linear-algebra

[cursive narwhal]

uh wot lol

[wintry steppe]

i wasnt even tired when i said that

[cursive narwhal]

(i still don't quite know what you said. The "every square matrix would be invertible if that were true" comment makes sense in context)

[wintry steppe]

they asked if every square matrix in rref is the identity. a square matrix is invertible iff it reduces to the identity. therefore if every square matrix reduces to the identity, every square matrix is invertible

so no, i was right in context

[cursive narwhal]

Yea exactly lol, it's not really copswing worthy because it makes sense in context

[wintry steppe]

yeah

i just woke up so i thought i was plain wrong, looking back

[cursive narwhal]

It's fine, it happens to the best of us

[wintry steppe]

thanks for catching that

i dont have the most keen eye right after waking up it seems

[cursive narwhal]

I don't think most people do lol. I've made stupid mistakes while doing math after having just woken up

But I've also given decently okay explanations even after just waking up. I remember giving that example to Chaf the other day after having just woken up lol

[wintry steppe]

i guess it really does depend on how you interpret their question, because if you interpret it how i did as "every square matrix's rref is the identity" then what i said is true, but if you interpret it as "every square matrix already in rref is the identity" then its definitely not true

[cursive narwhal]

Well, the person asked "Are all square matrices in rref identity matrices?". That's just the same as asking; if I take any random square matrix and do row reduction, will I get the identity matrix at the end?

So, i think you interpreted it correctly

[wintry steppe]

i think it's still reasonable to interpret it in the latter way, since if you remove the word "already" you have exactly their question

but this is a question on semantics and interpretation, not linear algebra, so it's probably best to keep it out of the channel now

[gritty frigate]

Why the definition of determinant works ??

I do not seem to find a proff on the internet

[limber sierra]

what do you mean by "works"

[gritty frigate]

Why do we do it by reducing the matrix in simplier matrixs

[spice storm]

So like a 3 x 3 matrix? why you have to reduce it?

[gritty frigate]

Yep

I know that it is because of a definition

But I do not know why it is defined that way

[spice storm]

Have you learn the propertiies of Determinants?

[gritty frigate]

Yep

I m right now giving a prove for them hahaha

In order to understand them well

Is because when I do row reduction I make that math with the numbers ?

[wintry steppe]

can you define "works"

[wintry steppe]

@gritty frigate the reason why we row reduce matrices is because each row corresponds to an equation

And we want them to be in the form x = .., y = .., z = ..

[ocean sequoia]

ok so i have to say i dont understand the idea of the dual basis at all

i keep seeing this idea thrown around and im not following

sorry was trying to type if out but im so bad at the TexIT

um the idea that

i think i can do it without it here hopefully its not to confusing

let f_i be a linear functional and v1,...,vn a basis of V

f_i(v_j) = 1 if j = i and f_i(v_j) = 0 if i =/= j

im so confused as to why we are going to 1?

so it takes every vector in the basis of V to 1?

i guess where im confused it that doesnt a linear functional just take something to F? for example x,y in R^2: x,y = 2x + 2y would be a linear functional right?

what exactly is f_i then?

ok hang on i think this is where im confused

given a basis vector in R^2,2x+ 2y would be uniquely determined on that basis but this where im getting lost

i dont understand why we have multiple f_i

yea because it will closure i can see that from that example

err it will satisfies the requirements for a vector space sorry

thats where im confused i think yea

LADR

ive really liked it up to this point

ok

how did you know that off the top of your head lol

ohhhh lol

oh i saw this proof too in a video

dim(V*) = dim(V)dim(F)

and because dim(F) is 1

ok so the basis for V* has to be multiple functionals

ok then why does it have to take basis vectors to one when i = j

is it just because 1 is the basis for F?

so since we know what it does for the basis we know what the transformation does everywhere?

[quartz compass]

the reason is convenience, if you imagine the vectors in a matrix then multiplying the two matrices gets you the identity matrix

in other words, they're matrix inverses

$f_i(v_j) = \delta_{ij}$

[stoic python]

Merosity:

[quartz compass]

see how this looks like FV=I

yeah

[ocean sequoia]

why does it take i =/= j to 0?

[quartz compass]

because they're orthogonal as vectors, think of it this way

F has the f_i as row vectors and V has the v_j as column vectors

making FV = I have the entries of every possible dot product of f_i and v_j

if i=j, the diagonal are all 1s

off the diagonal of hte identity matrix are 0s

[ocean sequoia]

wait fi (vj) is the dot product?

i thought it was saying to plug in

maybe im just a moron

ive heard of it

it is a no

actually no i dont know

yea

if v and u are orthogonal then v* u = 0

[1,0] * [0,1] = 1* 0 + 0*1

ok so wait why if i =/= j is f_i(v_j) = 0

honestly i havent felt lost at all till now

ok so just take it on faith?

im honestly ok with that answer

as dumb as that sounds

ok

same for why it takes it to 1?

[quartz compass]

yeah, it's just a choice that ends up being computationally convenient

it's not derived from anything

[ocean sequoia]

hm let me try this what does it mean that its 1?

like we are just trying to form a basis for the linear functional?

errr the dual space

which is the space of all linear functionals on V

so hang on what would be an example of the dual space on R^2?

2x, 2y?

lets just use the standard basis

[0,1],[1,0]

[stoic python]

slimvesus:

[ocean sequoia]

yea sorry

so do i need to specify that we take the y/x vector to 0?

2

so wait we know it takes x to 2x but because i havent specified anything for y?

ok what about just x?

and y then

because it has to have dim 2 right?

ok that helps alot to see

thank you

yep cause if you dot those you do get the identity like Merosity said cool

ok i think i feel better about this

neat

[wintry steppe]

write out what $$AA^{-1} = I$$ means

[stoic python]

TTerra:

[wintry steppe]

nice, you deleted your question

real there

[quartz compass]

@ them next time to shame them lol

I don't like to help cowards

[wintry steppe]

the question was "given an invertible $n \times n$ matrix $A$, show that the $ith$ row of $A$ and $j$th column of $A^{-1}$ are orthogonal for $i \neq j$, and that their dot product is $1$ for $i = j$"

[stoic python]

TTerra:

[wintry steppe]

you know, the definition of matrix multiplication

seemed like homework tbh

[gray dust]

rarely do i get stuff here that's NOT hw

[quartz compass]

yeah I was wanting to shame them for backing out of making themselves seem even remotely vulnerable for posting a question then deleting after they get an answer

I don't care if it's homework or not lol

[wintry steppe]

because you want the matrix to commute with its inverse

and that forces the matrix and its inverse to be square

[quartz compass]

I don't know about that

[wintry steppe]

maybe copswing worthy explanation

[quartz compass]

if the left and right inverse exist, then you can show they're equal and it commutes but

I would say if you want both a left and right inverse it has to be square

[wintry steppe]

oh right, for the multiplication to even make sense

brain fart

[quartz compass]

like for instance, imagine an inclusion mapping from R^2 to R^3, you're basically imagining your 2D plane to now be in 3D space

you can invert this operation by immediately forgetting that you did this, and everything is fine going back to R^2

but if you go from R^3 to R^2, no such way of undoing that operation so easily

[limber sierra]

more compactly:

$AA^{-1} = A^{-1}A = I$ doesn't make sense if $A$ isn't square

[stoic python]

Namington:

[limber sierra]

since then you cant multiply it by the same matrix on both sides (and produce a square matrix)

[quartz compass]

that just begs the question, why do you want a matrix to commute with its inverse?

[limber sierra]

because otherwise $A \neq (A^{-1})^{-1}$

[stoic python]

Namington:

[limber sierra]

which makes the notion of inverse much less useful

[quartz compass]

I don't think this answers the original question asked

"why must a matrix be a square to have an inverse?"

it only suggests why it's convenient

my response was just saying that I don't accept the premise, so that's why I didn't answer it 😛

[limber sierra]

well yeah your explanation is perfectly adequate if a "graphical" argument is acceptable

[quartz compass]

non square matrices can have inverses

[limber sierra]

left or right inverses

not two-sided

[quartz compass]

yeah that's why I had explained through an example already

but I think because @royal slate ran off this is further burying it

[limber sierra]

fair.

[wintry steppe]

suppose AB = BA for some matrices A and B. for these matrix multiplications to even make sense you need A to be m x n and B to be n x m for some positive integers m, n. then this is saying that the m x m matrix AB is equal to the n x n matrix BA, which only makes sense if n = m; i.e. A and B are square.

now, consider AB = BA = I. this means that B is the inverse of A. as the previous paragraph shows, A and B must be square. this shows that if you define "A is invertible" as "there is a matrix B with AB = BA = I", then that forces A and B to be square. it's much more common to define invertibility only for square matrices, but i just showed that there is no loss in generality in doing so. this is what i meant in my first message.

this should answer your question assuming i have interpreted it correctly as "if a matrix has an inverse, why must it be square?". if i have not, please let me know @royal slate

and looking back, this is basically what namington was saying. i just wanted to make it very clear what i meant by my initial answer

[cursive narwhal]

If you've learnt about matrices as equivalent objects to linear maps, then it's easier to think about why only square matrices can have multiplicative inverses (barring all talk of left/right inverses).

Let $A \in M(m \times n, \bF)$. Then, you can think of the matrix $A$ as being represented by some linear map $f: \mathbb{F}^n \to \mathbb{F}^m$.

Now, you can prove that two vector spaces are isomorphic iff they have the same dimension. That means that there exists a bijective linear map between two vector spaces iff they have the same dimension. If you want your matrix to be invertible, the linear map it is associated with has to be invertible as well. So, it has to be bijective. Thus, $m = n$. But that just means that $A$ is a square matrix.

This doesn't mean that it's always going to be invertible. Indeed, not all square matrices are invertible. I think that's a more useful way ot think about this stuff, though.

[gray dust]

rip latex though i think you've gotten almost obsessed with writing latex essays lately

[cursive narwhal]

Kek yea idk

[tame mural]

Is it wrong to say that a basis is the same as the smallest generating set for a vector space?

[shy atlas]

sounds about right

[tame mural]

hooray

[dusky epoch]

not really

"the" smallest implies uniqueness

which bases aren't

[tame mural]

I see

[shy atlas]

¯_(ツ)_/¯

[dusky epoch]

a basis is a MINIMAL spanning set

[tame mural]

I see, minimal is a better word

[dusky epoch]

meaning no proper subset of it still spans the space

[tame mural]

thanks!

[dusky epoch]

@shy atlas

[shy atlas]

no copswing on birthday

[dusky epoch]

it's my birthday.

[shy atlas]

yeah be cool

chill

[ripe hamlet]

happy birthday ann, what birthday is it for you?

[dusky epoch]

my 21st

[ripe hamlet]

ohh nice

[dusky epoch]

no no no @shy atlas chill only comes fourth in the order^tm

or third if you don't count prank

[shy atlas]

[clear sparrow]

What is the corollary saying that the theorem isn't already?

[dusky epoch]

nothing.

er

nvm

it's saying that the product of however many invertible matrices you want, not just two, is invertible.

[clear sparrow]

Ah 🤦‍♂️ ofc
Thanks for clearing that up!

[wintry steppe]

Makes sense

[strange obsidian]

thanks

[wintry steppe]

Is this right? The general solutions of the augmented matrix

[crystal oracle]

@crystal oracle I'm curious now, where does this appear in Linear Algebra Done Right?
@wintry steppe I used this to solve 3rd edition - section 6.B "Orthonormal bases" problem 13

[hollow finch]

is this correct? assuming the entries of B are functions of t, and A and C have only scalar entries
$$\int_{x_0}^{x}\left(AB(t)C\right)dt=A\left(\int_{x_0}^{x}(B(t))dt \right)C$$

[stoic python]

nix:

[hollow finch]

i guess same question for differentiation and indefinite integrals

[manic oasis]

Let's say that this holds, where $A$ and $B$ are real $3\times3$-matrices, $c,d\in\mathbb{R}$, and $P$ is a unitary $3\times 3$-matrix. Is $P$ real?

[stoic python]

gustavn64:

[eager burrow]

sure why not

set c = d and P = identity

@hollow finch yeah that's fine; matrix multiplication is just basically taking a bunch of linear combinations, and integrals are linear with respect to that

if you write out the matrix multiplications in terms of sums and matrix entries, it's easy to see that your relation holds

[hollow finch]

good point, that was kinda what i was thinking. so any linear operation should probably have some analogue for matrices, in this case pulling a scalar out of an integral.

[manic oasis]

@eager burrow I'm assuming that that holds for some values of $P,A,B,c,d$

[stoic python]

gustavn64:

[eager burrow]

yeah sure

oh whoops brainfart then

but no, this is not necessarily true; you could have P = i * identity, for example

[gritty frigate]

Guys

If I have a vector B so that |B| = 1

How can I get another vector with the same direction of B but such that |B| = X

[eager burrow]

multiply B with x

[gritty frigate]

Having the |B| = 1 vector, how can I get another with the same magnitude

You are right hhahahahaa

[gritty frigate]

I forgot that important property...

[wintry steppe]

@gritty frigate if you have B = (a,b) and |B| = 1 then sqrt(a^2+b^2) = 1

[gritty frigate]

I was thinking on doing pow 2 on both sides and them multiply by x

[wintry steppe]

Then xB = (xa, xb) giving us |B| = sqrt(x^2a^2 + x^2b^2) = sqrt(x^2(a^2+b^2)) = sqrt(x^2) = x

Oh yea that works too

[gritty frigate]

Well it did not worked D:

The first thing you said works perfect

But I do not what I m doing wrong that the method I proposed does not work

ax+by+cz+d = 0 is a line on R3 ??

[wintry steppe]

thats a plane in R^3

@gritty frigate

[gritty frigate]

I mean, if you know that a line in R2 is ax+by+c=0

Can I use ax+by+cz+d=0 on R3 ?

[wintry steppe]

If we rewrite this as ax + by + cz = d

And define n (the normal to the plane) = (a,b,c)

Then n * (x,y,z) = d

• is the dot product here

The dot product here essentially says that if we project a point along the normal, it should land on the plane defined by that normal if (x,y,z) satisfies the equation

And the magnitude of the projected vector yields d, the distance from the origin to the plane(which is a constant)

And if d = 0, the plane passes through the origin

[wintry steppe]

I'm trying to find the solutions from the augmented matrix, is this any good ?

I put the augmented matrix in rref, and I get that last step. Do I then interpret that final matrix to find the solutions ?

For example I'm trying to find for what values of "h" and "k" does the system have 1) no solutions 2) one solution 3) many solutions

<@&286206848099549185>

[gilded harbor]

i think this goes here

[wintry steppe]

For problem one I'm totally lost on how to set it up

I can do the LU and LDU factorizations but from there I'm lost

I would help y'all but I ain't got the time.

I'd say, for self help, try reading different sources.

If you have a link that explains that would be cool

but if your busy no worries of course

[keen patrol]

What do you have for the factorizations?

[naive skiff]

@wintry steppe I think this may help https://www.youtube.com/watch?time_continue=437&v=m3EojSAgIao&feature=emb_title

[wintry steppe]

Does anyone have some indicators for my problem? I'm still stuck 😅

[dry pulsar]

I have to prove the composition of two injective linear transformations is injective

I made this but i dont know if am i right

[wintry steppe]

a linear map sends the zero vector to the zero vector, regardless of injectivity

an injective linear map sends only the zero vector to the zero vector (this condition is equivalent to injectivity)

that is, if T is linear, then T is injective if and only if Tx = 0 implies x = 0

now try the problem again

prove the thing i just said if you haven't seen it before, it's a straightforward application of linearity

(this also just follows from T and S being functions....)

[dry pulsar]

Then can i assume TS is not injective then that implies kerS is different from 0? Because 0 of image of S is the 0 of domain of T

[wintry steppe]

you don't need a contradiction proof

you can either try a contradiction proof, or prove it directly (easier)

(ignore the deleted message, misread it)

[wintry steppe]

I'll have a look! 🙂 @naive skiff

[stoic python]

Abhijeet Vats:

[wintry steppe]

i assumed they knew that the composition of linear maps is linear

but, for their sake, they should know that this is absolutely necessary to know for using the thing i mentioned

you need to know TS is linear to be able to use the fact i mentioned after all

[cursive narwhal]

wot which fact?

[wintry steppe]

a linear map is injective iff kernel is trivial

it's so much easier to just say/prove that the composition of injective functions is injective, but for the sake of learning linear algebra it wouldn't hurt for them to prove it using that (and i am assuming that they are being implicitly asked to do so)

[cursive narwhal]

Urhh wait probably completely misinterpreted what you said lol

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Oh also, I got an analysis problem yesterday that's pretty spicy. It's pretty nice haha

[gray dust]

\ker format

[cursive narwhal]

oh nice

thanks

[gritty frigate]

$ A = (2,5,3), B = (0,0,2), C = (0, - 3,0)

Find the area of the triangle made by this points.$

[stoic python]

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

[gritty frigate]

$
Consider:
BA = (2,5,1)
|BA| = \sqrt{30}
BC = (0,-3,2)
|BC| = \sqrt{13}
CA = (2,8,3)
|CA| = \sqrt{77}

How would you find H, picking any magnitude as the base ?
$

Oh that hurts..

[stoic python]

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

[dusky epoch]

\\ for newlines is probably what you're looking for

$ BA = (2,5,1) \\
|BA| = \sqrt{30} \\
BC = (0, -3, 2) \\
|BC| = \sqrt{13} \\
CA = (2, 8, 3) \\
|CA| = \sqrt{77} $

[stoic python]

Ann:

[forest quail]

hey this is probably a very dumb question

is A an augmented or coefficient matrix?

like i genuinely don't know when matrices are coefficient or augmented and i feel rlly dumb

[dusky epoch]

it is neither

a matrix is a matrix

[forest quail]

wiat what

so it doesn't have a solution?

[dusky epoch]

augmented matrix vs coefficient matrix only matters when you're dealing with a system of equations.

[forest quail]

doesn't it also matter when u do rref and looking for the general solution?

[dusky epoch]

the general solution of a SYSTEM OF EQUATIONS

which you have not written down yet

but the question essentially asks you to solve Ax = 0.

for which A would be the coefficient matrix, if you so insist.

[forest quail]

ohhhhhhhhhh i finally understand it now!!!

dude thank uuu

[dusky epoch]

i'd rather not be called dude but you're welcome

[hushed mortar]

Can there be a Rank for a matrix A that is a set of F^(n*m)? I.e. are ranks for any spans or just spans of real numbers?

[forest quail]

im sorry! i just have a bad habit of saying dude all the time

[hushed mortar]

Dude

[dusky epoch]

johncarmack your question doesn't make much sense

please check your wording and try again

[hushed mortar]

You just called xir a dude. Check your privilege

[forest quail]

how can i do this without knowing what T is?

[dusky epoch]

Use the fact that T is linear

[hushed mortar]

Hmm, let me use latex. Is there a latex bot in this server?

[dusky epoch]

there is.

@forest quail you know Tu and Tv and that T is linear

that's all you need

[hushed mortar]

How should I invoke it?

[dusky epoch]

just write a message with at least one instance of mathmode anywhere in it and it'll render it for you like this: $T(u+v) = T(u) + T(v)$

[stoic python]

Ann:

[hushed mortar]

Deal, thanks

[dusky epoch]

(also my pronouns are she/her just for future reference)

waiting for feedback from soaringbear...

[gray dust]

[errant wyvern]

If matrix A is equal to some scalar x times some matrix B, what is the relation between detA and detB?

[severe magnet]

det A = lambda^n det B, n denoting dimension of the matrix

[wintry steppe]

if A is n x n, then det(cA) = c^n det(A)

lol

we literally said it at the exact same time

you dont need anything fancy to prove this btw, just multilinearity of det

[errant wyvern]

Yeah thanks just checking, just did an exam one of the questions was if det(A*B) defines a dot product, where A,B are 2x2 matrices. But it doesnt follow, in such dot product <xA,B>=/=x<A,B>

[wintry steppe]

ooh that's a nice question

[errant wyvern]

I think i got it right

[wintry steppe]

looks right

det((cA)B) = c^2 det(AB), not necessarily equal to c det(AB) as you'd expect from an inner product

even just addition wrt one component would go wrong, but scalar multiplication is definitely an easier way to see that linearity fails

[errant wyvern]

Im proud of my self, this year i have been strugling to keep up and adjust to uni, and then corona hit and my uni didnt adapt, so i was forced to study this alone without proper resources. Feels good i got at least that dumb question right

[wintry steppe]

if you keep doing questions like this (imo this is a good question because it tests understanding and isn't an immediately straightforward computation) then you'll get comfortable with the material

but math is hard, you should be proud

in the case of your question, n = 2

[stoic python]

TTerra:

[errant wyvern]

Yeah i see

Thanks

[limber sierra]

well satisfying that is equivalent to satisfying multilinearity

[wintry steppe]

sshhh

[limber sierra]

also thats unironically probably how i'd prove it if you forced me to, just because it takes up less space on the page

no writing necessary just a single block of mathmode

[wintry steppe]

i'd always thought of it as something that is just true for multilinear functions in general, so i've never thought of proving det(cA) = c^n det(A) in that way

Haha yea that seems to be a more natural approach to me too

the habit of writing out multilinear functions has been hammered into me by multilinear algebra

its just what ive gotten used to

i go through a lot of paper

[forest quail]

how do i prove this?

[wintry steppe]

what have you tried?

[forest quail]

do i just make up a matrix that follows the given

[pallid rampart]

Hint, if T is a linear transformation then what is T(0)?

[dusky epoch]

no, you do not make up a matrix for anything

would you prove that the product of any two even numbers is even by showing that 66 * 188 is even?

no you wouldn't

tterra don't thonk me i'm trying to demonstrate the absurdity of soaringbear's line of reasoning to them

[wintry steppe]

ok ok sorry

[dusky epoch]

@forest quail do you know what a linear transformation is?

and PLEASE do not ghost me like you did last time.

[forest quail]

after watching a few khan academy vids i finally figured it out! but thank u so much!!

[cursive narwhal]

@forest quail C'mon man, she even said please -.-

She literally told you not to ghost her

[forest quail]

oh wait i thought she was talking to tterra?

ohh nvm i just read the chat sorryy!!!

i have dnd on so i wasn't aware you responded

[wintry steppe]

all she told me was not to misuse one of the emojis lol

[cursive narwhal]

Listen. If you're going to ask questions, keep Discord open so you can immediately talk to the individual who's helping you. They'll usually ping you, like Ann did.

[forest quail]

i will!! i'm just not used to people responding so quickly

[wintry steppe]

time is a precious commodity

[shy atlas]

abhijeets mom comes cheap

[crystal oracle]

Is it true that for every field 𝔽, for every vector space V over 𝔽, for every vector subspace U of V, there exists a subspace W of V, such that U+W is a direct sum, and U+W=V?

[dusky epoch]

yes

[eager burrow]

Assuming the axiom of choice yeah, but if your vector spaces are not finite-dimensional, that space W might be awful

[dusky epoch]

imagine rejecting choice

[crystal oracle]

What about if there is no Hamel basis?

[dusky epoch]

wym no hamel basis?

every vector space has a hamel basis

[crystal oracle]

For example, if the the axiom of choice is false

[dusky epoch]

unless you're one of those people who reject cho-

oh.

[crystal oracle]

Well, I just want to know whether this statement follows from ZF

It's just that my textbook says that it can be shown "without the hypothesis that V is finite dimensional using considerably more advanced tools" and doesn't elaborate

[dusky epoch]

choice

zorn's lemma

[crystal oracle]

I see, thanks

I thought that maybe by "more advanced tools" he meant transfinite something

[shy atlas]

whats axiom of choice anyways

[shy atlas]

lat ouch

[eager kestrel]

When I do Au = .. I get:
Au= 2u^(T)u + vv^(T)u - Iu = 2u^(T)u + vv^(T)u - u
How do you get the first and second terms?

I only get why Iu=u

[shy atlas]

sorry i dont speak japanese

can u translate it

[eager kestrel]

How is the text relevant?

They just define A

and do A times u

and get the rightside, i dont know how the first and second terms are gathered

or mb its relevant

it says that v and u are orthogonal vectors in R5 @shy atlas xd

[shy atlas]

$u^Tv= u \cdot v$

[stoic python]

soαρ:

[shy atlas]

is that what you're asking about ?

[ripe hamlet]

Says Let u and v be two orthogonal vectors in R^5 and ||u||=2, ||v||=3, and let A = 2uu^2+vv^T-I

[shy atlas]

where did he go tho

@ripe hamlet is that swedish ?

[ripe hamlet]

yes

[hollow finch]

Does anyone have any resources for getting a geometric interpretation of a complex eigenvector (of a real matrix)?
I know it has to do with rotation, so is there some sort of way to find an associated ellipse or something?

[gray dust]

@eager kestrel if you treat u,v as column vectors and follow the rules of matrix multiplication then v^Tu gives the inner product (v,u)

[eager kestrel]

@eager kestrel if you treat u,v as column vectors and follow the rules of matrix multiplication then v^Tu gives the inner product (v,u)
@gray dust yeah i found it out

Although i need to read up on geometric and algebraic multipliciticy in relation to the spectraltheorm and defective eigenvalues

[delicate zealot]

If I send a problem and my work is someone able to help me with a linear algebra problem? It's about solving a matrix equation and then writing the solution parametrically, and when I got down to writing the solution parametrically I got some weird factoring problem. Hard to explain but if I sent my work and wrote some comments in it would someone be able to point me in the right direction at that step?

I think everything else about it is fine

[wintry steppe]

@delicate zealot

[delicate zealot]

I think the question is explained well enough but if there needs to be clarification I can provide it. Reducing the matrix is not the issue, rather it is the parametric definition that follows.

[ocean sequoia]

So if the dim of L(V,W) = dim(V) * dim(W) does that mean that an operator on V has a dim(V)^2? That just seems kinda odd

[wintry steppe]

you should be more precise: "the dimension of the space of linear operators on V (finite dimensional) has dimension dim(V)^2"

a nice way to think about this is to fix a basis of V, and then identify linear operators on V with matrices of size nxn. how many "basic" nxn matrices are there?

and with what linear operators on V do those matrices coincide?

results having to do with subspaces of linear maps are easy to intuit if you fix a basis and think about matrices (im sure that's not the only way)

[delicate zealot]

what do you mean by dim I've never seen that terminology before

[wintry steppe]

the dimension of a vector space

[ocean sequoia]

dim is dimension of a vector space

[wintry steppe]

the number of elements in a basis for the vector space. one can show that this makes sense (all bases have the same number of elements). unsurprisingly, R^n has dimension n, for example

if you accept choice, then all vector spaces have bases, infinite or not. those with finite bases are called finite-dimensional vector spaces, and one nice thing about them is that after fixing a basis, you can just think of the vector space as R^(dimension)

anyways

@ocean sequoia does what i said make sense?

that's my intuition for that result

[ocean sequoia]

no it does

[wintry steppe]

actually im pretty sure that's the proof for that result lol

[ocean sequoia]

im actually working atm but im pretty sure if did an example it would help

and R^2 shouldnt be too hard

[wintry steppe]

cause you can define an isomorphism sending the "all zeros except i,j" to the corresponding linear operator and then since isomorphic spaces have the same dimension you're done

an example? hmm

in R^2 you have a canonical basis, so all your linear maps are multiplication by a matrix

[ocean sequoia]

honestly dont give me on

one

i want to do it myself

ill send it later

[wintry steppe]

so the set of linear operators on R^2 is basically the set of real 2x2 matrices (yes, i know, they technically contain different objects, but they are so similar)

alright

(i hope that doesn't count as an example lol)

[ocean sequoia]

hahahha i wont count it as wone lol

oh

wait

duh

it would have to be the set of all 2x2 matricies

wow

ok well that also makes alot of sense

[wintry steppe]

yeah, and once you go to an abstract vector space, it's basically the same situation after choosing a basis

[ocean sequoia]

yep

[wintry steppe]

you don't actually need to introduce matrices to prove the result that dim L(V,W) = dim(V)dim(W), but you do need to pick a basis of each space. but at that point, you may as well do matrix magic

its a good way to visualize things (if that's even necessary)

[ocean sequoia]

it makes alot of sense in terms of matrix to visualize

which can help

yea

[wintry steppe]

yeah with a matrix it's easy to see how things work in the basis

now here's a related question: if V is finite dimensional and W is infinite dimensional, what's dim L(V,W)? (came up with this just now lol)

(you may assume choice)

[ocean sequoia]

i actually have never worked with infinite dimensional vector spaces : /

but let me try and think

id guess it would be

infinite

[wintry steppe]

yeah

(i haven't actually sat down and verified anything, but ive got a proof outline in my head)

[ocean sequoia]

that kinda makes sense but i know nothing about infinite vector spaces

[wintry steppe]

they are the spaces where rank nullity fails lmao

the most important examples of infinite dimensional spaces are probably function spaces

so you're probably more acquainted with them than you think

also dual space stuff goes wrong (basically anything where you would initially choose a basis goes wrong)

[ocean sequoia]

wait so like rank + null doesnt equal dim?

honestly i kinda was going to guess it was going to be a function-y

like it couldnt be injective or surjective

but i would guess it would still be unqiuely defined because V is finite?

[wintry steppe]

if V is finite dimensional, yes, rank T + nullity T = dim V for any linear map T : V to W

im not sure what you mean by "it couldnt be injective or surjective"

what is "it"?

[ocean sequoia]

Honestly nvm lol

I was trying to think of your example but I’m out of my depth

[wintry steppe]

dont be afraid to call me out if im saying things that go past what you know lol

also i thought about it a bit harder, and im not quite sure how to prove the question i came up with

maybe i can ask it here

[ocean sequoia]

i appreciate htat

btw just because i have always been shaky on my definition when people say uniquely determined on a basis means that each element of V for T(V) then each element gets taken to a unique element of W?

and the inverse must be true as well? So if we L(V,W) then each element of W can only be expressed in one way from V?

[wintry steppe]

do you mean in reference to a linear map @ocean sequoia

[ocean sequoia]

yes

i see that thrown around a linear map is unqiuely determined on its basis and i feel i sorta understand it

but not exactly

[wintry steppe]

do you want me to state the exact result

and explain how it implies that

[ocean sequoia]

yea

[wintry steppe]

ok one moment

if $v_1, \dots, v_n$ is a basis for $V$ and if $S,T : V \to W$ are linear maps (the dimension of $W$ doesn't matter here) then if $S(v_i) = T(v_i)$ for each $i$, then $S = T$

[stoic python]

TTerra:

[wintry steppe]

which can be interpreted as: if you have a linear map $T : V \to W$ and any other linear map $S : V \to W$ which agrees with $T$ on a basis of $V$, then $S = T$

[stoic python]

TTerra:

[wintry steppe]

as for constructing linear maps, you can do as follows: if $v_1, \dots, v_n$ is a basis of $V$ and $w_1, \dots, w_n$ are any vectors in $W$ (so long as there are $n$ of them, it doesn't matter if they're linearly independent or not), then define $$T(\sum a_iv_i) = \sum a_iw_i.$$ this can be shown to be the unique linear map sending $v_i$ to $w_i$

[stoic python]

TTerra:

[spice storm]

My linear algebra class over the summer isn’t going to cover change of basis. Am I going to miss anything important?

[wintry steppe]

the channel is kind of occupied

please wait a moment

anyways

[spice storm]

take your time. No rush. Take 20 days if you like.

[wintry steppe]

i didnt mean to come off as rude

@ocean sequoia this can all be taken as "if you know a linear map's values on a basis, you know it's values everywhere"

thonk me if i made a mistake

@spice storm change of basis is rather important, but there isn't a great deal to it so you can definitely cover it yourself

[gray dust]

if you covered linear maps and their matrix representations wrt some bases, then that's almost enough to cover change of basis. the change of basis matrix P from bases A to B is the matrix of the identity map wrt A & B, ie for any vector x, P acting on the coords of x wrt A produces the coords of x wrt B @spice storm

[wintry steppe]

wait, does the "wait until the channel isn't occupied" rule hold in the specific math channels as well? i know it's there for the question/help channels but i don't want to try to reference a rule that might not exist

[gray dust]

it IS a rule for all channels but it's one of the least followed and hardest to enforce in practice

[wintry steppe]

BUT please make sure that no one else is being helped before you jump in! if someone is already being helped, try another "questions" channel and ask your question there. 

as a general server rule

yeah

[gray dust]

rule2 specifically

[wintry steppe]

was just making sure, thanks

[gray dust]

i take that back waiting 15min before pinging helpers is the hardest to enforce

[gritty frigate]

Two R3 vectors can always exist on the same plane ??

[spice storm]

@gritty frigate did you end up proving your statement from yesterday?

[gritty frigate]

Oh which one ?

[spice storm]

Matrix det.

[gritty frigate]

i think I did

I saw some articles

Its the number we get if we dow Gauss-Jordan method

For what I understood

If I use common sense... Two vectors on R3 can always be considered on the same plane...

the problem comes when a third R3 vector comes into place. Am I right ?

[ocean sequoia]

@wintry steppe thank you that makes sense especially the S = T if S(vi) = T(vi)

[wintry steppe]

there's only a problem if you require three linearly independent vectors in the same plane

[ocean sequoia]

sorry for the late response i got pulled to help teach on the floor for a few minutes

[wintry steppe]

if you havent seen that theorem before, then you can try proving it yourself

its not a very tricky proof, basically just right from the definitions

(this is in response to brzig)

[ocean sequoia]

yea i just wanted to make sure i understood it correctly

i think i did awhile ago but i felt a bit shaky

and it seemed better to ask

[wintry steppe]

sitting down and trying to recall things you did way before is a really good way to review and remember stuff

[ocean sequoia]

yea and it seems pretty dang important

i see it mentioned all the time

[wintry steppe]

i think the other day you were asking about dual spaces

iirc that comes up here

[ocean sequoia]

yea which is why i wanted to refresh my self

[stoic python]

TTerra:

[ocean sequoia]

but to put this into my own words to make sure its correct

if v1,...,vn is a basis of V and w1,...,wn is a basis of W and S(vi) = wi then any transformation L(V,W) that sends vi to wi will be equal to S(vi)

sorry i just dont want to reguirate definition wanted to try and say it in my own correct way to make sure i get it

[wintry steppe]

remember that this result doesnt actually rely on W having a finite basis

so you can state it a bit more generally

BUT

what you've written seems alright, but i'd change the very last part to "equal to S" and i'd also write "any transformation in L(V,W)"

[ocean sequoia]

ok perfect thanks!

[opal osprey]

hello everyone

I want to write this matrix as PDP^-1

I understand I must use the eigenvectors as base, etc

but

is there any way I can calculate the eigenvalues without tackling this 4x4 matrix?

some line switching jumbo or something?

because lambdas will ruin the 3rd row and I was aiming for that one for the laplace determinant

[slow scroll]

maybe not the best solution, but I would personally reduce one of the columns and then do cofactor expansion

[opal osprey]

hmm

the top two lines are great for laplace

I'm not sure I know what cofactor expansion is

[slow scroll]

I'm pretty sure cofactor expansion is what you call laplace.

We know how row operations affect the determinant. In particular, adding a multiply of a row to another does not change the determinant. So I would reduce the first column w/ gaussian elimination, and then do laplace on the first column.

[opal osprey]

yeah, that's it!

trying it rn

thank you so much!

[slow scroll]

npnp

[pallid rampart]

I don’t quite understand what this sentence is saying: “In the case of a single dxd Jordan block with λ=0, these dimensions are ...”

[pallid rampart]

<@&286206848099549185>

[wispy cape]

can you help with these problems ?

[wintry steppe]

@wispy cape what do you need help with in particular?

[wispy cape]

@wispy cape what do you need help with in particular?
@wintry steppe question 3

[wintry steppe]

@wispy cape i assume that "equally inclined" here means that the plane has a slope that is equal respect to x, y and z axes

ill draw a diagram to make more sense of this

[wispy cape]

can you go DM

[wintry steppe]

sure

[wispy cape]

so I follow

[gritty frigate]

Whats the name of this:
AxB.C

A,B,C = R3

[empty copper]

a random sequence of symbols

[quartz compass]

scalar triple product

[gritty frigate]

scalar triple product
@quartz compass Thanks hahaha

[quartz compass]

which is just the determinant

[gritty frigate]

Yep

[quartz compass]

yeah you're welcome lol

[gritty frigate]

That formula is the area of a 3D figure created by the 3 vectors. What is its name (I do not speack english)

The area is made of the absolute value of it*

What is the name of that figure in english? hahahah

[shy atlas]

slanty slanty cube

or parallelepiped as the virgins call it

[pallid rampart]

@quartz compass do you have any idea what the sentence “In the case of a single dxd ...” means?

[pallid rampart]

Ah I think I got it

[ocean sequoia]

So since L(V,W) is a in fact a vector space does that mean there is a basis for for L(V,W)?

I didnt see that mentioned but given that a dual space is a linear transformation from V to F should and we can have a basis for the dual space would there be a basis for L(V,W) as well?

[pallid rampart]

Let $(v_1,\dots,v_n)$ be a basis for $V$ and $(w_1,\dots,w_m)$ be a basis for $W$ (when these vector spaces have infinite dimensions is pretty similar), we can let $$T_{a,b}(v_{i})=\delta_{a,i}w_b$$ then all the $T_{a,b}$, where $1\leq a\leq n$ and $1\leq b\leq m$, form a basis for $L(V,W)$

[stoic python]

Whoever:

[ocean sequoia]

im only wworking with finite spaces atm if that makes it easier

[pallid rampart]

Yeah it is finite dimensional

Also this didn’t come out of nowhere

[ocean sequoia]

no i figured there would be a basis

[pallid rampart]

So L(V,W) can be represented as matrices, and a natural basis for the set of mxn matrices is where we have 1 on a single entry and 0 everywhere else

So the linear transformation I defined there, when represented as a matrix with respect to the basis v and w, is exactly a matrix where we have 1 in one entry and 0 everywhere else

[ocean sequoia]

yea ok that makes sense

because any transformation for example from R^2 to R^2 could be shown as a linear combination of identity matrix

[pallid rampart]

[ocean sequoia]

maybe not?

hahaha

[pallid rampart]

There is only 1 identity matrix

So if every linear transformation is a linear combination of identity matrix then the dimension of L(R^2,R^2) is 1

Which is

Only slightly incorrect

[ocean sequoia]

yea cause it needs to be 4 right?

[pallid rampart]

Yeah

Wait are you talking about a matrix with 1 on one entry and 0 everywhere else?

Was that what you meant by “identity”?

[ocean sequoia]

no

i did mean what you said i was wrong

[pallid rampart]

ok

[ocean sequoia]

yea rip

[ocean sequoia]

sorry for the ping not sure if this is rude but when you said

Wait are you talking about a matrix with 1 on one entry and 0 everywhere else?
@pallid rampart this means that if we had four 2x2 matrices of a one in each corner that would be a basis for the operator?

this btw is not the same thing as finding a basis for the range of T right?

[pallid rampart]

Not a basis for an operator, a basis is always for a vector space, and in this case the 4 matrices form a basis of the set of all 2x2 matrices

And yeah it’s not the same thing as finding a basis of the range of T

L(V,W) is the set of all linear transformation from V to W

[ocean sequoia]

Not a basis for an operator, a basis is always for a vector space, and in this case the 4 matrices form a basis of the set of all 2x2 matrices
Isn't an operator a linear transformation so it is a vector space? and isn't that then a basis for L(R^2,R^2) b/c isnt every linear transformation from R^2 to R^2 a 2x2 matrix?

[pallid rampart]

Um

How is a linear transformation a vector space

The set of all linear transformation form a vector space

[ocean sequoia]

@pallid rampart isnt that what that is saying

or does it mean the set of all linear transformations is a vector space?

[pallid rampart]

L(V,W) is the set of all linear transformations

all

[dusky epoch]

from V to W

[ocean sequoia]

ok i read into that too much

[pallid rampart]

Ah yes from V to W

[ocean sequoia]

ok so I need to be more precise

but would the set of four 2x2 matrices of a one in each corner that would be a basis for the set of all transformations from R^2 to R^2?

[dusky epoch]

strictly speaking, no. because a transformation from R^2 to R^2 is not the same as a 2 by 2 matrix.

[ocean sequoia]

what would be the basis then

because if its a vector space it would have one

or could i have a hint to try and figure it out

ive tried googling it and going back through the book but i cant find anything

[dusky epoch]

"the" basis

ah yes

of course

[ocean sequoia]

a basis

sorry :/

[pallid rampart]

Well I gave you a basis

[dusky epoch]

well if you identify each 2 by 2 matrix A with the linear map (x ↦ Ax) : R^2 -> R^2

then your set becomes a basis

[ocean sequoia]

when you say your set you mean the set i gave above correct?

[dusky epoch]

yes

[ocean sequoia]

Ok thank you both

I appreciate the help

[shy atlas]

a fellow doer of axler

[pallid rampart]

I also did axler

But I didn't use the last edition

[shy atlas]

so yknow how this gives u the solution to the whole least square thing

why is this a much easier way of solving least square than $Ax = $ proj$_{\text{ran A}}b$

[stoic python]

soαρ:

[shy atlas]

just project it into the column space of A and solve

[ocean sequoia]

@shy atlas are you going through it on your own?

[stoic python]

AfterJack:

[slow scroll]

@shy atlas wdym? the A^T Ax = A^T b comes from Ax = proj_{ran A} b

[tame mural]

What does orthogonality tell you above independence?

[wintry steppe]

nonzero orthogonal vectors are linearly independent. the converse is very false

a counterexample for the converse should be easy to come up with

to prove the thing i said, just follow the definitions

[tame mural]

thanks!

[wintry steppe]

Linear independence requires the vectors to not be colinear

So if they arent colinear there is an orthogonal component

So orthogonality is a requirement for linear independence in a sense

[tame mural]

oh hmm I see

[wintry steppe]

you can think in R^2 and R^3 for intuition on orthogonality

you can say some things about orthogonality in infinite dimensional vector spaces (there may be a thing or two to be said about an orthogonal set of functions...)

but if you picture things in R^2, then, as robo said, orthogonality should "obviously" imply linear independence

[ashen terrace]

Hey yall, Im a little new to proofs within linear algebra and I ran across this problem:

[dusky epoch]

this is kinda hard to read

Suppose $m$ is a positive integer. Show that for each polynomial $q \in \mathcal{P}_m(\bR)$, there exists a polynomial $p$ such that $q = \dv[3]{x}( (x^3+2x^2+1)p)$

[ashen terrace]

Oh thank you!

My question is about the nature of this question.

[dusky epoch]

oh wait

that's a triple prime, not an m

jeez lmao

[stoic python]

Ann:

[dusky epoch]

wym by nature

[ashen terrace]

After a really long period of solving, this statement appeared to be false. I don't know why this exact problem is worded like this, especially if its statement is false. Did the problem maker make a mistake?

The error is a dimensional error

Sorry, im not sure if my question makes sense

[dusky epoch]

wym by dimensional error

the operator $T : \mathcal{P}_m(\bR) \to \mathcal{P}_m(\bR)$ defined by $Tp = \dv[3]{x} [(x^3+2x^2+1)p]$ can be shown to have trivial kernel, and since $\mathcal{P}_m(\bR)$ is finite-dimensional it'll follow that $T$ is an isomorphism and so has an inverse and then for your problem you'll have $p = T^{-1}q$

[stoic python]

Ann:

[dusky epoch]

the statement is true

[ashen terrace]

The one you just posted?

[dusky epoch]

the statement of the problem

what i wrote there is basically an outline of the solution lol

[ashen terrace]

Is it?! 😢

[dusky epoch]

yes

[ashen terrace]

Maaan

[dusky epoch]

it remains to show the following:

• T is linear
• T has trivial kernel

which are both things a linear algebra student ought to know how to do

[ashen terrace]

Yep! Im not quite there yet

ill have another go at it. Thank you!

[viscid kernel]

How come the rowvectors here span a 2d space instead of three ? I dont see any rowvector which is in the span of another rowvector.

[dusky epoch]

[2,1,5] = (8/7) * [1,2,4] + (3/7) * [2,-3,1]

[viscid kernel]

How did you find it ?

[dusky epoch]

just because your vectors aren't mutually parallel doesn't mean they form a linearly independent set

i found this by solving a linear system

[viscid kernel]

So, did you do this. ==> you transponed the matrix then you did gaussian elimination with your matrix. By setting your matrix equal to zero. You turned it into a reduced row echelon form and then you got the answer. Am I right ?

[dusky epoch]

no, i plugged it into WA bc i don't really feel like doing arithmetic while on the go

[viscid kernel]

Aight

Thanks

But the way Do you normally solve it the way I said ?

[dusky epoch]

eh

maybe? idk if i had some paper i would have probably just written out directly the system (row 3) = x * (row 1) + y * (row 2)

[viscid kernel]

This way

I made the rowvectors columnvectors bu transponing it

[half ice]

@viscid kernel
Don't transposition it

The way 3b1b wrote it was correct

[dusky epoch]

the thing is baklava i see no need to constrain myself to any particular way of arranging the numbers on a page

[viscid kernel]

@half ice yeah but is, is that Im looking if the rowvectors are independent not the column vectors. Thats the reason of transposition.

[half ice]

OH I apologize haha. Then yes, what you're doing makes sense

If you're just checking for independence, then the block of 0s is arbitrary. We typically leave it out

@viscid kernel

[viscid kernel]

@half ice alright, you scared me for a bit. 🙂

[flint cape]

any idea how to solve this please

If f(3) = 0.1 and f(3.1) = 0.35 for a function f then what is the slope of f at x = 3?

[half ice]

Know the "rise over run" formula for slope?

I'll happily answer this here this time, but in the future know that this isn't linear algebra, despite it being a question about algebra on lines. Math is fun like that

[flint cape]

apology for that

[half ice]

So do you know the formula?

[flint cape]

yes, I know rise over run

[wintry steppe]

How did my prof know to do a[2] - 4*a[1] ?

[shy atlas]

cuz u wanna get 0 under the 1

do u know how gaussian elimination works

[wintry steppe]

I thought it was already in rref

[shy atlas]

uhh no

do u know what rref form means ?

google the definition ig

[wintry steppe]

All non zero rows are above any rows of all zeroes .. check
each leading entry is in a column to the right of the leading entry of the row above ... check
all entries in a column below a leading entry are zeroes ... check

the leading entry in all non zero row is 1 ... check
each leading 1 is the only non zero entry in it's column ... check

Plus he used the command rref() - how would that not be in in rref form lol

[shy atlas]

each leading 1 is the only non zero entry in it's column ... check

theres a goddamn 4 in there

that dont look very 0 to me

[stiff frost]

I think you're talking about two different matrices and hence talking past eachother

[wintry steppe]

Ya I just noticed we are

My prof isn't using the rref(a) - he's using just a

So that means he only needed the echelon form (hence the a[2] - 4*a[1])

What's the rationale behind solving 8 - 4h = 0 ?

Why would he set it = 0 ? I thought it would be 8 - 4h = k - 8 ?

[ocean sequoia]

Hey so if we have a x that is a sub space of V such that dim(x) < dim(V) and we are trying to map x to w and we know dim(x) = dim(w). So let T:x->W be a linear transformation from x to W does this imply T:V->W?

No because we don’t know how T functions on a basis of V?

[wintry steppe]

@ocean sequoia yea its not the same mapping

Ex: take V = R^3, x to be a plane in R^3 and W to be another plane in R^3

[ocean sequoia]

@wintry steppe thanks yea! That’s what I said makes me so happy to know I got that right lol

Makes me feel like I might actually be starting to get my head around some stuff

[hazy gull]

is span, column space, range and image all just different ways to represent a similar idea?

[limber sierra]

uh... i mean they're all vector spaces (assuming you're using the convention where "range" and "image" mean the same thing)

and the span of a matrix's columns - i.e. its column space - is the image/range of the transformation it represents

but "span" by itself is a far more general concept

than this

the other 3 ideas are all essentially the same thing though yes

(at least in the context of matrices/linear transformations, that is)

(range/image are concepts that apply to all functions; for nonlinear functions, the concept of "column space" doesnt make sense - but this is #linear-algebra so i'm assuming you're working with linear stuff)

[hazy gull]

well I did have a chapter that went over range of nonlinear functions, even though it is a linear algebra class.

[limber sierra]

then again, in the context of arbitrary functions, the idea of "column space" doesnt make sense

[hazy gull]

but ya

[limber sierra]

whereas range/image does

but yeah this sentence summarizes the relation:

the span of a matrix's columns - i.e. its column space - is the image/range of the transformation it represents

[hazy gull]

thanks thats helpful

[limber sierra]

again all these concepts (except for column space) can be generalized

but in this specific context

that's the relation

[gray dust]

this is #prealg-and-algebra

[prisma smelt]

oh sorry

[humble oak]

hello, say for example from a characteristic equation i get (x-1)x^2, this does not have distinct eigenvalues right?

[umbral smelt]

Yeah

Because the multiplicity of eigenvalue 0 is 2

[shy atlas]

is the matrix gonna be diagonalizable tho

i feel it is gonna be diagonalizeable

[humble oak]

no

at least i don't think so 🤔

[shy atlas]

well i think so

[umbral smelt]

🤔

[eager burrow]

No, it doesn't need to be

[humble oak]

🤔

[eager burrow]

the matrix with entries (1,0,0),(0,0,1),(0,0,0) is an example

[humble oak]

ooh another quick question, if i was given a matrix and i had to show whether it is diagonalizable or not. it is fine for me to first convert the matrix to rref THEN determine if it's diagonalizable right?

[shy atlas]

@eager burrow hm so why is that matrix not diagonalizable ?

,w diagonalize {{1,0,0},{0,0,1},{0,0,0}}

ic

[stoic python]

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[eager burrow]

it's kind of a standard fact; this matrix is in jordan normal form, meaning that it's "as diagonalized as can be"

[shy atlas]

theres a lot of equivalent condition to diagonalizability. u can check them if u want

oh jordan form is like chapter 9 or smth

im at 7

[eager burrow]

ye you'll get there

once you know that, you can very easily give examples of matrices which aren't diagonalizable

and basically, every characteristic polynomial with a zero of multiplicity greater 1 can be attached to a non-diagonalizable matrix

[shy atlas]

ic thats p nifty

[wintry steppe]

"as diagonalized as can be" i love that phrasing

[wintry steppe]

It's easy to understand intuitively that (x,y) is being scaled linearly and all such points on a circle would become ones on an ellipse, but I don't see any obvious rigorous reason

Oh wait

(x,y) -> (ax,by), a^2x^2/a^2 + b^2y^2/b^2
Those are just different xs and ys in the book

[opaque laurel]

Hi guys, just a small question but the following transformation matrix only has shearing, correct?

[grim bridge]

that's a scaling and rotation matrix

[opaque laurel]

How did you tell?

Scaling because it's 2 I assume but rotation?

[grim bridge]

well visualise it

[gray dust]

Note how A acts upon the standard basis

[grim bridge]

it's simple, that's 2D

positive x ends up at positive y, positive y ends up at negative x

that's a 90° counter clockwise rotation

[opaque laurel]

are you sure about it being 90° though? Since if you would take (3,5) it ends up being (-10,6) which is not a 90° rotation

It would be a rotation but not a 90° one

[grim bridge]

that's a scaling and rotation

and yes it's 90°

a rotation can't be multiple angles at once

if you proved it for the basis vectors, if you prove a different result then you're wrong

notice that 3x-10+5x6 = 0

those two vectors are orthogonal

[opaque laurel]

So even when drawn and it's not a 90° angle the calculation is leading?

[wintry steppe]

Is parametric vector form === parametric equation ?

For example

[1 - 4t, -3/4 + 3t, -1/4 + t]

that's in parametric vector form ?

[cursive island]

Hey, could someone maybe explain what column space means to me?

Is it just the span of the vectors within the matrix?

[shy atlas]

span of the columns of the matrix, yes

[cursive island]

so if we have a the matrix

Is the column space 2d or 3d?

[shy atlas]

well we have 2 linearly independent vectors spanning the column space

what does that tell u about the dimension of the column space

[cursive island]

I would say its 2d

[shy atlas]

yup

[cursive island]

It just confuses me a bit because there are 3d coordinates

[shy atlas]

yeah i can see the confusion but just recall the definition of dimension ig

its just a minimal spanning list

[cursive island]

Yeah I think I kinda get it

[shy atlas]

it doesnt matter how many coordinates u have

[cursive island]

its like a 2d plane in a 3d system

[shy atlas]

you can reach any vector in the column space just using these 2

[cursive island]

Cool gotcha

Alright ty

[shy atlas]

np

[pallid rampart]

Let A,B be complex valued nxn matrices, then is it possible that AB-BA is invertible?

[wintry steppe]

here's an example i came up with

[pallid rampart]

Thanks

I might be too lazy to find matrices myself

But thanks

[wintry steppe]

the matrix with 1s on the antidiagonal and 0s elsewhere is a good one for coming up with examples

because its particularly easy to see what it does to matrices by left/right multiplication

left multiplication swaps all the rows and right multiplication swaps all the columns

might come in handy if you need a (counter-)example in the future

[pallid rampart]

Nice

[torn marten]

What does it mean to suppress a column vector from a matrix?

[wintry steppe]

@torn marten i think it means removing the column vector from the matrix

So A1 would contain all the column vectors from $a_{2}$ to $a_{n}$

[stoic python]

robo™:

[torn marten]

Thank you, I'll see if this proof makes sense with that interpretation

[wintry steppe]

How can I check that my vector equation is the right answer ?

Is this a difficult question ?

[wintry steppe]

what is the formal definition of direction in linear algebra

in R^2 and R^3

[limber sierra]

choose a basis (probably the standard basis); direction is the angle from those

"angle" here traditionally defined by using the standard dot product

[wintry steppe]

ok but

consider v = -w

do these have the same direction?

(where v, w are nonzero vectors)

they form the same angle

(or do they?)

[limber sierra]

do they??

[wintry steppe]

at least, they're on the same line with the same direction vector

i.e. if you formed a line with either vector

they'd be on that line

[limber sierra]

suppose v = [1, 1] and w = [-1, -1]; then

,w angle between {{1}, {0}}, {{1}, {1}}

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[limber sierra]

,w angle between {{1}, {0}}, {{-1}, {-1}}

[stoic python]

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[wintry steppe]

that's not (1;1) and (-1;-1)

?

[limber sierra]

what

[wintry steppe]

what

[limber sierra]

i'm saying that they form a different angle from a basis vector (1, 0)

if you want the angle between v and -w

,w angle between {{1}, {1}} and {{-1}, {-1}}

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[limber sierra]

but you could probably have guessed that from the two images

[wintry steppe]

oh wait, i see what you mean

so when we refer to direction

we do actually care about the actual, physical way they're pointing

because when we learned about projections

direction did not matter

as in, the way something was pointing

[limber sierra]

sure, i mean

direction is ultimately "arbitrary" in that it depends on your choice of basis vectors

but we usually choose the standard basis

ie {(1; 0), (0; 1)} or {(1; 0; 0), (0; 1; 0), (0; 0; 1)}

[hollow finch]

it also depends on the definition of the inner product, doesnt it?

[limber sierra]

sure but when referring to R^n i'm assuming we're taking the standard dot product

and hence are talking about euclidean direction

[hollow finch]

got it my b

[limber sierra]

nah its a valid question

if you change your inner product, you change the "spatial" relation between vectors

though that said

two vectors in the "same direction" in one inner product

will always have the "same direction"

in any inner product space

its just a matter of how the different "directions" are related, i.e. how we label the directions

that an inner product determines

to be more clear here:

"direction" of nonzero vectors is actually the equivalence class they fall into under positive scalar multiplication

[hollow finch]

when i was a course embedded tutor for linear algebra last spring, the professor demonstrated how the relationship between vectors changes by drawing the unit "circle" with different inner product definitions. a lot of people lost it when he was calling what we normally call a diamond or square a "circle".

[limber sierra]

if you have a normed space, then dividing a vector by its norm gives you a unit norm vector

which is usually how we canonically refer to direction

(this definition actually only makes sense over ordered fields, but it can be amended to work more generally)

see https://ncatlab.org/nlab/show/direction+of+a+vector

[wintry steppe]

@limber sierra what book do you recommend

LADR or LADW

or Strang's book

[limber sierra]

im not really qualified to give recommendations there im afraid

havent really used any of them

[wintry steppe]

why not what did you use

[limber sierra]

lecture notes

and the sections of artin on vector spaces for practice problems

[hollow finch]

so i got this as the solution to a problem i was working on and i was wondering what the heck this means geometrically

like how am i supposed to interpret this

it appears to form the ellipse which is the solution to the 2x2 first order system of differential equations with constant coefficients x'=Ax, where v is a complex eigenvector of A associated with a complex eigenvalue with a zero real part

but why would that be

[wintry steppe]

Suppose that $\textbf{u}\cdot \textbf{v} = 2$ and $\textbf{v}\cdot \textbf{w} = -3$. What are the possible values of $\textbf{u}\cdot \textbf{w}$?

[stoic python]

polynomial:

[wintry steppe]

rip kaynex

<@&286206848099549185>

any people

[limber sierra]

are these just arbitrary vectors? or do you have any other information about the space

[wintry steppe]

R^2

@limber sierra

[limber sierra]

ok then we're solving a system

[wintry steppe]

yeah i tried but feels like there's not enough info

[hollow finch]

no need for a system i dont think

[limber sierra]

theres alternate approaches

[hollow finch]

use the cosine definition of the dot product

[limber sierra]

thats what i meant

we have that $\abs{u}\abs{v}\cos \theta_{uv} = 2$ and simialrly $\abs{v}\abs{w}\cos \theta_{vw} = -3$, and we want to determine what $\abs{u}\abs{w}\cos \theta_{uw}$ could possibly take

[stoic python]

Namington:

[wintry steppe]

yes

[limber sierra]

but you can rewrite $\theta_{uw}$ in terms of $\theta_{uv}$ and $\theta_{vw}$

[stoic python]

Namington:

[limber sierra]

(think about how - use a geometric intuition)

you can then do some algebra on that

[wintry steppe]

uh

no i can't?

like there is casework or

[limber sierra]

this still works if the vector w is "between" u and v, btw

[wintry steppe]

yeah

that's what i meant by casework

i can't really do too much

[limber sierra]

at least modulo 360

note that the cosine of the yellow angle, and the blue+green angle, will be the same

since cosine has a period of 360 degrees (or 2pi radians)

so in any case, its safe to write $\cos(\theta_{uw}) = \cos(\theta_{uv} + \theta_{vw})$

at least for two-dimensional vectors

[stoic python]

Namington:

[limber sierra]

forgot cosines

bleh

[wintry steppe]

you also forgot mod 2pi

lol

[limber sierra]

no as i said

its fine since cosine "eliminates" 2pi terms