linear-algebra

[twilit minnow]

so my question is

[subtle gust]

Both sided invertible? 💀

[twilit minnow]

how can i go from rank(A) = n to injective

[lavish jewel]

wait what

let's run that again

[twilit minnow]

with the transformation T

[zinc timber]

epic/monic

[lavish jewel]

this is only true if m = n, gta

[zinc timber]

rank nullity

[subtle gust]

Isn't the def of an inverse AB=BA=In?...

[lavish jewel]

if m < n, you cannot have rank n

[subtle gust]

Why would it matter

[lavish jewel]

and if m > n, you have a nontrivial kernel

[twilit minnow]

if m > n, it is injective

[lavish jewel]

you need to rework this part

[subtle gust]

If we multiply by the inv from the right or the left

[zinc timber]

so may ppl so may topic lol

[lavish jewel]

if m > n and rank n, you don't have injectivity

just say rank(A) = m

[twilit minnow]

wait how?

[zinc timber]

?

[twilit minnow]

you have

full col rank

how isnt it injective

[lavish jewel]

what size is your matrix

[twilit minnow]

mxn where m > n

just any tall matrix

[lavish jewel]

the image you shared says n x m

that's what i was talking about

[twilit minnow]

B is n x m

A is m x n

[zinc timber]

that's the dim of B

[lavish jewel]

ffs i can't read

carry on

[zinc timber]

[twilit minnow]

i dont know if saying the rank nullity is 0 is enough

is it

[zinc timber]

just nullity, not rank nullity and yes it's enough

[twilit minnow]

okay then how do we go from injective to at least one left inverse

i read some answers online but i just got confused

[zinc timber]

I can give one informal construction

you can find a nxn matrix (say B') from im(A) -> R^n s.t. B'A = Id, now extend the rows till they have length m

[twilit minnow]

how would you do it in a formal proof

[zinc timber]

well your (A'A)^-1 A' is a left inverse

just show (A'A) is invertible when rankA = n

[meager harness]

<@&286206848099549185> 4b?

[lavish jewel]

please don't ping helpers immediately

what have you tried?

[gritty swift]

my solution: let $e_1,\dots,e_m$ be an orthonormal basis of $V$ with respect to $\langle \cdot, \cdot\rangle_1$. then we have
$$\langle v,w\rangle_1 = \sum a_jb_j$$
and
$$\langle v,w\rangle_2 = \sum a_jb_j |e_j|_2^2$$
now to show every $|e_j|_2$ is the same I derive a contradiction assuming they're different (i.e. if the sum is weighted I construct v,w such that the inner products disagree on orthogonality, which they can't). but this feels like an ugly solution and doesn't give insight. can anyone think of something better?

[stoic python]

uli

[spare widget]

why would <.,.>_1 = sum ...

It doesn't even say that the spaces are finite dimensional

[gritty swift]

they are

[spare widget]

Nor does it say that your inner product is the dot product

[gritty swift]

with an orthonormal basis it becomes the dot product

[spare widget]

It could very well be u^T M u

[spare widget]

depends on how they pick it

They can plug in any symmetric positive definite form

I think you should use only what they provide

[gritty swift]

ok I'll try without assuming finite-dim, the section is "orthonormal bases" and all the other exercises assume finite-dim so I guess I assumed it

[spare widget]

my point is not about finite dim

It's fine if they say it's finite dim

What isn't fine is you writing it as sum_i a_ib_i

[gritty swift]

it's correct because $\langle e_j, e_j\rangle_1 = 1$

[stoic python]

uli

[gritty swift]

the second one is the one that might be weighted (though you can show it's not)

[spare widget]

the following is a perfectly valid inner product <u,v> = u^T G v

[gritty swift]

I know

[spare widget]

G positive definite

[gritty swift]

we wrote $v = \sum a_j e_j$ and $w = \sum b_j e_j$ so $\langle v,w\rangle_1 = \sum a_jb_j$

expand it out

[stoic python]

uli

[spare widget]

They never define <v,w>_1 = \sum_i a_i b_i

[gritty swift]

that isn't a definition it's a consequence oml

[spare widget]

a consequence of what

[gritty swift]

$$
\langle v, w\rangle_1 = \sum_i \sum_j a_jb_j\langle e_i, e_j\rangle
$$

[stoic python]

uli

[gritty swift]

and the orthonormality of $e_i$

[stoic python]

uli

[spare widget]

^ they never define it like this

[gritty swift]

I defined e_i

[spare widget]

I can define an inner product in an orthonormal basis that includes a non-identity matrix

[gritty swift]

the orthonormal basis is picked with respect to <*, *>1

[spare widget]

So you picked the basis G^{-1/2}?

[gritty swift]

I'm considering the inner product space (V, <*, *>_1)

we aren't dealing with matrices here just linear maps

[spare widget]

It's isomorphic in fin dim

R^n to your V

[gritty swift]

yeah I guess, I'm not very familiar with the matrix definition of inner products

[spare widget]

but ok, I am starting to get what you mean

[gritty swift]

anyway the thing I was asking was is there a nice way to show $|e_k|_2 = |e_j|_2$

[stoic python]

uli

[gritty swift]

the way I showed it was assume $|e_k|_2 \ne |e_j|_2$ then exploit the unequal weighting to find $v,w$ such that $\langle v,w\rangle_1 = 0$ and $\langle v,w\rangle_2 \ne 0$ (a contradiction)

[stoic python]

uli

[gritty swift]

we haven't defined matrix sqrts or the matrix defn of inner products btw, though i'd be happy to hear a proof in that form if its intuitive

in matrix form I think the exercise would be: given that $x^TAy = 0$ if and only if $x^TBy = 0$ (where $A,B$ are positive definite) show that $A = cB$ for some scalar $c$

[stoic python]

uli

[zinc timber]

x'(A-B)y=0

[spare widget]

you plug in x,y the basis vecs

And you get the entries

Up to a multiplicative factor

[zinc timber]

^

[spare widget]

wait, does that work

[zinc timber]

[gritty swift]

it leads to the same thing I ended up with I think

this was kinda silly question its intuitive enough as it is that an unequal weighting implies you can make one zero and the other not

[zinc timber]

consider orthonormal basis wrt A

[spare widget]

I still think there should be a much simpler way

[zinc timber]

you get that the orthonormal basis is also a orthogonal basis of B, not necessarily of unit length

[gritty swift]

ooh

I had an idea

define a third inner product $\langle v,w\rangle_3 = \langle v,w\rangle_1 - \langle v,w\rangle_2$

[stoic python]

uli

[spare widget]

Afaik |ej|^2 = |ek|^2 wrt <>_2 is not true,is it?

[zinc timber]

it's not

i only said orthogonal not orthonormal

[spare widget]

Yeah so this doesn't work

[zinc timber]

that we have to prove separately

[gritty swift]

yeah nvm

its only an inner product if c /= 0 otherwise its identically zero

[spare widget]

I was thinking more along the lines of {v}^{perp} = {w : <v,w>_1 =0 } = { w: <v,w>_2 = 0}

i.e. the orthogonal complement to the span of v wrt both inner products is the same

And any vector can then be decomposed in this

I didn't think of how to continue this though

It should use somewhere that it's for every v

[gritty swift]

oh wait I'm dumb

[spare widget]

In order to get that the constant is tge same

[gritty swift]

I think you can use this somehow

maybe?

its from the same section

[spare widget]

idk how that helps you

[gritty swift]

true

I mean you could fix one of v,w then apply it to write both in terms of the same norm

it still assumes finite dim though (which is probably correct tho it spooks me he doesn't mention it since he usually does)

you can get $\langle v, w\rangle_1 = \langle v, u\rangle_2$ for all $w \in V$ using Riesz

[stoic python]

uli

[zinc timber]

say v1,..,vn orthonormal basis wrt A the these are orthogonal wrt B so a basis. now say Bv1 = ∑ ci vi then v1'Bv1 = c1

[spare widget]

You can try to pick v1,v2 such that <v1,w>_1 = c1<v1,w>_2 and <v2,w>_1 = c2 <v2,w>_2

And assume c1!= c2

Once you decompose v2 into a part parallel to v1 and orthogonal you can probably get a contradiction

[zinc timber]

let v2'Bv2=c2 similarly then v1'(Bv2-c1v2)=0

[spare widget]

I think the key is that the orthogonal complements to the span of v are the same

[zinc timber]

aah can't type

[spare widget]

Does the following work @gritty swift

Let <v1,w>_1 = c1 <v1, w>_2 and <v2,w>_1=c2<v2,w>_2

from this you have

<v1,v2>_1 = c1<v1,v2>_2 and <v2,v1>_1 = c2<v2,v1>_2 -> c1 = c2

from symmetry of the inner product

For the <v1,w>_1 = c1<v1,w>_2 part I think it suffices that you take a basis with 1 of the vectors being v1

And the others being in the orthogonal complement

[gritty swift]

it seems like it does

that's a more elegant way of showing it without a contradiction

you get <v1,v2>_1 = c1<v1,v2> from setting w=v2 and the other one from setting w=v1 right (the first line was for all w)

[spare widget]

yes

[gritty swift]

yep I think it works nice

[spare widget]

And v1 and v2 were arbitrary

So the c is unique

[gritty swift]

ooh true so it works in infinite dim

[spare widget]

Idk about inf

[gritty swift]

you never assumed you had a spanning list so it should extend

[spare widget]

This may or may not break for inf, idk

[gritty swift]

I mean you're setting c1 = <v1,w>_1 / <v1,w>_2

[spare widget]

Yes, but you have to show that it is a unique c1 for any w

[gritty swift]

so we just need <v1, w> /= 0 and same for c2 right

ah right

[spare widget]

which you should be able to do through picking a basis with v1 being 1 of the vectors

And the rest being in the orthogonal complement

So w gets decomposed in these two parts

The orthogonal part resulting in 0

[gritty swift]

yeah makes sense

[spare widget]

I haven't worked out the details though, there may be something missing

[viral olive]

guys if you have a linear function and a linear transformation from V ->V.
Is a set of linear independent lets call them (v1,v2,...,vn) still linear indepedent in the field.
(f(v1),f(v2),.....,f(vn))

[spare widget]

depends on f

[viral olive]

so if f is bijective

I assumed yes.
If f is not bijective its clear for me why it woulnt be the case.

[spare widget]

Yes if f is bijective it's all good

[viral olive]

Aight thanks Glados

[spare widget]

You're welcome

[spare widget]

You can use f(sum_i a_i v_i) = f(0) -> sum_i a_i f(v_i) = f(0) = 0

Vectors are lin indep if this holds only for (a1,...,an) = (0,...,0)

So the only way those can become linearly dependent is if f's kernel is nontivial

Because then let b != 0 be a vector from the kernel, and there exists (a1,...,an) != (0,...,0) such that b = \sum_i a_i v_i

But f(sum_i a_i v_i) = sum_i a_i f(v_i) = f(b) = 0

So by definition f(v1),...,f(vn) are then lin dependent

To be more specific, any vector that is in the kernel would become linearly dependent after the map since it goes to 0

Lin indep vectors not in the kernel are mapped to lin indep vectors

[meager harness]

i tried using the fact that A^k v = lamda^k v = 0 then lamda = 0

[viral olive]

perfect thank you

[oblique prairie]

just making sure, linearly independent subset here means that all vectors in the subset are linearly independent right?

[wintry steppe]

means that the vectors from S1 are linearly independent from S2, S3, ... Sk

and the union of all of them is a subset of linearly independent vectors of V

[bronze hill]

Can someone provide a straightforward proof for b) preferrably using latex?(I'm checking my solutions)

[spare widget]

$(C^TA^T){ij} = \sum_k (C)^T{ik} (A)^T_{kj} = \sum_k C_{ki}A_{jk} = \sum_k A_{jk}C_{ki} = (AC){ji} = (AC)^T{ij}$

[stoic python]

criver

[wise nebula]

hi guys i have an exam tomorrow and a really fast question about linear algebra

can someone come voice with me for a minute or two?

really don't know how to type that in the web

[spare widget]

I'd still suggest to try and write it

[bronze hill]

Oh thank you so much! This is basically what I wrote but I thought the dimensions of A and C would make a difference(e.g. the upper bound of sum)

[spare widget]

How so?

The multiplication is defined only if the inner dimension matches

so it's safe to omit it

[bronze hill]

Now I got it, thank you.

[spare widget]

The only thing I can think of to avoid any confusion is to write (D)^T_{ij} as (D^T)_{ij} instead, but I think it's understandable even as I wrote it

[wintry steppe]

$V$ and $W$ are two finite dimensional vector spaces, $dim V = n, dim W = m$. Could somebody help me understand why $dim \mathcal{L}(V, W) = n \cdot m$? My textbook gives a proof, but no intuition behind it.

[stoic python]

mate

[wintry steppe]

the set of all linear transformations from $V$ to $W$

[stoic python]

mate

[gray dust]

theres a correspondence between linear maps V->W and mxn matrices

so the dim of the latter gives the dim of the former

[oblique prairie]

probably just a little unnecessary nitpick, but why is this proof distinguishing between n_i and d_i as if they aren’t equal?

n_i = number of vectors in the ordered basis for E_{lambda_i}, d_i = dim(E_{lambda_i})

[wintry steppe]

they aren't

they haven't assumed the hypotheses of part (b) yet, they've only assumed that T is diagonalizable

they're defining \beta_i here using the same notation as part (b), but they're only proving (a)

[oblique prairie]

oh

ok then thanks

[wintry steppe]

👍

[oblique prairie]

i read it a bit more in depth (at least i think), only paying attention to the stuff from (a), and i still don’t see how those two aren’t equal

i mean, B_i is very obviously a basis for E_{lambda_i}

[wintry steppe]

why

[oblique prairie]

well, E_{lambda_i} is a subspace of V, so there must be some subset of the set of basis vectors that generate E_{lambda_i}

[wintry steppe]

i'm not convinced

{(1, 0), (0, 1)} is a basis of R^2, but no subset of it generates, say, the line y = x

[polar vessel]

how can i prove that for a matrix A and orthogonal matrix B, |AB|F = |A|F

so prove that the frobenius norm of AB equals the frobenius norm of A

[dark pier]

if you multiply two symmetric matrix, is the end result also symmetric?

[wintry steppe]

only if they commute

[wintry steppe]

this is sqrt trace(A^TA) or something like that, right?

that might help

[dark pier]

what does that mean

[wintry steppe]

AB = BA

[polar vessel]

we dont use trace it was never discussed

otherwise yeah it would be simple

[soft burrow]

maybe try studying B^tA^t instead and use that B^t, being an orthogonal matrix, should preserve the Euclidean norm of the columns of A^t

then relate that to the Frobenius norm

[polar vessel]

yeah thats what im trying right now

[soft burrow]

I think it follows more or less by definition

since the Frobenius norm is just the Euclidean norm on R^{n^2}

[polar vessel]

for all n?

[soft burrow]

yeah for n x n matrices

[soft burrow]

just write the sum in the definition carefully and use this

[polar vessel]

so for UC = [UC1 UC2 ... UCn] are C1 ... Cn the columns or rows of C

[tacit pelican]

could anyone help with this?

for the other questions, they were all based in \mathbb{R}^n so i could either create a matrix with the given vectors

and try putting it into rref

to see if it is generating or not

but i don't understand how to work with a space like $\mathbb{P}_3$

[stoic python]

sean

[hardy inlet]

Ok so if you were in Rⁿ, lets say R², how many vectors are needed (at a minimum) to create a spanning set?

[tacit pelican]

2

[hardy inlet]

Ok

So u need at least the dimension of the space itself

[tacit pelican]

yep

[hardy inlet]

(along with linear independence, but ignoring that)

Do you know what P3 means?

[tacit pelican]

yes, the space of polynomials of degree at most 3

but the dimension is kinda unclear to me, since the standard basis of the space is 1, x, x^2, x^3

so does that mean its dimension is 4?

[hardy inlet]

What is the definition of dimension?

(and yes its 4)

The dimension of vector space V is the number of vectors in a basis for V

[tacit pelican]

this actually hasn't come up yet in my book

thats the next section

this section is about analyzing the pivots of a matrix in echelon form

so i assume i have to compose some matrix or something to deduce if it is generating or not

,rotate

[hardy inlet]

hmm

[tacit pelican]

oh

[hardy inlet]

matrices

[tacit pelican]

so its just hte number of vectors in its basis?

[hardy inlet]

yes

[tribal willow]

you could make a matrix of the three polynomials and row-reduce them, no?

[tacit pelican]

the thing is, thats what i'm confused about

do i represent these polynomials as vectors somehow?

[hardy inlet]

the coefficients yes

ax^0 + bx^1 + cx^2 + dx^3

(using the ordered basis of the standard p3 basis)

[tacit pelican]

so for example $x^3 +2x$ is equivalent to the matrix $\begin{pmatrix} 0 \ 2 \ 0 \ 1 \end{pmatrix}$

[hardy inlet]

so x^3 + 2x would be 0 2 0 1

[tacit pelican]

oh wait

sorry typo

[hardy inlet]

i dont think u want them as columns

but its been a while since i've done raw matrix comps

[tribal willow]

pretty sure it's as rows

[tacit pelican]

oh ok my bad

so it would be

[tribal willow]

then you perform elementary row operations on the rows to row-reduce to echelon form

[stoic python]

sean

[tribal willow]

ye

[hardy inlet]

anamono would have to probably verify; but I think u put all 3 coefficient entries as rows in the matrix and row reduce; but you wont get the identity since u dont have 4 rows

[tacit pelican]

how do i combine them into one single matrix though? before, i was just putting the given vectors as the columns of a matrix A

[tribal willow]

^ do this

i dont think you need to have identity, you just show they're linear independent right?

[tacit pelican]

yeah

[tribal willow]

i forgot condition for span

[tacit pelican]

which means

i have to show that there is a pivot in each column

i can do that

[tribal willow]

tbh i have never worked with pivots

so i can't help you there

[hardy inlet]

this thing i'm reading has column vectors

[tacit pelican]

yeah i got that covered from my notes

wait one thing i'm still unsure about is why its a row vector this time

when working in $\mathbb{R}^n$

[tribal willow]

hmm

[stoic python]

sean

[tribal willow]

i row reduce my matrices when i construct vectors in rows

i wonder if it's something different for columns?

[tacit pelican]

they were $n\times 1$ column vectors

[tribal willow]

or something im just forgetting

[stoic python]

sean

[hardy inlet]

this thing used columns tho

[tacit pelican]

yeah for example, here i just stuffed the four vectors into a matrix so it became $\begin{pmatrix} 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \ 0 \ 0 \ 0 \ 1 \ 1 \end{pmatrix}$

[stoic python]

sean

[hardy inlet]

but that only shows linear independence in that example not span

but yeah probably that matrix

for 3.2

this thing claims u need a 1 in every row

so since ur last row on the P3 problem would be 0s its not spanning

[tacit pelican]

i can show that a system of vectors is spanning and/or linearly independent easily

but

i just need the matrix

to represent that system

so my confusion here is: why are these polynomials represented as row vectors here? whats different in $\mathbb{P}_3$ compared to $\mathbb{R}^4$

[stoic python]

sean

[hardy inlet]

i think we were just wrong about how to enter them into the matrix

so i think u do make them column vectors

so it'd be 4x3

(4 rows, 3 col)

[tacit pelican]

ahh alright

thank you

this is what i got

just need to row reduce now right?

[hardy inlet]

yea

which gives me

[tacit pelican]

got it

[hardy inlet]

so they're linearly independent but don't span because the last row doesn't have a pivot

and im pretty sure 3.2 row reduces to the identity

so that would be linearly independent and span; hence a basis

[tacit pelican]

yup

i got that as well

since theres a pivot in each column

they're linearly independent

alright thank you, that was a big help

[hardy inlet]

i'm only 90% sure on this stuff

and i think having pivots in the rows means spans

but idk

[tacit pelican]

yeah it does

[hardy inlet]

from my brief research

[tacit pelican]

pivot in every row = generating (spanning)

[manic crow]

Well, I'm new at this, I wanted someone to help me with a question, but seems like it's not going to happen yet, then, can someone tell me please how to solve this or where to find information to solve this?

[quartz compass]

gaussian elimination or reduced row echelon form with an augmented matrix are some keywords to help you get started

[humble oak]

could someone explain to me why $\Bar{v}^T v = ||v||^2$ where v is some non-zero vector

[stoic python]

Crown

[dusky epoch]

it works for v = 0 too, so...

the standard norm on $\bC^n$ is $\nrm{v} = \sqrt{\sum_{k=1}^n |v_k|^2} = \sqrt{\sum_{k=1}^n \overline{v_k} v_k}$

[stoic python]

Ann

[humble oak]

ah okay

i see thank you

[wintry glen]

So I've just gone over the section on eigenvalues and linear transformations and I have few questions.

Let T be a transformation from V to V and B be a basis of V. The book defined a B-matrix of T, but I don't seem to understand what the use of it is. Is it just so that we can represent any linear transformation on general vector spaces as a matrix, so as to have a systematic way of computing T(x) by first finding the B-coordinate vector of x, then multiplying by the B-matrix of T, then by converting back the B-coordinate vector of T(x)?

Also, they also discussed how if T is a linear transformation from R^n to R^n with standard matrix A, then if A is diagonalizable, then the B-matrix of T is diagonal, where B is the eigenvector basis given by A. What is the practical significance of this? Is it just saying that if we choose to work in the eigenvector basis that the effect of the transformation is the simplest?

[limber sierra]

The book defined a B-matrix of T, but I don't seem to understand what the use of it is. Is it just so that we can represent any linear transformation on general vector spaces as a matrix?
that's the idea, yes, at least as long as the vector space is finite-dimensional. representing transformations as matrices has its applications (if you've taken calculus, you may be familiar with the Jacobian matrix of a linear function, which tells you information about its local rate of change)

and being able to do that wrt a basis is essentially the way we "compensate for different coordinate systems"

What is the practical significance of this? Is it just saying that if we choose to work in the eigenvector basis that the effect of the transformation is the simplest?
essentially.

[wintry glen]

Hm, interesting. I don't fully get it yet, but I'll let it sit for a bit.

Thanks for the info regardless!

[gilded furnace]

isn't linear algebra about things like slope-intercept form and whatever

[dusky epoch]

no it isn't

[gilded furnace]

why does everyone make it seem so hard

then what

[limber sierra]

read the pins

[dusky epoch]

^

[gilded furnace]

wut

so hard

[limber sierra]

addendum: knowing that the eigenbasis always gives us a diagonal matrix is useful for decomposing matrices (we call it the "eigendecomposition")

there are various numerical reasons why decomposing matrices is useful

and this result (at least over ℝ or ℂ) eventually generalizes into the spectral theorem

[wintry glen]

Ah.

[limber sierra]

it's just one of many canonical ways to factorize a matrix

but if you happen to know the eigenvectors, or can easily compute them, it's a very convenient one

(and also the simplest)

[wintry glen]

Interesting how there's so many different factorizations with their own uses, like LU, PDP^-1, QR and SVD.

[limber sierra]

SVD is kind of like an eigendecomposition in that the principle is still trying to find "directions" (ie eigenvector coordinate systems) where the matrix "acts as a scalar"

[wintry glen]

I haven't learnt SVD yet. Maybe I'll understand when I get there.

[limber sierra]

on a more elementary level, this fact just gives you another way to think about diagonalizability

[wintry glen]

OK, I kinda get the ideas now. Thanks a bunch. The state of simultaneously understanding and not understanding is weird.

[limber sierra]

diagonalizability means that there exists an eigenbasis such that the matrix in that eigenbasis is diagonal

in other words, there exists a "coordinate system" that makes the matrix act "scalar" within that coordinate system

this isnt the most useful intuition like, computationally

but hopefully it makes "diagonalizable" seem a bit less... obscure

[wintry glen]

Ah, that does make sense.

[limber sierra]

"it's possible to simplify this transformation's action if we change how we measure"

whereas nondiagonalizable matrices are just fucked

and dont permit a nice simplification

[wintry glen]

I just understood diagonalizability as something that allows us to compute high powers easily.

[dusky epoch]

"linear"? you mean like coordinatewise scaling maybe?

[limber sierra]

yeah sorry

scalar

[wintry glen]

And this makes sense because repeated actions of multiplication would just be scaling, linking to my previous intuition.

[limber sierra]

essentially

you dont need to worry too much about the details here, it admittedly gets quite dense in definitions and computations

i find that that's when it takes longest for this stuff to set in

if it's just definitions i can parse that

if it's just computations i can see what the computations are doing

but eigenstuff and diagonalization involves both

so it takes a bit of practice to see the lines

you'll get the hang of it as you work at it

[wintry glen]

Yeah, I just wanted at least some understanding of why we bother with finding B-matrices of T and I've got that now, so I'm good. Thanks mate!

[broken notch]

I have a matrix u of size 3x12. The [x,y] values represent the coordinates of the lamps, [z] represents the height of the lamps. The equation for lightning each lamp, is lightning = x/d², where [d] is the distance from the lamp to the center of a matrix A of size 20x30. I want the lightning of each lamp to be as close to 1 as possible using the gram-Schmidt process and singular value decomposition. Does anyone have a clue? Please send help

[zinc timber]

wym by "distance from the lamp to the center of a matrix A of size 20x30"?

[broken notch]

or maybe someone can help me understand the question:

It is desired that the lighting level be as close as possible to 1.0 in all
squares. Use the least squares method to determine in python the brightness of each lamp using (i) 𝑄𝑅-decomposition via enhanced Gram-Schmidt, respectively. (ii) SVD decomposition.

because I am having difficulties understanding what they want and how to do it

[broken notch]

I have a matrix of size 20x30 and xy are the coordinates of the lamps in that matrix

[zinc timber]

how's this 20x30 then?

[broken notch]

then for instance the center of that matrix is xc=10, yc=15, if I have a lamp with coordinates for instance [4,20],

and that lamp has a height of z=8

then I have a distnace from [4,20,8] to [10,15,0]

[wintry steppe]

thank you. can we explain it without m by n matrices?

[gray dust]

yes, just find a basis of the former space

[lost temple]

someone help me

[coral flame]

1.The distance between (2, -1) and (3,y) is 1. Find the value of y.

how do I do this one?

[quartz compass]

do you know of any formulas for how to find the distance between two points?

[dusky epoch]

this barely qualifies as linear algebra

[broken notch]

can no one help me or throw me a bone

A parking space of 20m × 30m is illuminated via lamps placed in different places and at different heights, as indicated in Figure 1 page 3. The parking space is divided into a rectangular grid of 600 squares each of size 1m × 1m. The number 𝑦𝑗 indicates the illumination level squared 𝑗, for 𝑗 = 0,. . . , 599. Let 𝑥𝑖 indicates the strength of lamp 𝑖. We select units so that the contribution to the lighting in square 𝑗 from lamp 𝑖 is 𝑥𝑖 / 𝑑2 𝑖𝑗, where 𝑑𝑖𝑗 is the distance in ℝ3 from the lamp to the center of square 𝑗.

(a) State how the illumination level 𝑦 = (𝑦0,..., 𝑦599) and the powers 𝑥 = (𝑥0,..., 𝑥11) are related via a linear equation system. Set the coefficient matrix for the system in python.

(b) Make a heatplot showing the illumination level in each square when all lamps are lit with intensity 𝑥𝑖 = 20.0.

(c) It is desired that the illumination level be as close as possible to 1.0 in all squares. Use the least squares method to determine in python the brightness of each lamp using (i) 𝑄𝑅-decomposition via enhanced Gram-Schmidt, respectively. (ii) SVD decomposition.

I have done a and b

(I think)

but I dont know how to solve c)

this is the figure

can someone throw me a hint or help me understand what to do

[lavish jewel]

the part that takes effort is setting up the system of equations here

[broken notch]

I dont understand this "(c) It is desired that the illumination level be as close as possible to 1.0 in all squares. Use the least squares method to determine in python the brightness of each lamp using (i) 𝑄𝑅-decomposition via enhanced Gram-Schmidt, respectively. (ii) SVD decomposition."

[lavish jewel]

so, basically what you want is to set up a cost function to minimize

you have an ideal "lighting", let's say, which is a vector with 600 entries that are all 1

on the other hand, you have the actual lighting, which is given by some f:R^number of lamps -> R^600

(assuming you keep the intensities x_i fixed)

[broken notch]

R^600 lamps?

huh

[lavish jewel]

no, 600 squares you want to illuminate

[broken notch]

I thought I only want to illuminate the lamps

from the lamp to the center

[lavish jewel]

oh yeah, that's a lot easier and sensible lol, i misread

there, fixed

so what you want is the sum of intensities at each square to equal 1

i.e. $J_m = \sum_{n=1}^{12} x_n/d_{nm}$

[stoic python]

Edd

[lavish jewel]

this is the lighting at the mth square, where d_nm is the distance from each lamp to that square and x_n are the intensities of the lamps

i guess in your image you count from n=0 to n=11, but that makes no difference

what you have to consider is that the lamps are already placed and cannot be moved

this means 1/dnm is a constant

the variables are the x_n

with that in mind, you can notice that the sum can be rewritten as a dot product

[broken notch]

1/dnm varies from lamp to lamp

[lavish jewel]

let's say the distances from all lamps to the square m are put into a vector $\boldsymbol{d_m} \in \mathbb{R}^{12}$, and similarly, the unknown intensities are put into a vector $\boldsymbol{x} \in \mathbb{R}^{12}$

[stoic python]

Edd

[lavish jewel]

it does, but the quantities are fixed

you won't change them

each dnm is indeed unique, but also constant

now we notice that $J_m = \sum_{n=1}^{12} x_n/d_{nm} = \boldsymbol{d_m}^T \boldsymbol{x}$

[stoic python]

Edd

[lavish jewel]

and we want $1 = \boldsymbol{d_m}^T \boldsymbol{x}$

[stoic python]

Edd

[lavish jewel]

we notice we have exactly 600 of these equations, since there are 600 squares

we can set each d_m^T vector as a row in a matrix D. (btw in case it wasn't clear from the notation, the entries in each d_m vector are reciprocal distances, not just the distances)

this now gives us the matrix equation $\boldsymbol{1} = \boldsymbol{Dx}$, with $\boldsymbol{1} \in \mathbb{R}^{600}$ and $\boldsymbol{D}^{600 \times 12}$

[stoic python]

Edd

[lavish jewel]

now you want to solve this exactly if possible, or approximate it if not

so your goal is to solve $\min_{\boldsymbol{x}} \Vert \boldsymbol{1} - \boldsymbol{Dx} \Vert_2^2$

[dusky epoch]

boldsymbolx

[lavish jewel]

yes.

[stoic python]

Edd

[lavish jewel]

then we notice this is a linear least squares problem, which is convex (and strictly so if D has full column rank). this means you can take the gradient and set it to 0, then solve for x

that will give you a (possibly unique) global minimizer

if you have questions about this, it probably means you had parts a and b wrong

btw @broken notch if you could share screenshots of the original problem with me, i would appreciate it. seems like a cute problem to have my students practice their python

[broken notch]

the problem is in a diff language

[lavish jewel]

which language

[broken notch]

danish

[lavish jewel]

[broken notch]

I can translate it for you

if u want

[lavish jewel]

hmm nah, i think with the images and the description you already gave, it would be ok

[broken notch]

there are only two problems left

d and e

@lavish jewel 1.-Each row in D is a vector dm^T *x where dm= 1/dmn and x is the intensity vector, in our case is x=[0,1,2,3,4....11], right?

[lavish jewel]

each row in D is only a vector dm^T, without the x

we don't know what x is, and we solve for it

[broken notch]

2.-I don't understand how to solve min x||1-Dx||^2

how do I apply gram-Schmidt and sdv to solve the problem?

[lavish jewel]

well, from the other stuff i mentioned, you can take the gradient and set it to 0

the gradient would be D^T (1 - Dx)

or in other words, D^T 1 = D^T D x

and, for example if D^T D x is invertible, you get that x = (D^T D)^-1 D^T 1, which you should recognize as the solution to a least squares problem

(or well, even if not invertible, you can use the SVD method to get a projection onto the row space)

[spare widget]

In practice I use a CGNR or LSQR solver for this, but they are specifically asking for svd and QR here. Also the matrix seems to be dense.

To be physically correct they should use the squared distance too because of inverse square law but whatever.

Just fyi:

$\min_x |Dx - b| \implies D^TD x = D^T b$

[stoic python]

criver

[spare widget]

The equations D^TDx = D^Tb are called the normal equations

[broken notch]

how can i use x to apply the svd and the gram-schmidt?

[lavish jewel]

you don't use x

you solve for x

[spare widget]

you can google normal equations and you'll get a lot of info

You need do apply QR and SVD to the system matrix D^TD

The right-hand side is y = D^T 1

Having the svd you can compute (D^TD)^{-1}

Similarly having QR you can solve

QRx = y -> Rx = Q^T y

The Rx part is solved as you do triangular matrices (it's sequential)

[broken notch]

y = D^T 1 what is 1 here

[lavish jewel]

a vector with 600 1s

is this homework or did you just look it up?

i think you're missing a few things before you're able to solve it

[broken notch]

homework

[lavish jewel]

have you seen least squares in class?

[broken notch]

a little

I will read up on it again

it should be like this q,s,vt = svd(D^TD), right then we use (q)^-1 D^T =x, right?

@lavish jewel

it is the same with q,r = qr(D^TD), the we use (q)^-1 D^T =x, right?

[lavish jewel]

i don't think that's right

[spare widget]

For svd you should get U, S, V such that D^TD = U S V^T

Then (D^TD)^{-1} = V S^{-1} U^T

then x = (D^TD)^{-1} D^T 1

For QR I already explained what you should do

[lavish jewel]

you'll have to check how python returns the S matrix. i seem to recall that, by default, it provides a full SVD and S is a vector

i.e. it might contain zeros you need to remove, and then turn it into a matrix

and possibly truncate the columns of U and rows of V^T

[spare widget]

well, he'll keep the zeroes

He'll just take the reciproc of the nonzeroes

[lavish jewel]

that's more annoying to do than truncating

[spare widget]

He'll get zeroes if this is a rank-deficient D

i.e. if there is no unique solution

then the reciproc gives the pseudo-inverse

[lavish jewel]

i'm just saying that doing it this way requires you to iterate over the elements of S

whereas truncating is done by reference and without iteration

it's better practice for larger problems

[spare widget]

idk what by reference and without iter means

[lavish jewel]

exactly

the code is bad and slow

[spare widget]

I doubt

Both do the same thing

[lavish jewel]

nope

they give the same result

the computer does them differently though

[spare widget]

If it's slow it would be because one coded it slow

[lavish jewel]

that's exactly what i am telling you

[spare widget]

if you mean that you do some redundant ops because of 0s, sure, but that's definitely not the bottleneck in an svd

[lavish jewel]

no, but it's certainly a problem

this doesn't scale well

[spare widget]

depends on the problem

[lavish jewel]

you're adding an O(n) problem on top of the O(n^3) one

the behavior is surely dominated by that of the O(n^3) SVD and matrix multiplications, but the delay is there

[spare widget]

for instance I work with problems where some redundancy is better since it allows more efficient parallelization, but that's way out of scope for the current question

[lavish jewel]

yeah that doesn't make a difference here

[spare widget]

tbf, I think the person asking has an issue even with the simpler case

[lavish jewel]

sure, they need a way to find the inverse in the first place, given the output of the svd function

[spare widget]

And I would never use svd in practice for the above problem

CGNR/LSQR or Cholesky rather

[lavish jewel]

that's fine, but they are explicitly asked to use it

[spare widget]

Yeah, I meant wrt the perf comment

[wintry steppe]

Let $A$ be a bijective linear transformation from $V$ to $W$ and span $(a_1, a_2, \dots a_n) \leq V$. Why $dim A $span $(a_1, a_2, \dots a_n) = dim$ span $(A(a_1), A(a_2) \dots A(a_3))$?

sorry have no idea how to format this correctly.

[stoic python]

mate

[wintry steppe]

because A(span(...)) = span(A(...))

or are you asking why dim span (...) = dim span (A(...))

[spare widget]

because the map is bijective

Let (v1,...,vm) be a basis for span(a1,...,an)

by the definition of basis these vectors will be linearly indepedent

which means that sum_i b_i v_i = 0 only for (b1,...,bm) = (0,...,0)

now assume that A(v1), ..., A(vm) were linearly dependent

then there would exist (b1,...,bm) such that sum_i bi A(vi) = A(sum_i bi vi)

but if A is bijective, then it is injective, and then A has a trivial kernel

so you get a contradiction and it follows that dim(span(v1,...,vm)) = dim(span(A(v1),...,A(vm))

[wintry steppe]

thank you. i dont understand how did you get a contradiction, could you repeat please?

im lost after "then there would"

but i do understand why A injective iff A has a trivial kernel

there'd be a contradiction because at least one of the b_i's should be non-zero

and that sum should be equal to zero

that's where you get to use injectivity

[dusky epoch]

[sonic gulch]

Does a linear transformation T : V→W transforms V's basis to a W's basis ? Or there are any extra conditions ?

[wintry steppe]

not necessarily, take the zero transformation

a necessary and sufficient condition is that T be an isomorphism

[sonic gulch]

Oh

If Ker T = {0}, does it imply then ?

[distant schooner]

comes directly from $Dx = \text{P}_U(b)$, where $U$ is the column space of $D$ right?

[stoic python]

koala

[distant schooner]

i think this is the more "intuitive" version

[sonic gulch]

Sorry I'm not familiar with isomorphism of linear transformations

[wintry steppe]

bijective linear transformation

[sonic gulch]

Oh okay

Thank you

[distant schooner]

what is with the little leaf next to your name @wintry steppe ?

[wintry steppe]

it means i'm new

(i left for a month and came back)

[distant schooner]

right, but havent u been here for quite a while

oh i see

[zinc timber]

what leaf

[void path]

this is a nonogram, its a puzzle in which the numbers on each line represent how many blocks of continuous shaded pixels exist. There must be a white in between every blue "segment". When i go about solving them. Normally i solve them by thinking "hmm, 6 will need its middle 4 blocks colored no matter what so ill do that, then according to this column, that has to be a 1 so it should be white above and below" and so on...

But this feels like linear dependence and something to do with linear algebra, i just cant quite figure out how to apply it to solving these

this is a binary matrix, and im trying to figure out how to solve it mathmatically

my reasoning for wanting to figure this out, is that im building an operating system and this may be a better way of storing black and white photos

[spare widget]

It comes from taking the derivative wrt x and setting it to zero

Which gives you an equation for the stationary points

[spare widget]

There are research papers on this and it's far from trivial

I would instead suggest to use a readily available codec fir black and white images

[distant schooner]

ah okay yeah that too

i mean the projection is equivalent as well right?

ill try to show this

[spare widget]

That's how you derive it

[spare widget]

afaik you define the projection either through the x minimizing the L2 or as the x such that the error is orthogonal

the latter is typically called Galerkin formulation/condition

[subtle gust]

Can anyone explain linear subspaces to me?....

Specifically how you can prove that a given set is a subset of R^n

There's 3 ways and i only understand the first one

1. using the def of a subset (that i understand)
2. smthng related to homogeneous linear systems
3. smthng related to the span of smthng

[patent belfry]

For 3, the span of any subset of vectors in $\mathbb{R}^n$ is automatically a subspace of $\mathbb{R}^n$

[stoic python]

makatk

[wintry steppe]

I'm assuming by "subset" you mean "subspace"? These are different terms though so you shouldn't conflate the two. Here's what I think you mean by your "three ways". These are three equivalent definitions.

1. A subspace (of the vector space $R^n$ ) is a subset of $R^n$ which is closed under linear combinations (if $v_1, \dots , v_k$ are in the subspace, then $a_1 v_1 + \dotsm + a_n v_n$ is in the subspace, where the ai's are real numbers (scalars) ).

2. A subspace is a set of solutions to a system of linear equations: ${x=(x^1,\dots,x^n) \in R^n ~,~ f_1(x) = 0,~ \dots,~ f_k(x)=0 }$, where the $f_i$'s are linear functions $R^n \to R$, i.e. of the form $f_i(x) = a_{i1}x^1 + \dotsm + a_{in}x^n$. The above set will specify a subspace of dimension $n-k$.

3. A subspace is the "span" of a collection of vectors, where span means take the set of all linear combinations. This is nearly the same as my first definition.

[stoic python]

KS

[trim bear]

Going through my lin alg textbook and I wanted to show a cool proof I did

The first paragraph is the question

[stoic python]

Wyatt The Baguette

[round granite]

oh lol i just had this problem in an assignment

iirc the intuition was "S is an isometry so S* = S^-1 so its just a change of basis"

[last mist]

not sure if this is exactly "linear" algebra, but would anyone know a way to solve the following problem without brute force?

┌       ┐     ┌ ┐
│1 1 1 0│     │1│
│0 1 1 1│ x = │2│
└       ┘     └ ┘
where x_i ∈ {0, 1}
``` alternatively, if there is a better channel to post this in, please redirect me

[tacit pelican]

could anyone help me fix up this proof?

i'm pretty sure the first two paragraphs are correct (i hope)

but

idk how to prove that if dim V = dim W, V=W

i feel like something needs to be added in that space i have put

but idk what

maybe my approach itself was incorrect

ill look through mse to see if i can find anything

[native rampart]

(0,0,1,1) and (0,1,0,1)

Think of that as solving
x_1(1,0) + x_2(1,1) + x_3 (1,1) + x_4 (0,1) =(1,2)

You need the first co ordinate to be 1 so only one of x_1 ,x_2 or x_3 should be 1

You need the 2nd co ordinate to be 2 so exactly 2 out of x_2 ,x_3 or x_4 has to be 1

If you choose x_1 to be 1 you find you don't have a solution

[last mist]

yeah but wouldnt that amount to brute forcing?

[restive raft]

This is how I'd prove this: assume $V \neq W$ then $\exists w \in W$ and $ w \not\in V $ then say ${v_1,...,v_n}$ is a basis for $V$ then we also have that ${v_1,...,v_n,w}$ is linearly independent, else $w$ is in $V$, but we have an inequality given a linearly independent set of vectors in a vector space $W$ that is is less than or equal,
$|{v_1,...,v_n,w}| \leq dim(W)$ but if $dim(W) = dim(V) = n$ we would have $n+1 \leq n$ which is not possible, hence there can be no such $w$ and hence they must be equal.

[oblique prairie]

would distinct here mean that each eigenvalue of A has multiplicity 1?

[stoic python]

holazach

[restive raft]

no it just means no $\lambda_i = \lambda_j$ unless $i = j$

[stoic python]

holazach

[oblique prairie]

oh ok thanks, just making sure

[restive raft]

no problemo

[lavish jewel]

aren't those two things the same?

[restive raft]

huh

[lavish jewel]

the algebraic multiplicity is the power k so that (lambda - lambda_i)^k divides the char poly

if all of the eigenvalues are different, they have multiplicity 1

[restive raft]

I interpret statements like that as say the characteristic polynomial is $(\lambda - \lambda_1)^2(\lambda - \lambda_2)$ then the distinct eigenvalues are $\lambda_1, \lambda_2$ whereas you might say the eigenvalues are $\lambda_1, \lambda_1, \lambda_2$, if you have a n-dimensional operator with n distinct eigenvalues then it does follow each eigenvalue has multiplicity 1, but lacking that you can't say.

[tacit pelican]

alright, i think i fixed it

this is fine right?

its basically what you wrote

[stoic python]

holazach

[restive raft]

looks good

[tacit pelican]

thank you!

that proof took wayyy too long to do

[lavish jewel]

ah i reread what quantum posted, you're right. it doesn't say they all are distinct. that was my bad for skimming

[native rampart]

Brute forcing would be checking all 16 possibilities

Instead of 2^(4) we are checking 4-1 cases

Actually here's a much better approach

x_1+x_2+x_3=1
x_2+x_3+x_4=2

x_4-x_1=1
implies x_1=0 and x_4=1 implies x_2 or x_3 has to be 1

[last mist]

true

@native rampart
is there a common algorithm this follows? or are you going mostly off of just manipulating the equations directly?

[native rampart]

Just manipulating

https://mathoverflow.net/questions/187609/systems-of-equations-in-boolean-algebra

Maybe relevant

[oblique prairie]

could someone help me start the proof stated in the last line? i’m clueless on how to start it

[native rampart]

What is $x_1',x_2'...$

[stoic python]

BYE BYE!

[native rampart]

@oblique prairie

nvm derivatives

Prove x(t) is of that form first

Then try multiplying both sides
of
$\sum_{i=1}^k e^{\lambda_i t} z_i = 0$( Proving { $e^{\lambda_i t} z_i | i \in {1,2,3...k} $ } will work I think)
with $(A-\lambda_2)(A-\lambda_3)...(A-\lambda_k)$
This would cancel all the terms except $e^{\lambda_1 t} z_1$

The correct approach would be something along those lines

[stoic python]

BYE BYE!

[restive raft]

We want to prove a two way implication first start with if $x(t) = \sum_{n=1}^{k} e^{\lambda_it}z_i$ then
$x'(t) = \sum_{n=1}^{k} \lambda_i e^{\lambda_it}z_i$
and since $Az_i = \lambda_i z_i$ we have $x'(t) = Ax(t)$.
then going the other way given $A$ is diagonalizable the direct sum of each eigenspace, $E_{\lambda_1} + E_{\lambda_2} + ... + E_{\lambda_k} = V$ is the space so each vector $x(t)$ for every $t$ can be written as a linear combination of $z_i \in E_{\lambda_i}$ so $x(t) = \sum_{n=1}^{k}c_i(t)z_i$ then if we have $x'(t) = Ax(t)$ then
$\sum_{n=1}^{k}c_i'(t)z_i = \sum_{n=1}^{k}c_i(t)\lambda_i z_i$ given that this linear combination is unique it is a requirement that we have $c_i'(t) = \lambda_i c(t)$ which leads to a solution $c_i(t) =b_i e^{\lambda_i t}$ for some scalar $b_i$ but we can just transform $z_i -> b_i z_i$ and then the result follows

[stoic python]

holazach

[oblique prairie]

well, i already proved that those statements are equivalent

i was talking about the line at the bottom of the screenshot

oh wait

[vernal slate]

Hi how are you all doing, I'm just into linear algebra, I'm having confusion finding NPC (non pivot column)
For example,
[ 1 4 0 4 0]
[0 0 1 0 0]
There are no NPC's right?

[restive raft]

there are non pivot columns, columns 1 and 3 are pivot columns so 2 and 4 and 5 will be non pivot columns

[vernal slate]

Thanks man

[restive raft]

np

[broken notch]

Least squares via QR (𝐴, 𝑏)
Requirements: the columns of 𝐴 are linearly independent
1 Calculate a thin 𝑄𝑅 decomposition 𝐴 = 𝑄𝑅
2 Calculate 𝑄𝑇 𝑏
3 Solve 𝑅𝑥 = 𝑄𝑇 𝑏 via back substitution
what do they mean by thin

[lavish jewel]

this is what wikipedia shows

[zinc timber]

if $P \geq 0$ (entrywise) and $\rho(P) < s$ then $sI - P$ is non-singular M matrix (this part is easy to see) but how do I show that if $\rho(P) = s$ then $sI-P$ is singular (M matrix)

[zinc timber]

looks similar to stochastic matrix that if spectral radius is s then s is an eigen value

could find some help from there

[zinc timber]

https://en.wikipedia.org/wiki/Perron–Frobenius_theorem found something

[lavish jewel]

that's the one i had in mind when i suggested looking at stochastic matrices

but all i know is that it exists 😛

[languid sphinx]

P-F is a powerful theorem

IIRC the proof is not hard-hard, but tedious

[zinc timber]

I only need the statement for now

[zinc timber]

yes I stumbled onto this from a completely different context

to show that if H is non-singular M-matrix and B and BH^-1 both are Z-matrix then BH^-1 is non-singular M matrix <=> BH^-1 is non-singular M matrix

though again I only needed the statement

[broken notch]

@lavish jewel
im still stuck on this problem from yesterday perhaps you can tell me if I am correct?

import numpy.linalg as la
import matplotlib.pyplot as plt
import numpy as np

#Define the coordinate of the lamps
x=np.array([  4,  3, 19, 16,   9,   5,  18,   16, 12, 12,    2, 13])
y=np.array([ 20,  2,  4, 16,  28,  11,  13,   25, 12, 23,   15, 4])
z=np.array([2.8,  3,  3,  3, 3.4, 3.5, 3.6,  3.8,  4,  4,  4.5, 3.6])


#We define the center of the of the square, e.g. matrix A
xc=15
yc=10
zc=0

#We compute the distance from the center [xc,yc,zc] to [x,y,z] coordinates of each lamp
xxc = x-xc
yyc = y-yc
zzc = z-zc

#The ligthning square from each lamp is calculated by dividing the intensities xi 
# with the square of the distance in R³ to each lamp from the center.

xi=np.array([0.001, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11])

#Vector that contain the reciprocal distance from the lamps to the center.
dm = xi/(xxc**2 + yyc**2 + zzc**2)


b=np.ones(shape=(12,1))

#Create and populate matrix D of size 12x12 with the vector dm 
D = np.zeros((12,12))
for i in range(12):
    for j in range(12):
        D[i][j]=dm[j]
    
    
#Now we need to solve 1-D * x with:
# Least squares via SVD (D, 𝑏) 
# 1 Calculate a reduced SVD D = 𝑈Σ𝑉 𝑇
# 2 Calculate 𝑈 𝑇 𝑏
# 3 Solve Σ𝑦 = 𝑈 𝑇 𝑏
# 4 Set 𝑥 = 𝑉𝑦

u, s, vt = np.linalg.svd(D)

s= np.diag(s) @ np.diag(1/s)

utb = np.matmul(u.T,b)
sinv= la.inv(s)
y = np.matmul(sinv,utb)
x = np.matmul(vt.T,y)
Asvd=np.matmul(D[0],x)  
print(np.sum(Asvd))

[meager harness]

how to do 4b

[native rampart]

What have you tried

The solution is the minimal polynomial has to be either x or x^2(why)

In the former case,A is just 0 matrix

In the later case, pick any x such that Ax!=0

And x,Ax form a basis

Writing A in that basis gives you the answer

[wintry steppe]

Let k* be a linear form from the space of polynomials of degree less than 3 to R such that k* (q) = q(0). If B is a basis for that vector space, how to find B*?

[lavish jewel]

try writing a polynomial of the prescribed order and see what happens when you evaluate at 0

in particular, what happens to terms that have the variable present in them?

[wintry steppe]

they disappear

hm

if B = {b1, b2, b3} then we can write

q = Ab1 + Bb2 + Cb3

and i would expect that k* (b1) returns A, k* (b2) returns B, k* (b3) returns C

is that correct?

[lavish jewel]

i would expect k*(b_i) to be either 0 or the coefficient c_i

depending on what b_i is

can you show what exactly the problem statement is btw?

[wintry steppe]

so

k* : P2 --> R
k* (q) = q(0)
i need to write k* as a linear combination of vectors in B*. and B = {1, 2-x, 2-x^2}

[lavish jewel]

there we go, that makes it a lot easier that doing it in general

to give a hint

since it's a linear form, what rank does the linear transformation have?

what shape should the corresponding matrix have?

[wintry steppe]

i think rank = 1 because k* is not 0?

[lavish jewel]

yeah

and your matrix would be of size 1 x 3, just a row vector

what i would do is to split the problem up into easier problems

if the basis were nice and simple like the canonical basis {1,x,x^2}, then k* would be a lot easier to describe, right?

so we could do that first

and then, write a change of basis matrix that takes vectors in the B basis and maps them to the canonical basis

then the overall transformation is the composition of these two, i.e. apply k* to the vectors that have been transformed from the B basis to the canonical one

[dark pier]

If V = {x + y | x ∈ span{u}and y ∈ span{v}}, where u and v are two fixed
non-zero vectors in R3, then V is a subspace.

can someone help guide me on how to show if this is True or False?

[wintry steppe]

thank you for your time. i can express a2x^2 + a1x + a0 as a linear combination of vectors in B but not sure if that helps.

[lavish jewel]

that's precisely what the change for basis from B to the canonical one is 🙂

[wintry steppe]

i got

$a_{2}x^2 + a_{1}x + a_{0} = (a_{0} + 2a_{1} + 2a_{2}) + (-a_{1})(2-x) + (-a_{2})(2-x^2)$

is my matrix
$\begin{pmatrix} a_{0} + 2a_{1} + 2a_{2} \-a_{1} \ -a_{2} \end{pmatrix}$?

[stoic python]

mate

[humble oak]

hello there, i'm a bit confused about this example

how exactly did they determine the number of free variables?

nevermind, they are x^2 and x right?

[restive raft]

you can solve one of the variables in terms of the others say $ax^2 + bx + (-a -b)$ then the free variables are $a,b$

[stoic python]

holazach

[humble oak]

oh okay, thank you

[distant schooner]

is there a relationship between the geometric product and its reverse? such as $(AB)^\dag = B^\dag A^\dag$... im trying to prove or disprove this myself so any "hints" would be cool, if such a relationship exists (probably does)

[stoic python]

koala

[quartz compass]

well reverse(A) = (-1)^(grade(A)(grade(A)-1)/2) A so nothing super exciting using the fact scalar multiplication commutes

[short canopy]

so, i'm trying to understand this "the plane xy are all the points perpendicular to the Z axis. Therefore, we could say that they are the points (x,y,z) such that (x,y,z).(0,0,1) = 0, (0,0,1) being the normal to the plane "

i don't understand why (x,y,z).(0,0,1) = 0, isn't it (x,y,z).(0,0,1) = z ?

[lone flume]

The xy-plane consists of all the points (x,y,z) such that z = (x,y,z).(0,0,1) = 0.

[short canopy]

isn't (x,y,z).(0,0,1) = (x.0)+(y.0)+(1.z) = 0 + 0 + z = z?

[lone flume]

Yes, that's the point

The xy-plane consists of all the points satisfying z = 0.

[short canopy]

so, im supposed to assume z = 0 or something like that?

weird

[lone flume]

The point (x,y,z) lies on the xy-plane if and only if z = 0.

[quartz compass]

points without z=0 are the points that have floated off the xy plane

[short canopy]

right

[broken notch]

A parking space of 20m × 30m is illuminated via lamps placed in different places and at different heights, as indicated in Figure 1 page 3.
The car park is divided into a rectangular grid of 600 squares each of size 1m × 1m. The number 𝑦𝑗 indicates the illumination level squared 𝑗, for 𝑗 = 0,. . . , 599. Let 𝑥𝑖 indicate the power of lamp 𝑖. We select units so that the contribution to the lighting in square 𝑗 from lamp 𝑖 is 𝑥𝑖 / 𝑑2𝑖𝑗, where 𝑑𝑖𝑗 is the distance in ℝ3 from the lamp to the center of square 𝑗.

(a) State how the illumination level 𝑦 = (𝑦0,..., 𝑦599) and the powers 𝑥 = (𝑥0,..., 𝑥11) are related via a linear equation system. Set the coefficient matrix for the system in python. (You must estimate the coordinates of the location of each lamp from the diagram.)

please can someone help me with problem a) I am having a real hard time understanding what to do

figure

[restive raft]

$M_{ji} = 1/d^2_{ij}$ should be 600x12 matrix and then the system is $Mx = y$

[stoic python]

holazach

[broken notch]

where dij is the distance from the center to the cell right

[broken notch]

y would be 600x1?

[broken notch]

what about z

[restive raft]

Yes, Yes, what is z?

[broken notch]

z is the height

[restive raft]

the d_ij has the distance in R^3, wouldn't that already be included?

[broken notch]

import numpy as np
import matplotlib.pyplot as plt


center_x = 12/2
center_y = 600/2

M = np.zeros((600,12))
for i in range(600):
    for j in range(12):
        if(i==center_x and j==center_y):
             M[i][j]=1
        else:
            M[i][j]=1/((i-center_x)**2 +(j-center_y)**2)


fig, ax = plt.subplots(figsize =(1, 100))
plt.imshow(M, cmap = "hot")
plt.colorbar()

ax.set_xlabel('X-axis') 
ax.set_ylabel('Y-axis') 

plt.tight_layout() 
plt.show()    

looks weird, is it something like this?

I populate the 600 by 12 matrix

[broken notch]

do I populate the matrix M like this?

[broken notch]

what is y?

I will have to know y and A right

[sour flame]

Okay this is probably a super simple question but for life of me I can't figure out why S is not a subspace of W

like, I don't understand why you can't just for example do (0,1) * 3 and get (0,3) which is also an element of R^2 is there something very big I'm missing? Because my lecturer brought this up as an example for why it's not closed under scalar multiplication (unless I misheard him?)

[quasi vale]

@sour flame For S to be a subspace, we need to make sure the following conditions are satisfied: For any v in S, cv is in S, c is a scalar and for any v,w in S, v + w is in S. If you multiply (0,1) by 3, you get (0,3) and (0,3) is not in S. You can also do (0,1) + (1,1) = (1,2) which is not in S.

[languid sphinx]

Way too many things are not in S, so I feel like you're not getting the intuition right

[sour flame]

oh wait okay for some reason I was thinking about all linear combinations of the things in S was a subspace of W rather than S itself

thanks for the clarification

🙏 I was staring at this for 15 minutes for no reason

[dusky epoch]

do not confuse a set with its span

[broken notch]

@lavish jewel
will this be correct reasoning?

[spare widget]

We already went over this

Let (M^TM) =USV^T

Then (M^TM)^{-1}= V S^{-1} U^T

So the solution is x = (M^TM)^{-1}M^T 1 = VS^{-1}U^TM^T 1

You get the normal equations M^TM x = M^T 1 by differentiating |Mx - 1|^2 wrt x and setting the derivative to be equal to zero (i.e. the equations for the stationary points)

[broken notch]

@spare widget When you say (M^TM) =USV^T is it the same as: usv^t = sdv(M^TM)

in the equation VS^{-1}U^TM^T 1 , what is 1?

x will be the size 600x12, so by multiplying M^T*x^T I get Mx of size 12x12 so I guess if I take one row it is the solution to x, right?

[spare widget]

Yes it's the svd. 1 is a column vector of 1s. Idk what your last question is about.

[broken notch]

ok I think I figured it out thanks for helping 🙂

[wintry glen]

Hey, sorry if this is a bit of a stupid question, but for (c), do I just let $\mathbf{x}_0 = (1, 0)$ and compute $\mathbf{x}_2 = P^2\mathbf{x}_0$?

[stoic python]

PhenomPlasma

[elfin agate]

👋 i have a question regarding matrices corresponding to a map f:R^2–>R^3 . f is referring to two basis V={v,u} und W={x,y,z}

M(f)= ( 3 2, 1 0, 2 -1)

and the task is to transform M(f) so that it refers to the standard bases of R^2,3

to my actual question: i would like to find the map f from M(f) by having each column of M(f) as the solution of two equations

specifically: f(v)=f(2,7)= (3, 1, 2) and f(u)=f(1,4)=( 2,0,-1)

for the first column of f we had the system of equations like this:

2λ+7μ=3 and λ+4μ=2

leading to: λ=1 μ=-2

[zinc timber]

yes

[elfin agate]

if i do this for every column i get f(a,b)=(a-2b, 4a-b, 15a-4b)

is this form f unique??

[vale cape]

what do you mean by 2-digit rounding in solving systems of equations?

For example, use Gauss-Jordan Elimination method and 2-digit rounding in solving systems of equation

[elfin agate]

ups, i meant row and not column xD maybe thats clear now

so basically M(f)=(f(v),f(u)) ,f(v)/f(u) being column vectors with the values that you can see in the text above and i want to get f

in the way as i referred in the text above, will the result will be the right one?

[elfin agate]

or i have the totally wrong approach the hole time D:

[short canopy]

say i have this implicit equation for a line y = 2x + 1, how do i make it into a vectorial equation? my first approach was saying something like (x, 2x+1) = (x,2x) + (0,1) = x(1,2) + (0,1) but this is wrong when i compare it to some exercise from a book i'm using

[quartz compass]

that's correct

there's more than one way to do it though

so maybe that's why you think it's wrong

show some examples of what they give as the answer

for instance another way entirely is write it as -2x+y=1 and then you can recognize the dot product (-2,1) dot (x,y)=1

[elfin agate]

a quick understanding question: if k is an element of ker(f) with f being a homomorphismn would g(k), g being also a homomorphismn also be in the kernel of f?

if k=0 surely but im not sure if that statement holds for any k

[quartz compass]

nope, imagine f projects on the x axis and g projects on the y axis

[elfin agate]

damn, i want to show g(f(k))=f(g(k))=0 is there a trick im missing? g(f(k))=0 follows quickly since f(k)=0 and g(0)=0 but im not confident baout f(g(k))

[quartz compass]

is that true?

I'm thinking f projects on the x axis and g rotates by 90 degrees, that'd be a counterexample

maybe I'm not paying enough attention to something here

[elfin agate]

hold on, i didnt read the task correctly, i have to show the existence of a g, element of Hom(V=R^n) such that the equality holds

[quartz compass]

identity 🤣

[elfin agate]

i mean yes

am i tarded?

ah hold on again, g is unequal to zero

i thought they mentioned it so that you have to guess that you have to play with the kernel of f

hmm

[short canopy]

it's weird, i was doing some exercise where i did the same method and the result wasn't the same, i verified it with graphing software and it wasn't the same

this one #help-17 message

[little oriole]

is the det(0) = 0?

couldnt find a calculator that would work to confirm

[wintry steppe]

yes

[little oriole]

awesome

[wintry glen]

Thanks!

[trim bear]

How would I properly denote the fact that two vectors make up the columns of a matrix?

Col(A) = vec1, vec2?

[elfin agate]

i would use A=(u,v) and specifacally mentioning that the vectors u,v are columns by showing the dimension

[elfin agate]

a friend of mine shared his solution and i thought you might be interested, one can construct g such there are k_i =g(v_i) so that some of the k_i’s are the base of the kernel of f while others are in the image

[icy blade]

yo so kinda noob question

but after i find rref and find that the vectors are linear dependent

how can i find what combination of those vectors give zero vector

[hoary void]

Well you could solve the system of equations

Like say a1,a2,a3 are the vectors and b_n constants then: b_1a_1 + b_2a_2 + b_3a_3 = 0(vector) this would just make a 3x3 system that you could solve for b_n

[icy blade]

ok but i did solve them, now the problem is to actually get the b_ns

my final matrix is ([1 0 3 0] , [0 1 5 0] , [0 0 0 0]) where [] is a row

what do i do from here @hoary void

okay i got it nvm

[wise nebula]

this question is about a linear application

and asks for which a's the v vector is in the image of the application

but you shouldnt be supposed to multiply the matrix with the vector since its 4 columns 3 rows

i'm very confused about this and i have an exam tommorw

someone can help out?

(4 column matrix, 3 rows vector)

[pliant ore]

even the statement says the function takes vectors from R^4

[wise nebula]

yea but

how can you do the row x column product

if the number of the columns of the first doesnt equal the number of the rows of the second?

[pliant ore]

you cant

unless you add a 0 to the vector or something

[wise nebula]

then that problem makes no sense

since the application its literally

Av

but i can't do that

[wise nebula]

we agree or am i missing something?

because i think i am

[pliant ore]

well either the question is wrong or you're missing something

[wise nebula]

and that's for sure

but

you should notice

if i'm missing something

right?

idk this is just fucking my mind

the question shouldnt really be wrong

cause its a exam simulation for all class given days ago and nobody complained

its just so weird

they don't answer yet if youre wondering

[wintry glen]

Wait, what's wrong? v is a vector in R³ and it wants you to find the values of a such that v is in the range of the transformation. So you can just row reduce [A v] and find the values of a that make the system consistent.

[broken notch]

import numpy as np

x = np.array([-1, 1])[:, np.newaxis]
y = np.array([1, -1])[:, np.newaxis]
b = np.array([1, 1, 0])[:, np.newaxis]



A = np.array([[-1,0],[0,1],[1,-1]])




if ((A@(x+y)+b == A@x + A@y+b).all() and (A@(5*x)+b == 5*A@(x)+b).all()):
    print("It is linear")

is this correct?

I need to check if it is linear

[dusky epoch]

are you forced to check it for linearity with a program and forbidden from doing so by hand?

[broken notch]

I can do it by hand

but it is numerical linear algebra

that is why I am using python

bcs the exam will be in pytohn

[dusky epoch]

there's no need to use any programming here

L as written here is obviously not linear: L(0) is not 0!

and you don't need a computer to tell you that, surely.

[broken notch]

I need to check
#Check A(x+y) = Ax + Ay
#Check A(sx) = sAx

right?

[dusky epoch]

that's the definition of linearity yes

but you should know a thing or two about linear maps

such as: any linear map sends zero to zero

[broken notch]

hmm

[dusky epoch]

so a map that sends zero to something other than zero cannot be linear no matter what

[broken notch]

it is returning true

tho

im confused

[dusky epoch]

what is [:, np.newaxis]?

[broken notch]

just prints horizontally

I think

[dusky epoch]

...

you think?

so you are using things whose function you yourself are not fully aware of.

[broken notch]

I mean

[dusky epoch]

why, then, are you confused that you are getting bullshit results?

[broken notch]

new axis creates a new col

so instead of (3,)

[dusky epoch]

garbage goes in, garbage comes out.

[broken notch]

it is (3,1)

[dusky epoch]

also it looks like your calculation is bad anyway

[broken notch]

yea I thought so

bcs everything returns linear 😄

[dusky epoch]

you have L(x) = Ax + b, sure
and then L(x+y) = A(x+y)+b, but L(x)+L(y) = (Ax+b)+(Ay+b) not Ax+Ay+b as you wrote!

and once again!

this stuff shouldn't require a computer!

a computer can't prove linearity like that anyway!

the best you can do is demonstrate some examples of linearity being satisfied for some particular inputs!

do not rely on computers for that which can and should be done by hand!

[broken notch]

I need to use python

the entire course is numerical linear algebra in python

[dusky epoch]

what do you mean, "need"?

has your teacher said explicitly that EVERY SINGLE EXERCISE HAS TO be done in python and any non-python solution will be REJECTED OUT OF HAND?

even these?

[broken notch]

ok now it works

the teacher wont see this

these are just exercises

but I like to practice

in python bcs the exam is in python

[dusky epoch]

so why use python for things that don't require it...

you're not answering my question either

[broken notch]

"the exam is in python"

[dusky epoch]

has your teacher said this, yes or no

[broken notch]

I can do it in hand

np

but the exam is in python

[dusky epoch]

who gives a shit.

it sounds like you are just learning the very basics of linear algebra, and the 'numerical' part has not yet really come up.

if you're at the level where you're trying to verify whether or not a map is linear, you're not doing any numerics.

[broken notch]

I do

[dusky epoch]

it sounds like you are just learning the very basics of linear algebra, and the 'numerical' part has not yet really come up.

if you're at the level where you're trying to verify whether or not a map is linear, you're not doing any numerics.

you are NOT yet doing any numerics. you aren't trying to get any approximate solutions. you're not doing things like finding eigenshit numerically.

[broken notch]

I did the first one

by hand

and said cba

I dont see a reason to do it by hand if I know how to do it by hand

[dusky epoch]

this exercise can be done in one line

L(0) != 0

that's it

no need for any python shit.

it's going to take 10 times more effort to write a python program than to just do some observations yourself.

[broken notch]

yea but i am not well versed I dont see what you see

I just follow the formula

[dusky epoch]

and a program really will not help you with that at all.

i think trying to do everything programmatically will really muddle your vision even further.

ignore this advice at your peril, i suppose.

[quartz compass]

yeah that's kind of Ann's point, you're too busy distracted by programming stuff instead of the linear algebra stuff itself

[broken notch]

so what should I do :D?

not do the exercises or what I dont get it

[dusky epoch]

do things by hand unless there's good reason to do it programmatically.

[lavish jewel]

to do the logic on paper, and then code the logic afterwards

good code starts on paper

[dusky epoch]

if you need two dense 10x10 matrices multiplied fsr, obviously nobody is forcing you to do it manually. that's what computers are for. they're glorified calculators.

[broken notch]

doing it on paper would take way longer

and I dont have that kind of time

I also have other exercises and homework

[dusky epoch]

do you want to write shit code?

if you want to pump out thousands of lines of shit code you don't need any paper

but shit code, unsurprisingly, is shit.

[lavish jewel]

the logic, not the matrix multiplication

[dusky epoch]

if you are doing something that actually takes many computations like an iterative method of some kind, that's something for computers to do - but yeah, as edd said, first write out the formulas on paper.

[broken notch]

yea I know I usually do that

[dusky epoch]

well then do that

[broken notch]

but you dont need that to check if it is linear

[dusky epoch]

you do, as evidenced by you writing the linearity conditions incorrectly in your example.

[broken notch]

more of a typo

my question was about if it was correct to say Ax + b

bcs in the slides of my teacher he doesnt write Ax+b in his examples

[dusky epoch]

this only comes up now

your map is representable in that form, yes.

[broken notch]

I probably formulated the question badly

[dusky epoch]

yes you did.

[hard drum]

Anyway hello

[hushed hedge]

it keeps brining up calculus on the browser lol

[hard drum]

Lol dw

[hushed hedge]

for example let's say i have the point X = (-1,-2,0) in the plane of pi: x-y+z=0 and i want to solve for the orthogonal projection

so this was the problem

[hard drum]

Sure