linear-algebra

[flint nexus]

But the way he explains he use the kernel to show that the first row in the Integration of the Defferentiation( Av=w A =matrix i mean) A is all 0

thats what gets me confused

Im going to show a picture

Its in german but the drawings are there

A is the differentiate matrix and A^-1 is the integration matrix

he explains its 0 because in the matrixes top row of A^-1 of A is 0

And this is due the kernel being 1 so theres 1 constant

[spare widget]

so what is the question

[flint nexus]

I don get how he gets 0 with this explanation

[spare widget]

as K see it you're having matrices for differentiating and integrating polynomials

is T integration?

or is T differentiation?

T=A?

[flint nexus]

T is differentiation

No, T is Av

like f=ax

[spare widget]

well d1/dx = 0

[flint nexus]

ye

[spare widget]

then integral of 0 = 0

[flint nexus]

ye

that i get

[spare widget]

there should be a constant when integrating the lhs though

but assuming pseudo-inverse

with constant 0

[flint nexus]

But they explain with the kernel and that for all dim this is possible i believe

[spare widget]

idk, I can't read German

[flint nexus]

because the A-^1A has 4 dim and it still works

[spare widget]

T(x) = Ax?

[flint nexus]

yes

[spare widget]

ok

well then it's clear

[flint nexus]

But maybe in this case its saying that what the integration would be dependent on the dim

so ofc

[spare widget]

$A^{-1}A\begin{bmatrix} 1 \ 0 \ 0 \ 0 \end{bmatrix} = 0$

[flint nexus]

if it was v=x

this would end up in 1

[stoic python]

criver

[spare widget]

but what's your question?

[flint nexus]

They explain it thorugh here

i think i g et it now

[spare widget]

what is your question?

[flint nexus]

They explain its 0 with this

cuz of the one constant that you get with the kernel

I get it i think

This basically says it would give an identity Matrix but the dim 1 collum is 0

the one on the far left

Its matrix that describes the differentiatiation and then integration

and the thing they try to convey is that with the kernel you can certify that integral of differential of a constant its 0

Is what imsaying making sense or am I just waffling?

[spare widget]

idk what the question is

yes the matrix is not full rank

yes the matrix will spit out a zero for any vector (C, 0, 0, 0)

[ornate jetty]

hello

ive an answer to a q but i dont quiet understand

part D

thats the solution

Question is.

How is there an A squared on both sides?

[flint nexus]

ty

[spare widget]

A=A^2 since A is from U

so you can replace the A^2 by A

the point is that the matrix 5A != (5A)^2 = 25A^2 =25 A

unless you picked A to be 0

but you know that A = I is also from U

so you could pick that one

and see that 5I != 25I

[ornate jetty]

hold on repeat that again pls@spare widget

[spare widget]

I^2 = I then I is from U, but 5I != (5I)^2

hence U is not closed under scalar multiplication -> it's not a subspace

I being the identity matrix

[ornate jetty]

yep but in his solution he got 5A^2

[spare widget]

they show (5A)^2 = 25A^2 != 5A^2 = 5A

do a substitution for more clarity

B = 5A, and A is from U, then A^2 = A

now

check whether

B^2 = B

[ornate jetty]

ahhh ok perfect. got a bit confused

[spare widget]

B^2 = B only if A = 0

[ornate jetty]

yes

[spare widget]

but there are elements other than A=0 in this space

namely A = I

so you have your counterexample

[ornate jetty]

jesus this is a bit difficult to grasp

[spare widget]

what they provide as proof is technically not enough

since they don't show there are other elements than 0 in that set

(5A)^2 = 5A for A = 0

[ornate jetty]

only for the 0 yes

perfect spot on

thank you

[spare widget]

Turns out this is pretty standard in physics. Though I don't know how rigorous it is: https://physics.stackexchange.com/questions/188354/is-there-a-natural-suitable-definition-for-functional-derivative-in-curved-spa?rq=1

Namely point 8 in the answer

[thin spire]

need some help on my hw in algebra pls

https://i.imgur.com/UbwGDbs.png

[spare widget]

this is not linear-algebra

[wintry steppe]

Guys I would like to get help in math high school

Like post my questions

Im new here

[spare widget]

see #❓how-to-get-help

[fringe zodiac]

Q: prove that the subset of R4 are subspaces of E4 by writing them as the null space of a matrix.

I’m used to doing ones like this, but the wonky coordinate system is confusing me

[nocturne jewel]

what wonky coordinate system?

[fringe zodiac]

[a b a+b 2b] and the fact that there are no characteristic equations for a and b

Rather all a and b in R

Basically I’m not sure how I can make a homogenous system out of that

[nocturne jewel]

W=span{[1,0,1,0]^T,[0,1,1,2]^T}

Think about how you would solve Ax=0, then reverse engineer a possible A

Hint: Pick A st it's in RREF, it'll be easiest

[fringe zodiac]

^T meaning?

[fringe fjord]

Matrix transposition. In this case it's just a compact way to write column vectors.

[fringe zodiac]

Gotcha

I’ll write something up

[pulsar imp]

Hey nerds gimme your lunch money

[nocturne jewel]

yeah instead of typing out $\begin{bmatrix}1\0\1\0\end{bmatrix}$

[stoic python]

Mosh

[fringe zodiac]

So I got to this,

Do I take s and t as free variables, and try to make a matrix that works with that?

So I have 2 fvs and need 4 columns, meaning a 4x2 matrix is required

Could this work @nocturne jewel ?

[nocturne jewel]

looks like it'd work

though your augmented vector should be 2x1 not 3x1

[fringe zodiac]

Pardon?

[nocturne jewel]

you have a 2x4 augmented with a 3x1

[fringe zodiac]

Oh! Just noticed, thanks haha

Thank you for everything btw, I appreciate it.

[celest ore]

is there some relationship between between having a pivot position in every row to the solution space of the linear system?

[fringe fjord]

It just means none of your original equations were redundant.

[celest ore]

right, but that doesn't really answer the question

[fringe fjord]

Doesn't it? The solution space is not affected by whether you originally had redundant equations.

So you could have rows without pivots, or have pivots in all rows, no matter what the solution space is.

[wintry steppe]

when you have norm ^2 is that the same as taking hte quantity just square

[celest ore]

Right, so I've asked a dumb question.. i'm trying to justify how a linearly independent 3x3 matrix will span R3.

[wintry steppe]

so for example i want to do | | y - y_hat | |^2 and that quantity ends up being the column vector of 0 1/2 -1/2

[nocturne jewel]

By extremal properties, since dim(R^3)=3, any 3-tuple of independent vectors span the space

[grave garden]

Guys

Is orthogonal and perpendicular the same thing ?

[nocturne jewel]

Perpendicular usually only refers to R^2

orthogonal refers to perpendicularity, and the more abstract idea

[grave garden]

Im confused about normal, perpendicular and orthogonal now

Now im reading about Scalar Products and Orthogonality

[fringe fjord]

Don't worry, you won't go too wrong by thinking of them as synonyms.

Except for matrices.

An orthogonal matrix is a normal operator, but not the other way around.

[grave garden]

I see

[fringe fjord]

But that particular "normal" hardly has any association to right angles, anyway.

[grave garden]

What does non-degenerate mean ?

[fringe fjord]

About what kind of thing?

[grave garden]

The scalar product is said to be non-degenerate if in addition is also satisfied the condition...

If $v$ is an element of $V$ and $<v,w>=0$ for all $w\in V$ then $v=0$

[nocturne jewel]

[stoic python]

Potato

[grave garden]

Is there a latex for $<v,w>=0$ or just type it like this ?

[stoic python]

Potato

[fringe fjord]

THat looks like a definition of "non-degenerate" you have right there.

$\langle v,w\rangle = 0$.

[stoic python]

Troposphere

[grave garden]

I thought non-degenerate is sth, and then that condition make it so

[fringe fjord]

"is said to" is one of the more blink-and-you'll-miss-it ways to introduce a definition.

[grave garden]

I see, I'll keep an eye on it

[fringe zodiac]

The second pic defines V and W

And P2

How would I approach c?

[zinc timber]

take an arbitrary polynomial in P2 and show that you can write it in the form v+w

requires a little manipulation

[fringe zodiac]

Arbitrary poly being ax^2+bx+c

[zinc timber]

yes

[fringe zodiac]

Got that part, but v and w are not polynomials to begin with, are they?

Well they’re polys with conditions attached, but how would I write them out

[zinc timber]

why won't they be polynomials

try to separate out ax^2+bx+c = (...x^2+...x+0) + (...x^2+...x+c)

see that the constant part of the first term is 0

[fringe zodiac]

Which is derived from V

[zinc timber]

adjust the coefficients so that the sum of the coeffs of 2nd term turns out to be 0

[fringe zodiac]

2nd polynomial you mean?

[zinc timber]

ya

[fringe zodiac]

That brings me here

[zinc timber]

that does not work tho, remember you also need v+w = your original polynomial, i.e. ax^2+bx+c

do it step wise, say you split them like this (ax^2+bx) + (c) then it's still our original polynomial and (ax^2+bx) \in V but (c) \notin W

for that we need to add some -cx or -cx^2 to our 2nd term giving us (-cx+c) \in W ( u can also use -cx^2 or something else)

but now this dis balances our original polynomial, so to balance it again we add another +cx to our first term giving us (ax^2 + (b+c)x) + (-cx+c)

check this satisfies the condition

[fringe zodiac]

One sec

[zinc timber]

(also note that the representation is not unique)

[fringe zodiac]

Okay so we use V to make an arbitrary polynomial ax^2+bx+0 which works cause the only condition is c=0

[zinc timber]

ya

[fringe zodiac]

In W, we set a=0 for simplicity, although we could do it with b too

[zinc timber]

we are not setting a=0 but choosing a,b,c, is a way that a+b+c=0

[fringe zodiac]

Yes, and we choose b = -c right

We make our second poly 0+(-c)x+c

[zinc timber]

yes and a=0

[fringe zodiac]

The cs cancel and the condition is met

[zinc timber]

therefore P2 \subseteq V+W

[fringe zodiac]

Although any multiple/transformation on c is an alternative answer

[zinc timber]

I won't say any because it also has to be equal to our original polynomial

[fringe zodiac]

Any is too general yeah, a power, multiple would work though

[sweet willow]

maybe out of scope, but

how can I efficiently make two random matricies, g (1, N) and B (NxN, symmetric) s.t. B is indefinite and B = U'LU -> Ug_1 == 0?

also where B is indefinite and Ug_1 != 0

ping if responding please

[ionic stream]

verifying that a set with two operations is a vector space. could someone check parts e-g for mistakes? something tells me i’m supposed to show that all three axioms listed do hold, but instead i found that they don’t

[grave garden]

Guys, is it false to say that all vector space has a scalar product ?

[lavish jewel]

scalar product like the dot product? a map f: V x V -> F?

[grave garden]

$\langle v,w\rangle$

This thing

[lavish jewel]

mhm. the answer is yes, it is false

(remove the space before the $ btw)

[stoic python]

Potato

[grave garden]

Because in the book im reading, when they want to talk about it, they always assume that, that scalar product is defined

[lavish jewel]

it's not part of the definition of a vector space

so they're talking about inner product spaces instead

if you're working with R^n and C^n and subspaces thereof, though, yes

[grave garden]

Is the scalar product unique ?

[lavish jewel]

no

any operation that satisfies the definition of an inner product is an inner product

as useless as that sounds

for example in R^n, any symmetric positive definite matrix can be used to define an inner product

[grave garden]

I see

( i just learn about positive definite thing )

[flint nexus]

Hi, I am having some difficulties in understanding how to turn vector spaces with matrices

could someone help me understand plez and explanation would be helpful

the first simple example is

You have a standardbasis with v1=(1,0); v2 =(0,1)

And they turn along the angle theta

They ask for matrix

[spare widget]

https://en.m.wikipedia.org/wiki/Rotation_matrix

[flint nexus]

ty

Now they are saying imagine a standard basis of i and j so that the space the classic 2 d space

now project them to only fit in a line which is 45 degrees

[nocturne jewel]

Yeah..? and?

[flint nexus]

they want to project all vectors into this line

[nocturne jewel]

the span of that vector

yes

[flint nexus]

so i first thought for the vector in the middle

so the matrix would be

(1/2,1/2),(1/2,1/2)

But this matrix only works for one vector right?

[nocturne jewel]

wdym

[flint nexus]

I mean ye the vector span this grey line

So its Matrix transformation

So Av=w

[nocturne jewel]

L[v]=Av works for any v in V

[flint nexus]

and w is the grey line

[nocturne jewel]

w is a vector

W is the span of that vector

[flint nexus]

tru but i only want the grey line

so i need a matrice that describes only the gry lin

Do understand wht i mean

?

[nocturne jewel]

Not really

You found the matrix

[flint nexus]

But it only works for one point in the line

how immi supposed to get the whole line spanned

[nocturne jewel]

.

You have the entire span of [1,1]

[flint nexus]

ye

But I wanted so i dont get the entire span

i wanted to just span the line

and thats where the matrix comes in

[nocturne jewel]

That's contradictory

You both want and don't want the entire span

[flint nexus]

of course with the basis vector 10 and 01 i can span all in R^2

No i dont want the entire span

im saying the basis vectors here span R^2

[nocturne jewel]

Yeah

duh

[flint nexus]

but i only want the grey line

the one that connects bothe basis vectors

[nocturne jewel]

which grey line...?

[flint nexus]

ahhh

[nocturne jewel]

You have 4 black lines

[flint nexus]

ok lokk at the pic

the hypotenuse

[nocturne jewel]

Yeah, there's 4 black lines

[flint nexus]

the long side of the triangle is grey

[nocturne jewel]

It isn't but ok

[flint nexus]

click the pic itll get bigger then youll notice i think if not ill chnage the colour

bettter?

[flint nexus]

I dont want to be toxic but it is grey the other triangle

Just saying maybe you should do one of those colour tests there are a lot of people who dont see coulour that well

I think if your male its like 1 in 10

but in anyways

do understand what im asking now

I have 2 basis vector namely 1 0 and 0 1

I want to span with matrix only the orange line

So That Av=W and W= orange line

Is this making sense now?

I can get the midle vector with the matrix(1/2,1/2;1/2,1/2)

but the others i cant

I just found out about the help channels , Sorry for btohering you guys

[pastel merlin]

Any video or helping tool that explains matrices ( addition , subtraction , multiplication ) because I have an exam tmrw and I know nothing 😭

[oblique prairie]

https://www.mathsisfun.com/algebra/matrix-introduction.html

[pastel merlin]

thank you 🫂

[oblique prairie]

all the basic matrix stuff you need is either on that page or linked on that page

[zinc timber]

,av @oblique prairie

[stoic python]

quantum#6320's Avatar

Click here to view the GIF.

[zinc timber]

akame

[oblique prairie]

uwu

yes very cute

[wintry steppe]

Can we say that theory of matrices is evolved from theory of linear transformation . Coz for example if we consider matrix multiplication then the way it is defined is because of composition of linear transformation and that condition number of columns must be equal to number of rows in order to define matrix multiplication coz in order to define composition two inner dimensions must be equal ?

[nocturne jewel]

yes

Matrices arise from linear transformations

[wintry steppe]

Okay ty !

[grave garden]

Hii guys

I dun understand how they get the dimension for W

[fringe fjord]

Bacically "trace=0" is a single linear equation in the entries of the matrix, so enforcing it can at most reduce the dimension by 1.

Also, since symmetric matrices with nonzero trace do exist, enforcing trace=0 does take away something, so the dimension has to decrease by at least 1.

[grave garden]

I see

[exotic wedge]

If $T : V \rightarrow W$ and $\dim(V) = \dim(W)$. Then if $T$ maps basis to basis, this implies that it's an isomorphism right ?

[stoic python]

ru0xffian

[wintry steppe]

i need some help

Let$$f(x) = \frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$

[stoic python]

kaniii

[gray dust]

@exotic wedge try showing it

[wintry steppe]

is someone able to break down whats happening here?

[exotic wedge]

I am using it in a proof and i was trying to verify it. My initial idea is that if it is not surjective then it can't map basis to basis. If it is not injective then it can't be well-defined. But I didn't verify that rigorously.

[gray dust]

may be better to go into the details then

are V,W findim?

[exotic wedge]

yeah sure

They are findim

[wintry steppe]

i never know where to start

[gray dust]

pick a basis $v_1,\ldots,v_n$ of $V$ and a basis $w_1,\ldots,w_n$ of $W$

[stoic python]

RokabeJintaro

[exotic wedge]

Yeah and wlog you can say it maps v_i to w_i

[gray dust]

yes

show kerT={0}, conclude T is injective & surjective

[exotic wedge]

yeah i got it, thanks !

[gray dust]

no prob

[nocturne jewel]

Not really LinAl

But write out the 1st couple values for n

That's like always a good place to start w/ patterns

[wintry steppe]

ah yes

[nocturne jewel]

Had that not at all crossed your mind?

[wintry steppe]

i dont know if i realy find a pattern

[wintry steppe]

what was i supposed to find @nocturne jewel

[night wren]

is the most direct way to compute the matrix of a linear transformation for a nonstandard basis to use the change of basis matrix and the matrix for the standard basis?

[gray dust]

@night wren do u already have these matrices?

[night wren]

I calculated the matrix for the standard basis.

[gray dust]

then u can use the standard matrix to find the matrix wrt beta

[night wren]

So I'm looking for C^{-1}AC where C is the change of basis matrix from the standard basis to beta, yes ? and A is the matrix for the standard basis

[gray dust]

no, C should be from beta to standard

think of the order of composition

looking at the exercise, the alternative is faster tbh. take the coordinates, in beta, of the image of each basis vector

T is defined so that those coordinates are easy to find

@night wren

[night wren]

I'm not sure how to set this up

[gray dust]

say b_i is the i'th vector in beta

the i'th column of [T]_beta is the coordinates of Tb_i wrt beta

[night wren]

OK, I got it, thanks a bunch

so this is a more direct way, look at the image of the new bases and write it as a linear combo of the bases

[night wren]

could you say y this works?

[gray dust]

u can show that
$$[Tx]\beta=[T]\beta[x]\beta$$
for all $x\in V$ where $[x]\beta$ is the coords of $x$ wrt $\beta$

[stoic python]

RokabeJintaro

[night wren]

so is this method calculating the left side where x = b_i and then trying to deduce the coefficients for T_b so that you get the same thing?

[gray dust]

this essentially says [T]_beta enacts T, which is what we expect of matrix representations of linear maps

[gray dust]

this isnt the formula for finding coefficients, its a result u can show after u construct [T]_beta

[night wren]

But this technique you described uses this theorem, yes? You said to calculate T(b_i) wrt to beta so thats the left side. Then you try to write T(b_i) as a linear combo of beta which is the right side, kind of

[gray dust]

finding the coords of Tb_i wrt beta is its own matter

once u construct [T]_beta then the above is a result u can show (note its for all x in V)

[queen pagoda]

is it possible to find the rank of the matrix if you know that Ax = 0 and number of variables?

[haughty berry]

If you knew the number of linearly dependent vectors x such that Ax=0 and the number of columns (variables), then by the rank nullity theorem, yes

[wintry steppe]

so i know an ellipse can be represented as a 2 x 2 matrix

is it only a 2 x 2 matrix or can any square matrix represent an ellipe?

ah so its any matrix thats symmetrical

[nocturne jewel]

Refer to quadratic forms

[wintry steppe]

cool ty

additionally if i had a vector u in ker A^T and v in range A whats the best way to show u is orthogonal to v without using the kAT theorem?

i know i cant just do (A^Tu) w

and say that it would be 0 * w

thus being orthogonal but im not quite sure how to do it otherwise

[nocturne jewel]

That's not what v in range means

[wintry steppe]

i was just saying let Av = some vector w

for simplicity

[nocturne jewel]

$v\in \operatorname{Im}(A)\iff\exists w\in V$ st $Aw=v$

[stoic python]

Mosh

[wintry steppe]

coud i just do smt like this?

[nocturne jewel]

Im just correcting your error

[wintry steppe]

ah ok ty!

[fervent ridge]

So I’m confused on how to get specific solution from echelon form

I know you can use echelon form to see if it’s consistent or inconsistent but what if I want to find specific solution?

[nocturne jewel]

then you get it into RREF

[wintry steppe]

is this a valid operation?

[nocturne jewel]

and read off the "state" of solutions

Im gonna say no, cause you still have the error

[wintry steppe]

isnt this the same

Ax = some vector w given that x is in the range of n

[fervent ridge]

so you can never find specific solution with echelon form?

[wintry steppe]

and A is m x n

[nocturne jewel]

Yes

That's not what you have

[fervent ridge]

i just wanted to see if there a way without doing reduced

[wintry steppe]

lol i copied the problem though

[nocturne jewel]

v needs the form Ax

No

I'm not saying there's a problem with v in Im(A)

I'm saying there's a problem with the fact you then wrote Av=w

[wintry steppe]

so should it just be Av?

[nocturne jewel]

Are you reading what I'm saying?

Cause it doesn't seem like you are if you just asked that

[wintry steppe]

i dont know what lm is

[nocturne jewel]

Image

Range

synonymous

[wintry steppe]

oh

you're saying to say v is in R^n not range A

[nocturne jewel]

...

v is in range(A)

[wintry steppe]

Ok then i have no clue, i just copied the instructions for the problem and thats exactly what it had 😢

[nocturne jewel]

You're messing up applying the definition of range

[fringe fjord]

If v is something that A can output and A is not square, then v cannot possibly have the right shape for "Av" to be meaningful.

[wintry steppe]

So instead of multiplying v to get w its Aw to get v

[wintry steppe]

then i guess this method of proving they're orthogonal wouldnt work hmm

wait nvm u can

u can say u^T (Aw) where u is in the kernel of A^T and Aw = v

since u^T is a 1 x p matrix and Aw ends up being a p x 1 matrix

thus when u do this and take the full transpose u get

w^T (A^Tu)

A^Tu = 0 bc u is in the kernel of A^T

thus it 0

showing that the dot product of the two is 0 so they must be orthogonal

[feral iron]

I'm having trouble figuring out the eigenvectors for this. I took linear algebra a few semesters ago so I kind of don't remember what to do

work so far:

I was following an example where they did row echelon here but it seems kinda complicated with what I have

[quartz compass]

what I do is just put in variables like $\vec x = \begin{bmatrix}x \ y\end{bmatrix}$ and just solve for one of the variables in terms of the other, then factor it out since we only care about eigenvectors up to scalar multiples

[stoic python]

Merosity

[quartz compass]

this gets you 2 equations but you can just pick the easier one

$$\begin{bmatrix}1 & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix}x \ y\end{bmatrix}=\lambda \begin{bmatrix}x \ y\end{bmatrix}$$
then you get,
$$\begin{bmatrix}x+y \ x\end{bmatrix} = \begin{bmatrix}\lambda x \ \lambda y\end{bmatrix}$$

[stoic python]

Merosity

[quartz compass]

so obviously pick the second component as your equation, $x=\lambda y$. That makes your vector $$\vec x =\begin{bmatrix}\lambda y \ y\end{bmatrix} =y \begin{bmatrix}\lambda \ 1\end{bmatrix}$$ and you're done

[stoic python]

Merosity

[feral iron]

That is way easier just remembering its a system of equations. Thanks

[quartz compass]

yup you're welcome

[wintry steppe]

True or false : every vector space is the dual of some other vector space
Its false right ?

[fringe fjord]

In order to be interesting, that question must be interpreted as "is every vector space isomorphic to the dual of some vector space?"

It is false indeed (at least assuming the axiom of choice), but can you name a counterexample?

[grave garden]

Hii guys

How do they get this ?

[wintry steppe]

Every vector space is isomorphic to its dual

[fringe fjord]

That is true if the vector space has finite dimension.

[spare widget]

Try substituting with the above definition of w

it follows because the inner product is bilinear

[wintry steppe]

For counterexample P_3(R) is isomorphic to dual of R4

[wintry steppe]

I didn't get that parentheses part

[fringe fjord]

That's an example, not a counterexample. A counterexample to the claim should be a vector space that isn't isomorphic to the dual of anything.

[wintry steppe]

Ops my bad 😩

[fringe fjord]

I crossed out the parenthesized part, because I later concluded that we can find a counterexample without using the axiom of choice. (I may still be wrong about that, though).

[wintry steppe]

Okay

[wintry steppe]

As far as finite dimensions are concerned i dont think there is any counterexample .but for infinite dimension there may be a counterexample

[fringe fjord]

Correct.

One of the counterexamples is the space of sequences of real numbers where each sequence only has finitely many nonzero terms.

This space has countably infinite dimension.

It cannot be the dual of anything with finite dimension, because those spaces all have finite-dimensional duals.

[wintry steppe]

N how did u interpreted that que like how did u know "is every vector space isomorphic to the dual of some vector space "?

[wintry steppe]

Okay

[fringe fjord]

It cannot be the dual of anything with larger dimension either, because the dual of a space has at least the dimension of the space itself.

[fringe fjord]

That's mostly a matter of experience. It's a rather uninteresting fact that not every vector space is literally the dual of something else, because the elements of a dual space are always linear transformations, and we're free to define a vector space whose elements are represented by mathematical object that are not even mappings.
On the other hand I already knew that "isomorphic to the dual of something" has an interesting (and slightly surprising) answer, and I also know it is common to skip the "isomorphic to" wording, relying on the reader to understand it implicitly.

[fringe fjord]

Finally, this space cannot be dual to itself, because there's a linear functional for each map from N to R, without the restriction that only finitely many values are empty. And the space of such maps has uncountable dimension -- the set of functions of the form f(n) = a^n for some a in (1,2) is linearly independent, so the dimension has to be at least the cardinality of |R|.

[wintry steppe]

Oh okay

[wintry steppe]

Here dual of a space has the dimension of space itself . Idk how u r saying at least

[fringe fjord]

Again, the dimenions are only equal if the dimension of the original space is finite.

[wintry steppe]

Oh yeah correct .

Ty for ur precious time !

[wintry steppe]

lets say you have a c such that

c = (B^-1 A^-1)

c(AB) = I

can you say then that by associativity, that (AB)c = I as well or do u ahve to prove that separately?

i guess my question is

is there any property of matrices where i can trivially say since c(AB) = I then (AB)c = I too

from c(AB) = I the only thing associativity says is that (cA)B = I as well

sorry i meant commutative*

ik matrix multiplication isnt commutative

but this is the identity matrix so i was wondering if theres anyway we can trivially say (AB)c is the identity too

AB = I implies BA = I so long as all matrices involved are square

[fringe fjord]

If AB is a square matrix, then you know that the existence of either a left or right inverse means that AB (or really the linear transformation it represents) is bijective.

[wintry steppe]

they r squares

A and B are invertible n x n matrices here

[fringe fjord]

For a bijective map a left inverse is automatically a two-sided inverse.

[wintry steppe]

amazing tysm

[fringe fjord]

(And similarly for a right inverse).

[wintry steppe]

i ask bc this was a hw, and i got points off for not proving (AB)c bc i only did c(AB)

so im gonna ask my professor

bc he also made mistakes on two other problems so its not totally out of field that he messed up here as well

but i just wanted to clarify before i ask him for pts

[fringe fjord]

Just because it can be proved is not the same as you don't have to prove it ...

[sinful valve]

what is this notation exactly im confused

[wintry steppe]

i agree but for one, for example, our professor said we dont have to prove that a subspace is non empty only scalar multi and vector addition

and then he took pts off for not proving its not empty

even tho ik thats the definition

so i vaguely recall he said smt about this too but i wanted to clarify

https://cdn.discordapp.com/attachments/540211747613704221/945016917209546793/Screen_Shot_2022-02-20_at_11.59.19_AM.png

im still a bit confused on how this is the answer

[cold flint]

When doin the laplace transform of S(t-c) is the correct form e^(-cs) or (e^(-cs))/s?

[oblique prairie]

#odes-and-pdes

[cold flint]

I've seen these two forms but I'm not sure which is correct

[robust flicker]

can anyone help me out w this problem im not so sure where to begin

[teal grotto]

do you know what proj_a(v) ? thats where i would start

[robust flicker]

yeah ik that however im confused bc theyre asking for the magnitude of the projection to be 0 right??

[halcyon spindle]

Hmm I have an idea for it, think about orthogonal vectors.

[robust flicker]

ohh okay i think that gave me an idea thank u :))

@halcyon spindle can i dm u??

[halcyon spindle]

Sorry, I decline.

[robust flicker]

its alright dw

[gray dust]

@cold flint the functions are NOT the same. delta denotes dirac delta. u denotes the unit step function

[cold flint]

Thank you

[cunning pike]

if i got a given formula to get a matrix

they ask for the matrix where the reflection is in the plane x_3 = 0

how do i apply this?

[tired jungle]

Let $\mathbb{A}^m$ be an inner product space with tangent space $\mathbb{V}^m.$ We also have the inner product space $\mathbb{B}^n$ with tangent space $\mathbb{W}^n$.
We will use the standard euclidean metric d on $\mathbb{A}^m$ and $\mathbb{B}^n$, where $d(p_1,p_2)=|p_1-p_2|.$
Show that any isometry $F : \mathbb{A}^m \to \mathbb{W}^n$ is an affine map and its differential $$dF : \mathbb{V}^m \to \mathbb{W}^n$$ $$v \mapsto F(p+v)-F(p)$$ is an orthogonal map.

If I am asleep pls ping me

[stoic python]

fake neko

[spare widget]

Operator notation, D means differentiation and as defined (T(f))(x) = x^2f(x). It's demonstrating that the two operators do not commute: (TDf)(x) = (T(Df))(x) = (Tf')(x) = x^2f'(x)

(DTf)(x) = D(x^2f(x)) = x^2f(x) + 2xf(x) from the product rule

[robust flicker]

suppose the coefficient matrix of an m × n linear system has n pivots. is it true that the system has a solution?

if it does have a solution can someone explain why that is

[lavish jewel]

in general, no

if m > n, then the image of the matrix is only a subspace of the codomain

as an easy example, consider a 2x2 identity matrix, and append a row of 0s to it

making it 3x2

the image is the span of [1,0,0] and [0,1,0], so vectors spanned by [0,0,1] have no sol x so that v = Ax

[robust flicker]

im sorry to bother u but could u show the example thru here?? im having some trouble visualizing it

[robust flicker]

ohh i see

tysm :))+

[grave garden]

Guys, orthogonal basis = basis right ?

Because in orthogonal basis, its element are all perpendicular to each other, and for basis, all its element is linear independent, and that mean they all are also perpendicular to each other ?

[haughty berry]

An orthogonal basis is a basis by definition

But not all bases are orthogonal (for example {(1, 1), (0, 1)})

And not all orthogonal sets are bases or even linearly dependent (for example, {0})

[grave garden]

💀

So orhogonal basis can be called a basis, but a basis can't be called orthogonal ?

[haughty berry]

But any orthogonal set that doesn’t include the zero vector is linearly dependent, so it follows that any orthogonal set with n vectors (where n is the dimension of the vector space) that doesn’t include the zero vector is linearly dependent

[haughty berry]

Do you recall what the definition of an orthogonal basis is?

It’s a basis that’s orthogonal

So yes, obviously an orthogonal basis can be called a basis

And yes, like I said, not all bases are orthogonal

[grave garden]

Hmmm, so a basis imply orthogonal, but an orthogonal set is not a basis and could also be linear dependent

[haughty berry]

No

I just said, not all bases are orthogonal

So basis doesnt imply orthogonal

But yes, an orthogonal set can be linearly dependent (if and only if it contains the zero vector)

[grave garden]

Why zero vector make it linear dependent ?

[haughty berry]

If the zero vector is in any set, then the set is linearly dependent

Like suppose you have a set ${v_1,\dots,v_n,0}$. Then you can take the following sum:
[ 0\cdot v_1+\cdots+0\cdot v_n + 1\cdot0 = 0 ]
So you have a zeroing sum where one of the coefficients is not zero. So the set is linearly dependent

[grave garden]

Ahhh I see it now

[grave garden]

Glad that I ask this question, otherwise I would have failed the exam

Thank mate !

[haughty berry]

Np good luck!

[wintry steppe]

@haughty berry ?

[wintry steppe]

Show that every plane through the origin in R^3 may be identified with the null space of a vector in (R^3)^* . I didn't get this que how can they talk about null space of vectors of dual of R^3 without defining LT from dual of R^3 to some other vector space and what is meant by identified with null space ? Are they saying i have to find isomorphism ?

[spare widget]

It just means find a normal to the plane

And then $f_n(\cdot) =\langle n, \cdot\rangle$ is the element

[stoic python]

criver

[spare widget]

Since you know that $\langle n, x\rangle = 0 \implies x \in plane_n$

[stoic python]

criver

[wintry steppe]

Okay ty !

[spare widget]

I am trying to figure out how to incorporate an affine constraint into a projected gradient descent procedure.

The affine constraint can be turned into a linear constraint on the gradient of the form:

$\boldsymbol{1}^T \begin{bmatrix} \boldsymbol{I} & \ldots & \boldsymbol{I} \end{bmatrix} \boldsymbol{d} = 0$

[stoic python]

criver

[spare widget]

where the middle matrix is made of m identity matrices stacked next to each other

it's clear that one solution is:

to split d into m equal pieces

and then for each piece to impose the constraint

$\boldsymbol{1}^T\boldsymbol{d}_i=0$

[stoic python]

criver

[spare widget]

which would consist of projecting d_i on a normalized version of 1 and subtracting that from d

I don't think this is the projection of the steepest descent, since additionally one also wants to maximize the similarity with the initial negated gradient

$\max_{\boldsymbol{d}} \boldsymbol{v}^T\boldsymbol{d}$

[stoic python]

criver

[spare widget]

or potentially

$\max_{\boldsymbol{d}}|\boldsymbol{v}-\boldsymbol{d}|^2$

[stoic python]

criver

[spare widget]

with the caveat that d must satisfy the aforementioned constraint

[lavish jewel]

those two aren't really the same though

[spare widget]

I know

[lavish jewel]

what's the original problem

[spare widget]

I am guessing that |v-d| would be the equivalent of the projection in the sense of L2

while d dot v is just something I thought of

the original problem is constrained gradient descent with affine constraints

since I am minimising an L2 energy, it probably makes sense to prefer the L2 projection

[spare widget]

i.e. this

[lavish jewel]

gradient descent is usually not the original problem

it's the proximal of the 1st order taylor approx

what's the original cost func and constraints

[spare widget]

the original cost function is

$\min_c|u(c)-f|^2$

[stoic python]

criver

[spare widget]

where u depends nonlinearly on c

there are also the constraints on c I mentioned

I derived the gradient for the above

and I would like to apply projected gradient descent to it

in order to satisfy the aforementioned affine constraint on c

I basically have the negated gradient v

and want to project it onto the set

$X = {\boldsymbol{x} ,:, \boldsymbol{1}^T\begin{bmatrix}\boldsymbol{I} &\ldots & \boldsymbol{I}\end{bmatrix}\boldsymbol{x} = 0}$

[stoic python]

criver

[spare widget]

If the matrix with the Is was not there, then it's fairly simple as x_proj = x - (1 dot x)/|1|^2 * 1

i.e. removing the component parallel to 1

[lavish jewel]

well but, the way this constraint is written is separable

[spare widget]

yes I can do 1^T di = 0

but I don't think this will be the projection

[lavish jewel]

yes, that's exactly equivalent

[spare widget]

is it?

[lavish jewel]

do the linear algebra

[spare widget]

wouldn't I be able to trade-off different component

e.g.

[lavish jewel]

if you multiply the 1^T from the left, then 1^T replaces all of the I matrices

[spare widget]

say I have a d that I split into m blocks

[lavish jewel]

and vector equivalence is done elementwise

[spare widget]

and 1 dot d_i = 0

then let \sum_i d_{i,j} = k

now I can choose the different j differently

and this will still satisfy the constraint

however one would clearly have a higher energy

I think it's because [I ... I] is a rectangular matrix

so there are multiple ways to satisfy this

it takes us from a larger space (namely m times larger) to a smaller one

so there are multiple solutions that will seem equivalent, but I do not think they are equivalent under an L2 norm when considering |d - v|^2

[lavish jewel]

this is already captured in the 1^T d_i

each of those is a hyperplane

they're algebraically equivalent

[spare widget]

ah, is it because the spaces are orthogonal

when I split it into m pieces

each piece is orthogonal to the others

ok, that's great, thank you

[lavish jewel]

wait

lemme look at it again

yes

if it were a large I matrix, it'd just be 1^T d

but it's in blocks

[spare widget]

yes

[lavish jewel]

this chops up d into independent pieces

each one has its own hyperplane

if you know how to do the projections for a constraint of the type 1^T x, you just have to do it N times (however many I blocks there are)

[spare widget]

great, I was worried that the two may not be equivalent

but I guess because they are linearly independent

it can be split up

thank you

[lavish jewel]

hmmmmmm

actually

it's easier

it's just one long 1^T vector

i messed up the multiplication from the right

it's a single projection

[spare widget]

ah

you did

[lavish jewel]

1^T [I I I...] just makes one long 1^T vector

[spare widget]

1^T * [I ... I]

yes, you're correct

[lavish jewel]

this is even simpler, then

[spare widget]

I messed it up too, hence why I thought something was off

[lavish jewel]

😛

[spare widget]

I mean it makes perfect sense, it's just a different grouping of the terms 😂

[lavish jewel]

it takes 2 to tell each other they're dumb

[chilly hazel]

not sure in what channel I should ask this but why does the constraint Tr(X) = 1 increase numerical stability in semidefinite programming

[still lodge]

positive determinant implies that a linear transformation stays in the 1st or 2nd quads right

[spare widget]

positive determinant implies no odd number of reflections

e.g. change of handedness in 3d

idk what it means for a linear transformation to stay in the 1st or 2nd quads

A rotation matrix has determinant of 1 and can send an input vector to any quadrant

[still lodge]

i'll rephrase then

what does it mean for a linear transformation to preserve orientation

wait im stupid

basis vecs dont flip

[spare widget]

Well imagine an identity matrix with one of the 1s being a -1

That's a reflection along the axis

And changes the handedness

An even number of reflections yields you a rotation

An odd number yields you a negative determinant so a change in orientation

If you were to draw an example you can take a basis, and then negate one of the basis vectors

If you want you can also think of a hyperplane with normal n, flipping its normal changes its orientation, then you can extend this to manifolds too

e.g. think of flipping the normal of a sphere in R^3 to the inside

Basically det < 0

Also makes sense if you think of the determinant as computing the signed area of a high-dimensional parallelepiped with the edges being the matrix's columns

If you flip an odd number of the edges the sign and orientation flips

[wintry steppe]

is the inverse of a square matrix always itself?

its not

[slow scroll]

the easiest class of counter examples to consider are 1x1 matrices

[wintry steppe]

yea thats what i ended up doing, tysm!

what does it mean when someone says training data as i.i.d?

[finite karma]

independent and identically distributed

[keen siren]

What properties of matrix and vector algebra ensure that Ay = 0?

these are the theorems i have to choose from

[slow scroll]

this question doesn't make much sense. Assuming A is a matrix, Ay need not be 0 in general

[keen siren]

yea thats why im hella confused

[halcyon spindle]

If y was 0, you can definitely use a and b to show Ay = 0.

[slow scroll]

do you have a screenshot of the question?

[keen siren]

this is y

yeah

[slow scroll]

okay i guess we need to know what A is too lol

[keen siren]

^this is A

[slow scroll]

how is your understanding of property d? @keen siren

[keen siren]

not too well honestly

my teacher isnt the best

[gray dust]

it should be said that a_j denotes the j'th column of A

[slow scroll]

Okay, so if A is a m x n matrix, you can write A = [a1 a2 .... an] where a1, .., an are the columns of A. So its saying that Ae_i picks out the ith column of A

[gray dust]

d) can be found by direct computation

[keen siren]

ohhh i see

so then what does it have to do with Ay= 0 tho

[slow scroll]

well, this is kind of a weird question tbh, but the idea is that technically you can use properties a, b, and d to perform any matrix multiplication

maybe think about how

[keen siren]

oh i get what you're saying

[keen siren]

so like you're basically saying that is it able to be multipliable because it has the same amount of rows as a column vector

[timber gust]

How would I go about finding vectors in a span that aren't multiples?

For example,

[gray dust]

assign nonzero weights to each vector

[slow scroll]

nope. im saying that if you want to compute A(x,y,z) for example, then you can start by writing (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) = xe_1 + ye_2 + ze_3
Then by property a, A(x,y,z) = A(xe_1 + ye_2 + ze_3) = A(xe_1) + A(ye_2) + A(ze_3)
and then by property b, A(xe_1) + A(xe_2) + A(xe_3) = xA(e_1) + yA(e_2) + zA(e_3)
and then by property d, xA(e_1) + yA(e_2) + zA(e_3) = xa_1 + ya_2 + za_3
So A(x,y,z) = xa_1 + ya_2 + za_3 where a_1, a_2, a_3 are the columns of A

[timber gust]

Any non-zero?, I tried that but I get it wrong

[gray dust]

eg 1u+1v

[timber gust]

this the format

[keen siren]

ahhh okay this makes sense ty

[timber gust]

I keep getting that wrong, also thought it was 1

[gray dust]

did u compute 1u+1v

[timber gust]

Confused on how to do that

[gray dust]

see the definitions of vector adding & scaling

[timber gust]

The only thing I learned is that I can add/subtract the numbers within the brackets

Im also getting the 1 in front of the u and v wrong without even computing

[gray dust]

vector scaling is multiplying each component by the scalar

[timber gust]

im still confused, could you provide an example

[gray dust]

c(x,y,z)=(cx,cy,cz)

[timber gust]

Oh i did that before, but how would I apply that here?

[gray dust]

compute 1u+1v

find 1u & 1v then add the results

[timber gust]

IT WORKED

I LOVE YOU

Woudl it be the same process for the second part?

what would i be multioplying the u and v by in that part?

[gray dust]

any nonzero weights, just not the same as before

[keen siren]

So then based on what you're saying this would have the same answer as the previous one, right?

A remaining the same exact matrix and z being :

[slow scroll]

yea. Like i said before, these are kind of strange questions. It sounds like they want you to see some trick, but I'm not really seeing it. a, b and d are technically needed to compute any matrix multiplication by a vector though, so a, b and d are "not wrong"

[keen siren]

Alright thanks that's what I thought. The questions are pretty weird

[ancient patrol]

A linear combination of 1v + (-1)v = 0 holds right? so it's not linearly independent? Sorry if it's a dumb question but im new to this

[lavish jewel]

1v + -1v is 0v

to be linearly dependent, you need nonzero scalars

[ancient patrol]

yes so S is linearly dependent right? 1,-1 are nonzero scalars

[lavish jewel]

no

1-1 = 0

so it IS a 0 scalar

which definition of linear (in)dependence are you using in your course material

[ancient patrol]

wait...lemme show you

LI

[lavish jewel]

so in your case, this sum is of the form c v = 0

since you have a single vector v

[ancient patrol]

LD

[lavish jewel]

and you have written (1-1)v = 0

i.e. 0v = 0

[ancient patrol]

oh.....

im dumb

thank you so much

[zinc timber]

why is there a nptl logo

[forest frost]

Hey, I'm about to start linear algebra next week and was wondering if there was anything I should read or watch before I start the class

[zinc timber]

3b1b ELA is a good overview of things you'll be doing

[lavish jewel]

i would advise against that one, those videos make more sense to watch after you have seen the topics in class

[zinc timber]

It did help me tho, could be a different experience for others

[lavish jewel]

idk, the whole time he's referencing future knowledge

[zinc timber]

[haughty berry]

Yeah I agree, for me I feel like 3b1b videos are only good for giving a different perspective of the material after you already know it
I don't feel like his videos are so great at introducing new topics, but they are very good at giving intuition and different perspectives

Anyways, the point of a class is to teach you, so usually there isn't a reason to prepare for a class...

[spare widget]

analytic geometry is great to study before or along with linear algebra

gives most of the geometric intuition you would need

[halcyon spindle]

Do you understand what it means the line is normal to (x, b/(x+1)^2)?

Draw a diagram if that helps and try to think about it for a sec.

[spare widget]

I am trying to figure out what it means to take a derivative wrt the index of a Kronecker delta. In the continuous case with a Dirac delta it looks like the d_x operator if applied inside an integral

$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \int_{-\infty}^{\infty}\lim_{h\rightarrow 0} \frac{\delta(y-(x+h))-\delta(y-x)}{h}f(y),dy$

I am not quite sure how legal this is though

[stoic python]

criver

[spare widget]

Or what happens in the finite dimensional case with a kronecker delta

to me it looks like I'll get a 0 in the finite dimensional case

Is it just supposed to differentiate the other function and then multiply it with a dirac delta?

so in the finite dimensional case the analogy would be a discrete derivative multiplied by a dirac delta?

i.e.

$\langle\delta'{ij}, f\rangle = \langle \delta{ij}, f'\rangle = f'_i$

Where D is some discretisation of a derivative (e.g. forward differences)

[stoic python]

criver

[spare widget]

I think it should have a minus sign actually

[spare widget]

it makes sense because of integration by parts

$\langle \partial_x f, g \rangle = -\langle f, \partial_x g \rangle$

[stoic python]

criver

[spare widget]

if f or g vanishes on the boundary

[true jacinth]

is this right ?

$a_i\in A_i\iff A_i={a_i}\implies A_1\times A_2\times A_3\stackrel{?}{=}{a_1,a_2,a_3}$

[nocturne jewel]

the iff isn't always true

and your product doesn't have tuples, so no

[true jacinth]

oh ok

[stoic python]

DonutsenPLS

[true jacinth]

i don't understand how n-tuples work

[nocturne jewel]

$A\times B:={(a,b)|a\in A & b\in B}$

[stoic python]

Mosh

[true jacinth]

oh ok thanks

[zinc timber]

this thing looks a lot like distributional derivative to me

But I don't that's you meant

so I'll pass

[true jacinth]

so if $A_i={a_1, a_2,\dots ,a_n}$ then i don't understand what is $A^n$

[stoic python]

DonutsenPLS

[lavish jewel]

going by your previous messages, A_i is a set

A^n is the cartesian product of A n times

the difference is that A^n is an ordered list where you really want to consider a list with n elements at the same time, in a particular order, to do something with it

whereas with sets, you care about each element individually satisfying some property

so A_i^n would be a tuple of n elements taken from the set A_i

and you really consider all n elements at the same time

the tuple itself is a single object that may or may not have additional properties

[true jacinth]

ok thank you, i think i understand it now

[lavish jewel]

if you just run into {x,y,z} in the wild, there is no way to know which of the two it was without further context

but yeah, recall that a set is defined in set builder notation as a collection of individual elements satisfying some property

[true jacinth]

okok thank you again

[spare widget]

Turns out it's that, I looked it up, and it seems to match what I wrote. I guess it really depends of whether I want to treat the Kronecker delta as a discretisation of the Dirac delta or not, so the answer is likely application dependent.

[sleek sundial]

I have inner products classified as positive definite bilinear forms, how do I show that any two inner products are equivalent via linear transform on a vector space?

[dusky epoch]

is that even true for non-finite-dimensional spaces

[sleek sundial]

Only working in finite dimensions

[dusky epoch]

i mean ok like

i think you can just take two bases, one of which is orthonormal wrt the first inner product and the other orthonormal wrt the second, and have the linear transformation in question send one basis to the other

[sleek sundial]

ah yea I can see that happening

[wintry steppe]

can i please get help with this

linear equation systems

substitiution

[somber loom]

Pick one of those equations, solve for either x or y, and substitute into the second solution; that way you can solve for that particular variable, and then find the second variable value afterwards

[wintry steppe]

thank you

[wintry steppe]

here if t is in R

its a line through X_0 and parallel to t right?

and the (1-t) translates the X_0

[mystic frigate]

isnt (1-t) a scalar for x_0 should t be in R

[nocturne jewel]

"parallel to t"

[wintry steppe]

sorry parallel to x_1

[wintry steppe]

i think so

[still lodge]

in general, what kinds of linear transformations/matrices preserve the angle between vectors

[narrow egret]

3x + 6y =4
3x = 3 - 6y
x = 1 - 2y

2x - 10y = -9
2(1-2y) - 10y = -9

[ember vessel]

orthogonal transformations/matrices

[mystic frigate]

whats an orthogonal transformation lol

[still lodge]

what i mean is like what do those matrices look like

if that makes sense lol

[ember vessel]

there are alot of characterisations maybe the easiest one is that the inverse of a orthogonal matrix is the transpose

[ember vessel]

if you look at the column vectors of a matrix, the dot product of two columns is zero

but there is no general formula for them

[gray dust]

@ember vessel @still lodge orthogonal maps on R^n have a canonical (usual) representation. if A is orthogonal then there exists an orthonormal basis of R^n in which the matrix of A has, along its diagonal, a string of 1s, a string of -1s, and a string of 2x2 rotation blocks

ie in this basis, for some axes A does nothing, for some A reflects, and for some axis pairs A rotates by some angle

[spare widget]

Rotations

Reflections also preserve angles but they flip orientation

For orthogonal transformations you have R=R^T

And thus dot(u,v) = u^Tv = u^TR^TRv = (Ru)^TRv = dot(Ru, Rv)

it means that the columns/rows of R form an orthonormal basis

https://en.m.wikipedia.org/wiki/Orthogonal_group

[gray dust]

heres what a decomposition of an orthogonal map may look like

note that pairs of 1s (identity) & pairs of -1s (reflection) can be written like the 2x2 rotation blocks, theyre just rotations with respective angles 0,pi

[spare widget]

There can be a lone -1 somewhere if the matrix changes orientation

[viral magnet]

that just means time reversal

[gray dust]

sure, i meant take pairs of axes at a time when rewriting as rotations

[mystic dagger]

to a set A={v1, v2, v3} be a set of orthogonal vectors it has to have every dot product to be 0 or just one is enough?

[gray dust]

any vector dotted with any other vector must be 0

[mystic dagger]

ok thanks

[little frigate]

Oh gawd fuck latex I'm just gonna post a mere translation

Let {u1, . . . , a} and {v1, . . . , vn} two bases of E. Let P be the passage matrix from {u1, · · , un} to {v1, · · , vn}. If u ∈ E with A = P{u}{u1,...,un},
so P{u}{v1,...,vn} = P−1A.

Could someone explain me where the inverse is popping off from?

This is the diagram associated with that thing

Theorem 4.2.5 states that if {u_1, ..., u_n} is a base of E, v_1, ..., v_n in E constitute a base of E if and only if P^{v_1,...v_n}_{u_1,...u_n} is invertible

That's pretty much the only other place where inversion is mentioned around

[sleek sundial]

Is there such a thing as nonlinear algebra

Things feel so nice in linear

[wintry steppe]

"abstract algebra"

in this definition

would covariance go away only when x and y are independent? because isnt E(x-mu_x) = 0?

,rotate

[nocturne jewel]

this isn't linear algebra

[wintry steppe]

wait it is tho right bc im using this to figure out the EPE stuff

[nocturne jewel]

#probability-statistics

[wintry steppe]

im looking at the ESL text

ok

[slim thicket]

Your mom is independent

[wintry sphinx]

what if Cov(X, Y) is an inner product 🤔

[wintry steppe]

like this

dumb q sorry

[mystic frigate]

wait wtf is this

😆

[wintry steppe]

covariance matrix

[mystic frigate]

my undergrad brain is lost 😆

[wintry steppe]

relatable

relatable

what does it mean to write something as a linear combination of something else?

is it just literally when u have

a column vector = scalar[col 1] + scalar2[col 2]

[mystic frigate]

yes

[wintry steppe]

for however many columns u have

[mystic frigate]

pretty much

[wintry steppe]

ok cool bc

i found beta thru xtx-1 xty

and now its like

write y_hat as a linear combination of X

but thats literally beta??

so i was like

is there more to this lmao

[mystic frigate]

so from the pic above are the vectors linearly independent when there are no covariacne values

aka cov = 0

with the pivots as var(x_1)...var(x_k)

[wintry steppe]

im not sure

i think they must be independent

[grave garden]

HEyy guys

Is the number of element in orthogonal basis in V = dim V ?

[zinc timber]

hint: that's a basis

[gray dust]

@grave garden dimV is defined as the size of ANY basis of V

[grave garden]

I see 👁️

[sleek sundial]

wait I am not losing my marbles.. right?

31 is true just by linearity no?

and then 32 is the other way around, but how does 1-1 tie in?

assuming that T is a linear operator

[lavish jewel]

i think you could do it that way, yea. pick a v_i, express it in terms of the other v_js and use linearity in 31

for 32

if {T(v_i)} are lin dep

then there is one T(v_j) that can be written as a sum of the remaining T(v_k)

then use linearity to say that T(v_j) = T(v_k + ...)

then use 1-1 to argue that v_j = v_k + ....

so that the v_i are lin dep by definition

[sleek sundial]

wait for 31, how does that work exactly? say i get v1 = -a2v2 - ... - apvp/a1

so I take the transform of v1,...,vp, then what? i rewrite v1 with what I got?

[lavish jewel]

yeah

i'll give you an example, you can generalize it yourself

say {u,v} are lin dep

then u = cv for some c in F

then consider {T(u), T(v)}

where u = cv

so this is equivalent to {cT(v), T(v)} which is lin dep

[sleek sundial]

got it

thanks

[lavish jewel]

now use sigma notation for n vectors

[zinc timber]

https://tenor.com/view/sigma-gif-24293069

[grave garden]

Guys, if i have n linear independent elements, then the space generated by it has dimension = n right ?

[restive raft]

Sounds about right

[grave garden]

Guys, can you help me understand to get row rank + dim solution = n ?

I can't understand this part, if possible can you provide a linear map like how they solve it for the first part ?

[lavish jewel]

rank nullity moment

so the basic idea is to keep in mind the definition of the image and null space of a matrix

and to consider that a matrix vector product is a linear combination of the columns of a matrix, as they show there in the line L(X) = ...

the image of a matrix is all vectors y such that Ax = y

the null space of a matrix is all vectors x such that Ax = 0

so given an equation Ax = y, the number of solutions depends on the dimension of the null space

if the only vector x for which Ax = 0 is x = 0, then Ax = y has a single solution (at most)

if not, then you can use linearity of matrix mult to construct more solutions

let w be the set of all vectors for which Aw = 0

then if Ax = y, A(w+x) = y too

since A(w+x) = Aw + Ax = 0 + y = y

then you consider the space of all y generated by Ax, i.e. the space spanned by the columns of A

its dimension is the rank of A, i.e. the number of linearly independent columns of A

there's a few more steps to go from there, but i'll jump to your second point

regarding the row rank, consider that the alternative formulation of matrix multiplication is via inner products

in Ax = y, each entry in A is the inner product of a row from A with x

so that in order to obtain Ax = 0, x must be orthogonal to all the rows of A

the rows are of size n, and there are r linearly independent rows

by this definition, the set of x for which Ax = 0 is the orthogonal complement of the space spanned by the r linearly independent rows

you can go from there

[grave garden]

Ill let you know if i dun understand sth there

[grave garden]

Thanks Ed !

I think i could see why now

( but not really clear about it yet, but i think i will soon )

[spare widget]

If I have an overdetermined inconsistent system Ax=b, I can always find a solution minimizing it in an L2 sense: A^TAx = A^Tb. If I have an underdetermined system I generally have to add extra constraints, or drop some unknowns and columns, right?

That is, there are no other options (excluding minimisations wrt another metric for the overdetermined case), right?

[subtle gust]

find the inverse of A if 5A^3+3A^2-4A=0nxn

any idea how we do this problem...

[haughty berry]

Well, if you assume $A$ has an inverse:
[ 5A^3 + 3A^2 - 4A = 0 \implies A(5A^2 + 3A) = 4A \implies A(5A + 3I) = 4I]
[ \implies A^{-1} = \frac{1}{4}(5A + 3I) ]

[subtle gust]

Confused as to what we did in the third step

[haughty berry]

Multiply by A^-1?

or the step before that

oh wait

[lavish jewel]

you did the mult wrong though

[haughty berry]

sorry i dropped the 4

[lavish jewel]

in at least 3 places