linear-algebra

[lavish jewel]

your problem is convex

[spare widget]

it is

I went through Boyd's book

Couldn't find QP with conflicting constraints

[lavish jewel]

because that name is not used

idk where you found it

[spare widget]

what name is used for it?

[lavish jewel]

in engineering books like boyd's, this is a denoising problem

[spare widget]

Any references?

[lavish jewel]

not off the top of my head. it's anyways pretty easy because the target and constraint are both differentiable, you can set it up yourself

[spare widget]

set up what?

[iron harbor]

I'm trying to show that U is a subspace of R^2 but I'm not quite sure that this does the trick.

According to the material from my course, showing that U is a subspace can be limited to showing that both vector addition and scalar multiplication are closed and defined for U; is that really all I need here? What about the basis and dimension of U?

[lavish jewel]

how to solve it with gradient based methods without ever needing to find pinv or bases

since you said you dont want that

[iron harbor]

oh sorry didn't mean to interrupt

[lavish jewel]

this works, clock

[spare widget]

I would highly doubt this being possible, I went through some papers going over the literature for QP, and you generally either have to construct a basis, or do the projection, there's no escaping it afaik

The bonus of the conflicting case is that the projection is more nasty

[lavish jewel]

if you want the exact sol, yes

[iron harbor]

nice thanks

[lavish jewel]

if you're ok with an iterative alg that converges to it, no need

[spare widget]

I agree that it falls under convex optimization with projected gradient, but the projection is the efficiency-wise nasty part

[lavish jewel]

i never said projected grad tho

[spare widget]

then I am not sure what method you had in mind

interior point method?

Are you suggesting to relax the projection part?

[lavish jewel]

yep

[spare widget]

can't say much without a specific method, doesn't ring any bells except DS and DS*, but those were relaxations for doubly stochastic matrices

[lavish jewel]

otherwise you're looking at variations on kkt, so stuff like interior point methods or active set methods

[spare widget]

Yeah I was digging through kkt for a while

[lavish jewel]

you could pick an epsilon for Ax approx b and log barrier it, for example

[spare widget]

I found this: Optimization with Least Constraint Violation
Though it's wsy too complex for what I am solving

[lavish jewel]

all of these approaches tend to go with an extra hyperparameter and a sentence something like "it can be shown that for a correctly chosen value of the hyperparam, it solves the original problem"

but the proof doesn't tell you how to pick it 😛

[spare widget]

I've seen those 😂

dgfem ptsd

[lavish jewel]

my recommendation, depending on how strict the constraints are, is either log barrier, or to simply minimize f(x) + mu || Ax - y ||

where mu is in some obscure way related to || Ax - y || < \epsilon

this latter one should inherit the convexity flavor of f(x) in your case

[spare widget]

The first and simplest thing I would try is throwing cgnr at the singular system, if this proves too slow I will likely go with your suggestion of f(x) + mu ||Ax-y||

[lavish jewel]

so possibly strictly convex and positive def

[iron harbor]

given vectors a and b and scalars x and y, does this hold or have I lost the thread here?
x(a+b) + y(a-b) = 0
xa + xb + ya - yb = 0
(x+y)a + (x-y)b = 0
==> implies that x = y = 0 ??

seems like there's got to be something wrong with it that I'm not seeing

[lavish jewel]

there's no reason for this to be true unless you know a and b are linearly independent

[iron harbor]

right, yes. that's the connection I was struggling to make.

the proposition is: given linearly independent vectors a and b, are the vectors (a+b) and (a-b) also linearly independent?

[lavish jewel]

aha

[iron harbor]

but because that is the assumption of the proposition, it should hold right?

[lavish jewel]

yes

c = x +y and d = x - y

then we have ca + db = 0, which is a linear combination of a and b

and since a and b are linearly independent, we know that the only solution to this is c = d = 0

so you rewrite it as a linear combination of the original vectors and simply evoke the definition of lin indep

[iron harbor]

oh man I am relieved that it makes sense after all

[lavish jewel]

if they were lin dep, this would not necessarily be true

[iron harbor]

thank you for explaining that :)

I need to make flashcards with these definitions I think; that reasoning was only like 3/4 of the way formed in my brain until you wrote down basically the translation of what I was thinking

[lavish jewel]

it takes a while to get the hang of it

and linalg is usually one of the first proof-heavy courses people take

take your time to digest it

[iron harbor]

👍

[urban gale]

what can be the equation of negative y-axis?

[halcyon spindle]

What do you mean?

[lusty pumice]

Shouldn't this be 55?

[spare widget]

you're supposed to expand |u-2v|^2

expand it and then plug in the known factors

[lusty pumice]

oh mannnnnn

I'm going to smash my head against the wall

real quick

[spare widget]

use |x|^2 = x dot x

and the linearity of the dot product

x = u-2v in this case

[lusty pumice]

I used this

[spare widget]

you don't need that

[lusty pumice]

but didn't see the minus

[spare widget]

and the 2

use this:

[lusty pumice]

AND THE 2

[spare widget]

$|x|^2 = x\cdot x$

[lusty pumice]

mannnnnnnnnn

[spare widget]

and substitute x = u-2v

[teal grotto]

what?

[lusty pumice]

I give up on life

[spare widget]

then use linearity to distribute the terms

[lavish jewel]

you missed a square

[spare widget]

I missed a square

[teal grotto]

criver, this is exactly what ur describing

[stoic python]

criver

[spare widget]

except with a different argument

[teal grotto]

?

sorry, but that looks like the exact same argument to me

[lusty pumice]

No I just did the quiz

and I put 55

because I'm blind and dumb

and I didn't know why

but didn't see the minus and the 2v

My brain assumed it was | | u+v | |^2

[spare widget]

my brain also used to get fried at exams

I would walk out 5 steps out the door and it would start working

I think it's the sitting for extended periods of time, no blood circulation to the brain

[lusty pumice]

does it mean I need more practice?

I understand the concepts we're going through for the majority of the part

[spare widget]

no, that seems like you simply solved a different problem

try solving it right now with the correct thing

[lusty pumice]

but then I will literally burn my quiz,tests, and exams because of little mistakes I make

and since it's online school, we don't get marks for doing the work

we only get marks for getting the answer

[spare widget]

welp

[polar vessel]

i know the definition of an induced matrix norm (kinda) but idk where to start for this proof

i know that both the row sum and column sum norms are alpha so could i use that fact in the proof?

[torn stag]

@polar vessel Note that $A_{\alpha} = \alpha I$. The definition of induced matrix norm is $$\lVert A \rVert = \sup_{|v| = 1}|Av|.$$ Can you finish now?

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

i know that as well but i cant see how to construct the proof

[torn stag]

Compute the supremum

[polar vessel]

huh?

im sorry im lost

so Av is a vector

would the supremum be the largest value of the vector?

[torn stag]

sup is equivalent to max here (since we are in finite dimensions, due to compactness of the unit sphere)

So $|A|$ is the maximum value of $|Av|$ over all unit vectors v

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

what does "over all unit vectors" mean? i think understanding that will clear things up

[torn stag]

It means $$|A| = \sup{|Av| : v \in \mathbb{R}^2 \text{ and } |v| = 1}.$$

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

oh ok

[torn stag]

Note that each occurence of $||$ has a different meaning here

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

yeah thats another confusing part

[torn stag]

One is the induced norm, one is the norm on the domain of A, and once is the norm on the codomain of A. To be explicit I should have subscripted these differently

[polar vessel]

isnt v the domain and Av the codomain?

[torn stag]

Both domain and codomain are $\mathbb{R}^2$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

But they can have different norms.

[polar vessel]

im confused is that relevant to the proof?

[torn stag]

It says "any induced matrix norm"

[polar vessel]

ok

[torn stag]

That means that you have to show it is true for any norm on the domain and any norm on the codomain

[polar vessel]

ok

[torn stag]

Here is the more explicit specification: $$|A|{\text{induced}} = \sup{|Av|{\text{codomain}} : v \in \mathbb{R}^2 \text{ and } |v|_{\text{domain}} = 1}.$$

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

so im not really understanding this

can you give me an outline of the proof you have in mind because i have no idea where u are going with htis

[torn stag]

If $|v| = 1$, then $|\alpha Iv| = |\alpha||v| = \alpha$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Thats it

[polar vessel]

ok so the A = I * alpha part i understand but where does the assumption |v| = 1 come from?

[hardy inlet]

anyone know what the teacher means by "find two vectors having lots of square roots"?

[viral magnet]

Here

[hardy inlet]

oh wait nobody answer this I just got an idea and wanna see if it works first

[torn stag]

@polar vessel You want to compute $$|A| = \sup{|Av| : v \in \mathbb{R}^2 \text{ and } |v| = 1}.$$

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

is the answer a scalar?

is it not alpha?

[torn stag]

@polar vessel The most direct way to compute a supremum or maximum like this is to compute each element in the set and then find the maximum.

Here it happens that the only element in the set is $\alpha$.

[polar vessel]

how can i compute each arbitrary vector v

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

im confused

i want to calculate each Av? how do i do that

[torn stag]

just do it

let v be an arbitrary vector with $|v| = 1$. And compute $|A_{\alpha}v|$.

[polar vessel]

how lol. i feel like im having a mental lapse here

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

I showed it eariler. You use the norm axioms and definition of $A_{\alpha}$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Actually

There was a flaw in my work

We need the norm on the domain and codomain to be the same

Check your definition of induced norm to be sure

[polar vessel]

if |v|=1 then |Av| = 1*|A|

there is no mention of domain nor codomain in our definition

[torn stag]

@polar vessel It's implicitly $\mathbb{R}^2$ for both, since a 2x2 matrix acts on vectors in $\mathbb{R}^2$ and produces vectors in $\mathbb{R}^2$.

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

so now we are down to |A| = |I*a|

So |I| = 1?

I is identity matrix

i mean it makes sense to me. every norm i can think of has this property

[torn stag]

Yes $|I| = 1$ as long as we use the same norms on domain and codomain

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

otherwise its not true

[polar vessel]

Does this check out?

[torn stag]

no

[polar vessel]

why

[torn stag]

equality 2

[polar vessel]

i figured

[torn stag]

It's correct

[polar vessel]

huh?

[torn stag]

since $A_{\alpha} = \alpha I$.

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

what

[torn stag]

$$A_{\alpha} = \alpha I \implies |A_{\alpha}| = |\alpha I| = |\alpha| | I |.$$

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

right i have thatr

[torn stag]

yeah now basically the question is whether $|I| = 1$

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

well you said so

by definition

[torn stag]

Check yout definition of induced norm

It will depend on that

[torn stag]

Because if the norms on the domain and codomain are different, then $|I|$ might not be 1. For example, we can let the codomain have $10$ times the norm of the domain. Then $| I | = 10$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Yeah in your picture both norms are subscripted by v, so I guess they are the same

Then it's easy to see from the definition that $|I| = 1$.

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

yes so is my logic in my proof correct then

[torn stag]

I guess every equation is correct

the way you wrote it is just a bit weird (for me)

[polar vessel]

well how would you write it bc i dont fully understand it

im like 80 percent there

[torn stag]

I'd do $|A_{\alpha}| = |\alpha I| = \alpha |I|$ first. Then I'd show that $|I| = 1$. Then I'd conclude.

[stoic python]

IlIIllIIIlllIIIIllll

[polar vessel]

how would i show | I | = 1?

that would be equivalent to the proof itself

[torn stag]

for that use the definition of induced norm

[polar vessel]

oh it would be like |Ax|/|x| = |x|/|x| = 1, since A = I

So this?

@torn stag if this logic is correct then i can easily finish the proof

[torn stag]

yeah thats correct

[polar vessel]

woohoo

feels good

ty

[hardy inlet]

shoot it says orthogonal vectors in the theorem [editing in progress]

fixed*, is this completely correct?

[polar vessel]

Is this proof correct?

[polar vessel]

how to solve this?

the 2 norm involves eigenvalues and i know that eigenvalues of orthogonal matrix are +-1

[polar vessel]

This is the proof I came up with

[teal grotto]

why is lambda = 1?

[zinc timber]

why not just use the definition of ||A||_2?

[polar vessel]

eigenvalues of Identity matrix are 1

[polar vessel]

explain what you mean

[zinc timber]

your proof is fine

[hardy inlet]

Whats the complement bar here for? Isn't this only over the reals? or are we allowing complex arguments?

it has to be real since its restricted to the domain [0,1] i think right?

unless it has complex coefficients?

[zinc timber]

polynomials with real coeffs

[0,1] is taken so that the integral exists

[hardy inlet]

Why wouldnt it exist?

Like i guess u mean if its all of R its unbounded

[zinc timber]

yea that's what I mean

[hardy inlet]

Ok thanks. Me sleep and finish tomorrow with fresh brain like the light blue name said last week

[wintry steppe]

guys why is the 1st one right?

[gray dust]

its clearly linear, check that its bijective

[wintry steppe]

In a 3x2 matrix, why do we say that the column space lies in R^3 space? Shouldn't it be in R^2?

[dusky epoch]

$\bmqty{&\ &\ &}$

[stoic python]

Ann

[dusky epoch]

this is what a 3x2 matrix looks like

[wintry steppe]

Yes.

[dusky epoch]

its columns are of size 3

[wintry steppe]

Correct.

So, does the dimension change when we look at the individual column and the matrix as a whole?

[dusky epoch]

??

Col(A) is a subspace of R^3...

the colspace of any 3 by m matrix is a subspace of R^3

the columns of A themselves lie in R^3

[wintry steppe]

The above matrix has a dimension of 2 as there are 2 variables, but looking at the column and treating it as an individual vector I find it to be in 3d space.

[dusky epoch]

matrices do not have dimension

[wintry steppe]

In terms of geometric represenation?

[dusky epoch]

i have a feeling you're overthinking/overcomplicating it

Col(A) is a subspace of R^3. dim(Col(A)) can be 2, 1 or 0 depending on what A is

if you consider the matrix as representing a linear map, it'll be from R^2 to R^3.

[wintry steppe]

Suppose we've these equations:
1x + 4y = b1
2x + 5y = b2
3x + 6y = b3
When we represent them in matrix notation, why do 1 and 4, or 2 and 5, or 3 and 6 go with a different variable(x and y) if they lie in the same dimension?
What I mean is, 1 2 3 form a column vector in 3d space, and each component belongs to a specific dimension. If both 1 and 4 belong to the same dimension, why do they pair up with x and y, which appears to be different?

[dusky epoch]

??????

[wintry steppe]

Expected.

Let me illustrate through a diagram in your DM.

[dusky epoch]

with all due respect, what the hell are you talking about

no

don't DM me

post the diagram here

[wintry steppe]

Alright.

[dusky epoch]

i think you're trying to overcomplicate how matrix multiplication works

[dusky epoch]

sure... you can view this equation as $xv_1 + yv_2 = b$

[stoic python]

Ann

[wintry steppe]

1 2 3 is a vector in 3d, am I right?

[dusky epoch]

if you meant [1; 2; 3] then yes this is a vector in R^3

[wintry steppe]

Good. So, 1 2 3 each of them lies in a different dimension.
Similarly, 4 5 6 lies in a different dimension.
But, the dimension of 1 and 4 are the same. If they are in the same dimension, then why are we pairing them with a different variable, as in x + 4y = b1, etc.

[dusky epoch]

...you're misusing the word "dimension" a lot here.

that's all i can say.

[wintry steppe]

By dimension I mean the usual dimension we talk about. 2d, 3d, etc.

[dusky epoch]

you clearly mean something very different than the meaning of dimension in the context of linear algebra.

dunno about anyone else but i'm done here.

[wintry steppe]

✌

[wintry steppe]

could u pl elaborate

[vocal prairie]

To check if a given map is an isomorphism here, it suffices to check that it's

1. linear, and
2. bijective.
   As Rokabe pointed out, it is easy to see that it is linear, so you should verify if it's bijective.

[teal grotto]

in the case you know a priori that the two finite vector spaces have the same dimension, it suffices to check if the map is injective or surjective, since a linear map between two vector spaces of the same dimension is bijective iff injective iff surjective due to rank nullity

[zinc timber]

"finite"

[wintry steppe]

I don't know how do I express this in matrix form

[zinc timber]

do you know about quadratic forms

[wintry steppe]

I don't know how to express the k'(x1x2+x2x3) term

[wintry steppe]

can I include that term in the matrix?

[zinc timber]

you can, as a hint try to compute $\m{x & y} \m{1 & 2 \ 3 & 4} \m{x\y}$

[wintry steppe]

I just found a lecture note on quadratic forms and I haven't learnt it

http://www.rmi.ge/~kade/LecturesT.Kadeishvili/MathEconomics/Term3/Week3QuadraticLEC.pdf

[zinc timber]

@wintry steppe were you able to solve?

[chrome radish]

im a bit confused on step 2.

how did they get 1/3?

[lucid glacier]

1-2/3=1/3

[chrome radish]

I know they get -2/3 by doing $-\frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}=-\frac{2}{3}$

[stoic python]

John doe

[lucid glacier]

they're subtracting the vector (0,1,1) from (2/3,2/3,2/3)

[chrome radish]

yeah

[lucid glacier]

What's the problem exactly?

[chrome radish]

nothing now I just realized I oversaw the vector

u_2

[lucid glacier]

Ah

Alright

[iron harbor]

why is it possible to move the 1/2 outside of the scalar product here?

[zinc timber]

because inner products are linear( bilinear)

[iron harbor]

so is it an application of this rule that I can't see?

I can see how this rule is applied as the second step of solving it

[chrome radish]

isn't it this rule? <u, kv> = k<u, v>

k is any scalar

[iron harbor]

^ that's what it seems like to me but I can't find that rule and I'm not sure how to derive it myself

it seems right

it's a generalization of the associative property of multiplication isn't it?

because a * b is the same as b * a

[chrome radish]

yeah

[iron harbor]

alright then it makes sense and all is well... until the next exercise that I don't understand comes along ;)

[exotic cloak]

How to prove this?

[lucid glacier]

just set k=0 and you recover the rule <a,hc>=h<a,c>. The way stated in the picture is just more succinct

[lucid glacier]

also this is commutativity, not associativity

[iron harbor]

that's a very helpful observation, thank you

[iron harbor]

totally right; learning this in multiple languages doesn't really make it easier to remember which word it is 😂

[lucid glacier]

that's fair

[amber osprey]

import numpy as np

#(a)

v = np.array([[1, -1, 1, -1], 
              [1, 1, 1, 1  ], 
              [2, 0, -2, 0 ]])


dot_1 = v[0,:].T @ v[0,:]

dot_2 = v[1,:].T @ v[1,:]

dot_3 = v[2,:].T @ v[2,:]

I am applying the grammatrix, G = V.T*V right

It is supposed to be give 0'

but it doesnt

why

[lavish jewel]

why would they yield 0?

[zinc timber]

why should it be zero

[lavish jewel]

are you saying the vectors are orthogonal to themselves?

[amber osprey]

bcs

it is a problem

where the answer should be zero

when I take the grammatrix

[lavish jewel]

it seems you misunderstood the problem

[amber osprey]

but when I apply the formula it doesnt

[lavish jewel]

it will never work here

the dot product of a vector with ITSELF is equal to its 2-norm squared

the 2-norm is positive definite

[amber osprey]

but if I say V[0].T @ V[1] it gives me 0

[lavish jewel]

ofc

you misunderstood the question, please reread it carefully

the only vector for which V^T V = 0 is the 0 vector

[amber osprey]

V.T @ V doesnt give the norm?

[lavish jewel]

it does

the norm squared

[amber osprey]

ok then tell me how do I apply the grammatrix

[lavish jewel]

wdym "apply the gram matrix"

the gramian of a matrix with orthogonal columns will be diagonal

well, VV^T

V^T V has no special structure here

the cols are not orthogonal, the rows are

i think you mixed up several things at the same time

[amber osprey]

Make the Grammartrix for 𝑣0, 𝑣1, 𝑣2 and confirm that this collection of vectors is orthogonal.

[lavish jewel]

mhm

[amber osprey]

yea

I dont get it

what am I supposed to do

[lavish jewel]

i can see that

what does it mean for 2 vectors to be orthogonal to each other?

[amber osprey]

this grammatrix is confusing me

the dot product is equal 0

but

[lavish jewel]

mhm

[amber osprey]

what about the grammatirx

://///

[lavish jewel]

the gramian is A^T A

[amber osprey]

yea

but that is the norm

so

[lavish jewel]

what?

[amber osprey]

didnt you just say thart

[lavish jewel]

it isn't JUST the norm

it's a matrix

and each entry in the matrix is a dot product

[amber osprey]

huuh

[lavish jewel]

so you can use it check whether the columns of a matrix are orthogonal

i'm sorry, i can't help you

[amber osprey]

why

[lavish jewel]

you need to go review matrix multiplication

[amber osprey]

I know matrix mul

wtf

[lavish jewel]

can you write the product A^T A in terms of dot products of the columns of A?

or let's take a step back

you know how $y \cdot x = \langle y, x \rangle = y^T x$?

[stoic python]

Edd

[lavish jewel]

this is all you need

express the matrix product A^TA in terms of dot products of pairs of columns of A

[amber osprey]

ooh

so v0.T v1

v0.T v2

?

[lavish jewel]

mhm

and then you can see

if all the $v_j^T \cdot v_i = 0$ when $i \neq j$, then the vectors are orthogonal

[stoic python]

Edd

[lavish jewel]

and the gramian V^T V, where V has the v_i as columns, contains all of these dot products

[hardy inlet]

Are e1 e2 and e3 the orthonormal basis in this note example? Are is v1=1, v2=x, v3=x^2?

And these be the results of a different domain with same basis?

[zinc timber]

these aren't Legendre polynomials

they are normalized legendre polys

[hardy inlet]

normalized as in ||v|| =1?

[zinc timber]

yes

also why is your e1 = 1/sqrt(2)?

[hardy inlet]

thats what the teacher said the solutions were in class i thought

And i did something quite wrong here

oh i forgot to sqrt the norm but i think even fixing that its weird

[zinc timber]

you are supposed to get these

nvm

here

[urban gale]

like the y axis has the equation x=0 right?
so I want such an equation which is satisfied by only the points which lie on the y-axis. examples of such points can be (0,-1), (0,-3), etc.

[hardy inlet]

i can vaguely read what that python code's doing, but its still a bit over my head

So like what are the steps to solving it as a person? Or is it identical to the code

[zinc timber]

you don't need to understand the code

just read the polys

[hardy inlet]

yeah so they're the same as the ones in the notes

just with different roots

well different way of writing them

same value

so, knowing the values for that one; how can I manually compute e1

[zinc timber]

knowing the value of what

[hardy inlet]

the 1/\sqrt{2}

so i know what the answer should be

oh wait the norm is different i think i just realized

its <v1, v1>, but not the ordinary squaring I did before

sec

and repeat this right

[zinc timber]

ye

[hardy inlet]

ok i butchered the second one mainly because idk how to do this apparently

I tried (x-1/2)^3/3

as its F(x)

$\frac{1}{3}\left(x-1/2\right)^3$

[stoic python]

MattDog_222

[hardy inlet]

but i guess i'm having a lapse in calculus because the power rule isn't working

[zinc timber]

it will work

[zinc timber]

make sure you change the limits accordingly

[hardy inlet]

wait why would i have to change the limits. Isn't it still -1 to 1

[zinc timber]

ya I though you are using subs

[hardy inlet]

ok so I was right on that part, I just forgot symbolab expands sometimes

but i still got the wrong answer

nevermind I dont know how to use calculators 🤦‍♂️

forgot to wrap my 1/3 in parenthesis

so it does infact give the coefficient

wait no idk where I remember seing 7/6

I must have done something wrong with the numerator but i cant find it

this is getting 1x somehow

nevermind I forgot to square 😐

its the little things

so back to the matter at hand; I need to apply that concept and work to this interval, correct?

and so if it asked for "The complete orthonormal set on [-1,1]" would this be the "set"? I thought it was a basis 😐

[reef roost]

In general it’s nice to know that if you integrate any odd function from -a to a, it’s 0 for any a.

Saves work

[hardy inlet]

ah yeah that property, thanks for the reminder

[reef roost]

True

[zinc timber]

[hard vault]

say that I have a linear map F given by this matrix. Is there a "quick" way of finding a ON basis for R^3 consisting of eigenvalues of F? I know that I can first find the eigenvalues of this matrix, then find the eigenvectors, then choose 3 different eigenvectors corresponding to different eigenvalues and then use the Gram-Schmidt but this is very painful

[zinc timber]

use WA

[hard vault]

what is that?

[zinc timber]

wolfram alpha

[hard vault]

lmao

[reef roost]

Since this is real symmetric, all the eigenvectors are orthogonal for distinct eigenvalues. So you only need to graham schmidt on eigenvectors that have the same eigenvalues

So after you diagonalize, most of the work should be done already

Cuz if the eigenvalues are distinct you just normalize

[hard vault]

oh right

so I just find eigenvectors and normalize them right?

[reef roost]

Yes

Assuming there are no repeats

I can’t tell you off the top of my head if there are repeats but it doesn’t look like it to me

[hard vault]

I got that the eigenvalues are 3, -3 and -3 again

[reef roost]

Well then you gotta graham Schmidt the -3 eigenvectors

[hard vault]

right right

okay wait

[zinc timber]

finding spectral decomposition for anything >=3 is a pain

and not recommended

just use computers, if you are allowed

[reef roost]

Spectral decomposition above 3 is only ok if it’s separable

Preferably in powers of 2

Beyond that, 100% use a computer lol

[hard vault]

so for -3 I got that the basis for the eigenspace is (1, 0, -2), (1, -2, 0). So I guess I choose these two elements and gram schmidt these?

and then for the third one I just normalize

[reef roost]

Yeah

[hard vault]

okay okay

[zinc timber]

I still won't do it

[hard vault]

thank you so much!

[zinc timber]

except for cI_n ones

[reef roost]

I only do it for quantum gates because you know the common ones so well you can just write down the answer ;)

[zinc timber]

idk quantum gates

[reef roost]

On this note, Tokidoki you may be interested in the spectral theorem

It’s pretty sweet

[hard vault]

oh right I forgot about it

[amber osprey]

(c) Confirm that 𝑣3 := 𝑥 - 𝑃𝑥 is orthogonal to 𝑣0, 𝑣1 and 𝑣2.

v = np.array([[1, -1, 1, -1], 
              [1, 1, 1, 1  ], 
              [2, 0, -2, 0 ]])

u = np.array([3, 2, 1, 0])


counter =0

v3_check = set()


for i in range(0,3):
    x_min_Px = u @ v[i, :] - u @ v[i, :]
    v3_check.add(x_min_Px)
    for item in v3_check:
        if item == 0:
            counter +=1
            if counter == 3:
                print("v3 is ortogonal to 𝑣0, 𝑣1 og 𝑣2.")

(d) Use 𝑣0, 𝑣1, 𝑣2, 𝑣3 to determine an orthonormal basis for ℝ4.

But v3 is just 00??

[dusky epoch]

mind showing the problem?

[amber osprey]

(c) Confirm that 𝑣3 := 𝑥 - 𝑃𝑥 is orthogonal to 𝑣0, 𝑣1 and 𝑣2.

(d) Use 𝑣0, 𝑣1, 𝑣2, 𝑣3 to determine an orthonormal basis for ℝ4.

[dusky epoch]

this is part c of the problem

can you show parts a and b

[amber osprey]

a and b dont matter

[dusky epoch]

i think they kind of do.

[amber osprey]

b) Calculate the projection 𝑃𝑥 of

[dusky epoch]

it's impossible to tell what x is, what P is, what the v_i are

can you please show that

and also show where you're being instructed to use python

[amber osprey]

(a) create the grammatrix for 𝑣0, 𝑣1, 𝑣2 and confirm that this collection of vectors is
orthogonal.

[dusky epoch]

also this line

    x_min_Px = u @ v[i, :] - u @ v[i, :]

feels kind of suspicious since the value here will always be 0

[amber osprey]

it is numerical alg

[dusky epoch]

please just post the entire problem at once

[amber osprey]

the problem is in danish

I have to translate it

[dusky epoch]

please just post the entire problem at once it's very very very annoying to see it delivered piecemeal

i don't care if it's in danish

post it anyway

if there's anything that needs translating i'll tell you

[amber osprey]

ait

[dusky epoch]

so you want to project x onto the subspace spanned by v_0, v_1, v_2. yes?

[amber osprey]

#(b)


u = np.array([3, 2, 1, 0])

pr_v0_u = v[0,:]*np.dot(u.T, v[0,:]) / np.dot(v[0,:], v[0,:]) # [0.5, -0.5,  0.5, -0.5]

pr_v1_u = v[1,:]*np.dot(u.T, v[1,:]) / np.dot(v[1,:] , v[1,:]) # [1.5, 1.5, 1.5, 1.5]

pr_v2_u = v[2,:]*np.dot(u.T, v[2,:]) / np.dot(v[2,:], v[2,:])   # [ 1.,  0., -1.,  0.]


print(pr_v0_u,"\n",pr_v1_u,"\n",pr_v2_u)

#We have now found the points that are closest to the vector u

already did that here

I am trying to solve d

but I dont know what v3 is

[lavish jewel]

that's precisely the question you have to answer

review what you did in b and c

[dusky epoch]

what's u?

did you rename that from x?

[amber osprey]

yea

u = x

[dusky epoch]

okay

so you projected x onto v0, v1 and v2 individually... but not onto their span

[amber osprey]

I just used the formula to get the closest point

thats what I thought the problem asked for

idk

[dusky epoch]

that's for projecting onto one vector

or equivalently, that's for projecting onto the line spanned by one vector

you're projecting onto a subspace of dim 3

that's different

[amber osprey]

what

dim3?

[dusky epoch]

a subspace of dimension 3...

[amber osprey]

ait

[dusky epoch]

however, as your v_i are orthogonal (and you are asked to verify that in part a), it happens that the projection onto this subspace is the sum of the projections onto each v_i

obv this only works because of this orthogonality

[amber osprey]

huh

you mean pr_v0_u + pr_v1_u + pr_v2_u

[dusky epoch]

yes

that's your Px

should be your Px

[amber osprey]

what formula is that

I cant find it in my notes

sadge

[dusky epoch]

is your teacher so strict that they forbid you from using anything that isn't a formula that they explicitly taught you

[amber osprey]

probably

last time I tried to solve something differently that wasnt taught he said it was wrong 😄

[dusky epoch]

did he elaborate on why it was wrong? or did he just dismiss it out of hand?

[amber osprey]

he said "not the desired linear system"

even tho it gave the correct answer

[dusky epoch]

i mean

without seeing that problem i can't say for sure

but it may well be that you got the correct answer by sheer coincidence

[amber osprey]

nah

I made a drawing and everything

explaining the steps

and reason

[dusky epoch]

really would like to see the problem and your solution exactly as-stated now

[amber osprey]

ait

give me a sec

then

okay maybe I didnt explain the steps 😄

(d) Set up a linear system of equations for a function
𝑓 (𝑥) =

𝑝1 (𝑥), for 5.0 ⩽ 𝑥 ⩽ 8.0,
𝑝2 (𝑥), for 8.0 ⩽ 𝑥 ⩽ 10.0,
where 𝑝1 (𝑥), 𝑝2 (𝑥) are quadratic polynomials such that
(i) 𝑝1 (𝑥) passes through the data points at time 5.0 and 8.0,
(ii) 𝑝2 (𝑥) passes through the data points at time 8.0 and 10.0, and
(iii) in time 8.0, 𝑝1 (𝑥) and 𝑝2 (𝑥) have the same slope.

it is the last part

where I have to set the slopes to be equal

[dusky epoch]

the $16a_1 + b_1 = 0$ and $16a_2 + b_2 = 0$?

[stoic python]

Ann

[dusky epoch]

these?

[amber osprey]

yea

[dusky epoch]

yeah no you should have had $16a_1 + b_1 = 16a_2 + b_2$

[stoic python]

Ann

[dusky epoch]

you can't just assume out of hand that f'(8) will be 0 specifically

[amber osprey]

but the values of the coefficients are correct

at least I think so

since I made a plot

[dusky epoch]

hold on

this is what i'm getting with your values for the coefficients

[amber osprey]

try b_1 17.7778

I might have made a typo

[dusky epoch]

okay so you typo'd

hm.

hold up

[hardy inlet]

@zinc timber sry to @ but can you run that script on [0,1] and see if this is what u get

[dusky epoch]

were you told specifically that p1'(8) and p2'(8) had to both be zero or were you only told that they must be the same?

[zinc timber]

you can have the code if you want

[amber osprey]

" 𝑝1 (𝑥) and 𝑝2 (𝑥) have the same slope."

[dusky epoch]

so you were not told that this slope had to be zero. yes?

[hardy inlet]

idk python [dont have pycharm setup or numpi]

[amber osprey]

if they are both zero they have the same slope tho

[dusky epoch]

so you were not told that this slope had to be zero. yes?

please do not avoid the question.

[amber osprey]

wht

[dusky epoch]

yeah and if i have a million dollars then i'm a millionaire.

[amber osprey]

what 😄

[dusky epoch]

you do realize that it is very much possible for the two quadratics' common slope to be something other than zero, right?

the slope could be 1

or 2

or 42069

it could be anything

you can't just decree that it HAS to be zero simply because you decided on it.

[amber osprey]

yea but it just says the same slope

[dusky epoch]

IT SAYS THE SAME SLOPE

IT DOESNT SAY THE SLOPE IS ZERO

IT DOESNT SAY THE SLOPE IS ZERO

i don't know how else i can say this!

there is nothing in the problem that indicates the slope has to be zero!

you are essentially saying "A = B therefore A = B = 0"

[amber osprey]

alright

[dusky epoch]

also tbh your plot is bordering on garbage with how you're plotting 65 below 50

[hardy inlet]

this is going to be the plane and the perpendicular plane right?

[silver heath]

No, a plane is 3 points

this is a line and a perpedicular line

use grahm-shmidt process?

[dusky epoch]

yes

dim(U) = 2

[hardy inlet]

A hyperplane*

Or maybe its a plane

[amber osprey]

okay so how do I answer d?

[hardy inlet]

Vocab irrelevant, both both dim 2

[dusky epoch]

once you actually calculate Px and then x-Px properly, and once you verify that x-Px is orthogonal to v0, v1 and v2, you will have that {v0, v1, v2, v3} is an orthogonal set and your job will be to turn it into an orthonormal set

and if you know what both of these are, it should be rather obvious how to do it - hopefully without requiring a formula to be handed down on a silver platter.

[amber osprey]

what was the formula for px being the sum?

[dusky epoch]

...i guess you could look up the formula for projection onto a subspace

or projection onto the span of several vectors

[hardy inlet]

there has to be a better way that gram shcmit right? D:

[iron harbor]

seems like I should be able to figure this out but I'm stumped: given this fact: vector a is orthogonal to vector b when <a,b> = 0

how can I find actual examples of vectors that are orthogonal when I know one of the vectors but not the other?
a = (3,4)
<a,b> = 0
b = (x,y)
3x+4y = 0
x = (-4/3)y
=> ....
what now? 🤔

there will be ... infinitely many solutions, but what about just one?

[amber osprey]

?

[dusky epoch]

good enough, i suppose.

[halcyon spindle]

It very simple to calculate with gram-Schmidt since dim(U) = 2.

[hardy inlet]

but I need 2 bases

i need U and U perp

can i treat them seperately?

like i get what u mean about dimU = 2, but I think i need to do all 4 on an ordered basis of R4 such that the first 2 vectors are 1 2 3 -4, and -5, 4, 3, 2

[hardy iron]

55

[hardy inlet]

i would love to not do that [4x gramschmid] tho if its possible

[hardy inlet]

what have u done? It says to create an augmented matrix, so have u done that yet?

[hardy iron]

I don't understand the problem

I just needs someone explain to me to solve the others

[hardy inlet]

u set up an augmented matrix and use gauss-jordan elimination (RREF or row reduce)

[hardy iron]

Row

[hardy inlet]

?

[hardy iron]

Is this an application

[hardy inlet]

no its a method for computing the variables

Row Reduced Echelon Form

[hardy iron]

How to use it

[hardy inlet]

u had to have covered it in class if they're asking u to do it

R1 ← R1 + R2 for example

[hardy iron]

Oh

[amber osprey]

v = np.array([[1, -1, 1, -1], 
              [1, 1, 1, 1  ], 
              [2, 0, -2, 0 ],
              [0,  1, 0, -1]])



v @ v.T

would this be the answer to d?

[dusky epoch]

no

[amber osprey]

this is my reason

but I guess they are not of length 1?

[dusky epoch]

.

refer to this again

refer to your knowledge of the difference between the words ORTHOGONAL and ORTHONORMAL

[amber osprey]

an orthonormal vector is just when it is orthogonal and length 1 right

So I guess I need 1 in the diag

[dusky epoch]

there's no such thing as "an orthonormal vector"

orthonormality is a property of sets of vectors, not of individual vectors

but yes, an orthonormal set is an orthogonal set with the extra condition that all vectors have length 1

[snow totem]

What literature should one read to get into quantum communication with basic undergrad knowledge?

[hardy inlet]

so i have this, but im getting (-77)²+56²+39²+38² = 12030 which isn't a nice squareroot

[hardy inlet]

nvm thats correct but he said we coulda used calculators 💀

[elfin hawk]

What does this actually mean?

[fringe fjord]

It looks like it's the definition of a relation on R^2.

Not a particularly meaningful relation, mind you.

The condition simplifies to (a,b) R (c,d) iff (a+7)b = (c+7)d.

It's not clear that it is useful for anything except "find out if this is an equivalence relation" exercises.

[elfin hawk]

I see are there any particular properties that you can spot?

[tough veldt]

Think of a few pairs of numbers that are related to each other...

Figure out if this is an equivalence relation

If it is, try to see what the equivalence classes are

[fringe zodiac]

Any ideas on where to start?

Huh?

[torn stag]

@fringe zodiac Use the definition of linearly independent. This result is obvious because $U$ is an isomorphism.

[stoic python]

IlIIllIIIlllIIIIllll

[fringe zodiac]

We haven't covered isomorphisms yet, would there be any way to solve it without? @torn stag

perhaps using basis?

[torn stag]

@fringe zodiac You don't need to use isomorphism. This problem is easily solved directly using the definition of linear independence.

I'm just saying that the result is obvious from an algebraic point of view.

[fringe zodiac]

Alright, sorry for the confusion

Let me take a crack at it

Im a bit confused, working backwards i figured why T would be independent given B1...Bk's independence

But how can you know that B=UA is linearly independent if U is a matrix

[dusk fulcrum]

Am I allowed to row exchange in a 2x2 hessian? I need my hessian to be positive definite, but it is:

12 0
0 12

Sorry for dumb question

[wintry steppe]

why is yhe 3rd one right?

[zinc timber]

why do you need to exchange rows? just check the values as it will be a lot easier

[zinc timber]

how are you concluding 3?

[dusk fulcrum]

Oops I meant

0 12
12 0

[hardy inlet]

Pretty sure that its a change of basis operation and should be $B = C^{-1}AC$

[stoic python]

MattDog_222

[hardy inlet]

I'd recomment watching this vid https://www.youtube.com/watch?v=P2LTAUO1TdA&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=13

and when he says we're going from "our language to jennifer's language" thats essentially saying ur going from $\beta_1 \to \beta_2$ or vice versa

[stoic python]

MattDog_222

[wintry steppe]

It is the answet tht is given

[dusky epoch]

maybe they want <angle brackets> and not (parentheses)?

...

[wintry steppe]

ill show a new pic

[dusky epoch]

now you are cropping off the error messages

[dusky epoch]

why <c> now?

the c should not be enclosed in angle brackets

[wintry steppe]

ok let me try without the c

but it says an error for the top part too though

nvm

I got it you were right

thank you man

[pseudo cairn]

is it okay to start this question by taking the cross product of u and v to get p?

i.e, p = u X v

i just don't really know where to start with this T.T

[lavish jewel]

yeah that'd be a start

well, the determinant of that, right?

[pseudo cairn]

we use different language I think, im guesing what you call the determinant is what we call the magnitude?

since I don't have any numbers to work with, i'm guessing I'd have to do it using this formula :

[teal grotto]

nah, should just be the magnitude of the cross product

[lavish jewel]

ah i was mixing up two things

there's a different way to get the area using determinants

yes, the norm of the cross product

(magnitude)

[teal grotto]

right

[pseudo cairn]

alright, i've gotta do this on mathematica

[lavish jewel]

make them equal?

[pseudo cairn]

nvm i might have misunderstood

i think i better just go calculate the cross product first

[lavish jewel]

but yes, the next step would e to find the projections on the respective planes

[pseudo cairn]

and then the result might help with the intuition

yeah i mean initially i was really worried by the projections part because i don't know how i'd calculate that without any numbers

but we're also told that a projection onto the xy plane gives the coordinates a,b,0

and so on

maybe that helps?

[lavish jewel]

i would invite you to think of the planes as being spanned by subsets of the canonical basis of R^3

what you said is basically the consequence of what i just said

if you take the vectors [1,0,0] and [0,1,0] and do an orthogonal projection onto them,

[pseudo cairn]

I know those are linear alg terms but I can't remember them being defined properly in any of the lectures / readings

[pseudo cairn]

what is a "subset of the canonical basis of R^3" , I don't even really know what "span" means at this stage. maybe I could look at some linear alg videos on youtube for this?

[lavish jewel]

this is also used in the definition of your problem

i think you'll have to go review the definitions before you start

[pseudo cairn]

He has used the terms when talking about examples but he hasn't really defined them explicitly

[lavish jewel]

span is the set of all linear combinations of a set of vectors

[pseudo cairn]

I vaguely know what they mean because of videos i watched in preparation for this unit.

[lavish jewel]

e.g. any point on the xy plane can be expressed as a linear combination of the vectors [1,0,0] and [0,1,0], so these two vectors span the xy plane

[pseudo cairn]

we haven't really covered linear combinations or even matrices yet

all we've learned about so far is vector projections, area of paralellogram, area of paralellipid, and some geometry stuff in r^3

i could just peice the information together myself

[lavish jewel]

i guess that's all you need, then

[pseudo cairn]

let me have a longer think about what u said

and then i'll get back to you

if ur around ofc

[lavish jewel]

if you just accept that the projections onto a plane mean setting one of the coordinates to 0

[pseudo cairn]

I

[pseudo cairn]

when you say orthogonal projection, are you talking about

this

this type of thing?

the way i constructed it there is probably wrong

[lavish jewel]

yes

really weird notation but

that's the idea

[pseudo cairn]

im struggling with how to find the area M,N and O for the LHS

I have the value of the projection

to get the area, i'm guessing i'd need to take a cross product

but I don't really have another vector to take the cross product with

like for eg, for N I have the point (a , 0 , b) as my projection

but i don't know what to cross it with

[amber osprey]

v3 = u - Pu

#u - Pu = u - a1k1 + a2k2 + a3*k3

#w2 = u - w1

#w2 orthogonal to w1
would v3 be orthogonal to Pu?

[amber osprey]

when they are orthogonal you can always make them orthonormal right

[wintry steppe]

with Gram-Schmidt yes

and well sometimes it's only about dividing the vector by its magnitude

if they are orthogonal already

[amber osprey]

how do I get around this?

[burnt timber]

Hey im unsure how to do this question, I expanded out the rhs using the vector triple product rule, then the left hand part is perfect so I want the right hand part to equal 0, but im unsure as to why it does

[zinc timber]

'a' not defined

to use 'a' as a symbol, you need to make it one

use sympy, or better yet sagemath

[zinc timber]

why do you want rhs to be zero

that's not zero unless it's a straight line, you are calculating curvature

[amber osprey]

ty

[zinc timber]

I literally quoted the error message python gave you

[burnt timber]

Sorry hopefully my working will help @zinc timber

On the first line, if the second part equals zero it gives the form I want

[zinc timber]

my bad, that's not the curvature

anyway

[burnt timber]

Yeah sorry my message was confusing, I feel like what I have done is wrong but does that part equal zero somehow?

I can only think it would if r''=0 but that's not stated in the q anywhere

Or is there some other way of doing it?

[wintry steppe]

R^3 represents 3d space so what does p_2(R) represents ? And what about other vector spaces do they have any physical representation like R^2,R^3?

[zinc timber]

T is the unit tangent vector right? It's not arc length parameterized so why dT/dt = r''/|r'|?

[zinc timber]

P_2(R) is polynomials of degree ≤2 with coeffs in lR

[wintry steppe]

Yeah but whats their physical representation

[zinc timber]

if you choose a basis, they do act like R³

[burnt timber]

I assumed that differentiating this with respect to t would give that, is that wrong?

[wintry steppe]

My que is how to imagine p_2(R)

[zinc timber]

no the denominator is also a function of t

[burnt timber]

Ah okay, even if it's a scalar?

Nvm I realize my mistake now, it's not a scalar

[burnt timber]

@zinc timber What would the left hand side equal? Im struggling to know what to rearrange to

[wintry steppe]

guys the standard basis vectors are always the eigenvectors of a matrix ?

[lavish jewel]

of the identity matrix, sure

or did you mean any matrix? in that case, no

[wintry steppe]

thank u @lavish jewel

[karmic siren]

Unsure if this is the right channel, if not I'm sorry. Does anyone have an idea what I can change or add to this formula to get my force values to approximately equal each other? Only D changes as the towing angle increases. I will send someone a tenner if you sort me out, been trying for too long xD

[spare widget]

try minimizing $|p_1(t)-p_2(s)|^2$?

[stoic python]

criver

[spare widget]

this is some polynomial that is quadratic in t, and quadratic in s, taking the derivative with respect to t and equating to 0 and taking the derivative with respect to s and equating to 0 should a system of two equations. There may be an easier way though, using the fact that the shortest distance is along a line segment orthogonal to both lines.

[dreamy depot]

Has anyone seen this ? https://files.eric.ed.gov/fulltext/EJ923724.pdf Seems like the method indeed holds up and can be used for finding inverses of matrices

[spare widget]

here's a simpler solution: https://math.stackexchange.com/questions/210848/finding-the-shortest-distance-between-two-lines

[spare widget]

isn't this just multiplying the rows by the denominator, you're paying for the fractions with large numbers

I also have a better suggestions than:

Freed from unnecessary computational demands, students were instead able to spend more time focusing on designing an appropriate system of equations for a given problem and interpreting the results of their calculations.

Let your students use a computer to compute inverses after they are done practicing tedious matrix inversion. And stop at 3x3 or 4x4 matrix inverses.

[dreamy depot]

Yeah but the large numbers can be dealt by multiplying the reciprocal there was an optional step mentioned which that. I took Lin Alg a while back and remember doing a bunch of Gauss Jordan eliminations and asking whether there was a more eefficent method that had more speedup. The fractions don't come till at the end at the very last step

[spare widget]

more eefficent method that had more speedup
there is - a computer

in fact in numerics one avoids computing the inverse where possible

[dreamy depot]

Oh in my class we did everything by hand

[spare widget]

if you have to solve Ax = b, you do not compute x = A^{-1}b, you instead use a solver than doesn't have to invert A

[dreamy depot]

ahh kk makes sense

[spare widget]

that's useful only up till the point that you learn what the idea is, beyond that it seems useless

you're not going to be inverting matrices by hand, even if you have something that you need the symbolic inverse of

you'd use a symbolic math package for that (to simplify determinants or whatever)

it would be stupid to make someone invert a 10x10 matrix for instance

[dreamy depot]

in my class we had to do like an 8x8 for our final

[spare widget]

unfortunate

[dreamy depot]

Plus i'll have to reviewing for the gre so i've been looking at what stuff was missed in the undergraduate syallbus

We never did Linear Algebra over the complex plane nor talking about Jordan Canoical form and a grad student brtualized me for it

Can't believe undergrad maff dosen't go through everything 😢

[spare widget]

some books do, some do not

tests are monkey business anyways, I've taken exams perfectly only to forget everything 1 hour after the exam

if you're just studying for a test motivation's typically pretty low, and rightly so

[dreamy depot]

I was my DRP meetup and the head over there mentioned that there are things you would not see in an undergraduate course even in proof based courses

Huh relevant thread https://old.reddit.com/r/math/comments/tc7mjl/is_there_any_real_point_to_forcing_students_to/

[spare widget]

anything that starts with forcing, usually ends with forgotten 1 hour after the exam

I still remember all of the stuff that I studied because I found it interesting for which I picked exercises that I chose, conversely I don't remember even 5% of whatever I had to study for some exam

[dreamy depot]

I actually forgot a huge chuck of linear algebra while doing this thing in Rep. Theory (Representation Theory) I had the intuition from doing a bunch of linear algebra problems semesters ago and I was able to recall like basic things. But man I feel like I need to review linear

[pseudo cairn]

does the "span" of a plane represent its vertical and horizontal "sides" ?

[nocturne jewel]

no

span is the set of linear combinations of a set of vectors

[pseudo cairn]

i'm just trying to attempt this question

and i'm confused

i know that for the area of P, we can take the cross product of V and U

and take the magnitude of that

but I'm confused about how to get the area of everything else

idk if the fact that i don't really understand what it means for the vector v and u to span the paralellogram is contributing to this

[lunar sorrel]

apparently the first one is right... 2nd one is wrong...

i simply did 5^2 + 1^2 = c^2 and got sqrt(26)

[dusky epoch]

so you found the length of v?

[lunar sorrel]

i found the length of u

[dusky epoch]

in case it was not clear, i was asking if this was meant to be finding the length of v.

[lunar sorrel]

yes

[dusky epoch]

well this is not at all what you were asked for

u and v are the SIDES of your parallelogram, not its DIAGONALS whose length you were asked to find

[lunar sorrel]

ack ty ty

[bold ermine]

can someone help?

#help-10

[pseudo cairn]

is the 4th face just the grey shaded region?

[dusky epoch]

it's the face that's obscured by the yellow, purple and blue faces

[pseudo cairn]

mhm so the base right?

[pseudo cairn]

like, if i saw the corner as a pyramid, the base would be the 4th face?

[dusky epoch]

if you wish?

it's a tetrahedron, so any face can be considered as the base

[pseudo cairn]

yep okay, so all i had for part A of this question was 2 vectors that spanned a paralellogram and we're supposed to use this idea to answer part b. Would the area of the coloured faces be given by the cross product of each of the two vectors onto the xy, xz and yz planes?

[dusky epoch]

...

[pseudo cairn]

sorry i had to edit

what i wrote before

[pseudo cairn]

I changed what i wrote to try and make it a little clearer what i mean hopefully

[dusky epoch]

your message is still incomprehensible unfortunately

[pseudo cairn]

do you have any suggestion on how I could start or a prompt maybe?

I might just rewrite the message again

[dusky epoch]

no please don't

[pseudo cairn]

this is the question we did before

[dusky epoch]

i don't want to have to read more nonsense

[pseudo cairn]

and we're meant to be able to use what we did there

[dusky epoch]

perhaps if you want to go with the cross product thing

[pseudo cairn]

to do part B

[dusky epoch]

you can say that the area of the yellow face is 1/2 ||u × v||

and likewise for the other two colored faces

[pseudo cairn]

mhm, i don't actually have the vectors though, can I just write u_1, u_2, u_3, v_1, v_2, v_3 a and so on?

[dusky epoch]

............

you want to drown yourself in notation?

you want to deliberately give yourself nine whole variables to deal with and keep track of?

[pseudo cairn]

he wants us to use mathematica, so it might not be that bad using mathematica?

he said he deliberately made it so that it would be impossible to do by hand but i'm not sure that applies here. what would you suggest as an alternative?

[dusky epoch]

??

....

becoming more confused by the minute over here ngl

[pseudo cairn]

not 9 variables

just

u= u sub1, u sub2, usub3

u get me?

[dusky epoch]

you're introducing notation for the coordinates of u

if those are not variables then idk what is

am i to understand that you are specifically BANNED AND DISALLOWED from using such basic properties as the cross product being linear in both arguments, or that ||a+b||^2 = ||a||^2 + ||b||^2 when the vectors a and b are orthogonal?

or are you just being masochistic right now in your desire to have more letters to juggle

or what

i literally cannot think of any other reason why you would want to torture yourself like this

[pseudo cairn]

so would u and v correspond to a and b in this case?

[dusky epoch]

no

nothing "corresponds to" anything

you're seriously overthinking things a ton right now

and it's distressing

[pseudo cairn]

ok i see what you're saying with this

[spare widget]

This is way simpler, take f = v-u and g = w-u, then the area of the hidden face is 1/2 | f x g |

Note that f = u-v, g = w-v or f=u-w, g = v-w would have also worked

The only thing you need to know is |f x g| = |f| * |g| * |sin(u,v)| which is the area of the parallelogram with edges f,g, and the triangle is half that.

You don't need any extensions of pythagoras' theorem

You can of course do it for the 4 faces, but that's just extra work for no reason

[short stirrup]

how would i represent my skateboard as a 3D matrix? Im trying to apply a rotation matrix to my skateboard to represent a trick, but im not sure how to make my skateboard into a matrix i can apply the rotation to

[spare widget]

You would discretise your skateboard and then apply the rotation to every point of it

e.g. make a triangular mesh, then transform each vertex

[short stirrup]

Could you walk me through this? I'm in high school and this is something I am doing outside of the syllabus so im not sure how to begin

[spare widget]

what are you using to model your skateboard

[bold ermine]

can someone help me with this

[zinc timber]

what have you tried so far?

[languid sphinx]

Do non-symmetric positive definite (in the real sense) matrices define inner product spaces?

Positive-definite in that all eigenvalues are of the same sign

I'm not handling complex numbers

[spare widget]

inner products are symmetric

[languid sphinx]

Oh

oh yeah

So they do not define an inner product, but they can define a norm?

And we say this norm is not induced (not possible to be) by any inner product?

[spare widget]

x^TAy = (x^TAy)^T =y^TA^Tx

and this should hold for any x and y

so A=A^T

[languid sphinx]

Yes for inner products I get yo

you*

[spare widget]

how do you define your norm?

[languid sphinx]

In the same sense as an 'inner product'

x^T S x

Where the 2nd x would be an alternative vector if we tried to define an inner product

[spare widget]

you mean sqrt(x^TSx)

[languid sphinx]

Wait does it matter

[spare widget]

yeah, I think it works there

[languid sphinx]

I've not thought through this myself

[spare widget]

it does because p(sx) = |s|p(x)

for a norm p

[languid sphinx]

In this case $S$ does not have to be symmetric right?

[stoic python]

pepper

[spare widget]

I don't think it does, unless this somehow breaks subadditivity

(x+y)^TS(x+y) = x^TSx + x^TSy + y^TSx + y^TSy

[languid sphinx]

Hmm the cross term is worrying

I'm not sure if it's possible to find some cross term that breaks subadditivity

[spare widget]

it should look like this:

sqrt((x+y)^TS(x+y)) <= sqrt(x^TSx) + sqrt(y^TSy)

in the usual setting you just get 2x^TSy

I am not sure things will break

you can use Cauchy Schwarz to deal with the mixed term afaik

[languid sphinx]

In the usual case yeah

[spare widget]

I think you can just apply Cauchy-Schwarz twice

[zinc copper]

So we started bilinear forms in lin alg and my teacher started straight away by using forms of the type <u,v> = u^T Av for A a matrix. My friends and I guessed it was just because all bilinear forms are of this type and thinking about it this morning I proved this is true. However this means there is a natural identification between bilinear forms and matrices with respect to a basis, and so I’m wondering how we can see this as an identification between bilinear forms and endomorphisms

[spare widget]

this solves it: https://math.stackexchange.com/questions/87199/maximizing-symmetric-matrices-v-s-non-symmetric-matrices

[zinc copper]

Oop didn’t see this was still active sorry

[spare widget]

don't mind me

[spare widget]

for finite dimensional spaces sure

[languid sphinx]

Oooo thanks so much @spare widget

I'll think about this a bit more

[zinc copper]

Yeah I meant for finite dim

[spare widget]

you can write all your inner products as x^TAy where A is symmetric

or for the complex case x^TAy^* where A is hermitian and y^* is the conjugate of y

[zinc copper]

I’m talking about bilinear forms not inner products, so no necessary symmetry in A

[spare widget]

ah, sorry

[zinc copper]

And I’m mainly concerned with understanding how we can identify forms with endomorphisms

[spare widget]

I was still thinking about pepper's question

about inner products

[zinc copper]

Since both have a natural identification with matrices wrt a choice of basis

I’m thinking about maybe seeing the bilinear form as linear for a fixed u and seeing that you get an action on v like the dual of an endomorphism evaluated at u^T or something

And that endomorphism is A or A^T, something around those lines

[spare widget]

isn't the standard trick to feed it basis vectors and derive things from there

[zinc copper]

No I mean I’m not trying to prove anything rn

[spare widget]

https://www.mathematik.hu-berlin.de/~wendl/pub/connections_appendixA.pdf

[zinc copper]

Just see if there’s a connection between bilinear forms and endomorphisms

[spare widget]

A.2 would probably be relevant for you

[zinc copper]

What makes you say that ?

[spare widget]

As I understood it you want to show that any bilinear form can be represented through a matrix

[zinc copper]

I’ve already proved this 😂