather than having those who already know just tell you the answers

[storm python]

I suppose that's true, but it's true for almost everything else in math, but ty for explaining

[jovial sigil]

But matrix are special

A little more special

[molten pumice]

noncommutative multiplication guarantees that

[storm python]

What if I say

There is a saying " one can't be taught group homomorphism , one must see for himself what it is"

It still makes sense, that's what I meant

[molten pumice]

except that wouldn't be a quote/reference

so no joke undercurrent

[storm python]

True

[jovial sigil]

How many ways can u define matrix multiplication?

[limber sierra]

one

theres infinitely many ways you can phrase it but theyre all equivalent

[jovial sigil]

But each can be used in a different way

[subtle walrus]

is that a saying as well ??

[cursive narwhal]

How the fuck were you able to have your discord nickname in Hindi?

And how the fuck is someone supposed to be able to ping you?

Oh nvm

[wintry steppe]

Hi everyone

I need help with this problem

Find the parametric equation of the line that goes through the point (0, 1, 2) and is parallel to the plane x+y+z=2 and is perpendicular to the line (1+t, 1-t, 2t)

So far I found the direction vector of (1+t, 1-t, 2t) is [1, -1, 2] and x0,y0,z0 = (1, 1, 0)

After that I'm kind of lost.

I think I want to find a line that goes through (0, 1, 2) and is perpendicular to (1+t, 1-t, 2t) next ?

[hallow cliff]

Do you know about cross products?

You need to find a vector that is perpendicular to both the vector normal to the plane and to the vector that is the basis for the line

Which is basically taking the cross product with those two vectors

So find v = (1,1,1)x (1,-1,2)

And your answer will be (0,1,2) + tv

@wintry steppe

[wintry steppe]

if a vector is perpendicular to the vector normal to a plane it is parallel to that plane

yeah ?

[hallow cliff]

Yes but this is only true in R^3

[wintry steppe]

Okay I see

I think I will be able to do the problem now thx !

[hallow cliff]

no problem

[wintry steppe]

I hate that I can't put these ideas together

first time that it happens I've taken calc 1 and discrete math

linear algebra is the first one where I'm like... I have no idea how these concepts relate to each other so I can solve a problem]

[hallow cliff]

Nah you were almost there

Now that you saw how I worked out this problem I bet you'll be able to do any other problem of this type

[wise slate]

what is an eigenvalue?

[cursive narwhal]

A value that is eigen, obviously

[wise slate]

...

[subtle walrus]

the opposite of a fremdvalue

[wise slate]

a what?

[subtle walrus]

what kind of answer are you looking for?

[wise slate]

idk the definition i guess

[cursive narwhal]

Yes, a fredvalue

There are velmavalues and scooby values too

Just look for the definition in a textbook

[subtle walrus]

the eigenvalue a of a linear transformation L is a scalar, such that L(x) = ax for some non-zero vector x

[wise slate]

my brain blew up

[subtle walrus]

you're welcome

(fremd is the "opposite" of eigen; they're both german words)

(it's a good joke, i promise)

[thorn lichen]

What does AB = BA tell you about matrices A and B

[subtle walrus]

that they commute?

[thorn lichen]

Ight bet

[cursive narwhal]

It tells you that the matrices are in love with ABBA

[subtle walrus]

that exp(A)exp(B) = exp(A+B)

[thorn lichen]

Day 1 of calc 3 and I’m already mad because the homework submission for this worksheet I just did isn’t anywhere lmao

[jovial sigil]

@cursive narwhal there is something called copy and paste

[cursive narwhal]

Lmao

[jovial sigil]

Indian??

The name suggests

[cursive narwhal]

No, I'm German.

[jovial sigil]

Bengali??

[cursive narwhal]

I'm German.

[jovial sigil]

U live in germany or wht

[cursive narwhal]

No, i live in Italy.

[sick dragon]

is this correct:
e_1 is {1,0}
e_2 is {0,1}
e_n is {0, ..., 1}
where e_1, ... e_n = x_1, ... x_n

In the context of transformations

[cursive narwhal]

?

[cursive narwhal]

Okay, so you have a linear map $T:\bR^3 \to \bR^2$ and you want to find the $2 \times 3$ matrix associated with this linear map under the standard basis of $\bR^3$, yes?

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Is that what you want to do?

[sick dragon]

i just want to know if those definitions are correct

[cursive narwhal]

What do you mean those definitions?

e_1 is just the set containing 0 and 1. e_2 is the set containing 0 and 1. Not quite sure what e_n is supposed to represent but it is just a set.

These are not the standard basis vectors of R^n, if that’s what you were going for

[sick dragon]

thats what i was

[hollow finch]

in my linear algebra class we called e_i a vector with 0s in all entries except for a 1 in the ith entry (i.e. the standard basis vectors)

[sick dragon]

dope avatar

[cursive narwhal]

$e_1 = (1,0,0,0\ldots, 0)$ where it is understood that this is an n-tuple in $\bR^n$. That’s the correct notation

[stoic python]

Abhijeet Vats:

[cursive narwhal]

And yes, what nix said goes for $e_i$

[stoic python]

Abhijeet Vats:

[sick dragon]

so is e_1 {1, 0} and e_2{0,1} ? this notation is weird to me

[cursive narwhal]

That is set notation

[sick dragon]

is e_1 a vector

[cursive narwhal]

In that notation, e_1 = e_2. However:

$e_1 = (1,0)$ and $e_2 = (0,1)$ in $\bR^2$

[stoic python]

Abhijeet Vats:
Compile Error! Click the reaction for details. (You may edit your message)

[cursive narwhal]

????????? That is a standard notation for the canonical basis vectors. So I would assume that you’re talking about the vector e_1

[sick dragon]

OK, and i^N=R^N?

[cursive narwhal]

[thorn prairie]

Guys i have a question

[sick dragon]

i didnt mean i haha

i meant e^N is the number of columns

for the transformation

[thorn prairie]

Does A have same valuables with A^T and not same idiosyncrasies but A has same idiosyncrasies with A^-1 but different valuables?

[limber sierra]

$e_n \in \bR^n$

[stoic python]

Namington:

[limber sierra]

im not sure what e^n is supposed to mean

or how it's "equal" to R^n when it's apparently a number

while R^n is a vector space

@thorn prairie i have no idea what you're asking

[cursive narwhal]

I don’t see how e^N is the number of columns of the transformation, my friend.

[thorn prairie]

@limber sierra i translated. Im talking about λ and the vectors of λ

[limber sierra]

ah, eigenvalues and eigenvectors

[thorn prairie]

Oh lol

Thats the naming?

[copper mason]

does their exist an eigenspace per dimension that has a single element? can the element be an eigen vector?

[limber sierra]

yes, A and A^T have the same eigenvalues, and A and A^{-1} have the same eigenvectors

[thorn prairie]

Does A have same eigenvaluables with A^T and not same eigenvectors but A has same eigenvectors with A^-1 but different eigenvaluables?

Thats the correct question

Alright thanks

[limber sierra]

yeah, that sentence is correct

well, the eigenvectors MIGHT be the same between A and A^T

obvious example: if A is the identity matrix

but they don't HAVE to be

similar for eigenvalues of A^-1

@copper mason eigenspaces are vector spaces, and the only vector space with one element is the vector space of dimension 0, which only contains 0

but, since eigenvectors are, by definition, nonzero

this means that the answer to your question is "no"

there is no eigenspace that only contains one element, since then it would only contain 0, and the space must contain the eigenvector (so thus the eigenvector would have to be 0, a contradiction)

[copper mason]

That was I thinking but I was unsure

thank you so much

[thorn prairie]

@limber sierra sorry one more question. Is (AB)^T = A^T * B^T or the other way around?

[copper mason]

So is the eigenvalue in this scenario nonexistent then?

[limber sierra]

$(AB)^{T} = B^TA^T$

er

[stoic python]

Namington:

[limber sierra]

fixed

[copper mason]

@limber sierra So to be clear, is the eigenvalue in this scenario that you described above, nonexistent then?

[limber sierra]

there is no eigenvector associated with an eigenspace of one element.

since eigenspaces of one element dont exist

[copper mason]

so the eigenvalue wouldnt exist either then gotchu

[limber sierra]

eigenvalues dont really have much relation to eigenspaces, honestly

the connection is indirect

eigenvalues are related to eigenvectors, and eigenvectors determine eigenspaces

[copper mason]

If you were to assume the eigenvector can equal to 0, would the matrix of the eigenspace of a single element be an identity matrix?

[hollow finch]

theres no point to the concept of an eigenvector if it can be 0

and the identity matrix has infinitely many eigenvectors. in fact every single vector in the same space as it is an eigenvector with eigenvalue 1.

[copper mason]

in a identity matrix wouldnt the eigen vectors be a 0 vector ?

[hollow finch]

uh

no?

[limber sierra]

by definition, an eigenvector must be nonzero

review your definition

if an eigenvector can be zero then everything is an eigenvalue

which makes the notion kinda useless

[copper mason]

that's why I'm confused about the question because it asks me to give an example if the entire question isn't factual lmao

[sick dragon]

is there only one pivot here? column 1?

[limber sierra]

review your definition of pivot

13 is also a pivot.

[sick dragon]

is the Kernel of T's diimension the # of free variables?

[gray dust]

meaningless word salad

what are the standard basis vectors of P_2?

doesn't matter if you write a row or col vector for now, just show you know you can do it

that's it

no prob

[hexed steeple]

Just need clarification on what I actually need to do

[hollow finch]

@hexed steeple Probably verify which of them satisfy all conditions of being an inner product on R2

Your textbook probably has them in a box somewhere in the chapter introducing inner products

[hexed steeple]

@hollow finch Thought so. thanks

[hollow finch]

[cursive narwhal]

A list of vectors can span a vector space without being linearly independent

If they don't span R^3, then you can find a vector in R^3 that cannot be expressed as a linear combination of the vectors in your given list.

I can't think of an algorithmic way to do it off the top of my head. Suppose we had 4 vectors $v_1,v_2,v_3,v_4 \in \bR^3$. Then, you can express each of them as a linear combination of the standard basis vectors. So, if you simplify the linear combination of these vectors as a linear combination of the standard basis vectors instead, you can determine if they span $\bR^3$ or not. For example, if they simplify to something like $a_1e_1+a_2e_2$, then that linear combination does not span $\bR^3$.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

The above is just a sketch. In practice, I'd probably do something else depending on the problem

[celest bridge]

quick question (since I'm rusty on my linear algebra): det(A)*det(B) = det(AB) for upper triangular matrices, right?

[limber sierra]

det(A)det(B) = det(AB) in general

if multiplication makes sense

[quartz compass]

if determinant makes sense

[celest bridge]

oh, duh

what am I thinking about that only holds for upper triangular matrices?

[limber sierra]

determinant is product of diagonal entries, perhaps

(this also holds for lower triangular though, naturally)

[quartz compass]

AB can be square while A and B aren't

[limber sierra]

yeah true mero

[celest bridge]

mmm yes

[random crown]

can someone help me out real quick

7. a) What is the only Eigenspace per dimension that has a single element?

i cant figure this out

[limber sierra]

youre the second person to ask this

which is weird

note that eigenspaces are vector spaces, so they must contain 0

but because they're eigenspaces, they must contain the eigenvector

so if they only have one element, the eigenvector must be 0

but this is contradictory (an eigenvector is nonzero by definition)

so the question is poorly-posed, in that no such eigenspace exists

[random crown]

so no answer exists?

[limber sierra]

image is a separate concept, and doesnt apply to vectors

rather, it applies to functions

(or matrix representations of functions)

its the set of all possible "outputs" you can get from your function

you might've called this the "range" of a function in high school algebra

as mentioned, it coincides with the span of the matrix representation's vectors.

thats a different thing

different meaning of "image"

where did the definition you gave come from?

[wintry steppe]

so it just wants illustration

just google

[limber sierra]

well, no

the "image" in this context just means

"what you get after applying the transformation"

the definition you gave is an entirely different meaning of "image"

[wintry steppe]

so just the standard matrix

?

[limber sierra]

no.... it wants what vector you get after applying the standard matrix

oh whoops

wrong row

[wintry steppe]

so vector is [w1,w2,w3]

alright that makes sense

[limber sierra]

no, the vector is (-2, 1, 2)

[wintry steppe]

no i mean the new one

[limber sierra]

uh

ok sure

that works

[wintry steppe]

so the answer is just (xcostheta - ysintheta, x sin theta + y cos theta, z)

?

[limber sierra]

what is x? what is y? what is z? what is theta?

[wintry steppe]

bruh

theta is 90 degrees

x is -2, y is 1, z is 2

[limber sierra]

cool, so plug those in and simplify

and that's your result

[wintry steppe]

alright i got another :/

i know it involved sheering

although it goes from R4 to R5

so its not a square matrix anymore

[cursive narwhal]

Sure but you can still find a matrix. In particular, you're looking at a 5 x 4 matrix

@wintry steppe

The columns of this matrix will just be the images of the standard basis vectors in R^4

[wintry steppe]

hm

is everything except the main diagnol gonna be 0

and would the numbers be like

New Variable divided by Old Variable

heres what i have in mind

[cursive narwhal]

[wintry steppe]

wrong dimensions but am i on the right track

[cursive narwhal]

No.

Your columns are the images of the standard basis vectors of R^4

So, T(e_1), T(e_2), T(e_3) and T(e_4)

[wintry steppe]

uh

whats e

[cursive narwhal]

The standard basis vectors of R^4. So:

e_1 = (1,0,0,0)
e_2 = (0,1,0,0)
e_3 = (0,0,1,0)
e_4 = (0,0,0,1)

[wintry steppe]

oh ok

T(e_1) means image of e_1?

[cursive narwhal]

Yes, it is the image of e_1 under the transformation T

[wintry steppe]

ohhhhhhhhhhhhhhhh

alright hold on

?

[cursive narwhal]

Let's see:

T(1,0,0,0) = (0,1,0,0,1)
T(0,1,0,0) = (0,0,0,1,0)
T(0,0,1,0) = (0,0,1,0,-1)
T(0,0,0,1) = (1,0,0,0,0)

So it appears that you are correct

That is the matrix associated with this linear map

[wintry steppe]

did i get dimensions wrong

do i need to transpose

oh wait no

columns

alright great

[cursive narwhal]

Yeap

[wintry steppe]

thanks a lot ✅

[cursive narwhal]

You're welcome.

[wintry steppe]

also uh

just to get a second opinion

so the answer i got according to what the guy said is (-1, -2, 2)

he said image was just a vector but

it looks like a rearranged version of the original one

so isnt it T(x,y,z) -> T(-y, x, z)

and i write that as an image somehow

[gusty needle]

Hey guys can anyone assist in understanding a question I have for a slightly more advanced lin alg class?

[pale coyote]

So what's the definition of the transpose

Of an operator

[dusky epoch]

oh my bad

you two have the same default green discord pfp

[pale coyote]

word

[dusky epoch]

if either of you were to change it that'd be great

[pale coyote]

NAAH

[wintry steppe]

pls

tags wrong person

[gusty needle]

Ok so ive made progress on it, if i get stuck again ill pop you a question @pale coyote thanks for offering to help

[nocturne oracle]

Gonna review my LA before reading through an algebra book, anyone know the differences between these

or does it not really matter

[half ice]

Applied books tend to be less technical for the return of being easier to read, and quicker to be useful. If you've already done LA, I'd suggest checking out Axler

[nocturne oracle]

Ok sounds good, I just want a quick refresher of anything I might have forgotten

[strange obsidian]

yeah, LA done right gets a recommendation from me

[gritty frigate]

Why do we now that the determinant of a 2x2 matrix is a11a22-a12*a21?

[torn silo]

its the area

[cursive narwhal]

The definition of the determinant of an n x n matrix yields that result when applied to 2 x 2 matrices

@gritty frigate

[half ice]

I think that you're actually trying to ask "wtf is a determinant?"

[gritty frigate]

Nono I know what it is

But I want to know how

Expression

And why if an A has its A-1 the system is defined

[torn silo]

then its the area shouldn't surprise you

[gritty frigate]

area of what

I think I get it now...

Lest me see if my process is correct

Let*

I dont know how to use the math bot...

That would make it easier

Considering A, B two 2x2 matrix and I, identity matrix

[eternal finch]

If you are simply interested in the 2-by-2 case...

Linear algebra began with the study of systems of equations. A natural question to ask is if the system has solutions.

For a 2-by-2 homogeneous system whose coefficient matrix is {{a, b}, {c, d}}, when you do row reduction, you get {{a, b}, {0, d - bc / a}}.

The system has a nontrivial solution if d - bc / a = 0.

Another way to write this is ad - bc = 0.

And so our determinant appears. It’s called a determinant because it determines if the system has a nontrivial solution.

Sorry for text bombing. Just thought another perspective would be interesting.

[cursive narwhal]

And why if an A has its A-1 the system is defined
This makes absolutely no sense, as far as I can see.

[gritty frigate]

Nono I havent finished