linear-algebra

[magic light]

and this is a different base

B

[spare widget]

B1 = (1,1,1,0,-1)

[magic light]

a different basis

[spare widget]

Then this B1 is a coordinate reprrsentation of some vector b1

Is it true that $B_1 = [b_1]_e$?

[stoic python]

criver

[spare widget]

Where _e indicates the canonical basis

[magic light]

well what is [b1]_e? it means that b1 = a1 * v1 + a2 * v1... where v_i is the standard base

[spare widget]

yes

It means (a1, a2,...,an)

wrt e1,e2,e3

[magic light]

sure then, I guess? it changes nothing basically right?

[spare widget]

e.g. $B_i \ne [b_i]_C$

[stoic python]

criver

[magic light]

hmm

yeah

[spare widget]

Ok so B_i are b_i wrt e

And I is the matrix such that

[magic light]

please don't tell me to sandwhich xd

[spare widget]

$I [v]_B = [v]_C$

[magic light]

wait

[stoic python]

criver

[magic light]

what is $$[b_i]_B?$$

[stoic python]

blackmamba[they/them]

[spare widget]

(1,0,0,0,0)

[magic light]

yeah

So

I brings me from B to C

don't I need to multiply by (1, 0, 0, 0, 0)

and then I get $$[b_1]c = (1, 0, 0, 0, 0)$$

[stoic python]

blackmamba[they/them]

[magic light]

and what does that mean

it means b1 = 1c1 + 0c2 ....

so b1 = c1

$$[I]^B_C*[b_1]_B = [b_1]_c$$

[stoic python]

blackmamba[they/them]

[spare widget]

I think that's incorrect because while $[b_1]_B = [1, 0, 0, 0, 0]$

[stoic python]

criver

[magic light]

and this my matrix

[spare widget]

b_1 is not a contravariant vector

It is a basis vector

[magic light]

that's the definition of the transition tho

[spare widget]

So it requires a transformation like a covariant vector

[spare widget]

For contravariant vectors, not covariant ones

[magic light]

do you agree this property holds for all w?

[spare widget]

The basis vectors transform in the opposite way

I'll give you an example

[magic light]

wait, one thing at a time

I'm not sure what contravariants are, but this is partly what the solution shows so it is probably correct...

[spare widget]

if w is a contravariant vector then sure, if v is a covariant vector then no

Let me give you an example

[magic light]

according to my sheet it holds for all though

🤔

[spare widget]

Say you're in 1D

And you do a transformation that does [v]_C = 2 * [v]_B

That is, the coordinates become twice larger

So the vector coord [v]_B=2 becomes the vector coord [v]_C= 4

[magic light]

the transformation matrix is the identity one here

[spare widget]

it's not it's 2

[magic light]

I mean in my example

its the identity matrix

[spare widget]

So it looks like your vector became twice longer in my example

[magic light]

I get your point w/ the issue

the thing is I wrote it to be true for identity transformations

[spare widget]

But it is the same vector, so it should be the same length

Now let the basis vector b1 = 1 in the canonical basis

In C b1 is 1/2

In order to compensate for the coordinates becoming twice larger

From there you see that the vectors (i.e. contravariant vectors) transform inthe opposite wsy of the basis vectors

for the identity, the opposite is the identity

So nothing changes

But for other transformations its the reverse

You have to be careful how you transform vectors and covectors

The rules for transforming them are different

And the rules are such so that length and angle measurements of the same vectors agree between different coordinate systems

[magic light]

alright

fk, so annoying this material.

[spare widget]

https://youtu.be/A1h_eucHFW4

[magic light]

if A is anti-symmetrical(A=-A^T) does that mean that over R all of its eigenvalues are 0?
because:

$$Av = xv => -A^Tv=xv => (-A^Tv)^T=(xv)^T $$
$$=> -v^TA=xv^T => -v^TAv=xv^Tv => $$
$$-v^Txv=xv^Tv => x=-x => x=0 over R$$

[stoic python]

blackmamba[they/them]

[spare widget]

it's 0 or imaginary only eigenvalues

[magic light]

question is my proof correct or did I mess up the matrix multiplications

[finite karma]

is the Jacobian limited to 2 dimensions?

[spare widget]

no

[nocturne jewel]

Jacobian is just abs value of the derivative

whatever size the derivative is matrix wise doesn't change

[finite karma]

I have a vector function and want to take the matrix derivative

[spare widget]

well some people call the first derivative matrix the Jacobian

[finite karma]

is the result some form of jacobian?

[nocturne jewel]

yes

[spare widget]

by vector function you mean (x(u,v), y(u,v), z(u,v))?

[nocturne jewel]

Jacobian is by definition a derivative matrix

it's abs value of the det

[spare widget]

you can get a non-square matrix e.g. for surfaces in 3D

but then you can take sqrt(det[J^TJ])

[finite karma]

but is the Jacobian still 2D?

[magic light]

$$Av = xv => -A^Tv=xv =>transpose (-A^Tv)^T=(xv)^T $$
$$=> -v^TA=xv^T =>{"multiply by v"} -v^TAv=xv^Tv => $$
$$-v^Txv=xv^Tv => x=-x => x=0 over R$$

[spare widget]

what do you mean is the Jacobian 2D?

[magic light]

fk

[spare widget]

you can define it for an arbitrary dim vector function

[finite karma]

if I have M vector derivative of N vector

[magic light]

can anyone check this?

[finite karma]

the jacobian is MxN

[magic light]

I transposed it and in the second row multiplied both sides by v

[spare widget]

so you have: (x_1(y_1, ..., y_M), ..., x_N(y_1, ..., y_M))?

sure you get a matrix, that will not be square if M!=N

[finite karma]

sure, thats not my issue

[spare widget]

whether the Jacobian is defined as MxN or NxM depends on convention

it depends on how you like to order you element

[finite karma]

now if I have a PxQ matrix derivative of R vector function, is the Jacobian PxQxR

[spare widget]

wdym O vector functions?

[finite karma]

length, bad naming

[spare widget]

I still don't understand what that means

[finite karma]

a function that outputs a vector of length R

[spare widget]

but that was already the case:

x(y) = (x_1(y_1,...,y_M), ..., x_N(y_1,...,y_M))

you feed this an M-dim y vector

and it spits out a N-dim x vector

so wdym vector of length R?

isn't N the R that you are referring to

to be sure, x(y) is a vector with N elements

so taking the M different derivatives of the N elements yields you the NxM or MxN Jacobian matrix

[finite karma]

I see, let me rephrase. Jacobian of a function with matrix input and vector output

[spare widget]

matrix input can be vectorized

m = vec(M)

then you're back to the original thing

but by all means, you can form a holor with a 3rd dimension if you want

[finite karma]

so the jacobian is PQxR

holor?

[spare widget]

a 3-d array

sure you can produce something that is NxMxR

or some permutation of those

it's just a derivative

[finite karma]

its not a jacobian then though right

[spare widget]

$\frac{\partial x}{\partial M}$

[stoic python]

criver

[spare widget]

I haven't seen it anywhere, but you can generalize jacobian to mean that I guess

I don't think it matters does it?

[finite karma]

thanks, the reason im asking this is because im trying to generalize the gauss-newton algorithm

[granite crow]

@spare widget can you help me with regular algebra

Wrong channel

Sorry

[spare widget]

At the end of the day Jacobian just refers to

$\frac{\partial x}{\partial y}$

[stoic python]

criver

[spare widget]

it's just terminology, it shouldn't have any bearing on whether you're able to generalize an algorithm

as mentioned you can vectorize the matrix

then you can probably apply the original algorithm as is

[finite karma]

thats a good idea, thx

[spare widget]

$(\lambda+\overline{\lambda})\langle x, x\rangle = \langle \lambda x, x \rangle + \langle \overline{\lambda} x, x \rangle = \ \langle Ax, x \rangle + \langle x, \lambda x\rangle = -\langle x, Ax \rangle + \langle x, Ax\rangle = 0$

[stoic python]

criver

[spare widget]

@magic light

in the above x is an eigenvector corresponding to the eigenvalue lambda (e.g. we can assume x!=0), that is lambda * x = A * x, then from the above it follows that lambda + conjugate(lambda) = 0

but that's possible only if lambda is 0 or imaginary

the inner product is <u,v> = v^H u

for the first equality I have used linearity, for the second I have used that lambda * x = A * x if x is an eigenvector, and I have also used the conjugation property of the inner product when moving scalars between the two slots, for the third equality I have used that the adjoint of A is -A, and that lambda * x = A * x if x is an eigenvector and lambda is its corresponding eigenvalue

[magic light]

x is eigenvalue

[spare widget]

eigenvector in my case

[spare widget]

this thing

[magic light]

o

but how did we get inner products? 🤔

[spare widget]

$\langle u, v\rangle = \overline{v}^Tu$

[stoic python]

criver

[magic light]

oh I see

makes sense, yeah

[spare widget]

it's the standard inner product for C^n

[magic light]

another way is to simply see that the characteristic polynomial is only solveable for eigenvalues 0(in R)

[grave garden]

Heyy guys

What is U in here ?

[wicked palm]

some other vector space

[grave garden]

Are the square and parallelogram both can be written in the form of $t_1E_1+t_2E_2$ ?

[stoic python]

Potato

[gray dust]

@grave garden no, points in the square take the form t1E1+t2E2. the solution shows that points in the image of the square take the form t1(1,1)+t2(-1,2)

[grave garden]

Ohh

I see

[grave garden]

Given a vector space V, can we know how many linear map are there ?

[zinc timber]

a map is completely determined by where you send the basis, now think through

[grave garden]

So they are the same number as the maximal set of linear independent ?

[dusky epoch]

those are definitely some words coming out of your mouth

...are you like, working with a vector space over a finite field or something?

[grave garden]

Yes

[grave garden]

Is it wrong ? 💀

[dusky epoch]

this message here?

it's nonsensical.

[grave garden]

I see

Im having a hard time with linear mapping

What are we trying to do when talking about describing the image of a subset under a mapping ?

Are we trying to find $W(F(x,y))=1, (W(x,y)=x^2+y^2)$?

[stoic python]

Potato

[proper cradle]

How to find number of solutions of linear system of equations over finite field?

[zinc timber]

by solving 🙈

[grave garden]

Give a system of equations, say 3 with 3 variables, it has a unique set of solution when the determinant of matrix of constant ( i forgot its name ) is ≠ 0?

So $a≠\frac{3}{2}$ and b belongs to R ?

[stoic python]

Potato

[stark acorn]

I am confused on how matrix P is generated

Figured it out!

[gaunt quiver]

Can anyone recommend me good book for vector space. I am not able to get it

[vocal prairie]

A few books you could look into:

1. Linear Algebra by Friedberg/Insel/Spence
2. Linear Algebra Done Wrong by Treil, available for free from the author's webpage
3. Linear Algebra Done Right by Axler; this one is a bit infamous for unconventional presentation of determinants

[gaunt quiver]

Ok

[grave garden]

Guys, (2y,y) belongs to y=2x ?

That would be y=4y right ?

[dusky epoch]

have you already put this system in RREF?

[wintry steppe]

matrix representation of L_A should be equal to A right ?

<@&286206848099549185>

[wintry steppe]

but when i substituted some value coordinate matrix of L_A(v)=Av wrt gamma is equal to [L_A]_gamma * (coordinate matrix of v wrt gamma) . the thing i am not getting is matrix representation of L_A is not A but still i get the same ans

[proper cradle]

Yes i got two equations but idk how to solve them

[dusky epoch]

show your equations

[proper cradle]

$x_1+x_2+x_5=1 & x_3+x_4+x_5=4$

[stoic python]

BLツ

[dusky epoch]

\\ for a newline

okay so x_1 and x_3 are pivot variables and everything else is free

[proper cradle]

$x_1+x_2+x_5=1 \ x_3+x_4+x_5=4$

[stoic python]

BLツ

[proper cradle]

Yes

[dusky epoch]

how many values can each of the free variables take?

[proper cradle]

5 and 5 for x_1 and x_3 respectively

[dusky epoch]

i said free not pivot

you mentioned precisely the variables you shouldn't have mentioned

and you could have even just said "5" and nothing else

[proper cradle]

By solving this both two i got $x_1+x_2+4x_3+4x_4+3=0$ so if I choose any three of them then i got remaining one right?

[stoic python]

BLツ

[dusky epoch]

you're overthinking it

you're overthinking it

massively

[proper cradle]

Number of free variable are 3 then?

[dusky epoch]

yes, there are 3 free variables.

or

as you would like to put it,

Yes, Number of free variable are 3

[proper cradle]

So we need to check possibility of values for free variable

[dusky epoch]

you're overthinking it again

any assignment of values to the free variables, of which there are 5^3 = 125, will lead to exactly one solution of the system

[proper cradle]

Ohh for every choice we get unique solution

I mean assignment of values to the free variable

Then number of solutions will be 125 right

[dusky epoch]

yes

[magic light]

if two vectors are linearly dependent can I say u = xv? for u, v dependent?
x is some scalar

or is that wrong 🤔

[spare widget]

it's correct

[magic light]

great - so I solved this question really differently than the professor using this fact

if the norm of u, v, w = 1 and <u, v> = <u, w> = <v, w> = 0, can I say that u, v, w are linearly independent?

because, if I assume they are not(for example u and v are not) then u = xv and then <xv, v> = 0 => x<v,v>=0 but since the norm of v is 1, then <v, v> = 1 as well, meaning x=0 => u=v=0 which is a contradiction to the norm being 1?

[spare widget]

Since a * u+ b * v = 0 for (a,b) != (0,0)

Without loss of geeraity assume a!=0, the u= -b/a * v

[magic light]

yeah I thought so

any ideas on the proof?

[spare widget]

Let me see

[magic light]

it's proof by contradiction(i obviously simplified)

[spare widget]

But that's 3 vectors

[magic light]

yeah I just wrote for 2 here

[spare widget]

Then the assumption must be u = a * v + b * w

[magic light]

well if u is not dependent on v and v is not dependent on w that means u is not depenednt on w

[spare widget]

that's not how it works

thereexists u

Such that u != a *v and u != b * w but u = a * v + b * w

The definition for linear independence

Is that sum_ia_i* v_i =0 <-> a_1 = ... = a_n = 0

[magic light]

well yeah, but if we show that u is independent of v, and v is independent of w, then u is independent of both v and w no?

[spare widget]

But it's the same idea

[spare widget]

No

u and v ay be linearly idepedent on their own

v and w may be,and u ad w may be

But that doesn't mean that u,v,w are

So the subsets being independent does't mean that te whole set is

The logic should go: let u,v,w be lin dependent

Then there exists (a,b) != (0,0)

such that

u = a * v + b * w

Now 0 = <u,v> = a * <v,v> + b * <v,w> = a * |v|^2

And the inner product with w too

You get that either v is 0, or a is 0

[magic light]

hmm

[spare widget]

And from the second that either w is 0 or b is zero

Let's do that for n vectors

Suppose they are linearly dependent

Then there is an a_i !=0 such that

[magic light]

I didn't know proving that all subsets of (u, v, w) are independent except the set itself isn't enought o show the entire set is independent

[spare widget]

$\sum_{i=1}^n a_i v_i = 0$

[stoic python]

criver

[spare widget]

Now assume that $\langle v_i, v_j\rangle = 0, i\ne j$

[stoic python]

criver

[spare widget]

Then

$\langle v_i, \sum_{j=1}^n a_j v_j \rangle = a_i \langle v_i, v_i \rangle = 0$

[stoic python]

criver

[spare widget]

So either a_i is 0 or v_i is zero

If v_i is zero, then sure,it islinearlydepenfent withevery vector

[magic light]

thanks

[spare widget]

If v_i is not zero however thena_i = 0, and we assume a_i !=0 -> contradiction

So te only way for $v_1, ..., v_n$ to be liearly dependent with $\langle v_i, v_j\rangle = 0, i\ne j$ is for one or more of the vectors to be zero

[stoic python]

criver

[spare widget]

If you remove the zeroes, then you get a linearly independent set

[spare widget]

The geometric intuition is that you can have 3 vectors lying in the same plane, with none parallel, but they are linearly dependent since they are in the same plane and every one of them can thus be written as a linear combination of the other two.

If two are parallel then clearly they are linearly dependent. But if none are parallel and they are all in the same plane , then any two of those form a basis for the plane, so the third one can be expressed as their linear combination.

[magic light]

Got it.

Another question where I did something differently than the professor

actually it's wrong

now I noticed

lol it's so long

[magic light]

if I show DimV=n and v1, v2... vn are linearly independent, can I say that they span the space V by definition?

or is it not enough?

acutally just thought of a counter example to that... damnit

[subtle walrus]

what is the counterexample?

[magic light]

say V is the space of anti-symmetrical matrices 3x3, dimV=3

but the 3x3 matrices
1, 0, 0
0....

0, 1, 0
0...

0, 0, 1
0...

don't span V

even though they're linearly independent

[near quiver]

could be more of an English language question, but I simply don't understand this 😂 I don't even know how to ask.

F^n is a set of n lists ..... of elements of F?

[magic light]

i think it's like F^2 = {(0), (0, 1), (0, 1, 2)}

oh

maybe not

it's actually

{(0, 1, 2), (1, 0, 2)...}

all the permutations of {0, 1, 2}

[near quiver]

It's very beggining of linalg book. I just came back to it, because I want to properly understand linalg. In uni I just knew how to solve majority of problems and it carried me. But it was more like mechanical skill rather than understanding theory/proofs

[subtle walrus]

those matrices aren't skew symmetric to begin with

[magic light]

exactly, they are linearly independent of the same dimension but don't span the space

[subtle walrus]

i mean, they aren't even in the space...

[magic light]

yeah...

[subtle walrus]

i think you have to adjust your original statement

[spare widget]

F^n is a set of n-tuples with coordinates in F

[subtle walrus]

to be about subspaces of a larger space

[magic light]

would it be true for a sub space?

Like

[subtle walrus]

because any dimV linearly independent vectors in V do in fact span V and are a basis

[magic light]

if v1... vn are from a subspace of V

[spare widget]

Set F = R then it's R^n vectors, set F = C, then it's C^n vectors

[subtle walrus]

your statement is something like "if U is a subspace of V of dimension n, then any n linearly independent vectors in V span U", which is wrong, because if you take a vector not in U, then its instantly wrong

[magic light]

I realized the statement is wrong after writing it, I just thought if I take any n independent vectors (v1... vn) then they span any space of dimension n which is clearly wrong

[near quiver]

I'm trying to rephrase it somehow for myself, so I can verify I understand the concept, but I can't. I'll give it few more minutes

[subtle walrus]

my point is you have to specify where your vectors are from

from the space? from some ambient space?

[magic light]

hmm

well, originally this came from a question so perhaps that'll give some context

[spare widget]

What are elements of R^3?

[subtle walrus]

please show the question

[near quiver]

I know answer is {(x,y,z): x,y,z E R}.
So.... all possible combinations of x,y,z which are within R? x,y,z combinations being tuples

[magic light]

T:V->W is bijective(I think that's the word), v1... vn are in V
prove:
v1... vn are a basis in V iff T(v1)....T(vn) are a basis in W

So lets assume v1... vn is a basis and look at

a1T(v1)... anT(vn)=0
if we prove a1...an = 0 then T(vi) are linearly independent

by transformation properties:
T(a1v1... anvn) = 0
and since it is bijective then a1v1... anvn = 0 and thus the only solution is a1=a2=...=an=0

[spare widget]

The only difference with F is that the field doesn't have to be R

[magic light]

that's the first direction

[spare widget]

e.g. it can be C or Q

[magic light]

@subtle walrus

well, actually that just shows they're linearly independent

I'm not sure why they span the space

I guess because T(v1) are in W

[subtle walrus]

because any n linearly independent vectors inside a space span it

[magic light]

yeah, and DimW=DimV

that's important as well

because it's bijective

[spare widget]

Write the definition of linear independence

[subtle walrus]

otherwise, you could find one vector not in the span and extend the set to make a set of n+1 linearly independent vectors

[spare widget]

Then use linearity

To pull out the sum and constants

out of the map

[near quiver]

Ok thanks. I was very unclear with wording and firstly didnt understand if it means there are 3 tuples, or tuples consist of 3 elements. Although I could reverse engineer it from R^3.

[spare widget]

A basis is a set of linearly independent vectors that span the space, if you have a bijection then you need the same number of vectors, and it remains to be shown only that they are lin indep.

[spare widget]

A n-tuple is just an ordered list of n elements: (a_1,...,a_n)

[magic light]

the other direction though:
assume T(v1)...T(vn) are independent
and look at
a1v1...anvn=0
T(a1v1... anvn)= T(0) = 0
a1T(v1)...anT(vn)=0 => a1=a2...=an=0

so v1... vn are linearly independent, and because T(v1)...T(vn) span W then DimW=n and DimW=DimV=> DimV=n and thus a1v1... anvn also span the space, right? @subtle walrus

[subtle walrus]

more or less

you need to know the dimension is n before you can conclude that T(v_1), ..., T(v_n) span your space

alternatively if you have one direction, the other is exactly the same but you replace T with T^{-1}

[magic light]

sorry, I assumed T(v1)...T(vn) span the space

or more specifically, that they are a basis

so they are a minimal span, so the dimension is n

[subtle walrus]

i think you have all the right ideas, but you should maybe think about this again tomorrow or so

[magic light]

tomorrows the test 🙂

[subtle walrus]

damn

good luck then

[magic light]

thx

[mellow saddle]

Bro njose

[wintry steppe]

hey all

part a in specific

if anyone has tips that would be amazing

[magic light]

QUICK technical question:

$$T(x,y) = (x-y, 7x-3y)$$

[stoic python]

blackmamba[they/them]

[magic light]

do I write the matrix as
1 -1
7 -3
or
1 7
-1 -3

[gray dust]

the convention is to write the image of ur basis as the columns

this way Tv=Av

[magic light]

so the 2nd one?

andn does it matter?

[gray dust]

whats the image of the standard basis?

[spare widget]

it matter if you don't want Tv to mean v^TA^T

I'd suggest sticking to Av, it has the added benefit that it hints that v is a contravariant vector

[magic light]

T(1, 0) and so on

so T(1, 0) = [1, 7]

so [1, 7] would be the column

oh that's the first one

[gray dust]

yes

[magic light]

I thought I wrote it the other way around

my bad, thanks.

[gray dust]

ur welcome

[magic light]

so if I'm ever confused I just do T(1, 0...) and that is my first column

^^

[gray dust]

the main reason for doing that is

this way Tv=Av

ie if ur matrix A is made this way then applying T is the same as left multiplying by A

[magic light]

Right.

[night wren]

Since diagonalizing can be seen as the characteristic polynomial having a root, can you always diagonalize a matrix in a field extension?

[gray dust]

@night wren diagonalizable doesnt mean that. it means an eigenbasis exists

[fringe fjord]

Concretely, consider $\begin{bmatrix}0&1\0&0\end{bmatrix}$. Is characteristic polynomial is just $\lambda^2$ which definitely has a root, but it's not diagonalizable.

[stoic python]

Troposphere

[night wren]

But if a matrix has n distinct eigenvalues, then it is diagonalizable, right? If I have a characteristic polynomial of degree n with n-1 roots and is not diagonalizable, can I consider the matrix in a field extension where the polynomial has n roots (suppose they are distinct)?

[fringe fjord]

Yes, if there's any field extension where the characteristic polynomial has as many distinct roots as the side length of the matrix, then it will be diagonalizable over the extension field.

(It can never happen that there is exactly one root missing, though).

[wintry steppe]

whats the best way to formally prove this? https://cdn.discordapp.com/attachments/935224727746805792/941797584656298044/Screen_Shot_2022-02-11_at_2.46.08_PM.png

for part a

[sleek sundial]

If we have V as a euclidean vector space, and T an orthogonal transformation on V, how do I show that T has at least one real eigenvalue without using a characteristic poly/complex conjugate argument when dim V = 3

[grave garden]

Guys is linear map always injective ?

[wicked palm]

nope

[grave garden]

Is "linear map" and "linear transformation" the same statement ?

[fringe fjord]

Yes.

[grave garden]

I misread about kernel

[wintry steppe]

if anyone has any advice on part a that would be amazing ❤️

[nocturne jewel]

T is injective iff ker(T)={0}

[magic light]

and its surjective too if T:V->V 🙂

[nocturne jewel]

no

Being an operator doesn't make it surjective

T is surjective iff Im(T)=V

for T: V->V

[slow scroll]

I think they were trying to say an injective operator on a finite dim space is surjective

[zinc timber]

one-one operator on finite dimensional vector space => surjective as welll

[slow scroll]

Jinx

[zinc timber]

🙈

[mortal isle]

How do I present part b

[sleek sundial]

bump

[grave garden]

dim{0}=0 or 1 ?

[slow scroll]

0

[grave garden]

I see thanks !

[slow scroll]

the empty set is a basis for {0} 🤓

[grave garden]

I thought 0 is the basis for {0} so its dim is 1

[slow scroll]

nah {0} is linearly dependent

[grave garden]

Ohhh

[zinc timber]

char poly will have degree 3, now what can you say about the root

[nocturne jewel]

"without using char poly"

[grave garden]

Guys, say $x_1+2x_2=b$, what would the mapping L be ?

[stoic python]

Potato

[grave garden]

Do you still consider the same mapping ?

[zinc timber]

yeah ok I didn't see that, the argument will be a little involved tho @sleek sundial

one thing you can try is show det(A-xI) is a continuous function on x from R -> R and then show that large enough -ve value of x will give you -ve det and large enough +ve x will give you +ve det so from continuity it must be zero somewhere

(still uses the char poly but implicitly )

[grave garden]

I can show that $v_0+u$ is indeed the solution, but how do I prove that there is no other solution that is not in the form $v_0+u$ ?

[stoic python]

Potato

[zinc timber]

if v0+v is a solution with v not in the kernel, then L(v0+v) = w+(some non zero vector) != w therefore can't be a solution

[grave garden]

Ahhh I see

[ionic stream]

here, i was asked to prove that identity written in the last line of the second image. red markings are from my professor. what makes this invalid? (apologies for low quality btw, can get a better version if need be)

should i just have been more clear with the whole cofactor expansion thing in the beginning?

[sleek sundial]

I can’t use that argument for this unfortunately

[sleek sundial]

Thx I’ll try that

[gray dust]

@night wren

But if a matrix has n distinct eigenvalues, then it is diagonalizable, right?
yes, try proving this
can I consider the matrix in a field extension where the polynomial has n roots (suppose they are distinct)?
yes, an example is the pi/2 rotation matrix which isnt diagonalizable over R but IS over C (eigenvalues i,-i)

[brave hound]

how do I solve 40?

39 was straightforward because the first 2 columns are multiples, no clue about 40 though

[dusky epoch]

perhaps you could apply the definition of linear (in)dependence

[brave hound]

i'm trying to reduce it to RREF and see if that helps

but i'm having trouble with fractions mixed with variables

[wintry steppe]

if you concatenate the vectors of a linearly dependent set into a matrix, its determinant will be 0. think u can do it from there?

[brave hound]

we weren't taught what a determinant is

😓

[dusky epoch]

fractions mixed with variables?

can you show your work

[brave hound]

sure

[dusky epoch]

i have a feeling you're taking a suboptimal route if you are going with row reduction

[dusky epoch]

yuck!

[brave hound]

yea its bad

I have an exam tomorrow

[dusky epoch]

i mean first off why decimals

[brave hound]

Not lookin good

[dusky epoch]

ok so like

do you explicitly want to do matrixbashing on this

or are you ok with a method that is not matrixbashing

[brave hound]

Decimals to make (3,1) = 0

i would like to do it the smarter more efficient way

[dusky epoch]

your equations should look like this:

-2x + y - z = 0
z = 0
x - 3y + rz = 0

that's the equation x * (first vector) + y * (second vector) + z * (third vector) = 0

do you understand how this came about?

[brave hound]

yeah

[dusky epoch]

do you understand how to proceed with solving this system of equations?

||you should get that it only has the trivial solution no matter the value of r||

[brave hound]

oh wait I can plug in z

right?

Wait nvm that gets rid of the r

So does that mean that there is no value for which it is linearly dependent?

[dusky epoch]

yes

your set is always LI regardless what r is

[brave hound]

Ok thank you @dusky epoch

[lime zinc]

any hint .

[grave garden]

That x•v/v•v thing looks like the component of x on v

[grave garden]

Hmmm $v=P(v)+Q(v)$ ?

[stoic python]

Potato

[grave garden]

By the given $P(v)+Q(v)=I$

[stoic python]

Potato

[grave garden]

[dusky epoch]

P(v) + Q(v) = I(v)

I(v) = v

[grave garden]

Ohh it is an identity mapping 💀

I take it as identity element

Is the third condition optional ?

I didn't see they use it

[hybrid gulch]

6/2 (1+2) =

[wintry steppe]

why is the product of elementary matrices a regular matrix?

[dusky epoch]

define 'regular matrix'

[wintry steppe]

invertible

non-singular

[dusky epoch]

every elementary matrix is invertible

the product of invertible matrices is invertible

[wintry steppe]

thanks Ann :)

[zinc timber]

T is housholder reflection

you can assume v=e1 and the see how a T looks like

[lime zinc]

V is arbitrary ,i tried with v=e1 , i got the answer, but i want to know the proper solution by considering v as arbitray

[lime zinc]

Yes but denominator is misssing

[zinc timber]

proper way would be to see that T² = I

meaning only possible eigen values are +1 or -1

now -2(v'x)/(v'v)x means we are subtracting twice the component along v

means GM of eigen value is -1 and so rest of the egen values are 1

[lime zinc]

Along x?

[zinc timber]

therefore trace = n-2, det = -1 and orthogonal as well (check symmetric)

[zinc timber]

no we are flipping along v

in other words reflection about the hyperplane perpendicular to v

[lime zinc]

I have solution ,but i am not getting one line

Let me send that

[zinc timber]

still waiting

[lime zinc]

I have open my laptop

Give me some time

[grave garden]

Is the last option the answer ?

[zinc timber]

there is another way tho, we know that eigen values of $M= \frac{vv^T}{v^Tv}$ are one 1 and rest 0. so now T=(1-2x)•M so eigen values will be one -1 and rest 1

[lime zinc]

yes

[zinc timber]

which line?

[lime zinc]

Span v

[zinc timber]

all vectors in span(v) has eigen value -1 not 1

they even showed it, probably a typo

[lime zinc]

How?

[zinc timber]

T(v)=v-2v=-v

[lime zinc]

I get this part

[zinc timber]

[lime zinc]

But that one

Above that

[zinc timber]

yeah than eigen value is -1 not 1

[lime zinc]

Just find the ....

[zinc timber]

which one exactly 😐

[lime zinc]

Just find the eigenspace

Line

[zinc timber]

oh "ust find the eigenspace"

they showed eigen space of -1 is span(v) and eigen space of 1 is span(v) \perp

[lime zinc]

Yup

[lime zinc]

This is what i am not getting

[zinc timber]

the 2nd part?

ok are you able to see why span(v) perp has eigen value 1?

like just plug in x•v=0

then every vector perpendicular to v is in the eigen space E_1, so dim E1= dim v\perp which is n-1 (here n=3 hence 2)

[lime zinc]

Why they considering thete are two eigenspace on is sapn V and other sapnv perpendicular?

[lavish jewel]

maybe an alternative approach helps

we can rearrange T(x) to write it in matrix-vector form in the canonical basis, and maybe this helps you out

[grave garden]

Guys, is showing bijection enough to show that a mapping is isomorphism ?

[lavish jewel]

notice that $T(x) = x - 2 \frac{x \cdot v}{v \cdot v}$ can be rewritten as $Ix - 2v \left( \frac{x \cdot v}{v \cdot v} \right)$

[stoic python]

Edd

[lavish jewel]

so far we've only added in an identity matrix and moved around a scalar, so nothing funny there

then, we notice that $x \cdot v = v^T x$

[stoic python]

Edd

[zinc timber]

lol no

[lavish jewel]

so that $T(x) = Ix - 2 \frac{v v^T x}{v^T v}$

[stoic python]

Edd

[lavish jewel]

and now we factor out the x from the right

[zinc timber]

definitely this is the approach I would recommend

[lavish jewel]

$T(x) = (I - 2\frac{vv^T}{v^Tv})x$

[stoic python]

Edd

[zinc timber]

u need to show linearity as well

in the context of VS

[lavish jewel]

now you notice that $I - 2 \frac{vv^T}{v^Tv}$ is a symmetric matrix

[stoic python]

Edd

[lavish jewel]

and this means it is diagonalizable in an orthogonal basis

now pair this together with what ryu already told you regarding v being an eigenvector with eigenvalue -1

[grave garden]

I see, i've been doing just bijection for past several exercise

[zinc timber]

[grave garden]

Do I need to show linearity of $(GF)^{-1}$ as well or just $GF$ ?

[zinc timber]

no

[stoic python]

Potato

[zinc timber]

bijectivity and linearity is enough

[grave garden]

I see thank you !

Is the isomorphism mapping of two vector space unique ?

[zinc timber]

no

[grave garden]

Just wondering, can one determine how many of the mapping are there ?

[lime zinc]

Wait can we use the fact that eigenvalues of I- A matrix is 1- eigenvalues of A?

[lavish jewel]

if you know why that is true, yes

if not, you have to show it first

[lime zinc]

I know this fact

[lavish jewel]

but do you know why?

(it's easy to show if you read what i wrote above)

you noticed that vv^T is a symmetric matrix diagonalizable in some orthogonal basis, so let's say c vv^T = QDQ^T

since Q is orthogonal, QQ^T = I

so I + c vv^T = QQ^T + c QDQ^T = Q(I + cD)Q^T

this also immediately establishes you'll be working with orthonormal eigenvectors, one of which is v

so what ryu says follows

[lime zinc]

I have to solve this on my notebook first. Thank you for the help both of you

[spare widget]

I need someone to check whether my logic is correct because there's something bothering me when solving a problem.

I have the minimisation problem

$\min_x -v^Tx, , 1^Tx = C, , x\in[0,1]^n$

[stoic python]

criver

[spare widget]

I take the negated gradient: v

And project it onto the plane with normal 1/sqrt(N)

$\vec{d} = \vec{v} - \frac{1}{n}(\vec{v} \cdot \vec{1})\vec{1}$

[stoic python]

criver

[spare widget]

Thus if I have an initial guess, for example x^0_i = C / n, the gradient descent step will look like

$\vec{x}^1 = \vec{x}^0 + \gamma \vec{d}$

[stoic python]

criver

[spare widget]

And to take into account the [0,1] constraint I do

$\vec{x}^1 =max(\vec{0}, min(\vec{1}, \vec{x}^0 + \gamma \vec{d}))$

[stoic python]

criver

[spare widget]

What bothers me is whether x^1 satisfies 1^T x^1 = C

[lavish jewel]

if you did it right, yes

[spare widget]

After the clamping*

It's clear that it satisfies it before the clamping

[lavish jewel]

since you're adding a contribution orthogonal to 1^T

[spare widget]

On the other hand my argument for the clamping is that the clamping is equivalent to projecting out the e_i component

of d

[lavish jewel]

this is a valid formulation of projected gradient descent, yes

[spare widget]

What bothered me is that if I bring gamma large enough

Then I would get a solution of 1s and 0s except where d has a 0 component

1s and 0s don't necessarily sum up to an arbitrary constant C in [0,1] however.

So is it always the components where d is 0 that make up for the fractional part?

Or is there some flaw in this clamping that in fact violates the 1^Tx = C constraint

[lavish jewel]

hmmm

[spare widget]

Notably I would get 0s where d_i < 0, and 1s where d_i > 0

And x^0_i = C/n where d_i = 0

That seems to be the optimum according to the above

[lavish jewel]

but that's part of the problem, right? by doing gradients you are now considering a proximal form

and it may very well be that the distance you can move is 0

this parameter gamma cannot be arbitrary

it can only be as large as the feasible region allows

[spare widget]

I think it can, since v is constant

[lavish jewel]

yes but you have constraints

equality and inequality

so no

otherwise the solution would just be infinity in the entries

inf* v

[spare widget]

i.e. the gradient doesn't change, except when it meets a [0,1] constraint, but at that point this is equivalent to removing the i-th component of the gradient

So that's where the clamping comes from

[lavish jewel]

it doesn't remove the element though, if you simply do that you change the direction

and it's no longer the direction of the gradient

[spare widget]

here's how I think about it

Let's say I do the largest possible step gamma

such that one [0,1] constraint comes into play

without loss of generality let that be x_i = 0

now I project away this from the gradient

but that's equivalent to setting d_i = 0

Since $d' = d - (d \cdot e_i)e_i$

[stoic python]

criver

[lavish jewel]

ah, indeed

but yes, this also means this is no longer a direction of steepest descent

or possibly no longer feasible in the other constraint

[spare widget]

with respect to the constraints it is the steepest descent direction

[lavish jewel]

what makes more sense here is to instead use this result to cap gamma

with respect to the inequality constraint, yes

but it is no longer a feasible point in the first place

since as you noted, you are ignoring the equality constraint

[spare widget]

so while all of the above seems plausible to me, I was wondering whether I am overlooking something regarding this 0,1 solution with occasional C/n terms that supposedly satisfies 1^T x^* = C

[lavish jewel]

yes, that you're ignoring the 1^T x = C entirely

you can add it to the original cost function with a lagrange multiplier

or you can use the inequality constraints to cap gamma

instead of truncating

because ATM you're projecting onto the feasible set of the ineq constraint completely ignoring the equality one

[spare widget]

lagrange multiplier doesn't help here, I tried it, linear independence of constraints doesn't hold for kkt

[lavish jewel]

then simply cap gamma

[spare widget]

So min(max is indeed not equivalent to doing the steps by capping gamma

[lavish jewel]

the new x^i at each iter can be treated as a ray that intersects a hyperplane given by one or more of the inequality constraints

you can find the intersection of this ray and hyperplane analytically

and parametrize it in terms of gamma

[spare widget]

I am still trying to reason about it

Since dot(d, 1) = 0 even if I set some components of d to zero

So maybe they are equivalent after all

My idea is that I should not be able to escape the plane 1^Tx = C, if x^0= C/n is on the plane and 1^Td = 0

But 1^Td = 0, even if I zero out some components

[lavish jewel]

that much should be true, as long as d remains orthogonal to 1^T, yeah

so, as long as you are taking gradients only on the hyperplane with normal 1^T

[spare widget]

So afterall the min(max should provide the minimum?

I can't shake off the feeling that I am overlooking something

[lavish jewel]

hmmm

[spare widget]

I'll try to get a proof by induction and come back later I guess

[lavish jewel]

i'm not sure what you meant about not satisfying lin indep for kkt btw

[spare widget]

My worry arose from the optimum being made up of 1s and 0s and potentially some C/n

Then their sum is numel(sgn(d)>0) + numel(d==0) * C / n and that must be equal to C

So either the $max(0,min(1, x^0 + \gamma d))$ is an incorrect expression or the above indeed sum up to C.

[stoic python]

criver

[lavish jewel]

what about the kkt stuff

[spare widget]

The constraints are linearly dependent, it doesn't give anything useful.

[lavish jewel]

what i mean is that you can take the gradient of -v^Tx + c 1^Tx

and adjust c as needed

[spare widget]

I tried this, but you get v = -c

And it doesn't account for the other constraints

I also tried it with a quadratic term, it makes things harder as then I need to slide along geodesics on a hypersphere

I'll focus on trying to prove/disprove more formally the min max thing for now

[lavish jewel]

that's not quite right, c is a scalar and v is a vector. you'd need to do an orthogonal projection of v onto the vector of 1s

and yes, another term would be needed for the inequality constraints

[spare widget]

c us a scalar but c * 1 is a vector

When you add all the terms they are linearly dependent

[lavish jewel]

but you could get around by using boundary functions

but why would you add them up

[spare widget]

Same thing happens in the quadratic ||x||^2=C^2 case

It's essentially a linear programming problem in the linear case, but my issue is not with that it's with proving/disproving the min max optimum. I'll work on it and try to post something more coherent when I've worked out the details.

[lavish jewel]

at any rate, yeah, doing the projection as you did doesn't appear to work in general

take an example in R^3 where you start with x0 = 1/3[1,1,1]^T, with C = 1

you can built an orthogonal basis for the hyperplane with normal [1,1,1]^T by using the basis vectors [-1,1,0]^T and [1/2, 1/2, -1]^T

[wintry steppe]

I need help to find this equation. m = 2/3, passing through (-4, -6). I checked the answer and it is in general form (Ax + By + C = 0). I am unsure as to how I'm supposed to get from m = 2/3, passing through (-4, -6) to Ax + By + C = 0 . <@&286206848099549185>

[lavish jewel]

and let's say you add the second vector of this basis to x0. after clampling, the vector no longer satisfies the equality constraint

[lavish jewel]

please check #prealg-and-algebra

[wintry steppe]

ty

[lavish jewel]

on the other hand, one can make a sort of trust region by instead checking that for a given step size, you are still within the feasible region

i think that makes more sense here

enforce the ineq through the step size instead of through projections

(i'm not sure this converges to the true solution btw, but your current method has very simple counterexamples showing it doesn't work)

[spare widget]

solution according to min max is 1,1,0 which sure enough doesn't satisfy the constraint

[lavish jewel]

yep

[spare widget]

So there is indeed something faulty with the logic

I am up to the point of

[lavish jewel]

my geometric interpretation of this is "how parallel can we get to v"

with some constraints on needing a specific component in a given direction and not being able to add any other component that is too large

so there must also be some way to reformulate it as a convex combination of v^T and 1^T

[spare widget]

$\gamma^{i+1} = \inf {\gamma : x^i_j + \gamma d^i_j = 1, d^i_j > 0} \cup {\gamma : x^i_j + \gamma d^i_j = 0, d^i_j < 0}$

[stoic python]

criver

[spare widget]

because if I take steps iteratively then I assume I can show by induction that it will stay in the plane, it's just that this process is not equivalent to the min max thing I had

[spare widget]

Yes we want to maximize the dot product with v after all, but the constraints make this harder

[lavish jewel]

well, i already presented one approach to you

[spare widget]

You're referring to the iterative?

The gamma I wrote above?

[lavish jewel]

do you have any example for which you know the true solution btw?

[spare widget]

Not really, no.

Thanks a lit for the counterexample though

At least I won't be stuck trying to prove something that isn't true

[lavish jewel]

i would really try a little longer with kkt and replace the ineq by a barrier function

or try the gamma stuff

[spare widget]

That feels off, since this is a linear programming problem, so there are more efficient ways to solve it than through penalizers

I think that this iterative gamma thing is a special case of the simplex method

[lavish jewel]

well sure, you can always use the tried and true solution approach

should be able to transform the equality constraint into an inequality one or backwards with some extra variables

but this is pretty much the same as doing KKT

[spare widget]

I guess I am just deriving a simple version for the special case that I have yes. Thanks a lot for the counterexample again. I'll just stick to picking the largest gamma that hits a constraint, descent, update gradient, then rinse and repeat.

[wintry steppe]

is anyone here able to help me rq, itll probably be easy for you

if not i understand

[lavish jewel]

i'll give you a hand cuz the other channel seems dead

[wintry steppe]

ty

[lavish jewel]

notice that in slope intercept form we have y = mx + d

[wintry steppe]

yes

[lavish jewel]

and in standard form we have ax + by + c = 0

[wintry steppe]

yep

[lavish jewel]

so let's do some arithmetic on them and see

first, we can write -mx + y - d = 0

this is already very close

what sometimes is done is that one notices that m = rise/run, let's call them p/q

so that -p/q*x + y - d = 0

and we can multiply both sides of the equation by q

so that now we have -px + qy - qd = 0

which is of the form ax + by + c = 0

[wintry steppe]

holy shit your smart

[lavish jewel]

this only works when q is not zero, so when we don't have a vertical line

[wintry steppe]

okay

[lavish jewel]

if you didn't understand any of the details, you can also try opening a help channel, maybe you'll have better luck there

check out #❓how-to-get-help

[wintry steppe]

i think i can work with this, thank you this helps a ton

[spare widget]

I realized where my mistakes was working through your example btw. Zeroing out a component of the gradient can make it so that 1^Td != 0 as you said. So after each zeroing out I have to project the gradient back onto the plane with normal 1. Now everything makes sense. Thank you.

It was my gradients leaving the feasibility plane that was the issue, so I had to project back.

[lavish jewel]

i thought about that as well, but it's possible that projecting back can make the inequality be violated again... i think?

you'd have to check this

it may or may not be necessary to iterate projection -> clamping -> projection ->...

[spare widget]

Yeah it needs to be projected onto the lower dim hyperplane orthogonal to both ei and 1

otherwise it gets stuck

[lavish jewel]

but you'll notice that this depends on the length of the original gradient vector, too

so in some sense this is similar to directly picking gamma

(in a very handwavy sense )

[spare widget]

Basically at step k, my gradient needs to be in the hyperplane orthogonal to $1, e_{i_1}, \ldots, e_{i_k}$

[stoic python]

criver

[lavish jewel]

i'm not sure i see why

[spare widget]

where the index i_j is the index of the constraint that comes into effect at step j

[lavish jewel]

ah

yes

ok

[spare widget]

So it projects into a smaller and smaller space until I get a single point, or it will stop before if d has some zeros at the very beginning

The point is that d^k eventually becomes 0, so no step is possible

[lavish jewel]

sounds ok

[spare widget]

Should lead to a global minimum for convex sets I believe

and a hyperplane cutout by a hyperube should be convex afaik

[lavish jewel]

intersection of convex sets should be convex, yea

[spare widget]

The only nasty thing is that the set $1, e_{i_1}, \ldots, e_{i_k}$ is not orthonormal, although very close to it

[stoic python]

criver

[lavish jewel]

right, but it shouldn't be all that complicated to handle in code

start with 1 in your set, and add in new vectors with gram schmidt as needed

[spare widget]

I can always just fix the 1 I guess

Since the rest are orthonormal

And the orthogonal complement remains unchanged

[lavish jewel]

actually, i don't think this is needed. since all you need is to take away components parallel to these vectors.

yeah

[spare widget]

I wonder whether this orthogonalization will mess up with the length of the projected vector though

i.e. $v_{\perp} = v - v_{\parallel}$

[stoic python]

criver

[spare widget]

Won't the length of v parallel be affected

[lavish jewel]

shouldn't be the case

as long as you return to the canonical basis

hmm not even

[spare widget]

yeah, I don't think it should matter provided the projection to parallel has the proper denominators, or the basis is orthonormal

[lavish jewel]

as long as the vectors are orthonormal

yea

[spare widget]

Ok that's great. Thanks a lot again. It was very helpful.

[lavish jewel]

cool

[brave hound]

could someone explain this?

[lavish jewel]

what definition of rank does your book use

[brave hound]

number of nonzero rows in RREF / number of basic variables

[lavish jewel]

and do you know how this related to the number of linearly independent columns in a matrix?

(e.g. through the so-called "pivots")

[brave hound]

if it's linearly independent, the nullity is 0

[lavish jewel]

i don't think that's the correct thing to consider here

[brave hound]

ok

[lavish jewel]

unless you can say how the nullity relates to the number of (non)zero rows in RREF

[zinc timber]

why do I have a bad feeling about this

[brave hound]

lol

Ik that nullity is n-rank but I don't think that helps here bc there are no values

[zinc timber]

Edd correct me here but $\m{1 & 0 & 1 \ 0 & 1 & 1}$ has LD columns and rank = 2 but it's equal to m=2?

[brave hound]

wait rank also refers to the amount of pivot columns

if that's what you were asking

[lavish jewel]

oh ur right, idk why i thought m was the columns

is this a true/false question?

[zinc timber]

fuu thought I was tripping because you weren't reacting

[lavish jewel]

my brain is fried

[lavish jewel]

is this a true/false question? cuz otherwise...

[brave hound]

wait i have an idea

Yeah it's T/F

[zinc timber]

ffs

[lavish jewel]

then ryu already gave you the answer

[zinc timber]

[lavish jewel]

that moment when your linalg book be trippin

[brave hound]

the book says this

[lavish jewel]

everything there is true

[brave hound]

so it implies that it would also be linearly dependent if the rank of A is GREATER than m

Therefore the question is false

[lavish jewel]

what do you mean by this?

[brave hound]

oh wait

wrong row

one sec

ok nvm I'm confused again

ignore what I said

alright I get what Ryu is saying

Thanks @lavish jewel @zinc timber

[spare widget]

@lavish jewel Turns out my minimisation problem had a trivial solution actually not requiring gradient descent.

If I sort the v_i in ascending order, and let C = m + delta, where delta in [0,1). Then I need to set x_i = 1 for the i corresponding to the m largest elements of v. And x_j = \delta for the m+1 largest. And then everything else should be 0.

This satisfies the constraint, maximizes the energy. And if there are other optimal solutions it is in the case where some v_i are equal.

Reminds me of m-term best function approximation wrt L2 in an orthonormal basis but with quantisation constraints [0,1] and total sum constraints.

i.e. I think $\min_{x\in[0,1]^n, , 1^Tx = C} ,|v - x|^2$ has the same minimum

[stoic python]

criver

[spare widget]

except for when some v_i are equal

[lavish jewel]

aha, i see

[spare widget]

The m-largest v in the L2 energy need to be >1 though, and the rest <0, and something else for the m+1st. So I guess it's not exactly the same.

[burnt hearth]

The question says to "Find the general solution of the system in parametric vector form" so is final answer correct??

[glass harness]

hello

[gray dust]

,w rref[[1,1,1,1,1,1],[1,2,3,4,5,6],[1,0,-1,-2,-3,-4]]

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by RokettoJanpu#6002

[burnt hearth]

Yeah I got that but I'm wondering the part of parametric vector form

[gray dust]

the vector for x3 should be (1,-2,1,0,0)

[burnt hearth]

Ok i fixed it

[gray dust]

its good beside that

[burnt hearth]

So vector x with the columns is my final answer correct?

[glass harness]

i hace a question

The moment of a force F is defined as M = F * r. A force given by the expression 4i + j N is applied at point (2,1)m. Find the moment with respect to the origin

since vector r is (2,1) and the force is (4,1), we cross product the 2d vector, which formula is (a,b)*(c,d) = (ad-bc)k. vector r * vector f = (2,1) * (4,1) = (2-4)k = -2k.
thus, the moment of force with respect to the origin is -2k?

[gray dust]

yes

[glass harness]

wait

wrong server

[burnt hearth]

Ok thank you

[glass harness]

please help me with a vector question

Given the vectors u1 = 2i-j+k; u2 = i+3j-2k; u3 = -2i+j-3k; and u4 = 3i+2j+5k, find the value of the scalars a, b, and c in which u4 = au1+bu2+cu3

[wintry steppe]

plug them in

just like the following

2ai-aj+ak+bi+3bj-2kb-2ci+cj-3ck

then you need to get a common feild out like the following

i(2a+b-2c)+j(-a+3b+c)+k(a-2b-3c)

and then you just create three equations

2a+b-2c = 3

-a+3b+c = 2

a-2b-3c = 5

and solve them

[lavish jay]

Is there an easy way to invert a matrix ??

[haughty berry]

There are a couple of algorithms here:
https://en.wikipedia.org/wiki/Invertible_matrix

[zinc timber]

use python

[haughty berry]

Wolfram alpha?

[oblique prairie]

no, why use something free when you can go out of your way and use python?

sorry ryu you know i had to do it

[zinc timber]

python is also free

I wanted to write MATLAB ngl

[oblique prairie]

i should have said simple

[reef sierra]

Hey quick question regarding definitions. If say A is a linearly dependent set of vectors but A is also a subset of B, does that mean that B is also linearly dependent?

[fringe fjord]

Yes.

Because a linear relation between vectors from A is then necessarily also a linear relation between vectors from B.

[reef sierra]

Okay great, thank you for the explanation

[solar steppe]

If i have a strong positive autocorelation how do i make adujstments to my lag? and where to I modify parameters?

[grim cliff]

Is an eigenspace a subspace of the range space of a square matrix?

[somber slate]

wow the first questions of LADW are already killin me

[gray dust]

@grim cliff yes, u can show this

[grim cliff]

Let V be a linear subspace of $C^n$ and t an automorphism of $C^n$, and $E_{\lambda,t}$ be the eigenspace of t associated to the eigenvalue $\lambda$.

I read somewhere that
$V\cap t(V)\cap E_{\lambda,t}=0$ iff $V\cap E_{\lambda,t}$=0

Does anybody have an idea why this is?

[stoic python]

fajitas

[lavish jewel]

the if direction is easy, since intersection is symmetric and associative

in the only if direction, think first of what V cap t(V) is

[grim cliff]

See Im confused cause couldn't it be V\cap t(V)=0?

[lavish jewel]

what is an automorphism?

[grim cliff]

Well it's an invertible endomorphism

[lavish jewel]

what does this tell you about its image and kernel

[grim cliff]

The kernel is trivial

But I'm not sure about the image

[lavish jewel]

rank nullity theorem

[grim cliff]

I know the image has the same dimension

[lavish jewel]

what does it tell you

well

[gray dust]

invertible maps are surjective

[grim cliff]

Is the conclusion I'm supposed to get that t(V)=V?

[gray dust]

yes

[lavish jewel]

yes

still, you raise a valid point in that the problem should be tackled separately when t(V) cap V = 0, as this means V is the 0 vector space

but you quickly reach that the statement holds

[grim cliff]

Ahhh okay I think I see what you mean

Thanks so much guys,

If I can probe one last thing, what about t(V)\cap E?

Is this nontrivial as E\subset Image(T)?

[lavish jewel]

sure, but i think you don't need need to worry about that

since you can rearrange the expression to check first V cap E

[gray dust]

idt this needs to be done separately?

[lavish jewel]

it doesn't need to, no, but if they were concerned

it is easy to show it still works

[grim cliff]

Okay thank you so much

[gray dust]

it p much goes
$$\brc0=V\cap tV\cap E=V\cap V\cap E=V\cap E$$

[stoic python]

RokabeJintaro

[gray dust]

w/o having to think whether any caps are 0

[grim cliff]

The only thought that messes me up is imagining V as a line through the origin. Then t(V)=V iff V is an eigenspace, right?

But not every line going through the origin is an eigenspace

[lavish jewel]

all subspaces contain the origin

if they are 1D, you can imagine them as lines passing through the origin

also no, because an eigenspace can be associated to the 0 eigenvalue

then t(V) = 0

or are you talking only about automorphisms

[wintry steppe]

Where have I gone wrong?

Ignore the stuff under the question btw that’s old

Oh my the angles are around the wrong way

Smh

[lavish jewel]

careful with the magnitudes @wintry steppe

notice they tell you the direction and magnitude of the vectors separately

so double check whether they need to be scaled before you add the forces

[lavish jewel]

ah nvm i just noticed you had already corrected that lol

[brittle gyro]

quick one: does $\lambda \in \lambda(A) \implies \lambda \in \lambda(A^T)$?

[stoic python]

Sydd

[errant mist]

What other differences are there between positive symmetric matrices and stochastic matrices other than the entries of the stochastic matrices are probabilities between 0 and 1?

[haughty berry]

Yes:
\begin{multline*}
p_{A^\top}(x) = \abs{xI - A^\top} = \abs{(xI)^\top - A^\top} = \abs{(xI - A)^\top} = \abs{xI - A} \ = p_A(x)
\end{multline*}
So the characteristic polynomial of $A$ is equal to that of its transpose.

You can actually prove that $A\sim A^\top$ but that’s more complicated.

[brittle gyro]

How do we know $\det(M^T)=\det(M)$ without using the result I want to prove?

[stoic python]

Sydd

[haughty berry]

It’s a pretty well known identity

[zinc timber]

you can use the definition of det, $$det(A) = \sum_{\sigma \in S_n} \epsilon(\sigma) A_{1\sigma(1)}A_{2\sigma(2)}\cdots A_{n\sigma(n)}$$

[haughty berry]

\def\sign{\operatorname{sign}}
Like:
\begin{multline*}
\det(M^\top) = \sum_{\sigma\in S_n} \sign(\sigma)\cdot\prod_{i=1}^n[M^\top]{i, \sigma(i)} = \sum{\sigma\in S_n} \sign(\sigma)\cdot\prod_{i=1}^n[M]{\sigma(i), i} = \
= \sum{\sigma\in S_n} \sign(\sigma)\cdot\prod_{i=1}^n[M]{i, \sigma^{-1}(i)} = \sum{\sigma\in S_n} \sign(\sigma^{-1})\cdot\prod_{i=1}^n[M]{i, \sigma(i)} \\sum{\sigma\in S_n} \sign(\sigma)\cdot\prod_{i=1}^n[M]_{i, \sigma(i)} \
= \det(M)
\end{multline*}

[brittle gyro]

Makes sense, thanks

[haughty berry]

Ignoring all of the mistakes I made, that’s the idea

[haughty berry]

That’s a funky font

[brittle gyro]

I am stuck in something slightly related: if $A$ is square and symetric is clear that $|A|2=\sqrt{\lambda{\max}(A^T A)}=\sqrt{\lambda_{\max}( A^2)}= \lambda_{\max}(A)$

I think this should be true for unsymetric A as well, but I couldn't rigorously prove this

Or it isn't?

[haughty berry]

This is the norm of A? So like sqrt(tr(A A^T))?

[brittle gyro]

the 2-norm

sorry

[brittle gyro]

what does the trace has to do here?

[haughty berry]

Different norm, ignore that

Is $\lambda_{\max}$ the maximum eigenvalue?

[brittle gyro]

yes

[stoic python]

Sydd

[haughty berry]

Well, I’m a bit confused why this is right, because if the spectrum of A is say {-1, -2} and the spectrum of A^2 is {1, 4} the sqrt of the maximum eigenvalue of A^2 is 2, but the maximum eigenvalue of A is -1

[brittle gyro]

You're right, I think that's bad... But the should the 2-norm of A be the biggest eigenvalue in module?

[haughty berry]

It should be the maximum absolute value of A’s eigenvalues I think

Anyways, regarding your original q, why should this be true for non symmetric matrices?

[brittle gyro]

It is true that $|A|_2= \sigma_1$ in general, where this is the first (biggest) singular value. The thing is I am a bit lost on how the singular values and eigenvalues of $A$ are related (when $A$ is square)

[stoic python]

Sydd

[haughty berry]

You should probably wait for someone who knows more than me to answer, but I’m not sure if there always is a relation like this

Like you have one if A is symmetric (which you showed)

[brittle gyro]

this here is bad because of what you mentioned about negative eigenvalues etc

but something like that

like this maybe

[haughty berry]

Yeah

[brittle gyro]

trying to figure out if this is the same as the 2-norm in general