linear-algebra

[clear spoke]

How would you solve for x

x is a scalar

when I try to find x i don't get a scalar result...

[dusky epoch]

what is this

[clear spoke]

vectors

[dusky epoch]

... wat

[clear spoke]

vector algebra

[dusky epoch]

what is this notation

[clear spoke]

WHat exactly do you mean?

[dusky epoch]

i don't know what the notation $5 + 5\angle{90}$ is supposed to refer to

[stoic python]

Ann:

[dusky epoch]

sorry for the late reply

@clear spoke

[obsidian jackal]

how to solve the second one?

I tried solving it by doing the dot product then multiplying the dot product to the vector

the program though is asking for a scalar not a vector

[gray dust]

program broken. (u dot v)u is a vector

[obsidian jackal]

oh

well there goes my 1%

thanks for the answer

[gray dust]

your "wrong" ans is right so don't thank me

[obsidian jackal]

okay

[south wadi]

No idea where to start

can't find any youtube videos for this

[gray dust]

The image of u under T, is written as T(u)

By T’s definition, T(u)=Au

[south wadi]

Rokabe

what do i do with the v vector given

am i solving for T(x) = v

[gray dust]

The image of v under T is written as T(v). By T’s definition, T(v)=Av

[south wadi]

then i solve the matrix?

bro im so confused

this is the only thing im given

[gray dust]

do you know how to read T(x)=Ax?

[south wadi]

im just going off this'

no i dont

[gray dust]

go back to algebra

do you know how to read f(x)=2x?

[south wadi]

dude its just a linear function

yes

i know how to read it

[gray dust]

the idea is similar to reading T(x)=Ax

[south wadi]

ok

[gray dust]

in this case x is not a real number but a 2d vector

[south wadi]

right

[gray dust]

did you master matrix multiplication beforehand

[south wadi]

yeah dude im a master

[gray dust]

The image of a 2d vector x under the transformation T is written as T(x). By T’s definition, T(x)=Ax

[south wadi]

so are wwe doing

[2 0][a]
[0 2][b]

[gray dust]

that's T(v)

[south wadi]

We want T(x) = Ax

ive already solved for T(u)

[gray dust]

you showed me how to get T(v). do it

[south wadi]

holy fk my messages just dont load

my discord is broken

test

tes

a

and got [2 ]
[ -6 ]

t

finally

o

WHAT THE FUCK

?

im good

a

and got [2 ]
[ -6 ]

a

t

[gray dust]

result isn't right

[south wadi]

wtf

g

a

[gray dust]

result isn't right

stay put til discord devs fix this

[south wadi]

wtf

[gray dust]

stay put til discord devs fix this

[south wadi]

o

wtf

[gray dust]

stay put til discord devs fix this

[south wadi]

am i back

yeet im back

@gray dust

where were we

[gray dust]

T(v)

[south wadi]

@gray dust

sorry back

so t(v) = Av

t(v) = 1 -3
3 5 * x1 x2
-1 7

so i just mutliply in?

t(v) = x1 + 3x1 - x1 -3x2 + 5x2 + 7x2

also why is there no youtube vids on this shit

fk worgn question

a matrix is wrong

@gray dust

so is it just A matrjx time vector b

[gray dust]

you're making this more complicated than you need

T(v)=Av, aka matrix A times vector v

[wintry steppe]

is anyone available to help with this

[gray dust]

go to #real-complex-analysis

[wintry steppe]

kk thx

[south wadi]

yeah i got it now rokabe

[junior nacelle]

wait

is complex vars

a new channel?

[gray dust]

no

[junior nacelle]

huh

[deep swan]

What the fukc is a pivot

What does it actually mean

[south wadi]

rokabe

do u u do square brackets or parenthesis

[dire bluff]

tensors

if we have a tensor T :V x V -> R

i can see that it could be decomposed into symmetric and anti symmetric parts

[vast torrent]

That's true but not trivial

Oh wait

Is it just uhh (B-B *)/2+(B+B *)/2

Let me google it

Yeah that's it

Okay continue

[dire bluff]

so we can decompose it likethat

by some manipulation that doesnt take into account any basis

[vast torrent]

I'd hope so

[dire bluff]

i think T(x,v)+T(v,x) would be the symetricc kinda stuff and whatever

that is your B* here?

[lilac rampart]

Me wishing I remembered my linear algebra rn. Figured I'd stop here to ask a quick question if that's cool. Is [-1, 0, 1] invertable? I fear it isn't but I'm kinda screwed if that is the case.

[dire bluff]

so, i saw in a book that the symmetric part could also be further decomposed into a "traceless" part and a pure trace part

[vast torrent]

B* is B with the arguments switched in my statement

[dire bluff]

but that doesnt make much sense to me because theres no trace for tensors liek this one right?

[vast torrent]

Not sure what that means, traceless part

Take a photo?

[dire bluff]

like, a tensor with trace 0 plus a tensor with any trace

not any trace but the trace of the original tensor

more likely

[vast torrent]

Well if you have

[dire bluff]

the book is a physics book so im a bit skeptic

[vast torrent]

Gonna assume for simplicity the vectors in each argument are all the same size

[dire bluff]

you can't define a trace for a tensor that takes in 2 vectors like this one right?

[vast torrent]

Riesz says that any bilinear form is characterized by a matrix B(u,v)=<u,Av>

= u^T A v = u^T(A1+A2)v

Where A2 pulls out the diagonal entries and A1 has 0s on the diagonal

[gray dust]

@south wadi i used square a lot back then, nowadays curvy parens

[dire bluff]

if it helps, this discussion comes up when discussing SO(n) groups, so we should have some kind of natural bilinear form to rely on?

[vast torrent]

Doesn't what I said give you what you wanted?

[dire bluff]

i dont know. would that trace survive a coordinate change?

[vast torrent]

I don't think it does

Because edit:fuck, beaten

Well.

Let's use the other definition of trace

Sum of eigenvalues

Now A1 and A2 are well defined if we insist on using jordal normal form

I'm not happy with this answer though

[dire bluff]

hmm. i mean, you can't define eigenvectors for a tensor like this one?

[vast torrent]

This is a bilinear form

[dire bluff]

because the tensor acting on a vector gives you a linear functional, not a vector

right?

[vast torrent]

Nevertheless, by Riesz, there's a matrix A such thst B(u,v)=u^T A v

I really don't want to define anything in a way that isn't basis invariant

[gray dust]

@lilac rampart one may talk of left/right invertibility of squares/nonsquares, but one can only talk of general invertibility of just squares

[vast torrent]

So i don't like my solution of using the JNF and then splitting into a diagonal matrix + an upper triangular nilpotent matrix

Which is why I am passing the baton over to rokabe

After giving him a hearty pat on the back

[gray dust]

wait what no i'm only here to give terse answers to buried questions

[dire bluff]

this is part of the stuff im talking about

[vast torrent]

hides

[dire bluff]

lol

thanks anyway lips

[gray dust]

throws baton back to fauxpas with incredible force

[dire bluff]

i kind of see it this way. if the vector space has anything to do with the group SO(3) for example, we need to have a bilinear form to talk about the size of vectors and so on, right?

and such a bilinear form would give us a kind of isomorphism from V to its dual?

im not sure what properties the bilinear form would need to have for an orthogonal group to be possible

but a basis on the dual space would kind of give us a way to switch from a tensor that takes 2 vectors to a tensor that takes a vector and a dual

and those have a trace right?

[dire bluff]

is this the place to ask (very basic) questions about representation theory?

[pallid rampart]

Linear transformation from vector spaces V->W require V and W to be vector spaces on the same field right?

🤔 I have never thought about this

[dire bluff]

i think that if you need L(a*v) = a L(v)

for linearity, then a must make sense in both the V and W vector space

so it should be the same field?

[pallid rampart]

Yeah

That's what I thought

It wouldn't make sense to have L(a*v)=a*L(v) if V and W have different scalar multiplication

[dire bluff]

Hoffman defines it like this:
Definition. Let V and W be vector spaces over the field F. A linear
transformation from V into W is a function T from V into W such that (yadda yadda)

[sonic osprey]

@dire bluff sure

or #groups-rings-fields if you want

[dire bluff]

ok, thanks! i'll drop em when they come to me

right now i think i did it all wrong because i did everything in reverse.

i think a representation of a group like SO(n) on any old vector space would in turn give me a bilinear form (at least up to a unit) if i define angles between elements of V by checking which rotations map which vectors into one another

[restive shuttle]

im confused with how to even come up with a solution space for this

is it just a straight line?

how would i represent it and then how would i get the basis for it?

nvm

i think ive figured it for the most part

[gray dust]

is it just a straight line?
so whatcha think?

[pallid rampart]

If subspaces are straight lines, then what is the line formed by the subspace of polynomials with degree lower than m 👀

[restive shuttle]

idk

but i said straight line

cus i kinda felt like it

[gray dust]

you said you got it so i'm curious

[restive shuttle]

i got it in the sense i have an answer that i think ill go with

so it's likely still very wrong

[agile gyro]

It will be the xy plane

I think

[gray dust]

It will be the xy plane

[restive shuttle]

reee

[leaden ermine]

Hi, if I have a term a(b)(b_transpose)a(transpose) where a and b are vectors, and I take the gradient with respect to b, I am getting 2 (a_transpose) (b_transpose) (b). is that correct?

[dusky epoch]

$abb^Ta^T$?

[stoic python]

Ann:

[leaden ermine]

hello, yes.

if i take the gradient of that with respect to b what should I get?

[restive shuttle]

how does one reason through part a

is that just R2 to R2

i have a terrible grasp of invertability in general

[clear otter]

Is S invertible?

Is it possible for S to be invertible

@restive shuttle what do you have troubles understanding with invertability?

[restive shuttle]

uh

lemme check my theorems really quickly

im not sure

if S can be invertible or not

how can i tell if a transformation is invertible

does it just ask if there is another transformation that can undo the transformation that S does, and if so, it's invertible

im guessing yes

[clear otter]

@restive shuttle so when we go from a vector space to a smaller vector space, you'll basically get "collisions"

For example, look at the map from R^2 to R^1 by projecting onto the x axis

[restive shuttle]

oh so if you go back to r2, you lose out on all that stuffs

[clear otter]

With this projection map, [2,1] and [2,10] both map to [2,0]

So if I give you [2,0] can you determine its preimage?

[restive shuttle]

right, but if [2,0] were to map to r2, then you can get a lot of different solutions

3 i mean

i wouldn't be able to get its preimage i think

[clear otter]

So it's not invertible

Now if you go R^2 to R^3, you have a slightly different thing happen

Basically you can make every vector in R^2 map to something unique in R^3. So it's injective

Or one to one

Meaning if I give you something in R^3, you can determine its preimage (if it exists)

So that means it's invertible

Does that make sense?

[restive shuttle]

yes i think so

[clear otter]

So in your example, you have S o T

And want to know if it's invertible

If it is invertible, its inverse would be T^-1 o S^-1

So T and S both need to be invertible

Is it possible for both to be invertible?

[restive shuttle]

and that's not true, since S is not invertible

[clear otter]

Do you understand why S isn't invertible?

[restive shuttle]

yes

[clear otter]

They suggest thinking of S and T as matrices

[restive shuttle]

so same logic applied to T o S, and it isn't invertible either for same reason

[clear otter]

If you think of them as matrices, you get non square matrices

[quaint pelican]

Please help me ! Is a polynomial a vector ?

[pallid rampart]

[tranquil junco]

[dusky epoch]

[tranquil junco]

hai ann

[pallid rampart]

Hello sloth and Ann

[dusky epoch]

hewwo

[tranquil junco]

mrowr

hi whoever

[quaint pelican]

Help me plz =((

[pallid rampart]

Hello sloth

So what have you been up to?

[dusky epoch]

@quaint pelican your question cannot be answered in a way you would find useful

[tranquil junco]

@pallid rampart sumac

application

i wrote an essay at 3 AM lol

[quaint pelican]

I have an problem which is "Given vectors S = { 1 + t, 1 - t, 2} in P1(t)" and i have to prove that S is linear dependent. And i really have no idea whether a polynomial is a vector or not.

[pallid rampart]

Lol

sumac

So vectors are just elements in a given vector space

So you have the vector space P1

So these polynomials are in P1 and are thus vectors

But it doesn't matter

What you gotta do is find a1,a2,a3, not all 0, such that a1(1+t) + a2(1-t) + a3(2) = 0

[dusky epoch]

@quaint pelican P1(t) ought to be the vector space of all polynomials with degree at most 1

in linear algebra, you abstract away from the idea that "vector" must always refer to a geometric object

[quaint pelican]

in linear algebra, you abstract away from the idea that "vector" must always refer to a geometric object
@dusky epoch Okay i'll try because i always stick to the idea that "vector" is a geometric object

[dusky epoch]

that's your downfall really

vectors are just things you can add and scale

[quaint pelican]

vectors are just things you can add and scale
@dusky epoch Okay, tks a lot !

[dusky epoch]

you didn't need to ping me

[quaint pelican]

Okay sorry

[tranquil junco]

lol

[turbid valley]

Zero matrix is the only matrix that is both symmetric and skew symmetric. Right?

symmetric: matrix is equal to its transpose
skew symmetric: matrix is equal to negative of its transpose

[dusky epoch]

well if A^T = A and A^T = -A then A = -A surely

[turbid valley]

Okay thanks

[clear spoke]

Can someone explain to me how to solve for x

$x \angle 135 = 5 \angle 0 + 5 \angle 90$

[stoic python]

<3 Hoodie <3:

[clear spoke]

I know that

$5 \angle 0 + 5 \angle 90 = \sqrt{50} \angle 45$

[stoic python]

<3 Hoodie <3:

[dusky epoch]

$5 \angle 90$

what does this notation refer to

[stoic python]

Ann:

[clear spoke]

90 is the angle and 5 is the module of the vector

It's a way of writing vectors

[dusky epoch]

oh jesus. polar coordinates.

okay well your equation doesn't have any solutions then

[clear spoke]

Some things don't make sense to me

It has

I will show you

[dusky epoch]

no it doesn't

[clear spoke]

Look

$x \angle 135 = \sqrt{50} \angle 45$

[stoic python]

<3 Hoodie <3:

[clear spoke]

$x\angle135 = -x\angle45$

[stoic python]

<3 Hoodie <3:

[dusky epoch]

no

[clear spoke]

no?

[dusky epoch]

these two vectors are at 90° angles to one another

[clear spoke]

Which two vectors

[dusky epoch]

$x \angle 135$ and $-x \angle 45$

[stoic python]

Ann:

[clear spoke]

Oh shit

I failed

That makes sense

[dusky epoch]

$x \angle 135 = \sqrt{50} \angle 45$ does not have a solution

[stoic python]

Ann:

[dusky epoch]

no value of x makes it true

[clear spoke]

You're right

I had my head exploding because of this

Thank you

[hollow finch]

Is it possible to have a nilpotent matrix of a given size that has any given index? Like could we have a 2x2 with index 10?

[shrewd slate]

If you have finite dimensional subspaces U1, U2, U3,

Isn’t it the case that we can let U4 be a new subspace = U2 + U3
dim(U1+U2+U3) = dim(U1+U4) =
dim U1 + dim U4 - dim (U1 ∩ U4)
= dim U1 + (dim U2 + dim U3 - dim(U2 ∩ U3)) - dim(U1 ∩ U2 ∩ U3)?

[sonic osprey]

check your signs here

[shrewd slate]

Ok hol up

Now?

[sonic osprey]

Okay yeah, but the last term is still wrong

[shrewd slate]

How come?

[sonic osprey]

because you get U1 ∩ (U2 + U3)

[shrewd slate]

O right

But it’s still negative right

?

[sonic osprey]

Yeah the signs are okay

[shrewd slate]

So you could claim that that’s smaller than or equal to dim U1 + dim U2 + dim U3

[sonic osprey]

what is smaller

[shrewd slate]

What I typed

(<=)

I hate that that looks like an implication arrow

[sonic osprey]

yeah sure

[shrewd slate]

Ok cool

Thanks

[clear spoke]

Any idea?

[dusky epoch]

knowing a thing or two about scalar products might help. or perhaps the parallelogram law if one feels so inclined

[tulip galleon]

hey,
if i have 3D vector and roll, pitch and yaw
how can i calculate the forward vector ?

[wintry steppe]

Why isn't b a basic solution?

it solves ax = b, yet the answer key says it's not a basic solution

[half ice]

The question is asking "which vectors x can you sub into Ax in order to make it the same as b"

Note that the multiplication Ax only makes sense if x is a vector of height 5

[wintry steppe]

it is

[half ice]

OHH I thought you meant the vector b, not the answer (b)

[wintry steppe]

yeh

In linear programming, the basic variablesin (b) are 2, -5, and -1. the non basic are the two 0's

and the vector is used as x, which solves Ax = b

which i thought is the definition of a basic solution: a vector that satisfies Ax = b

and has m LI columns

[half ice]

,w matrix multiplication {{2,3,1,0,0},{-1,1,0,2,1},{0,6,1,0,3}} times {{0},{2},{-5},{0},{-1}}

[stoic python]

Display results online and refine query

Requested by Kaynex#7520

[half ice]

Yeah strange, you're right that should work

[wintry steppe]

hi

can anyone help me

why does {x_1,x_2,...,x_k} generates W?

[empty copper]

What does Theorem 1.7 say

[wintry steppe]

@empty copper

oh i think i kind of see ow

another vector in w will bne in the span of the set

@empty copper why couldnt they use this

let G = B_V, let L= B_W then |B_W| <= |B_V|

so dim(W) <= dim(V)

[lilac kindle]

how can I reduce this matrix to a row echelon form? pls @ me

[south wadi]

byt getting 0 in the corner

Hi can someone help me with this

I don't understand why T(e_2) would be 1 / root 2, 1 / root 2

please explain 😦

@ me please

[lilac kindle]

@south wadi i know that, but i tried doing
r2 = r3 - r2
r3 = r3 - r2
r3 = ar3 - r1
and i've got
a 1 1 4
0 2b 0 1
0 ab-1 a-1 3a-4

[south wadi]

honestly i cant be bothered

i just need help with my question

[native river]

@south wadi what is the questions?

@south wadi The transformation in this case will be $T(x,y)=(-sin(\theta),-cos(\theta))$

[stoic python]

joseph2531:

[native river]

Hi, i have a question, usually we denote an field if it is commutative has a additive and multiplicative inverse and identity $Z_{p}$.
Is it possible for to have a negative field i.e: $Z_{-2}$?

[south wadi]

@native river

I don't understand why T(e_2) would be 1 / root 2, 1 / root 2

[native river]

@south wadi can you post the full question?

[south wadi]

thats the full question

[native river]

I am not sure i will say that $e_{i} = \theta_i= -\frac{\pi}{4}+(i-1)*\frac{\pi}{2}$

[south wadi]

wat

we havent delt with omplex number

[pallid rampart]

Suppose that $U$ and $V$ are finite-dimensional vector spaces and that $T:U\to V$ and $S:V\to W$ are linear transformations. How do I prove that $$\dim\text{ker}ST\leq\dim\text{ker}S+\dim\text{ker}T.$$

[stoic python]

Whoever:

joseph2531:

[native river]

@south wadi the i is this case does not refer to the imaginary number (bad use of letter on my part), think of $i \in \mathbb{N}$ where N is the natural number

[stoic python]

joseph2531:

[native river]

Hi, i have a question, usually we denote an field if it is commutative has a additive and multiplicative inverse and identity $Z{p}$.
Is it possible for to have a negative field i.e: $Z{-2}$?<@&286206848099549185>

[stoic python]

joseph2531:

[dawn fractal]

where is the unity in Z-2 tho?

[native river]

unity?

[dawn fractal]

assuming you mean Z-2 as {-1,0}

[native river]

yes

[dawn fractal]

the multiplicative identity

surely you've heard of it

(-1)(-1)=1, which is not in Z-2

and (0)(-1)=0

so there is no unity

i.e. no elt. a in Z-2 exists such that a multiplied by any elt. b in Z-2 equals b

also, your question should've been posted on #groups-rings-fields

[native river]

sorry , what is elt?

[dawn fractal]

element

[native river]

@dawn fractal what if we define z_{-2} = {-1,0,1,2}, does it have a unity?

[sonic osprey]

you need to tell us what are the operations on this

[dawn fractal]

how would you define addition modulo then

for example in Z_2, addition modulo 2

moreover, you'd also have the problem of defining multiplication modulo

[native river]

@dawn fractal @sonic osprey Thank You!

[pallid rampart]

Suppose that $U$ and $V$ are finite-dimensional vector spaces and that $T:U\to V$ and $S:V\to W$ are linear transformations. How do I prove that $$\dim\text{ker}ST\leq\dim\text{ker}S+\dim\text{ker}T.$$

[stoic python]

Whoever:

[pallid rampart]

@sonic osprey

👀 can you help me

<@&286206848099549185>

[sonic osprey]

I'm here

[pallid rampart]

I posted a while ago

Ok

[sonic osprey]

uh isnt this

the kernel of ST is a subspace of the kernel of T

[pallid rampart]

Isn't it the other way around?

Because if a vector is in the kernel of T, then it's obviously in the kernel of ST

But it might not work the other way around

[sonic osprey]

oh yes im dumb

[brittle juniper]

Have you tried factoring the restriction of T to ker ST?

[pallid rampart]

?

I'm not sure I understand what you're saying

But here is what I have tried

(I'll just use ker for dimension of ker, and im for dimension of range)

[brittle juniper]

you just need to find an injection Ker(ST)/Ker(T) -> Ker(S)

(a linear injection)

[pallid rampart]

ker ST + im ST = dim U = ker T + im T <= ker T + dim V = ker T + ker S + im S

🤔

What's Ker(ST)/Ker(T)?

Oh quotient group stuff 🙃

Not sure how to do that

[vast torrent]

@pallid rampart a common shorthand is nu and rho respectively

[pallid rampart]

for dimension of kernel and image?

[vast torrent]

Yeah

Nullity, rank

[pallid rampart]

ok ill keep that in mind

[sonic osprey]

so don't you just have to show that im ST \leq im S?

[vast torrent]

It's more fun to draw rho than to draw nu

[sonic osprey]

and isn't that just true

[brittle juniper]

,tex
\begin{tikzcd}\Ker(ST)\arrow[d, "\pi"]\arrow[r, "T"]&\Ker(S)
\Ker(ST)/\Ker(T)\arrow[ur, dashrightarrow, hook]
\end{tikzcd}

[stoic python]

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

[vast torrent]

[brittle juniper]

Fuck this

[pallid rampart]

Ah

Diagrams

[brittle juniper]

,tex
\begin{tikzcd}\Ker(ST)\arrow[d, "\pi"]\arrow[r, "T"]&\Ker(S)\
\Ker(ST)/\Ker(T)\arrow[ur, dashrightarrow, hook]
\end{tikzcd}

[vast torrent]

Commutative diagrams in discord is ambitious

[stoic python]

Tuong:

[vast torrent]

Nice!

[pallid rampart]

I

Don't understand diagrams

[sonic osprey]

did u see what I said

[pallid rampart]

yeah i can show that but

i can't just subtract them from both sides

[sonic osprey]

oh I don't know how inequalities work lmao

[pallid rampart]

Fuck it

I'll skipping this one

[brittle juniper]

D:

[pallid rampart]

[sonic osprey]

i'll think about it when I get home

[brittle juniper]

Big sad, I put so much effort into that diagram

[pallid rampart]

I'm sorry but I really

Don't understand diagrams

:(

I mean yes I understand that T=(some other function)π

[brittle juniper]

you don't really need to be able to read the diagram actually

[pallid rampart]

and my job is to find that other function

[brittle juniper]

you just need to know that T(Ker ST) is a subspace of Ker S and that there's a theorem that gives you an isomorphism between (Ker ST)/(Ker T) and T(Ker ST)

[pallid rampart]

uh what's (Ker ST)/(Ker T)

[brittle juniper]

the set of equivalence classes for the relation R on Ker ST defined by
aRb iff a-b in Ker T

[pallid rampart]

Alright I think I got a proof

But with um

Very primitive approach

(T: U->V, S:V->W)

So we noted that ker T \subseteq ker ST

So we can have a basis for ker T

namely (v_1, ..., v_n)

extend that to be a basis for ker ST

namely (v_1, ..., v_n, ..., v_m)

then nu T = n, nu ST = m

we can prove that T is injective on v_{n+1}...v_m

then since v_{n+1}...v_m is in ker ST, it follows that ST(v_{n+1}), ..., ST(v_m)=0

so T(v_{n+1}) ,..., T(v_m) are in the null space of S

so m-n <= nu S

and m = nu SY <= nu S + n = nu S + nu T

omfg

i hate this

[sonic osprey]

there has to be a basis free argument

[brittle juniper]

you can directly take a supplementary space of Ker T in Ker ST

and the restriction of T on that supplementary is injective

[pallid rampart]

Ah

I have been thinking about another problem

For a really long time

And during dinner

And turns out

It's just a trivial consequence of a theorem that I forgot about

Fuck

me

[hardy blaze]

is the jordan canonical form

the same thing as solving for A = PDP^-1

cuz im in diff eq 2

& we r learning it the jordan canoncal form way which just seems way harder

😐

[gray dust]

please explain what letters mean what

[hardy blaze]

P is the matrix formed from the basis vectors, P-1 is the inverse of that, D is a diagonal matrix with the eigenvalues

its the diagonalization of a matrix basically

[gray dust]

P's cols are eigenvectors corresponding to each eigenvalue in D

[hardy blaze]

yes

isnt D just the jordan can form

[gray dust]

in the context of ode, jordan decomp is useful if A is not diagonalizable, ie if A is n by n and A has less than n linearly independent eigenvectors

follow the algorithm for A's jordan decomp and you'll get A=SJS^-1 where S is loaded with A's eigenvectors (generalized eigenvectors if A isn't diagonalizable) and J is A's jordan normal form which, if A isn't diagonalizable, definitely won't be a diagonal matrix

[stoic python]

Whoever:

[pallid rampart]

So this fact is like saying, the center of GL(n) is the set of diagonal matrices such that the diagonal elements are all the same

Is that right?

[sonic osprey]

No lmao

There are elements of \mathcal{L}(V) that aren't elements of GL(n)

[pallid rampart]

Oh right

xD

[cold topaz]

For this problem, I added the third row to the first row. But slader swaps the first and the third rows.
Is my solution correct, as well?

[nimble egret]

Sure

[cold topaz]

thanks!

[prime nebula]

How do I calculate the angle between two vectors based on the dot product?

[half ice]

v•u = |v||u|cosθ

Solving cosθ in this gives the angle between the vectors

[prime nebula]

Can you do an example

[half ice]

Give me two vectors in R²

Or, two 2D vectors

Or, two 2D vectors

[prime nebula]

[ 0 1 ]
[ 4 6 ]

(Pretend those are stacked vertically)

@half ice

[half ice]

The two vectors are (0,1) and (4,6)?

I'll let it be
u = (0,1) and v = (4,6)

[prime nebula]

yea

continue

[half ice]

u•v = 6
|u| = 1
|v| = √52

[prime nebula]

Wat

Im very new to Linear Algebra what does |v| and ||v|| mean

err

||v||

[half ice]

That's the magnitude of v

If
v = (a,b)
Then
|v| = √[a² + b²]

The magnitude of a vector is a real number

Like, the length of the arrow

[prime nebula]

ok and what is the
||v||

[half ice]

The magnitude of v

I'm just writing it as |v|

[prime nebula]

Oh so ||v|| = |v| In linear algebra?

[nimble egret]

I mean

If it's clear v is a vector

Yes

[prime nebula]

Because my textbook as used both in the first section and it has been confusing

[wintry steppe]

ok wtf

[nimble egret]

Both are acceptable notations

[prime nebula]

Ok

anyways

[wintry steppe]

is it equality or inequality

what the actual fuck

IS MIT WRONG

[sonic osprey]

uh, do you not know what the symbol $\leq$ means?

[stoic python]

Zopherus:

[wintry steppe]

yes but the two posts don't agree

why does this book say standard form is using equality constrants when the internet that i've googled so far say it's canonical form

[prime nebula]

How do I calculate the angle between two vectors based on the dot product?
for the vectors
( 0 1 )
( 46 )

[vast torrent]

https://proofwiki.org/wiki/Angle_Between_Vectors_in_Terms_of_Dot_Product

[prime nebula]

Ok Cool

thanks!

[vast torrent]

Follow links if you want to see the proof that

The regular formula for dot products

[prime nebula]

Doing that atm

[vast torrent]

Gives you cosine

[prime nebula]

I am unfamilar with arccos

[vast torrent]

It's very clear, the person who made those pages must be very smart

[prime nebula]

does arc[trig function] just undo a trig function?

[vast torrent]

Click the link defining arccos then :) yes it does but you need to limit the domain otherwise cosine has no inverse

Doesn't pass the horizontal line test,if you've heard of that

[prime nebula]

Alright thanks

[gray dust]

checked the proof revision history. just as suspected

[vast torrent]

[pallid rampart]

Oh man, I binged through 4 chapters of linear algebra in 3 days

and I completed all the exercises like an idiot

But it was fun anyways

and rewarding ofc

now it's time to learn about

eigen vectors and eigen values

https://cdn.discordapp.com/emojis/473701113359499284.gif?v=1

[prime nebula]

@gray dust What did you suspect

[gray dust]

fauxpas wrote the proof

[prime nebula]

lol

@gray dust Which one?

[gray dust]

the one he linked @prime nebula

[sonic osprey]

damn nice

@pallid rampart learning fast

[pallid rampart]

I mean

I'm planning to binge through linear algebra

This week

[prime nebula]

Same

[pallid rampart]

So that @gray glen can stop laughing at me

[sonic osprey]

nice

I'll still laugh at you though

[pallid rampart]

@sonic osprey shut up you weeb

[nimble egret]

I'll be laughing at all of you

[sonic osprey]

with what knowledge tho

[pallid rampart]

Don't you love it when (U_1+U_2)\cap U_3 is not (U_1\cap U_3) + (U_2\cap U_3)

[sonic osprey]

did I do this today

wait no you did this

[pallid rampart]

Lol

and arch

[viscid kernel]

What is the dimension of a matrix?

How do you calculate it or how do you see it ?

[vast torrent]

@viscid kernel if you mean the dimensions of a matrix, it's the number of rows and the number of its columns

If you mean dimension of the span of columns, you can't generally tell just by looking, but you'll learn ways to tell

[dense saffron]

my current exercise asks: what are the advantages of hermite vs bezier interpolation

this confuses me as to my understanding those are pretty much the same thing

just a little different in their input parameters

[vast torrent]

I just looked them up and it seems like bezier interpolation gives splines

@dense saffron rather than polynomials

So those are very different types of things

[dense saffron]

they both result in a polynomial

the bezier interpolation matrix is just the hermite interpolation matrix multiplied with a correction matrix

[vast torrent]

When i learned bezier curves they were used for splines, idk

Never learned hermite

It's locally a polynomial except at the knots

The way we used it in my class

[dense saffron]

a spline is just multiple polynomials chained together

[vast torrent]

Yes , so.the resultant curve won't be C^infinity like polynomials

[dense saffron]

but thats the case in both

C1 on the connecting points and C3 everywhere else

[vast torrent]

Unlike Newton interpolation, Hermite interpolation matches an unknown function both in observed value, and the observed value of its first m derivatives.

Well that's a bit of a tall.order, knowing derivatives of the underlying relation

Isn't it?

[tidal kettle]

big brain time

approximate the derivatives of the underlying relation so that you can use Hermite interpolation

[vast torrent]

[dense saffron]

tbh i might have messed up by not providing proper context, maybe hermite interpolation can be used to interpolate X points

but in the context of my lesson plan its only ever used to compute splines

so its this https://en.wikipedia.org/wiki/Cubic_Hermite_spline

[vast torrent]

You still need approximations of the derivative it says

[dense saffron]

yeah and bezier calculates those automatically based on your control points

[vast torrent]

I'm asking you

When would you know derivative data for a curve if you don't know what the curve is?

It sounds pretty useless

[dense saffron]

when you want a curve that looks pretty

for computer animations

instead of being exact

[vast torrent]

But how would you get data for the derivative?

Bezier curves look pretty and you just need control points

Not data about the derivative

[dense saffron]

i guess thats the answer, you dont know the derivatives anyway so why not use 2 of the points to estimate them anyway

[vast torrent]

And you told me the disadvantage of bezier curves

That they don't go through all data points

No big deal if you're drawing vector graphics

But certainly there's an advantage to an interpolation that actually matches f(x)= approximated f(x)

For x known empirically i mean

It's called residual error i believe

The difference between f(x) and approximated f(x) iirc is called residual error

[dense saffron]

thanks for the help !

[vast torrent]

Np

[keen mirage]

if P^T = 9P^-1 is P orthogonal

@vast torrent

i have a matrix M that is symmetric and want to diagonalise using spectral theorem

P is eigenbasis matrix

[vast torrent]

Why are you tagging me

[keen mirage]

i just wanted a quixk yes or no

[vast torrent]

Use the helpers tag if no one answers after 15 minutes,im a helper

[pallid rampart]

Why in linear algebra we don't use parenthesis for Tv unlike literally all other field of math

[sonic osprey]

probably matrix multiplication things

[pallid rampart]

Ah

I can see that

[gray dust]

@keen mirage P is orthogonal if P^T=P^-1

[keen mirage]

is there a name for P if P^T = kP^-1?

@gray dust

[gray dust]

everything needs a name...

[pallid rampart]

I got lost at the sentence "Because v≠0, the coefficients a_1, ..., a_m cannot all be 0, so 0<m≤n"

How does v≠0 imply that not all a1, ..., am are 0

[dusky epoch]

i think they meant a_1, ..., a_n?

or sth

uh

yeah

they want to ensure the existence of such an m

[pallid rampart]

yeah i got it

it's a1,...,am, not a0, ..., am

so basically they don't want the polynomial to be a constant polynomial

[dusky epoch]

[pallid rampart]

so that we can factor

[dusky epoch]

oh uh.

yeah

[vast torrent]

The usual proof of this is with determinants

Oh i see

This avoids picking a basis

[pallid rampart]

But what's the point of not choosing a basis? It's finite dimensional

This is from axler

Who avoids determinant

🙃

[vast torrent]

Because then you don't have to prove eigenvalues are basis invariant

[pallid rampart]

What does that mean?

[vast torrent]

Means changing the basis doesn't change the eigenvalues

[pallid rampart]

Wait um

Eigenvalue in "linear algebra done right" is introduced on linear transformations

Sooo there is no basis involved

I'm still not sure

[vast torrent]

To use determinants you need to pick a basis

And make a matrix

[pallid rampart]

Ah

Yeah

I see now

Thanks

[vast torrent]

With this proof you never use matrices

[lilac kindle]

can someone explain what does T^(2)_(0,0) mean ? (at the bottom)

[smoky lagoon]

brain stuck

I know there's a zero row in v1,v2,v3,v4 in r5 because theres more rows than collumns, but I'm not sure how to use that and the properties of other vector to show what i'm trying to prove

[subtle walrus]

well, what's the definition of span?

[smoky lagoon]

you can make a linear combination of the vectors that equal the vector in the span

so v1+v2+v3 does NOT equal v4

oh hold on

so if I know that v4 is linearly independent, then I know that v1v2v3v4 is linearly independent

because v1,v2,v3 is

or im missing something

[subtle walrus]

span is the set of all linear combinations of the vectors

[smoky lagoon]

so I know that v1,v2,v3 doesn't span r5, but i knew that already

[subtle walrus]

you know that v_4 is not a linear combination of v_1, v_2, v_3

because otherwise it would be in the span

[smoky lagoon]

mhm

so I can't add the vectors in some combination of vectors and scalars to get v4

so I know that v4 is independent from v_1, v_2, v_3

[subtle walrus]

yes

now convince yourself that this is enough

[smoky lagoon]

and the problem presupposes that v1, v2, v3 are already linearly independent

sigh

cool

thanks man

[lilac kindle]

i dont understand how the hessian matrix of the f(x,y) given becomes
6x 6y
6y 6x

i calculated it to be
6x 3
3 0

[vast torrent]

I think you're right

Looks like a mistake on their end

[lilac kindle]

@vast torrent aah right, thanks, it really confused me

[pallid rampart]

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

So it had been established that if T has dimV number of distinct eigenvalues then it will have a diagonal matrix with respect to some basis

But this proposition made it seem like it might not be the case if the eigenvalues are all distinct

I don't know if I'm interpreting this correctly

Oh wait, it doesn't need to have dimV number of eigenvalues

Lmao, m ≠ dimV

Ok

[obsidian jackal]

i have a question

in this picture, my professor mentioned that we need v1 and v2 orthogonal to v

where v1 and v2 are the vectors that are parallel to a plane

if v1 and v2 is orthogonal to v, doesn't it mean that they are parallel?

my second question is how did he get the vectors?

i did get 1 of the vectors by doing the dot product of the plane and the vector v but my other vector is different from his answer

[gray dust]

in most contexts, perpendicular and orthogonal are interchangeable

& your prof most likely spent 30 secs trial'n'erroring to find 2 vectors (which are not scalar multiples of each other) whose dot prod w/ v is 0

[obsidian jackal]

ooohh

so that is (t1v1 + t2v2) . v = 0?

since it passes through the origin

[gray dust]

sure

[obsidian jackal]

one thing still confuses me though

he mentioned that v . v1 = 0 and v . v2 = 0 in class

that means both vectors are orthogonal to v

and it means that it is parallel, right?

or am i mistaken?

[dull nexus]

I think for orthogonal the scalar must be 0

[gray dust]

in most contexts, perpendicular and orthogonal are interchangeable
do you get this?

[obsidian jackal]

yes

[gray dust]

then you can't confuse ortho & parallel

[obsidian jackal]

hmmm

maybe i really dont understand what orthogonal means

[dull nexus]

orthogonal means perpendicular to another

[gray dust]

for clarity, two things are perpendicular if they form an angle of pi/2

[obsidian jackal]

yes

[gray dust]

in most contexts, perpendicular=orthogonal

[dull nexus]

[obsidian jackal]

so does that mean v1 and v2 is perpendicular to v?

[gray dust]

yes

perpendicular=orthogonal
just drill this into your head

[obsidian jackal]

so does that mean that they are parallel?

v1 and v2

[gray dust]

do you know the difference between parallel and perpendicular

[dull nexus]

Lmao

[obsidian jackal]

not v

i meant v1 and v2

[gray dust]

we have no guarantee of that

can be parallel, perpendicular, or neither

[obsidian jackal]

oh

[dull nexus]

Do you know the inner product?

[gray dust]

also i'd prefer if just 1 person answered q's

[obsidian jackal]

maybe because it's 3d?

[dull nexus]

Whats with askew ? @gray dust

[gray dust]

there's an entire plane full of vectors that are ortho to v. there very clearly exist many pairs of vectors in the plane that are neither parallel nor orthogonal

[obsidian jackal]

oh okay

[dull nexus]

@obsidian jackal are maybe from Germany:D

are u*

[obsidian jackal]

now i get it

[gray dust]

Whats with askew ?
wdym

[dull nexus]

skewed vectors

[gray dust]

you mean skew lines?

[obsidian jackal]

lol no@dull nexus i come above the land of the free

[wintry steppe]

North Korea?

[dull nexus]

Oh ok

@gray dust isnt that also possible with vectors?

[gray dust]

lines can be skew. not vectors

to me, vectors always "start" at the origin

[dull nexus]

Oh. My bad .. the local vector...

[gray dust]

wdym

[dull nexus]

(x,y,z) - (0,0,0)

Sry if my english is bad

[gray dust]

i still don't know wdym

[dull nexus]

Hmm

Wait

I mean isnt it that they all have a start at the origin bc of the local vector?

[gray dust]

what's "they"? what's a local vector?

[dull nexus]

In german its called " ortsvektor"
So they are vectors with the distance of (x,y,z) - (0,0,0).

So Im not sure how its called in english

I was googling and they showed me the word "local vector"

[wintry steppe]

hi, is every basis of C^N also a basis for R^n?

[gray dust]

never heard of local vector

[wintry steppe]

basis of $C^n$ also a basis of $R^n$ ?

[stoic python]

nxue:

[wintry steppe]

Is that complex and reals?

yes

i know its mathbb but im lazy

$\mathbb{R}$

[stoic python]

nxue:

[wintry steppe]

C is two dimensional

Over R

C can be n dimentional

No but C by itself

what i mean is if i have basis of C^n
is it also a basis for R^n?

Over what field

And generally no

same n bruv

ill ask it another way:

No you have to define a vector space over a field

To multiply scalars with

If you define R over Q it is infinite dimensional for example

But R over R is 1 dimensional

[gray dust]

@dull nexus still got no idea what that's about if it helps you i think of lines as the set generated by constant vector + parameter*direction vector

[stoic python]

nxue:

[wintry steppe]

this ^

How has your course defined vector spaces?

Do they assume the underlying field is just R?

what do you mean?

The scalars you multiply with

Because that is essential for when you want to find a basis

coefficients over C (because over R won't work here)

So C as a C-vector space

C over C is one dimensional so adding a dimension as the standard space will add just 1 dimension

so does $C^2$ spans $R^2$ also?

So C^n has the same dimension as R^n

[stoic python]

nxue:

[wintry steppe]

If R^n is an R-vector space

They are not the same

They just have the same dimension

ofcourse, one has another field of numbers but it also includes R because

$RCC$

[stoic python]

nxue:

[wintry steppe]

dunno the symbol

Wdym

wait ill find the symbol

[stoic python]

nxue:

[wintry steppe]

Yes

[vast torrent]

kind of

[wintry steppe]

so C does span R

[vast torrent]

it's an embedding

[wintry steppe]

What do you mean by span

span{v1, v2, v3 .. etc}

https://en.wikipedia.org/wiki/Linear_span

You have defined them over different fields

[vast torrent]

R is embedded in C but tat's slightly different than saying it's a subset

[wintry steppe]

Well the typical definition of C is R^2 and then define multiplication and addition accordingly

R is a subset

because

[stoic python]

nxue:

[vast torrent]

what does "+" mean

[wintry steppe]

But the span doesn’t work either because for R you used the field R for coefficients and for C you used C

[vast torrent]

$\mathbb R \not \subseteq \mathbb R \times \mathbb R$

[stoic python]

gfauxpas:

[gray dust]

yay minor notation abuse

[vast torrent]

but the isomorphism $x \leftrightarrow (x,0)$ is close enough to equality to be effectively the same in most contexts

uh what's the two headed arrow

[wintry steppe]

\mapsto ?

I think that is just an injection

\Rightarrow ?

[stoic python]

gfauxpas:

[wintry steppe]

Homomorphism

[vast torrent]

there

[wintry steppe]

https://www.codecogs.com/latex/eqneditor.php

great website tho

i use it all the time

[vast torrent]

so we can say $\mathbb R \times { 0 } \subseteq \mathbb R \times \mathbb R$

[stoic python]

gfauxpas:

[vast torrent]

and $\mathbb R = \mathbb R \times { 0 }$ by abuse of notation

[stoic python]

gfauxpas:

[vast torrent]

set of imaginary numbers is also a copy of R but people don't talk about that as much

[wintry steppe]

i have another question 😅 :
is: $[I]^E _B$ the transformation matrix ? because I don't know what's the I doing there

[stoic python]

nxue:

[wintry steppe]

or it's just a notation

and it could be A , B , K

? 😟

What’s E what’s B what’s I

in general it could be transformation matrix from A to B (up to down)
in this one its E to B where E is the standard basis and B is some basis

I = ? the matrix itself i suppose

So you want to switch a basis from R to C?

no no they are over the same field

F

Field?

the bases span the same field

Do you mean vector space?

yes.

Okay

So what’s the question

is: $[I]^E _B$ the transformation matrix ? because I don't know what's the I doing there

[stoic python]

nxue:

[wintry steppe]

There are a lot of notations for this

if you havent heard of transformation matrices then you'll be clueless

No no I know what they are

😟

But I cannot know what notation your course uses

it's the general notation

My course used $S_{B,E}$

[stoic python]

N/𝔄:

[wintry steppe]

So no

There is no general notation

but how come you havent heard of transformation matrices yet know the notation?

never seen that notation

Because I don’t study in english and wasn’t sure what the term meant initially

[hardy blaze]

http://prntscr.com/r5fjvj

is there only one for this one? 😐

i got one form

[wintry steppe]

no there are 2 eigenvalues

$|A- \lambda I| = (1- \lambda)(1- \lambda)(a- \lambda)$

[stoic python]

nxue:

[wintry steppe]

oops didn't read the question my bad lol

[hardy blaze]

its ok

ik those r the eigenvalues

[pallid rampart]

So I did this by assuming U is not V, and create a basis for U, then define an operator that is not invariant on U and concluding that U is {0}

But

Can I do this without basis

[lilac kindle]

i need help with this one

[pallid rampart]

<@&286206848099549185>

sad life man

[sonic osprey]

I think the answer is no

but I have no proof of it

[pallid rampart]

Ok

🤔 probably some axiom of choice stuff again

[wintry steppe]

Hi , i need help about eigenvalues/eigenvectors
If it is given that
p(x) = |xI - A|= x²-3x+5

I need to find det(A) and trace(A)

I dont have a clue, i though the sum of the eigenvalues is the trace but there are no eogenvalues for this equation... (And the answers say the trace is not 0)

[vast torrent]

@wintry steppe eigenvalues are the roots of the polynomial

[wintry steppe]

Yes i know

There are no roots

[vast torrent]

It has complex roots

[wintry steppe]

Yes. But i dont need to find the eogenvalues

Eigen"

I need to find , given the polynom , trA and detA

[sonic osprey]

You said yourself, the sum of the eigenvalues is the trace

[vast torrent]

And the product of the eigenvalues is the determinant

[wintry steppe]

@sonic osprey what about det?
And do i sum the complex eigenvalues?

@vast torrent
Can eigenvalues be complex?