linear-algebra

[native rampart]

I mean,not a basis in general

{$e_1,e_2...e_n$} is orthonormal means
{$\phi_{e_1},\phi_{e_2}...\phi_{e_n}$} would be a basis of that space

[stoic python]

DrunkenDrake

[native rampart]

I don't think it would span (R^n)* at all

If the basis weren't orthonormal,This wouldn't be a "dot product"

A basis for which the inner product is a "dot product" is called orthonormal basis

idk just Intuition

How do you know there is not another x' Which satisfies this $\phi_{x}=\phi_{x'}$?

[stoic python]

DrunkenDrake

[wind pasture]

how do i find a vector v thats in span(a1,a2) and span(b1,b2)?

[dusky epoch]

0

[wind pasture]

nvm

i figured it out

(non zero)

[nocturne jewel]

Good cause you didnt give us much to go on with that question

[dusky epoch]

and i gave the maliciously compliant answer

[wind pasture]

was it that vague?

[nocturne jewel]

a1,a2

So vague we dont know what vector space you're working in

[wind pasture]

ok mb

i got another question

if A and B are linearly dependent matrices does the product AB have to be linearly dependent as well

[lavish jewel]

do you mean the matrices are linearly independent from each other or each one has linearly independent columns (i.e. full rank)?

[nocturne jewel]

I think they mean cA+dB = 0 for non zero c and/or d

[lavish jewel]

then no

[wind pasture]

i mean the columns of each matrix are linearly dependent

[lavish jewel]

then yes

[wind pasture]

the columns of AB must be linearly dependent?

is there a theorem for this?

[lavish jewel]

yes

you can easily show it yourself by doing operations on the vectors

you can also read wikipedia

oh oops, i meant to circle the property below this one

[wind pasture]

thanks

im just confused about how rank and linear independence are connected

oh wait

rank = # of pivot columns?

nvm i got it

[spice dock]

Hello again! A is a plane in R^3 with normalvector = n = (1, 5, 2). S = R^3 to R^3. Calculate reflectionmatrix S. Any tips on where to start? I've already googled for hours and checked my notes etc without luck..

[wary lily]

a reflection matrix changes n to -n?

[spice dock]

Yepp.

The way I was thinking was, if i find 3 orthogonal vectors to n, and find the reflection of those?

[lavish jewel]

the plane is spanned by 2 orthogonal vectors v1 v2 on that plane. v1, v2 and n span all of R3

you can do the reflection by expression vectors in R3 in terms of v1, v2, and n, and swapping the sign of the n component

[lavish jewel]

you can find 3 vectors, or as many of you want, that are orthogonal to n, but they all lie on a plane and are in a 2D subspace

there's no need to find more than 2

without thinking too hard about it, and probably with mistakes from my side, you're looking for a similarity transformation that looks kinda like this. if M = [v1 v2 n], you'll end up with MTM^-1, where T is probably something like diag{[1 1 -1]}

[spice dock]

I see, and I have to be honest, that doesn't make much sense now but I'll look at it and maybe in a few minutes I'll understand. I'm really struggling to understand linear algebra, even though I'm having no difficulty with any calculus math.

[lavish jewel]

surprisingly enough, linear algebra is more straightforward

[spice dock]

Hehe, so they say..

But thanks, I'll keep on it. I'll probably come back with more questions :). Have a nice day!

[novel hamlet]

is there a nice proof of scalar triproduct? v · (w × u) = w · (u × v) = u · (v × w) where v, w, u ∈ R3×1

[lavish jewel]

you can write out how the result looks and compare them

if you wanna be fancier, you can write the cross products as matrix multiplications

[quartz compass]

I just write it as a determinant and see that these are just manifestations of determinant rules

[wary lily]

since it's the volume of a parallelpiped, I guess there is a nice and intuitive geometric proof. Tho I'm very new to this stuff and haven't properly learned it yet.

[quartz compass]

but maybe that's cheating

what I said lol

[novel hamlet]

proof, is proof

[wary lily]

lol

[novel hamlet]

no matter how cheating it is

[quartz compass]

well if you don't know what the determinant is, you'd have to at least at some point in your life see that you can manipulate the determinant in that way without it changing, that's all I really mean

[novel hamlet]

thanks

i think i just calculate the corss producst open and get the desired result

on different problem i have troulbe on creating transformation matrix from set of values

i have polynomial space p4 with derivation as operation and i try to create fp->p transformation matrix

i have derived all functions on form a4X^4+a3x^3+a2x^2+a1x+a0 but im not sure how to get the transform matrix

[lavish jewel]

what's fp

you want a matrix that maps an order 4 poly to its derivative?

[novel hamlet]

F:P->P

yes

[lavish jewel]

ah

[novel hamlet]

ffs.... the photos never come on correct direction

[lavish jewel]

well, the easiest way is to see what happens to each of the basis vectors as you transform them

[novel hamlet]

im stuck on that part

[lavish jewel]

let's say we start with the vector [1,0,0,0,0], which corresponds to x^4, yeah?

that's one of the canonical basis vectors for 4th order polys

[novel hamlet]

That should return 0 as there is no x^4?

[lavish jewel]

no

it should return a vector isomorph to 4x^3

that's the derivative of x^4, and that's what your transformation matrix should do

i'll also teach you a trick of matrix multiplication before we go on

[novel hamlet]

So 1,0,0,0 should return 0,4x^3,0,0?

[lavish jewel]

if you have M = [m1 m2 m3], where m1, m2, m3 are column vectors

then the operation Mx yields x1m1 + x2m2 + x3m3

in other words, each element in the vector x is multiplied as a scalar to the corresponding column of M

also no, Tx, with x = [1,0,0,0,0]^T should return [0,4,0,0,0]^T

x = [coef of x^4, coef of x^3, ..., constant]^T

[novel hamlet]

so basically 1,0,0,0 -> 0, 4x^3, 3x^2, 2x, a, 0?

[lavish jewel]

1,0,0,0,0 but yeah

well

no

where did x^2 and x come from

what's the derivative of x^4

[novel hamlet]

4x^3

[lavish jewel]

why did other stuff pop up then

we're looking at a single basis vector, just x^4

[novel hamlet]

if i have polynomial P4 isnt it same as A4X^4+a3x^3+a2x^2+a1x+a0?

[lavish jewel]

yes but you already said you didn't understand how to work with that, so we are working with one basis vector ata time

[novel hamlet]

just misunderstood this then

[lavish jewel]

just look at one basis vector at a time, as i said at the beginning

[1,0,0,0,0]^T corresponds to x^4

[novel hamlet]

yes, and when you derive it you get 4x^3

[lavish jewel]

right

so [1,0,0,0,0] -> [0,4,0,0,0]

yeah?

[novel hamlet]

that sounds reasonable

[lavish jewel]

ok

now look at what i wrote above

[novel hamlet]

and by that logic 0,1,0,0,0 -> 0,0,3,0,0?

[lavish jewel]

if you have M = [m1 m2 m3], where m1, m2, m3 are column vectors
then the operation Mx yields x1m1 + x2m2 + x3m3
in other words, each element in the vector x is multiplied as a scalar to the corresponding column of M

yes

so do all of those, one at a time

0,0,1,0,0 -> 0,0,0,2,0

[novel hamlet]

and finally 0,0,0,0,1

[lavish jewel]

0,0,0,1,0 -> 0,0,0,0,1

and last, 0,0,0,0,1->0,0,0,0,0

yeah?

[novel hamlet]

yes

[lavish jewel]

ok

so this is the result of having done Mx for specific choices of x

this M is a transformation from R^5 to R^5, since we are inputting vectors with 5 elements and we are getting vectors with 5 elements out of it, yeah?

[novel hamlet]

yes

[lavish jewel]

ok

[novel hamlet]

oh wait no,

we get 4 elements out

[lavish jewel]

well, we left the 0 in front

[novel hamlet]

since we lose x^4 on derivation

[lavish jewel]

it's equivalent in the sense that the output is a subspace of R^5 where the first element is always 0

you can either make a 4 x 5 matrix or a 5 x 5 that will have an extra 0

[novel hamlet]

5x5 is easier

[lavish jewel]

that's what i thought as well, so let's go with that

now, let M = [m1 m2 m3 m4 m5]

it has those 5 columns

and if we multiply it by some x = [x1,x2,x3,x4,x5], we get x1m1 + x2m2+x3m3+...

the x_i are scalars, the m_i are vectors

so far so good?

[novel hamlet]

yes

[lavish jewel]

wait til we're done please

[keen coyote]

my bad

sorry

[lavish jewel]

aight, so we know that if x = [1,0,0,0,0] we get that Mx = [0,4,0,0,0], yes?

it's what we did above

[novel hamlet]

yes

[lavish jewel]

but this also means that [0,4,0,0,0] = x1m1 + x2m2 + x3m3 + x4m4 + x5m5

[novel hamlet]

so when we do this enough we should get

[lavish jewel]

but x1 = 1, and the rest of the x_i are 0

so [0,4,0,0,0] = 1*m1 + 0

[lavish jewel]

precisely

[novel hamlet]

something that looks like htis

[lavish jewel]

you see that the first row is 0s, so this matrix is rank defficient

which is the same as both of us said: i said it spits out a subspace of R5, you said it spits out polys of order 3

same thing

[novel hamlet]

so this is my transformation matrix M?

[lavish jewel]

yeah

test it out and convince yourself if you want

feed it random polys and compare with your own derivatives

[novel hamlet]

oh well that was kinda easy now that you walked me trough how this works

now to do the easy part

[lavish jewel]

if you can't see it clearly, go for inspection of the basis vectors

[novel hamlet]

calculating eigenvalues

[lavish jewel]

as someone else suggested up there

[novel hamlet]

I'll be bback later with more problems i dont really understand chain theory but I'll finish this first

[dawn mesa]

Let m = number of clubs, and let n = number of people in whole clubs.

1. Assume in any club there's an odd number of people registered in that club.
   2)In any 2 club, the sum of registered people from each club is even.
   Prove that m <= n .

[novel hamlet]

basically you have at least M students (number of clubs), so they can have more so you have to show that m<=m+1

[keen coyote]

anyone have any suggestions how to do this other than calculus

[lavish jewel]

the projection should be straightforward

you could do v3 = v1 x v2, find the coordinates of this vector x in that basis, and find the length of the vector v3 * its coefficient

(which still uses projections)

[keen coyote]

yeah I couldn't think of any way to calculate it without projections that would "simplify" the calculations so idk

cause projection isn't really that bad

[lavish jewel]

wow i just noticed it's R4, no cross products, but gram schmidt instead, my bad

[tame mural]

yes but gram schmidt is projection too, I think people understood you

I think it's hard to think of a non-projection way

[keen coyote]

yeah I mean calculus with partials isn't exactly simplifying

i was thinking maybe somehow decomposing v into a vector in the span of v1 and v2

and the complement of the span

[tame mural]

but decomposing is projection

lol

[keen coyote]

but again thats not really simpler

yeah

well maybe something with like the kernel

of a matrix made of v1 and v2

[lavish jewel]

that is also projection

[keen coyote]

rip

[lavish jewel]

project onto the null space

[tame mural]

I'd be curious to hear the answer

I love simplifications

[keen coyote]

same

[lavish jewel]

i can't think of anything either

[keen coyote]

thats why im stuck on it

unsatisfying to move on

otherwise

[lavish jewel]

the book doesn't give any fancy solutions?

[keen coyote]

nah

no solutions other than this random guy who posted some partial ones for some questions

its Linear Algebra Done Wrong

chapter 5 section 3

pdf is online in case you’re curious

[lavish jewel]

everything is happening at v1 = 1, maybe you can make a pretty 3D drawing

idk

idk what they mean by "simplify the calculations", it's literally 2 dot products

https://tenor.com/view/oh-well-elmo-shrug-gif-11295857

even drawing it in 3D is a projection onto lower dimensions

[keen coyote]

i feel like they must just mean partial derivatives

literally the only thing i can think of

[tame mural]

hmmm

[novel hamlet]

I got question, i got Γ = (V, E)

i need to find transformation matrix for laplace operator for Γ with ∆: F(V ) → F(V ), with base (χp1, χp2, . . . , χp8)

[lavish jewel]

i must admit i have no idea what i'm looking at. let's see if someone else helps you out and i learn something by looking

[autumn haven]

Can someone help me with the motivation of $\mathrm{dim}({\vec{0}})=0$

[stoic python]

Mr. Stewart

[keen coyote]

assume not

well actually I guess a better explanation id

Is

Every vector space has a zero vector by definition

Of a vector space

So let's take R^2 for example

[graceful vortex]

the dimension of vector space which has a finite generating family is the cardinal of a basis

The empty set is a basis of the zero vector space

because the empty sum is zero

Hum that's not convincing

The empty set is trivially a linearly independant family

since there's nothing to check

and it's a generating set : since every vector space contains 0

(cont'd), the smallest sub-vector space containing the empty set is the zero vector space

i.e. the span of the empty set is the zero vector space

so the empty set is both linearly independant and generating : it's a basis

and it has cardinality 0

[prime drum]

Can i potentially set up an appointment with someone to help me figure out where my lack of knowledge is for my linear algebra class? I have a homework assignment that ive been working on that has me questioning everything

[limber sierra]

it might be a little hard to find someone to give one-on-one tutoring here, i'm afraid; does your class have office hours you can attend? can you maybe book a session with your instructor/a TA?

[novel hamlet]

its question about graph theory

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

[prime drum]

I'm going to attempt that aswell namington, i just am very lost and was just wondering if someone here would potentially be willing to help me later is all 🙂

[nocturne jewel]

Ok I asked this before but I dont fully understand still

If I have 2 linear operators S,T on finite dimensional V, then ST is invertible iff S and T are
Why does that break/not hold if V is infinite dimensional / what's an example?

[nocturne jewel]

But matrices arent linear operators?

or are those the matrices for the operators?

[limber sierra]

they represent linear operations in the vector space R^infty

[nocturne jewel]

which is R^n but inf?

[wintry steppe]

can i get help with matrix algebra?

[limber sierra]

yes; intuitively you can think of it as the set of all real sequences

[im handwaving a bit here]

R^infty certainly has a basis

(1, 0, 0...), (0,1, 0, ...), (0, 0, 1, ...), ...

[nocturne jewel]

Yeah last time I asked someone (I think Terra) mentioned infinite sequences

[limber sierra]

uh

then wtf does "infinite dimensional" mean to you

these are not standard definitions

[nocturne jewel]

But im not seeing the jump from those matrices to operators / sequences

[haughty lagoon]

I think they might mean finite linear combinations? is R^infty supposed to have finite support here

[limber sierra]

you dont need finite support here

you can view those matrices as representative of linear functions

but if this layer isnt quite satisfying

see https://math.stackexchange.com/questions/349807/does-invertibility-of-product-imply-invertibility-of-each-term-of-product

(the proving-these-actually-represent-linear-maps is the part im handwaving away)

[haughty lagoon]

I think this is a hilbert basis here instead of a hamel basis?

[limber sierra]

oh i see your confusion mirza

you dont actually need that tighter notion of bases here

to map lienar operators to matrices

the map just isnt invertible but we can ignore those issues

[nocturne jewel]

Is a hilbert space just a vector space?

[limber sierra]

that said, we still have that hamel bases exist

by the axiom of choice

every vector space has a hamel basis

it just usually wont be constructible

in this case though, it will be constructible i believe?

wait no brain fart

it isnt

you need choice to show it exists

and so it isnt constructible

again though that isnt quite necessary in this case

to map between matrices and linear maps it suffices just to consider hilbert bases

as long as those matrices are infinitely large

(im hiding some details here, dont worry about them)

[nocturne jewel]

Ok I think I digested the stackexhange post

[novel hamlet]

are matrix colums interchangeable?

[limber sierra]

what do you mean by that?

[novel hamlet]

ie. i have colums lets say i have 3x3 matrix wiht colums W,V,U and i want to calculate det(w,v,u) is it same as det(u,w,v) or det(v,w,u9?

[limber sierra]

its not the same, no, but swapping columns affects the determinant in the same way swapping rows does

in particular, swapping two columns multiplies the determinant by -1

so det(w, v, u) = -det(v, w, u)

but also det(w, v, u) = -det(v, w, u) = -(-det(u, w, v)) = det(u, w, v)

[novel hamlet]

i was thinking about this since i had to prove triscalar product

that it is commutative

and since V(UxW)

is essentially this

and from course book i know that

this is what i mean by interchangeable colums

[lavish jewel]

swapping columns changes the sign of the determinant

[novel hamlet]

but for triple scalar it is W(UxV)=V(WxU)=U(VxW)

[wintry sphinx]

note that these are always two column swaps away from one another

[novel hamlet]

yeah i know that

but im not sure how to prove that

[proper trout]

can someone help

Im pretty sure this is linear algebra not sure lmao

[limber sierra]

it is not; try #prealg-and-algebra or any of the ten #questions-_ channels instead

see the pins for a description of what linear algebra is.

[compact relic]

I'm familiarizing myself with Invertible matrix and its theorem right now. In one of the problem of a true/false statement, it's asking " Let A be a 3x3 matrix, for some vector v b in R3, Ax =b has more than one solution. would Ax = b consistent for every b in R3?

My answer is that it would be false, since Ax =b in this case is not necessarily an invertible matrix ( because the problem stated some vectors not all?)

[half karma]

Is this proof written properly? It doesn’t need to be rigorous. Just wanna see if it’s correct

[half karma]

<@&286206848099549185> can someone help me plz ^

[sonic osprey]

I'm not really sure what you're doing, I don't think its correct

I'm not sure what your first line even means? What are these two vectors?

[half karma]

I’m trying to generalize for this theorem

[sonic osprey]

These are uh, pretty different situations. The second picture is talking about a specific example of a subspace.

[half karma]

I think my variables might be wrong but it’s how my professor writes the proof for condition 1 and 2

[half karma]

Oof

[sonic osprey]

You should focus on the first picture more

[half karma]

Would it be correct if I replaced x with u and y with v? In the first part I’m showing u+v=0

[sonic osprey]

why are you showing that?

[half karma]

Isn’t it the case that if it contains the zero vector then U is a subset of W?

[sonic osprey]

uh what

what is W in your problem?

[half karma]

Oh sorry I actually forgot to include that in my question

Just a second

[sonic osprey]

Just look at

[half karma]

Oh W is the non-empty set

[sonic osprey]

uh

yes?

The notation in your question is different though be careful

In your question U and V are subspaces of W

[half karma]

Oh so I’m my case the non-empty sets U and V of the vector space W is a sub space of W if it satisfied U+V is also in W

[sonic osprey]

No

U is a subspace of W

V is a subspace of W

They're both nonempty subsets of W that satisfy the conditions to be a subspace

[half karma]

Ah so would I need to show that they contain the zero vector?

[half karma]

Ok that U+V is also in W

[sonic osprey]

You need to show these two things for U \intersect V

No?

[fervent gulch]

can anyone help me with lin alg? My dms are also open.... I got long questions

[sonic osprey]

U + V doesn't make sense here. U and V are not vectors, they're subspaces of W

[half karma]

Oh I was thinking like showing U and V are subspaces of W then their intersection must also be a sub space of W

[sonic osprey]

No that's what you're trying to show

You're given that U and V are subspaces of W

and you're trying to show that their intersection must also be a subspace of W

[half karma]

Oh

I think I’m getting it now

Bear with me plz I haven’t taken a proofs course

[fervent gulch]

anyone?

[half karma]

How would I show that u+v exists in U⋂V? Intuitively I’m thinking u+v exists since U⋂V is the subset of all the vectors in U and V

[sonic osprey]

what are u and v there?

[half karma]

Plz @fervent gulch this room is occupied

[half karma]

They are vectors in U⋂V?

[fervent gulch]

Alright ping me when u done so Ik when I can go.

[half karma]

Ok

[sonic osprey]

Okay now think about what you know about U and V

use the fact that U and V are subspaces

[half karma]

If U and V are subspaces of W then U⋂V is also a sub space of W since:

1. u+v exists in U and V,
2. cu is also in U and vice versa for V.

[sonic osprey]

yes

[half karma]

Does that prove it?

[sonic osprey]

yes

[half karma]

The last portion is correct right? @sonic osprey

[sonic osprey]

yes

[half karma]

Thank you so much!!!

@fervent gulch I’m done now

[fervent gulch]

yass

[half karma]

Repost your question

[fervent gulch]

I did

[half karma]

[1 0 0 -3] row 1
[0 1 0 4] row 2
[0 0 1 -3] row 3

Notice the bottom two rows cancel out. But idk if this is correct since the matrix isn’t square

[fervent gulch]

what happened to the other matrix

[limber sierra]

they asked for an LU factorization, not a row reduction.

[hollow wharf]

lowkey need some guidance here

[fervent gulch]

U can say she/her

[hollow wharf]

well, um am i supposed to like do one by one and see if all of them work or not?

or do i stop after one disproves it?

yeah well i dont know instantly lmao

this is the first problem like this i was given

so im trying to learn it

show all ok, sheesh thatll take a while

yeah i will fs

thanks

[fervent gulch]

Y'all left me on read...

[hollow wharf]

lmaooo

idk how to do dat lolz

[fervent gulch]

Anyone Ping me if ya know?

[lavish jewel]

@fervent gulch notice w = u - 3v

[fervent gulch]

okay but how does that go with T(w)

like do i do t(w)= t(u)-3t(v) @lavish jewel

[lavish jewel]

precisely, since T is a linear transformation

[fervent gulch]

alright i got it right. I can figure out most things but I just need lil guidance

I got a similar question to this

[lavish jewel]

same as above

[fervent gulch]

i'll get back to u in 10 mins if i got this right

[lavish jewel]

i'll be sleeping so dont @ me haha

[fervent gulch]

Ah I got it done

[compact relic]

how do you simplifies when you evaluating a determinant inside | |

say |A x B X A^-1 |

can you move the A^-1 in front to simply it out to Identity Matrix, is it not restricted by the order?

[nocturne jewel]

$|AB|=|A||B|$

[stoic python]

moshill1

[dusky epoch]

please do not use the letter x, and especially capital X, for multiplication

[hollow wharf]

does anyone know how to do #4?

[wintry steppe]

write in set notation
show your steps

[hollow wharf]

#4?

[wintry steppe]

$$S = {A \in M(3 \times 2, \bR) : \text{condition goes here}}$$

[stoic python]

(T*(Terra), -dτ)

[hollow wharf]

yeah but how does the matrix look?

im confused on how to set up #4, i finished #3

[wintry steppe]

in general, it'd look something like

hold on let me type it

[hollow wharf]

okay

[wintry steppe]

$$\begin{pmatrix} a & c \ b & d \ -a-b & -c-d \end{pmatrix}$$

[stoic python]

(T*(Terra), -dτ)

[wintry steppe]

it's a 3x2 matrix

and the elements in each column all sum to 0

i guess by "show your steps" they meant you're supposed to write out a matrix with entries like a, ..., f, and then show that it has to take on this form?

anyways, for the next part

verify subspace criteria

there is nothing more to it

[hollow wharf]

hmm

okay well thanks, ill try my best

'

can i write it like that?

and say that a+c+e=0?

and b+d+f=0?

[wintry steppe]

that's what it means for the entries in each column to sum to zero, yes

[hollow wharf]

ah ok

how would i write it in set notation tho?

[digital bough]

Like terra wrote above and then state your condition. One way would be just using indexing :

$$S = {A \in M(3 \times 2, \bR) : A_{ij}=0 , i\leq3 , j \leq 2}$$

[stoic python]

rts

[hollow wharf]

hmm

[digital bough]

Or Well, i and j are not less than 1 and they are integers

[wintry steppe]

huh

[hollow wharf]

im like having the biggest brain fart of my life

soooo confused

[wintry steppe]

one way to start would be to write $$S = \left{ \begin{pmatrix} a & b \ c & d \ e & f \end{pmatrix} : a,\dots, f \in \bR : a+c+e=b+d+f=0 \right}$$

[stoic python]

(T*(Terra), -dτ)

[wintry steppe]

looks shitty

but you know e = -a-c and f = -b-d, yeah?

so you can "eliminate" e and f

[digital bough]

Ah, my bad he wanted to sum all Columns

[wintry steppe]

columbus

[digital bough]

Ty ultra you the best

Terra*

[hollow wharf]

got it @wintry steppe

so

for substitute -a-c and -b-d for e and f?

[wintry steppe]

sure

[hollow wharf]

what does that help with tho?

[wintry steppe]

well the thing i just wrote is perfectly fine "set notation" for S

but presumably you're supposed to simplify

"simplify"

[hollow wharf]

oh okay

i see

so should i?

[wintry steppe]

sure

worst case ask your grader lol

[hollow wharf]

okay, i mean its prolly better to simplify

[wintry steppe]

ya

[hollow wharf]

ighty

and that alters the S, for every e and f, there is the -a-c and -b-d?

[wintry steppe]

if you "eliminate" $e$ and $f$ from the previous "set notation" you should get something like $$S = \left{ \begin{pmatrix} a & b \ c & d \ -a-c & -b-d \end{pmatrix} : a,b,c,d\in\bR \right}$$

[stoic python]

(T*(Terra), -dτ)

[hollow wharf]

do i need to write that they add up to 0?

[wintry steppe]

well they already do, dont they?

[hollow wharf]

or is that already explained in the a+c-a-c

[wintry steppe]

i'd think so

[hollow wharf]

yeah sounds about right

,rotate

[hollow wharf]

thats what i got rn

so thats the first part right?

now i have to fine the subspace of V=M3x2(R)

right?

[digital bough]

Ye, you can use your set notation to see if it is at least closed under standard addition or scalar, if it isnt then uh... idk.

[hollow wharf]

lmaoo

ighty imma give it a shot and submit it, its been a long day

[wintry steppe]

meme answer: it's the kernel of a linear transfornmation M(3 x 2, R) -> R^2 so it's a subspace

[hollow wharf]

hehe 🙂

[wintry steppe]

namely, the linear transformation whose first, second components are the sums of the first, second column entries of the matrix, respectively

[hollow wharf]

ight i finished

[silk ether]

I wonder, why can't Gauss-Seidel method solve every system of linear equations? (like sometimes values would go to infinity)

[dusky epoch]

well if you have zeros on the diagonal then you have a problem

[silk ether]

But umm, the values given in the diagonal don't have any zero

The solutions are still getting larger, approaching infinity

[dusky epoch]

maybe your matrix isnt diagonally dominant?

can i see your matrix

[silk ether]

Umm wait

This was the system that I tried on

I was trying to solve using excel

[lavish jewel]

is there any special reason you're using that specific method btw?

[dusky epoch]

might be that they'll get sent to gulag if they don't use it

[silk ether]

Just try for fun 😂

[dusky epoch]

anyway, yeah, the coefficient matrix for this system is $$\begin{bmatrix} 1 & 2 & -3 \ 2 & -5 & 4 \ 5 & 4 & -1 \end{bmatrix}$$ which isn't diagonally dominant and doesn't look positive-semidefinite either

[stoic python]

Ann

[dusky epoch]

wait what am i talking about

it's not even symmetric, forget the pos-semidef thing

[silk ether]

I need to check this out later 🤔

Thank you very much :)

[lavish jewel]

for a quick test of your method, you can multiply both sides of the equation by M^T (the matrix you have there)

that should be solvable with your alg if it's working correctly

[novel hamlet]

you guys got a easy cheat to do 8x8 matrix determinant?

[native rampart]

Block matrices or smt?

Gaussian can take up to a max of 28 steps if you are ok with that?

[novel hamlet]

my textbook jsut gave hint to sum 7 first colums

[native rampart]

Yea,It depends on the kind of matrix

[novel hamlet]

this is the matrix i got

[native rampart]

Yea,This one's easy

[novel hamlet]

i summed firstt for colums and got 1

not sure what im supposed to do with that result

ah nvm sum is 0

det should be 0

but im not sure how to "calculate" it

[native rampart]

Yes

operations of form R_1->R^1+aR_2 preserve determinant

[novel hamlet]

how do i do trace for it

just sum diagonal?

[native rampart]

Sum the diagonal of the original matrix

[novel hamlet]

that is the original, so i get 16

thanks

just read the final question instructions: "you may use calculation software or calculate eigenvalues and eigenspaces by hand, in either case show your reasoning"

im kinda curious what is the reasoning for i put values on eg. wolfram alpha and trust it count better than me

xD

[mellow solar]

what is below the y-1 and 3-z, do i need to divide everything by two or is it 1?

this is line

[dusky heart]

if I have a matrix A = [4 -1; 5 2] and it's asking for (the L looking things are capital I's as in igloo) 3I_2 - (3I_2)A

im not sure what is meant by I

above the question it also says: **In the rest of this exercise set and in those to follow, you should
assume that each matrix expression is defined. That is, the sizes of
the matrices (and vectors) involved "match" appropriately. **

[urban egret]

maybe the I stands for identity matrix?

so I_2 is (1 0; 0 1)

does the notation O(n) for an orthogonal group mean that the matrix is orthonormal? or just orthogonal?

[dusky heart]

oh I see what u mean thanks

[wintry steppe]

Could someone explain to me what through the plane x_3 = 0 means?

[quartz compass]

imagine the plane like a mirror

whatever point is on one side of the mirror reflects an equal distance to the other side of it

[wintry steppe]

I'm not entirely sure how to do this, can't really find the explanation in the book

The book references an example solution where a transformation is the multiplication of a vector by a scalar, and the explanation of how this is a linear transformation

But that does not seem very similar to this problem

[fast vector]

Do you have a definition of „linear transformation“?
It should be defined by saying that one or two equalities must hold.
Then try to verify these equations for the map T: ℝ³→ℝ³, (x1, x2, x3)↦(x1,x2,-x3).

[wintry steppe]

Yeah, the two claims would be

1. T(u + v) = T(u) + T(v)
2. T(cu) = cT(u)
   I actually understand the answer to this question (how to prove it is linear), the only issue I have is that I don't understand what through the plane x_3 = 0 means

[lavish jewel]

it is telling you up there what it does

it flips the third component of the vector

if x = (x1,x2,x3), T(x) = (x1,x2,-x3)

[wintry steppe]

Oh man I feel so dumb lol

Thank you, sorry about that

can someone please help me with the inversions of permutations

i'm losing my mind over it because there are like two "definitions" and i dont know what is going on

first my notes say that (i, j) are an inversion if i < j and p(i) > p(j)

where p is some permutation

but then i saw in so many places that if (a b c d e f) is a permutation

the inversion is when some number is larger than a number thats more right

so what's going on here?

[limber sierra]

those are the same definition

[wintry steppe]

how about (1234)

[limber sierra]

there are no inversions there.

[wintry steppe]

wait a moment

hmm but thats not the same as if i > j and p(i) < p(j) right?

ok lets look at (2314)

is (2,1) an inversion here?

[limber sierra]

yes

2 occupies the 1 slot, and 1 occupies the 3 slot, so 1 < 3 but p(1) > p(3)

im assuming this is the second row of a permutation map and not cycle notation

[wintry steppe]

ohhhhh okay that was my problem

i thought that this second rule is for cycles

i never saw a notation of just the second row

[limber sierra]

ah no

[wintry steppe]

so i assumed that must be a cycle but now i think youre right

ok thank you so much i was losing my mind over that

[limber sierra]

cycle notation is a bit harder to see the inversions directly

[wintry steppe]

yes because if it was a cycle notation (2,1) wouldnt be an inverse right?

because 2 isnt even smaller than 1?

[limber sierra]

well it wouldnt fit the strict definition

that would actually still be an inversion if we consider it as a two-row matrix:
(1 2)
(2 1)

but it wouldnt fit the definition laid out above

[wintry steppe]

if (2314) is a cycle?

then if i understand you i could write it as (1423)

[limber sierra]

yeah (2314) and (1423) are the same cycle

[wintry steppe]

and then it would be an inversion?

[limber sierra]

but they would have different numbers of inversions

[wintry steppe]

but (1,2) would be an inversion right?

not (2,1)

[limber sierra]

er

other way around

$\begin{pmatrix}1&2\1&2\end{pmatrix}$ has no inversions but $\begin{pmatrix}1&2\2&1\end{pmatrix}$ has one inversion

[stoic python]

Namington

[wintry steppe]

ok yes i agree

thank you so much!!!

[limber sierra]

so youre right that if you tried to apply this definition

to cycle notation

youd run into issues

since (1 2) and (2 1) would then have a different amount of inversions

but they represent the same permutation

[wintry steppe]

yes you have to work with the strict definition when you have cycles

[limber sierra]

yeah, you want to convert to the two-row matrix representation first

ideally

(alternatively you could break it down into a product of 2-cycles)

[wintry steppe]

yes ok thanks i get it now

[worthy plank]

what would be the easiest/quickest way of finding the 4th column of C^-1?

[lavish jewel]

you can invert L pretty easily

inversion commutes with transpose, so once you have L^-1, you're pretty much set

also, since the matrix in the middle is diagonal, the 4th column will be the 4th column of L^T^-1 scaled by the sum of the elements of the 4th column of L^-1 scaled by 1/elements of the diagonal matrix

[sleek spruce]

hello good people

maybe i ask why

a matrix A, that is 5x6, with rank 3

can not span R^3?

the cols of A

[wintry steppe]

the columns of A don't even live in R^3

so no

[sleek spruce]

could you elaborate

if it's a 5x6, the dimensions are 5

and if it's rank 3

it has 3 pivots

so shouldn't it span R^3

[wintry steppe]

the columns have how many entries?

and elements of R^3 have how many entries?

[sleek spruce]

5 entries

and 3 entries

but the cols of A

reside in a 3d subspace

so why doesn't it span R^3

my imagination isn't working

my brain clearly shows the cols of A spanning a 3d space in a 5d

[wintry steppe]

because the columns literally don't lie in R^3. they span a 3 dimensional space, yes, but they don't lie in R^3

that three dimensional space need not be and will not be R^3

[sleek spruce]

could you differentiate what it means to lie on a R^n?

[wintry steppe]

is a member of

[sleek spruce]

oh

OH

ok

bruh

tyvm

so just to get my understanding

it spans a 3d space, but it's not in a 3d space

[wintry steppe]

yeah that sounds good

[sleek spruce]

tyvm

[wintry steppe]

the span will be a 3 dimensional subspace of R^5

Before we defined the determinant we defined multilinear transformations in the following way: F: U^n -> V is n-linear if all maps U -> V, u -> F(u_1, ..., u_n) are linear for all i ∈ {1, ..., n} and all vectors u_i ∈ U

The professor said that in order to do that we have to assume that 1 + 1 isn't 0 in the field that we're working with

my question is why is that condition necessary?

probably just so you don't have to deal with nonsense involving characteristic 2 fields

1 + 1 = 0 is bad

what is characteristic 2

i never heard of it before

it means a field in which 1 + 1 = 0 (a field is of characteristic p if p is the smallest positive integer for which 1 summed p times equals 0)

slightly better answer:

some aspects of the theory of bilinear forms (and by extension multilinear forms) become a little strange when you're working over a field in which 1 = -1, if i recall correctly

https://kconrad.math.uconn.edu/blurbs/linmultialg/bilinearform.pdf
this will probably elucidate that better than i can, see the table of contents first paragraph

e.g. any alternating (B(u, u) = 0 for all u) bilinear form is skew-symmetric (B(u, v) = -B(v, u) for all u, v). the converse is true outside of char 2 (because B(u,u) = -B(u,u), and since 1 is not equal to -1, B(u,u)=0), but false in char 2

oh okay thanks so much

we haven't yet covered bilinear forms so i'll wait until then for more understanding

but i think this thing with antisymmetry is the main thing

ya

because we have used that and apparently it doesnt work with char2

[wispy pewter]

does anyone know how to smoothly write the inner product of a bra and ket in LaTeX?

like this

[gray dust]

$\ip{x}{y}$

[stoic python]

RokabeJintarou

[gray dust]

physics package @wispy pewter

[wispy pewter]

ah ok that looks good, how about the lineing up a row vector and column vector

[gray dust]

bmatrix

[stoic python]

EmilD

[wispy pewter]

is there no way to line them up?

also dam that LaTeX command for the inner product didn't work in overleaf

[gray dust]

import physics package

[limber sierra]

what is meant by "line them up"

what are you looking to do

[gloomy arrow]

When you want to do multiple rotations all at once, do you add the rotation matrices or do you multiply them?

[limber sierra]

what do you mean by "at once"?

like rather than rotate by 90 and then again by 90 [by applying a rotation matrix twice], rotate by 180 [by applying a rotation matrix once]?
you would multiply, since linear transformation composition is the same thing as matrix multiplication

[gloomy arrow]

^^^ This is in fact what I meant

Thank you

[stable kindle]

is there any geometrical intuition for why Ax(BxC) = B(A.C) - C(A.B)?

like for the scalar triple product being cyclic it's like any way you do it you get the volume of the parallelepiped which is quite nice

[wintry steppe]

$$\begin{bmatrix}x\y\z\end{bmatrix}=xe_1+ye_2+ze_3.$$ use linearity.

[stoic python]

(T*(Terra), -dτ)

[wintry steppe]

(presumably "transformation" here means "linear transformation")

[fervent gulch]

Hold on

[fervent gulch]

so how do i obtain that and I can resize the matrix

[wintry steppe]

i don't understand

[fervent gulch]

I want to get t ( x y z) , I'll do the calculations but I do need guidance

[wintry steppe]

$$T\left(\begin{bmatrix}x\y\z\end{bmatrix}\right)=T(xe_1+ye_2+ze_3).$$ apply linearity and simplify

[stoic python]

(T*(Terra), -dτ)

[wintry steppe]

you know how T acts on e_1, e_2, and e_3

that's it

[fervent gulch]

ok gimme 5 mins and i'll let you know if i understood

ty Terra

[fervent gulch]

I obtained -22, 17

[fervent gulch]

My bad I realized what when wrong

Ok I got it right, now next question

[wispy pewter]

does anyone know the command to have wolfram alpha compute an innerproduct of a row vector and a column vector?

[nocturne jewel]

basically solve Au=b, so either gauss jordan or find the inverse of A if A is invertible

[fervent gulch]

hold on i'll let u know if i got it wrong

I got this for gauss jordan

[nocturne jewel]

yeah, did you do it w/ the augmented matrix?

[fervent gulch]

eh i haven't only inserted a for gauss jordan

[nocturne jewel]

Ok so either solve the augmented matrix or find the inverse

[fervent gulch]

this is inverse of a

[nocturne jewel]

Ok.. so solve Au=b

[fervent gulch]

wait but I don't have u

[nocturne jewel]

Yes but you found A^-1

[fervent gulch]

how do i use the inverse to lead me there

[nocturne jewel]

left multiply both sides by A^-1

[fervent gulch]

mulitply b by A^-1?

[nocturne jewel]

[fervent gulch]

both sides of what moshill1

[nocturne jewel]

Au=b...

[fervent gulch]

Maybe visuals would help, I am getting there

[nocturne jewel]

$A^{-1}Au=A^{-1}b \to u = A^{-1}b$

[stoic python]

moshill1

[fervent gulch]

for A^-1 times b

[nocturne jewel]

im guessing you're just getting a program to do the calculation for you from the screenshots, but sure

[fervent gulch]

y so salty? I am getting the easy things done manually by hand, i would do so if I had time to spare this evening

He/She/They left the groupchat, anyone willing to guide me

<@&286206848099549185>

I am not sure I feel like i prob left out something

The previous person said to multiply by both sides I only did invertible A times b

#linear-algebra message

I am not sure how to do the rest that all

I feel like i did a portion but I want to make sure, u said that" computing A^{-1} b gives you u" so I thought u meant I was done

I did read it but I only understand partially

so like after i did the A^{-1} b gives you u, what do i gotta do next. did i get the complete answer or was there more required.

#linear-algebra message

nvm I got it wrong.

yea no worries

[hollow finch]

anyone have an example of a LA proof that uses the contrapositive?

[nocturne jewel]

proof by contradiction?

[hollow finch]

i mean just a theorem or fact in linear algebra that has a proof by contrapositive

[nocturne jewel]

what's a contrapositive?

[lavish jewel]

off the top of my head, vandermonde matrices are shown to be full rank by contradiction, if that's what you mean

[hollow finch]

$(P\implies Q )\iff (\neg Q \implies \neg P)$

[stoic python]

nix

[nocturne jewel]

Havent learned formal logic yet lol

or wherever those types of statements show up

[hollow finch]

P=the matrix M is invertible
Q=zero is not an eigenvalue of M

If the matrix M is invertible, then zero is not an eigenvalue of M
is equivalent to
If zero is an eigenvalue of M, then the matrix M is not invertible

for example

[nocturne jewel]

[lavish jewel]

that looks like a proof by contradiction

[hollow finch]

i was just giving an example of what the contrapositive is

[fervent gulch]

u busy?

[lavish jewel]

my bad, contrapositive is a more special case

[nocturne jewel]

Probably getting more into proof/logic, but what's the distinction b/w contradiction and contrapositive then?

Proving contradiction is conditional and contrapositive is biconditional?

[lavish jewel]

in a contradiction, assuming p leads to q and not q at the same time

[nocturne jewel]

contradiction is start with the opposite of P and show an obvious false-ness?

[lavish jewel]

mhm

the two can sometimes be the same

[hollow finch]

sounds like Modus Tollens

[lavish jewel]

if p then q and you can show that q implies w and not w, then not q and therefore not p

ins't modus tollens contraposition?

[hollow finch]

$P\implies Q$

$\neg Q$

$\therefore \neg P$

[stoic python]

nix

[hollow finch]

personally not quite seeing the contraposition. i thought this was like contradiction because you usually do those by assuming Q is false

[lavish jewel]

oh, you mean to show that not q then not p, not just not q

in what you wrote one has to show q is false, not assume it

you start with knwledge that if p then q

you show that not q, therefore not p

in contradictions you show that if p then q and not q, which doesn't make sense

modus tollens is the law of contrapositive

but still this is all tangent to what you asked, sorry. i can't think of any examples at 3 am lol

[hollow finch]

tfw im an idiot

[lavish jewel]

anyway yeah

do you mean to prove p then q by showing that not q then not p?

or to start with knowledge that p then q, and show that not q therefore not p?

the vandermonde matrix example i gave starts with the assumption that the columns (for a certain vandermonde mat m x n) are lin dep, which means they have a nontrivial linear comb that gives a 0 vector

you show that the latter is false, amd therefore so is the former. this uses a contradiction as well tho

someone will give you a better example and correct my mistakes while i sleep

[pseudo thicket]

Can I check for this slide, after reducing the matrix A, I got :

A=[
1 0
0 1
0 0 ]

There is a zero row. That means it does not have a solution for every vector b, right?

But the span/column space of A is a 2D plane since we only have 2 Linearly Independent column.

[wintry steppe]

the columns of A can't span all of R^3 since there are only two of them

so yes, you are right, there will be some b for which Ax = b has no solution

you're also right that the span is a two dimensional plane, since the columns are linearly independent

everything you've said checks out 👍

[pseudo thicket]

Thank you @wintry steppe !

[wintry steppe]

[silk ether]

How do y'all solve matrices larger than 4x4 👀

Is that done manually?

[native rampart]

Matlab or some other program,ig

Assuming you mean solving a system of linear equations in more than 4 variables by "solve"

[limber sierra]

it's not really any harder

just more tedious/time-consuming

(and more opportunities to make mistakes)

[vocal prairie]

\textbf{Problem. }Let $V$ be a vector space having dimension $n$, and let $S$ be a subset of $V$ that generates $V$. Prove that there is a subset of $S$ that is a basis for $V$.\begin{proof}If $V$ is the zero vector space, then $S=\emptyset$ is the only possible spanning set, and $\emptyset$ is a basis for $V$, thus proving the claim. Consider now that $V$ is not the zero vector space. Suppose to the contrary that no subset of $S$ forms a basis for $V$, that is, any linearly independent subset of $S$ has at-most $n-1$ vectors(a linearly independent subset of $V$ can't have more than $n$ vectors by the replacement theorem). Call this linearly independent subset $A$. Now, all the vectors in $S\setminus A$ can be represented as a linear combination of vectors in $A$, which means $\text{span} (A)=\text{span} (S)=V$. Thus, $A$ is a linearly independent spanning set of $V$, and is therefore a basis by definition. But $A$ has at-most $n-1$ vectors, which contradicts the fact that $V$ is $n$-dimensional. Hence, $A$ must have $n$ vectors as well. This proves the claim.\end{proof}

[stoic python]

Ted

[vocal prairie]

Does this proof feel shaky?

The bit about A having n vectors...doesn't sound right. I think I'm either missing something, or adding some redundancy.

[native rampart]

That shows A has >=n vectors

So,There would be one more step

[vocal prairie]

But I already ruled out A having more than n vectors by the replacement theorem

[native rampart]

Then, It's fine

[vocal prairie]

[vocal prairie]

Is contradiction redundant in the above proof?

Can I not just take the largest possible linearly independent subset, claim that it spans the space, but can't have more than/less than n vectors due to replacement theorem and its corollaries?

[native rampart]

how do you know it spans the space?

[vocal prairie]

Same reasoning as above

[native rampart]

a better approach would be to take the minimum subset which spans the space and conclude it is a basis

[vocal prairie]

Any linear combination in terms of vectors of S can be reduced to a linear combination of vectors of this largest linearly independent subset

[vocal prairie]

How do you formalise this idea for a proof?

[tame mural]

perhaps u just remove 1 vector at a time

[vocal prairie]

S could be infinite

The process need not end

[tame mural]

it says dimension n

right?

[vocal prairie]

Yeah, but it doesn't mean S has to be finite

You could keep picking out vectors forever

[tame mural]

ic

[native rampart]

start with zero add vectors as long as the new gen set is LI

[vocal prairie]

Hmmm, okay

[tame mural]

but u could also choose the wrong vectors forever

[vocal prairie]

¯\_(ツ)_/¯

I'm more insistent on the contradiction approach for this reason

I still don't usually get the "choose one vector, choose another..." approach.

[tame mural]

it's just that i remembered axler had a proof with removing vectors until it was the minimal span

i shall perhaps look it up

[vocal prairie]

Thanks!

[tame mural]

ah

Axler's proof is for finite lists which are still spanning

Ah

Axler's approach is to prove that every finite-dimensional vector space by definition has a spanning list that is also finite

[vocal prairie]

So that's just saying every finite dimensional vector space has a finite basis

I don't think that helps with this problem, does it?

[tame mural]

but once you prove the existence of such a list

you can start subtracting in a finite way

You can bluntly define a finite-dimensional vector space as implying the existence of such a set

[vocal prairie]

I see. For now I'm just going with the idea of picking the biggest possible linearly independent subset of S and then proving that it is indeed a basis; I'll try your approch as well. Thanks!

[sudden egret]

can somebody explain how did they get the det A^-1=19?

[pseudo thicket]

In a homogenous equation where mxn matrix,m>n

Is it right to say we can state these points:
there could be infinitely many solutions or no solution
But there can never be just 1 solution?

[marble lance]

No. It has 1 solution or infinitely many solutions. 0 is always a solution, so it must have at least one.

[pseudo thicket]

but there could be no solution since there is a zero row

[marble lance]

What do you mean by a homogenous equation? Ax = 0?

[pseudo thicket]

yes

[marble lance]

A(0) = 0 is always true. It doesn't matter what A is.

So x = 0 is always a solution

[pseudo thicket]

okay so at least 1 solution

But how does infinite solution come about?

[marble lance]

If A is not invertible

[magic light]

Is there a way to get from the base (2,1,1), (-1, 1, 0), (-1, 0, 1) to the base (1,1,1), (-1, 1, 0), (-1, 0, 1) without zeroing (2, 1, 1)?

Basically turn (2,1,1) to (1,1,1)

[lavish jewel]

what exactly do you wanna do

[pseudo thicket]

thanks, so it either has only the trivial solution or infinitely many solutions (in addition to trivial solution)

[marble lance]

Yes @pseudo thicket

[pseudo thicket]

thank you

[marble lance]

Np

[pseudo thicket]

I'm trying to create a sample question to get the infintiely many solutions, how should I craft them?

[marble lance]

Create a matrix with zero determinant

Then it won't be invertible

Easiest answer: the zero matrix.

[magic light]

I made a mistake in the test, I was asked to orthonormalize the eigenvectors of some matrix
after fixing a mistake in one area I didn't fix it in another and got the base with (2, 1, 1) instead of (1, 1, 1) - rest of the vectors were accurate/correct.
I was wondering if I can argue that (1, 1, 1) is a combination of (2, 1, 1) and the rest of the base and thus I still deserve some points despite the last vector being "wrong"

[lavish jewel]

well

,w {{1,-1,-1},{1,1,0},{1,0,1}}^-1*{{2},{1},{1}}

[stoic python]

Results provided by WolframAlpha

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Requested by Edd#5182

[magic light]

nice, so it is one

[lavish jewel]

ofc, both of the matrices are full rank transformation matrices

[magic light]

why the reverse though?

[lavish jewel]

my guess is you will get no credit, though, because finding a basis is easy. they specifically asked you for an orthogonal basis

[magic light]

I got an orthonormal basis, but with the wrong 3rd vector

the rest of it was correct

[lavish jewel]

yours isn't orthogonal, and even less orthonormal

[magic light]

I know, ok:

I had a matrix A
I was asked to find an orthonormal basis for eigenvectors of A^TA
I found eigenvalues(correct)
I found eigenvectors(2 correct, 1 incorrect) <- my question here
I did gram schmidt on the 3 vectors I found
I normalized all the vectors
in the end I had 2 normalized vectors and 1 incorrect vector

[lavish jewel]

if you did gram schmidt correctly, the result will be a set of 3 orthonormal vectors

[magic light]

Yeah, but they are not the eigenvectors of A^TA

I also think I did a mistake in gram schmidt at v3

though I still didn't figure out what it is

[lavish jewel]

that is more likely

[magic light]

I still didn't get the results, I just realized I made a mistake halfway with the vectors and was wondering if I can argue that the step is (technically) correct since I still had an eigenvector

in case I lose points

[lavish jewel]

i mean, it's not an eigenvector

it'll change directions

[magic light]

yeah.

[lavish jewel]

the matrix is full rank, so your argument won't hold. you can do what i did up there for any vector in R3

you'll get credit for those 3 vectors and for trying G-S, most likely, but that's about it

[magic light]

well (2, 1, 1) is linearly independent

so it's not just any vector

also at the end, end result
two out of three vectors are correct

Since I used (2, 1, 1) as v_3

only the last vector is not an orthonormalized eigenvector

[lavish jewel]

they'll probably notice that

[hushed dock]

For tensor products, does $$(A \otimes B)(A' \otimes B')=AA' \otimes BB'$$ where A,A',B,B' are matrices?

[stoic python]

init0

[lavish jewel]

yes

[hushed dock]

proof?

[lavish jewel]

you can expand the product block by block and check

[hushed dock]

could you explain?

what do you mean by expanding it block by block

[lavish jewel]

kronecker products have a block structure

[magic light]

@lavish jewel
If Av = xv
then isn't it true
3Av = 3xv => A3v = 3xv => 3x is an eigenvalue as well as x?

for any number really?

except 0

[lavish jewel]

what you wrote is correct, but the conclusion is wrong

[magic light]

isn't 3v it an eigenvector of 3x?

[lavish jewel]

if 3x is an eigenvalue, then the corresponding vector is scaled by 3x

[magic light]

yeah

so 3v is an eigenvector as well

[lavish jewel]

no, because 3v is scaled by x

that'S why you got 3xv

not 9xv

[magic light]

Ok, but by definition can't we say x' = 3x is an eigenvalue because there exists Av' = x'v?

with v' = 3v

[lavish jewel]

look at your equation

that's not the definition of an eigenvector

it's not the same vector on the other side

[magic light]

Right, it's two different vectors with two different values

but aren't they both eigenvectors and eigenvalues, albeit differently? or am I misunderstanding something

[lavish jewel]

you are misunderstanding something indeed

[magic light]

Av = xv for non-zero vectors, means that x is an eigenvalue and v is an eigenvector

A3v = 3xv where x' = 3x and v' = 3v means that x' is an eigenvalue and 3v is an eigenvector because Av'=x'v

[lavish jewel]

that's wrong

[magic light]

hmm

[lavish jewel]

you're saying

Av = xv means v is an eig vect with eig val x

[magic light]

yeah

with you so far

[lavish jewel]

so Aw = uz makes u and z an eigenvector z and an eigenvalue u

??

[pseudo thicket]

hold on, wouldnt there be only inifinitely many solutions?

since there are free variables!

[magic light]

w is an eigenvector of u eigenvalue

[lavish jewel]

no

because you got z on the other side, not w

what the fuck

it has to be the SAME vector on both sides

SAME

[magic light]

okay, don't get angry with me! 😦

I see what you mean now

A3v = 3xv => A3v = x3v
so x is an eigenvalue of 3v?

now we don't have infinitely many eigenvalues, just vectors I guess

[lavish jewel]

right

that's why one usually normalizes the eigenvector

any vector parallel that direction will get scaled by the corresponding eigenvalue

so only the direction is important

[magic light]

right, which is why (2, 1, 1) is a bad answer nonetheless S:

even though it is a combination of other eigenvectors

[lavish jewel]

yep

[magic light]

sucks

hopefully they won't notice, the answer is 2 pages long and the result is still orthonormalized vectors lol

at least partial points

it's a 20 point question 😦

[lavish jewel]

you'll get at least partial credit

[magic light]

hopefully

[lavish jewel]

a merciful teacher would check separately that you did G-S correctly with the vectors you picked, so you'd get most of the credit

[magic light]

I checked it with a calculator and didn't find any issues, but the result of G-S in an online calculator seem different

actually I should just try to inner product them and see

[pseudo thicket]

in a matrix where m<n, where the equation is homogenous, then there will be inifinitely many solutions only, right?

[lavish jewel]

yep

[hushed dock]

if a complex matrix is invertible, is it also unitary?

[native rampart]

No

[quartz compass]

unitary is like the complex version of an orthogonal matrix

it has determinant |z|=1

[stable kindle]

okay, asking one more time, is there any geometrical intuition for why Ax(BxC) = B(A.C) - C(A.B)?
like for the scalar triple product being cyclic it's like any way you do it you get the volume of the parallelepiped which is quite nice

[native rampart]

I don't think cross products work nicely

[quartz compass]

hmm well the vector BxC is perpendicular to B and C, so when you cross A with BxC you get something that lies in the plane spanned by B and C, so you can look at the projection of A on B and C with those dot products I guess

I'd have to think a bit more about why it has that exact form but at least that seems reasonable enough intuition for why you end up with the dot products A.C and A.B at least

[novel hamlet]

Edd, I managed to figure out the math of you are intrested

[stable kindle]

hmmm

[lavish jewel]

what math

[novel hamlet]

The one I asked few days ago about the net

And you said you dont know howbto do it

With the funny circle imagine and needed to make matrix out of it

[lavish jewel]

oh

yeah after you said it was a graph i realized

the graph laplacian is a difference matrix

[novel hamlet]

Yeah I managed to count it

[lavish jewel]

aight

GG

[novel hamlet]

But it took quite long

[lavish jewel]

it's a big matrix even for only a few nodes, yeah

[novel hamlet]

Before I realised for each Xpn when pn is same then 1

Othervise 0

[lavish jewel]

if you rotate [1,0] by 4/3 pi, what does it become?

[fervent gulch]

hold on, i'm doing it

(4/3pi, 0) is what I got

@lavish jewel

[lavish jewel]

nope, the rotation should change both coordinates

draw a unit circle and check

[fervent gulch]

I am checking and I still have trouble @lavish jewel

[limber sierra]

where does a rotation of 4pi/3 land on the unit circle?

it's shown somewhere in that image

[fervent gulch]

(-1/2, -sqrt3/2)

[limber sierra]

right (assuming the rotation is counterclockwise)

[fervent gulch]

Ok so how do i proceed w/ that

[limber sierra]

well we know $\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}1\0\end{pmatrix} = \begin{pmatrix}-1/2\-\sqrt{3}/2\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

this isnt quite enough to solve for a, b, c, d

but if we also check where (0 1) ends up upon a rotation by 4pi/3

we can do the same thing and combine these data to get a system of linear equations to solve

though

idk if youre expected to rederive this