linear-algebra

[versed topaz]

yeah, I'm gonna think about this harder

[sleek helm]

lmk if u think of a counterexample

[violet sage]

Is the result true in char 2?

[versed topaz]

I don't know

[sleek helm]

its not true in general

[versed topaz]

But it feels true for F4

[sleek helm]

it might be true for larger fields

[mystic sentinel]

Oh so I do have to take it one step further to get to the contradiction

I guess because my hypothesis wasn't about 2u, it was about u

[sleek helm]

Well your instinct was correct

yeah exactly

[mystic sentinel]

So I was onto something ... I just was getting lost because I thought I had to work with three vectors

[sleek helm]

yeah tbh I just figured it out by trying to like think about it as a game

like you are trying to make such a counterexample

but math stops you

so like, first thing I thought about was why your 2-space proof didn't immediatley work for 3

so i was like where does v+w end up

and then I figured that 'felt' like a lot of leverage

[mystic sentinel]

I still struggle with that kind of intuition sometimes --- sometimes I get it but sometimes I really hit a brick wall

[vital tapir]

To show that a transformation T is a linear transformation, I only have to show that it satisfies the addition and scalar multiplication properties for a linear transformation right?

[violet sage]

It's a theorem that if U, V, W are subgroups of G whose union is G, then (among other things) G/(U n V n W) is isomorphic to Z2 x Z2. We let G = U + V + W in this case, and take U, V, W to be subspaces of some vector space L. We have dim (U + V + W)/(U n V n W) = 2. We still haven't invoked the assumption on the incomparability of the subspaces.

(L is over F2).

This should imply that we only need to check a two dimensional space over F2.

Actually no

[wet finch]

can we modify to work in any field except for F_2 by like

let a and b be nonzero elements of F whose sum is also nonzero

and consider (au + v) and (bu - v) or something

if u is in U but not V or W the same will be true of au and bu

then both of au + v and bu - v must be in W as they can't be in U or V

and then their sum (a+b)u is also in W, and hence u is as well since (a + b) \neq 0

@sleek helm @versed topaz @mystic sentinel

so I think it's not even that "2 = 0" in F_2 that's the problem

but rather that F_2 just doesn't have enough scalars, period

whereas F_4 does

[versed topaz]

ahh I think that makes sense!

This fits my intuition better

I was thinking of like

Writing down a polynomial or linear equation in the scalars

And finding that it has to be nonempty

But didn't get further

[wet finch]

yeah

it was just weird to me that this felt like it should be true in other fields of char 2 but not in F_2

so somehow the issue couldn't just be that 2 = 0

[versed topaz]

Yeah haha

I think it's like

An intuition about size

In fact, I wonder if we can get more precise about the size of a possible counterexample over F2

[wet finch]

well there is a counterexample in (F_2)^2

[versed topaz]

That's what I meant, sorry

[wet finch]

oh sorry I see

[versed topaz]

Like we used lines

Right?

[wet finch]

like, "over F_2 such a thing can only happen with lines"

yeah

[versed topaz]

Yeah exactly

So the full proof is

Take u in U, v in V, w in W where each element is outside the union of the other two subspaces. Choose nonzero a, b in the scalars such that a+b ≠ 0. Then u+av+bw is not in U cup V cup W?

[wet finch]

sorry what are you doing with u + av + bw?

[versed topaz]

I lost track of what the actual proof was, sorry

[wet finch]

so my new idea was look at the vectors au + v and bu - v. These are in U cup V cup W because that's assumed to be a subspace. however, they aren't in U or V, so they must be in W

but then their sum is (a+b)u is in W, and therefore u is in W as well since a+b \neq 0

contradiction

since u, v, w were assumed to be in only their respective subspaces

(which is also why you can't have au + v in U or V)

your typing and backspacing is making me nervous

[versed topaz]

Sorry haha

Just thinking and then answering my own questions

[wet finch]

so actually yeah about the sizes of things

let F = F_2 here

[versed topaz]

Right

[wet finch]

assume U cup V cup W is a subspace and assume none is completely contained in any other

and assume dim U > 1

oh wait i need one more assumption

hmm

so what I want to say is

let u1 and u2 be elements of U such that u1 + u2 \neq 0

and neither u1 nor u2 is in V or W

then repeat the proof with u1 + v and u2 - v

since U cup V cup W is a subspace, those are both in U cup V cup W, but they can't be in U or V

therefore they're in W, so their sum is in W

but their sum is some nonzero element of U

which I really want to not be in W hahahaha

[versed topaz]

Yeah, I see

Hmm

ah right so like

We take u1 = a u and u2 = b u in the previous case, right?

If I'm translating right

[wet finch]

yeah

[versed topaz]

I see

[wet finch]

and as long as u is not in V nor W, then no multiple of it will be

so that's why you're fine with like (a+b)u

[versed topaz]

Yup

[wet finch]

I want to sort of say like, "maybe it wasn't that our field didn't have enough scalars, but really that our subspaces just didnt have enough elements"

which was your idea too I think

[versed topaz]

Yeah, I see

Yup

I feel like there's a really cute proof hiding somewhere thst involves the homology of a complex haha

Hopefully I can think about this tomorrow

Going to log off and prep for my analysis final rn though

Thanks for thinking about this with me!

[wet finch]

actually wait no this is still false over F_2 I think, in (F_2)^3

you can have 3 planes

which are distinct but union to the whole thing

grrr

[versed topaz]

Oh hmm

[wet finch]

okay good luck!!!

:)

If I come up with anything I'll ping you sometime tomorrow

[versed topaz]

Please do!

[brazen sentinel]

som1 plss help i have a math test tmr and im struggling on these linear equation word problemss

[mystic sentinel]

Thanks for the help by the way. 🙂 That's definitely a technique I think I'll be able to use in the future.

[dusky epoch]

@hybrid elk do you still need help w/ this?

[hybrid elk]

yeah

[dusky epoch]

would it help if you were to write the actual system (equations and all) represented by this matrix?

no

unless you want to spend an infinite amount of time doing that

$\begin{cases} x + 3y - 4z = -7 \ y - 2z = 3 \ (a+3)z = b-5 \end{cases}$

[stoic python]

Ann

[dusky epoch]

here's your system

do you understand that the first two equations can be used to express both x and y in terms of z?

and since the first two equations don't involve a and b, any conclusions made from them are independent of the values of a and b?

so the equation that requires detailed analysis - the one everything hinges on - is the third equation

(a+3)z = (b-5)

notice that if a+3 ≠ 0 then this equation is guaranteed to have a unique solution

if a+3 ≠ 0 then dividing both sides by (a+3) gives you the solution

z = (b-5)/(a+3)

but that only works as long as you aren't dividing by zero

...

i meant like

shit like don't plug in a=4 and b=22 just because you have nothing better to do

if you didnt care about things being legal or not then you would just divide both sides of (a+3)z = (b-5) by (a+3) and get your solution

but obviously you need to think about what could stop you from doing that

and what could stop you is division by zero

find all pairs (a,b) for which the system has infinitely many solutions

[dusky epoch]

yeah?

what about this

Like if you were to say x is -5 and y is -3 it wouldn't be > 0
this just means (-5, -3) ∉ U

nothing more

the way to go about any question that goes "is this set a subspace of <vector space>"

is to check it against the definition of a subspace

i.e. to ask yourself three questions

• is the set nonempty?
• if you take two elements of the set and add them, is the sum still in the set?
• if you take an element of the set and multiply it by a number, is the product still in the set?

if the answers to all three questions are YES then your set is a subspace, if the answer to even one is NO then it is not a subspace

?

yes??? it's literally the definition of a subspace

phrased in a way that lends itself well(ish) to plug n chug

"larger than x" what

...

no?

the elements of your set are vectors in R^2

[lavish jewel]

wdym by that

[quartz compass]

what are you thinking so far

[lavish jewel]

are you familiar with the terms there though? like what they mean by "span all of R3"

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

do you know what a linear combination is?

that's right

hmm i would put plus signs instead of commas tho

[limber sierra]

theres a theorem in linear algebra:

all linearly independent spanning sets of a space are the same size

do you know what "linearly independent" means?

so an easy way to show a set of vectors is not spanning is to show that, if you "force" it to be linearly independent (say by row reducing + removing 0 rows), there are less vectors than the space's dimension

[lavish jewel]

i don't think they have taken linalg yet,

[limber sierra]

an alternative way is to find a vector that's impossible to express as a linear combination of the other vectors

ah

then the "alternative" way involves less machinery

though often more clever arguments to show that it's actually impossible

[kindred yacht]

how do you do this?

[lavish jewel]

you can say that a set of linearly independent vectors that span a subspace are a "basis" for that subspace

[limber sierra]

also i dont think this is quite right

or at least is phrased ambiguously

we say a set of vectors is linearly dependent if one of the vector is a linear combination of the other vector_s_

note the plural

[lavish jewel]

i'll leave it to you nami, i have no experience teaching at highschool level, cheers

[limber sierra]

so $\left{\begin{bmatrix}1\0\0\end{bmatrix}, \begin{bmatrix}0\1\0\end{bmatrix}, \begin{bmatrix}1\1\0\end{bmatrix}\right}$ is linearly dependent

[stoic python]

Namington

[limber sierra]

no, that would prove it does span R^3.

yes, a way to show it doesnt would be to show the set is linearly dependent

since we know the size of a linearly independent spanning set of R^3 is 3

(we call this the space's "dimension")

so if you have a set of 3 vectors and its NOT Linearly independent

then it cant possibly be spanning

since you can remove one of the vectors without affecting the span (since that vector is a linear combination of the others!), and then its not "large enough" to be spanning

i gave one above

its linearly dependent since we can write (1, 1, 0) = (1, 0, 0) + (0, 1, 0)

and hence it cant span R^3

although in this case its "obvious" it cant span R^3

since there's cleaerly no way to add those together to get the vector $\begin{bmatrix}0\0\1\end{bmatrix}

[stoic python]

Namington
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[limber sierra]

because of the last entry

[dusky epoch]

missing $ at the end

[limber sierra]

im aware

not worth fixing

a less trivial linearly dependent set might be:

[\left{
\begin{bmatrix}3\2\0\end{bmatrix}, \begin{bmatrix}1\2\5\end{bmatrix}, \begin{bmatrix}-1\2\10\end{bmatrix}
\right}]

[stoic python]

Namington

[limber sierra]

this is linearly dependent because:[
-1 \cdot \begin{bmatrix}3\2\0\end{bmatrix} + 2 \cdot \begin{bmatrix}1\2\5\end{bmatrix} = \begin{bmatrix}-3 + 2\ -2 + 4 \ 0 + 10\end{bmatrix} = \begin{bmatrix}-1\2\10\end{bmatrix}
]

[stoic python]

Namington

[limber sierra]

the prototypical linearly independent set for $\mathbb{R}^3$ is just[\left{\begin{bmatrix}1\0\0\end{bmatrix}, \begin{bmatrix}0\1\0\end{bmatrix}, \begin{bmatrix}0\0\1\end{bmatrix}\right}]

[stoic python]

Namington

[limber sierra]

of course this is quite uninteresting

you can certainly make more complicated examples:

[\left{\begin{bmatrix}1\\sqrt{2}\0\end{bmatrix}, \begin{bmatrix}0\15\-\frac{\pi}{2}\end{bmatrix}, \begin{bmatrix}3\-3\1\end{bmatrix}\right}]

[stoic python]

Namington

[limber sierra]

should be linearly independent unless i did my mental math wrong

go ahead.

[lavish jewel]

they got you buttered up

[limber sierra]

are you familiar with the criteria for when a linear system has one/zero/multiple solutions?

if i use the jargon "free variables" and "zero rows"

do you know what i mean

because we have the following result on real linear systems in row reduced form:

• A linear system has exactly one solution if it is consistent and has no free variables.
• A linear system has infinitely many solutions if it is consistent and has free variables.
• A linear system has zero solutions if it is inconsistent.

okay so clearly, no matter what a, b are, the first two rows of this system are unaffected

so we only care about the third row

[dusky epoch]

are you sure

[limber sierra]

wait

holy

brain fart

sorry im a dumbass LMAO

[dusky epoch]

you only get inconsistency when a+3 = 0 but b-5 isn't 0

[limber sierra]

yeah

my bad

apologies

i just had a stupid moment

LMAO

ann's right

[dusky epoch]

when you have 0z = (something other than 0) the equation is contradictory

when you have 0z = 0 the equation has every real number as solution

[limber sierra]

^

and in any other case you have exactly one solution

yes.

a slightly more detailed framing of ann's solution:

first let's suppose a + 3 = 0, so a = -3

then the last equation becomes 0z = b-5

obviously 0 * anything = 0

so if b - 5 is NOT equal to 0, then this equation (hence the system) has no solutions

hence we have no solutions in the case:
a = -3, b ≠ 5

[dusky epoch]

it's stronger than just "infinitely many" but the distinction doesn't come up in linear systems

[limber sierra]

meanwhile, we have infinitely many solutions in the case:
a = -3, b = 5

why is this? well, because then we have 0z = 0

and so ANY value of z we pick will work

okay, so that deals with the case where a + 3 = 0

now let's assume a + 3 is NOT zero (so a ≠ -3), and therefore we can divide by it

we can then rearrange the last equation to 1z = (b-5)/(a+3)

and so there's exactly one solution for z

which we can compute exactly by computing (b-5)/(a+3)

[limber sierra]

at least, not linear systems of spaces with infinitely many elements

if your vector space has finitely many elements, of course youll only have finitely many solutions

but that wont come up for ℝ^n

[dusky epoch]

yeah

[limber sierra]

anyway, to summarize:

infinitely many: a = -3, b = 5
none: a = -3, b ≠ 5
exactly one: a ≠ -3

does that make sense?

also btw i didnt really explain this at first but

from the beginning i was thinking "linear systems are easier to study if we're looking at simply z = [stuff] rather than [stuff] * z = [stuff], so i want to make the entry with a+3 into a 1"

"the easiest way to do this is to divide by a+3"

"but of course, that only makes sense if a+3 is not 0"

thats why i started off by assuming a+3 = 0

i wanted to deal with that case separately

and then i could deal with a+3 ≠ 0 later

if that helps explain why i took the approach i did.

[since it might seem that i just considered a+3 = 0 randomly; this was my reason.]

you might be tempted to consider an alternate approach, where instead of trying to divide by a+3, we instead subtracted a+3 by a+2

in order to make it 1

and indeed, that seems like it would work

if we subtract both sides of a+3 = b-5 by a+2, we get 1 = -a + b - 7

the problem, of coures, is that we're forgetting about the "z"

we dont ACTUALLY have a+3 = b-5

we have (a+3)z = (b-5)

so we cant actually get rid of the a+3 in such a simple way

im just explaining why i defaulted to division instead of subtraction

subtraction wouldnt work!

(this is why "add/subtract an entire row by some number" isnt a valid operation when row reducing/doing gaussian elimination, BTW)

(i.e. why we only have multiplying and dividing, or adding rows to each other)

anyway, i should get going but feel free to ask your question

someone else might be able to help

just a tip: whenever you look at systems of equations like this, a good strategy is to think "how can i get this into RREF?"

working with RREF matrices is much easier than REF

(im assuming you know what RREF means)

but you just have to be careful not to divide by 0 in this process

hence youll often have to deal with separate cases where things youre dividing by are 0s, if they involve variables.

thats basiclaly what i was doing above, i was just doing it informally.

a more algorithmic phrasing of what i did is just:

if we assume this is nonzero, we can put it into RREF with no free variables [besides the solution column], hence 1 solution; meanwhile if we assume it's 0 we either have a column without a pivot [hence free variable] or an inconsistent row [hence no solutions]

and then reverse-engineer what the values should be in each case

that might help give you a "Strategy" for more complicated forms of this problem

anyway, i really have to leave, good luck!

[humble oak]

how would i prove

if $A^{H}A$ is hermitian then $AA^{H}$ is hermitian?

[stoic python]

Crown

[humble oak]

that timing

[lavish jewel]

take the hermitian transpose 😛

[humble oak]

of A^HA or AA^H?

i don't really know how these two parts are linked

why is it necessary for A^HA to be hermitian in order for AA^H to be hermitian

[lavish jewel]

i don't actually think they are

[humble oak]

huh

interesting

[lavish jewel]

you can show it separately for both cses using only properties of the hermitian transpose, afaik

[humble oak]

this is what i got by hermitian transposing the AA^H

[lavish jewel]

you have to swap the places before you transpose

[humble oak]

swap the places of the A's?

[lavish jewel]

yes

(AB)^T = B^T A^T

[humble oak]

ah i see

and since $(AA^H)^H$ = AA^H$, $AA^H$ must be hermitian?

[stoic python]

Crown
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lavish jewel]

indeed

[humble oak]

thank you very mucho

[acoustic path]

How are these two spaces isomorphic when the first one has length 2 and the other has length 5.

Isn't a requirement for an isomorphism that both spaces have the same dimension

[lavish jewel]

what do you mean by length

the only thing this is saying is that cartesian products of R2 and R3 are kinda like R5, in very poor and unprecise words

or that a list containing a list of 2 numbers and a list of 3 numbers behaves like a list of 5 numbers

[acoustic path]

R^2xR^3' elements are (x1,x2) and (x3,x4,x5), that's only dimension 2

R^5 has x1, x2, x3, x4 and x5

[acoustic path]

He claims it is an isomorphism tho

[lavish jewel]

they said nothing about dimension though

the isomorphism they're mentioning is really just "take the 5 elements and put them in a list"

[limber sierra]

@acoustic path i think the thing youre missing is that

this isnt an isomorphism between R^2 and R^5

or R^3 and R^5

this is an isomorphism between R² × R³ and R⁵

R² × R³ denotes the product of vector spaces

which is the vector space made up of elements from R² "fused with" elements of R³

(inheriting coordinate-wise addition and multiplication from their respective spaces)

in the finite-dimensional case of spaces over the same field, the dimension of the product is the sum of the dimensions

so dim(R² × R³) = 2+3 = 5 = dim(R⁵)

as expected.

[acoustic path]

Ah

[lavish jewel]

there is more than one problem though. even if it was a map from R2 to R5, you can make an isomorphism by just embedding the lower dim space into the higher dim space

so that argument was also false

[limber sierra]

...huh?

thats... not true

dimension is an isomorphism invariant

[lavish jewel]

i used dimension wrong, my bad

[limber sierra]

there is no isomorphism between R² and R⁵.

(as vector spaces)

[lavish jewel]

not as vector spaces, no

[limber sierra]

there is a bijection in that both have cardinality beth_0.

[lavish jewel]

that's the one i meant

i ignored the very last line of the image they shared cuz i'm dumb

[limber sierra]

so you assumed "isomorphism" meant in Set in #linear-algebra

??

[lavish jewel]

apparently

[acoustic path]

thx Edd, Nami

[lavish jewel]

i'm gonna go return my degree lol

[humble oak]

if W is a mxn complex matrix where $W^HW = I_n$ is there anything i can determine about what $WW^H$ is equal to?

[stoic python]

Crown

[dusky epoch]

W is unitary

so WW^H = I_n too

[humble oak]

thanks 😄

[violet sage]

They are isomorphic as Q-vector spaces

[dusky epoch]

Q-vector spaces are basically just divisible abelian groups

[lavish jewel]

i just reread this

[lavish jewel]

this isn't necessarily true, is it?

the matrix isn't square

[dusky epoch]

oh

[lavish jewel]

WW^H is an ortho projection matrix

[dusky epoch]

fuck

[lavish jewel]

not necessarily I

[dusky epoch]

youre right

blurgh

[lavish jewel]

summon the person

[humble oak]

hi

[lavish jewel]

heyo

WW^H is hermitian, not necessarily I

[humble oak]

uh oh

[lavish jewel]

if the columns are orthonormal, it is also an orthogonal projection matrix

[humble oak]

how do i know if the columns are orthonormal?

nvm

but i don't know anything about W's columns

[lavish jewel]

then it's just hermitian

ah, you also know that n >= m

[humble oak]

hmm, by the end product i'm supposed to show (WW^H)^2 = WW^H

not too sure how i can put this all together

[lavish jewel]

is all of this dealing with the same W? all the questions from before

because none of this implies that on its own

oh

yeah, no

[humble oak]

ya mb misstype

[lavish jewel]

wait

no yea its ok

WW^H WW^H

the product in the middle is W^H W = I

[humble oak]

wait a second

[lavish jewel]

which is what they told you

[humble oak]

i can do that

??

that's crazy

[lavish jewel]

wdym

[humble oak]

so W(W^H W)W^H?

[lavish jewel]

associating the terms?

[humble oak]

ya

[lavish jewel]

sure thing

you missed an ^H at the end

[humble oak]

ah okay i believe i got it now

thank you very much

[lavish jewel]

i guess it was a projection matrix after all

nice

[ornate loom]

Not sure if I'm using the right channel, but I was reading a proof that proves the existence of an evolutionary stable strategy for all non trivial 2x2 matrices

and I was just wondering how they worked out that we need this value for p

[marble lance]

What's an ESS?

[ornate loom]

Evolutionary stable strategy- I will get the mathematical definition, one moment

[marble lance]

Oh

[ornate loom]

p and q are probability vectors

[marble lance]

Yeah, idk this stuff but can't you just put it into those inequalities to verify with an arbitrary q

[ornate loom]

i tried i think, would you have to take E(p, 1-p)

where p'=(p1,p2)

and q=(1-p1,1-p2)^T

[marble lance]

Huh? p is the vector (p, 1-p) and q seems to be any vector

[ornate loom]

Ahhh I see, thats probably why I was getting some weird stuff then hahaha

I will try that

and then i suppose solving for p you would get the value of p required

[marble lance]

Oh, I just mean you should verify that is an ESS first

I'm not sure how they got that it is one

[ornate loom]

yeahh i don't really understand this proof

[marble lance]

Repost question so someone else who knows this can help

[ornate loom]

thank you

I was just wondering how they worked out that we need this value for p

[wheat prairie]

does anyone know why does the A here equals to this summation?

[lavish jewel]

the one at the bottom?

since S is diagonal, the product USV is equal to the sum of weighted outer products of the columns of U and V^H

it becomes more evident if you notice that a product AB can be written not only as inner products, but also as outer products

then just shove an I matrix between A and B

[nocturne jewel]

Kinda stuck on how to show ii, mainly cause not sure how to prove something doesn't hold

[sonic osprey]

Just give an example

where it doesn't hold

[nocturne jewel]

ok so I do that for all n greater than equal to 2

[sonic osprey]

yes

[nocturne jewel]

there's kinda infinitely many n

[sonic osprey]

okay?

[nocturne jewel]

might take me a while to write out all n examples

[sonic osprey]

even doing part (i) needs you to show things for infinitely many vectors

there are infinitely many complex numbers, so how did you do part (i)

[nocturne jewel]

yes but you're saying give an example, not prove it

[sonic osprey]

What's the difference? You're proving that it doesn't hold.

[nocturne jewel]

I was trying to prove it doesn't hold given the properties

[dusky epoch]

disproof by counterexample is very common in math

[nocturne jewel]

$||v+w||_1^2 \leq ||v||_1^2+2||v||_1||w||_1+||w||_1^2$

[stoic python]

moshill1

[sonic osprey]

I don't think this will work, you have to use something specific about the norm since there are other norms that do satisfy the parallelogram identity

So you can't only use the properties

[nocturne jewel]

yeah ik it doesn't work, cause if I assume the identity holds for contradiction, I get a true ineq

[sonic osprey]

So just give a counterexample for all n

it's not hard to give an example for all infinitely many n

for example maybe you could use (1, 1, ...., 1)^T where there are n ones

[ornate loom]

Gonna ask again since I'm not sure if people saw this as someone else put a question immediately after:

I was just wondering how they worked out that we need this value for p

[pseudo thicket]

can someone explain why yhat or projc w y is just y?

is that because y is in the space spanned by u?

i'm quite confused

[lavish jewel]

that would be it, yes

if y is already in W

projecting it onto W shouldn't change anything

[pseudo thicket]

makes sense, thank you

however i am trying this formula

i converted each u to unit vector and applied the formula to get the weights

and also this:

however i got different answers for the weights

am i not supposed to get the same weights?

the former formula requires it to have unit length 1 hence I converted it to unit vector..however i'm getting nowhere similar answer when applying formula 2 which seems to be correct

[lavish jewel]

the two are equivalent

those 2 images show the same thing

y dot u/ (u dot u) is a scalar, so you can move it to the right of the vector u

then rewrite y dot u as u^T y

that leaves you with u u^T y / (u dot u), which normalizes u u^T

[pseudo thicket]

i'm getting a weird value though by using the first formula

[pseudo thicket]

unit vector of u2:

[pseudo thicket]

then i dot product of that vector with y, as in this formula to get the weight

[pseudo thicket]

however that weight is not the same as using this formula

[lavish jewel]

lemme matlab it up, i'm too lazy to do it on paper

[pseudo thicket]

so sorry to trouble you, i've recalculated it 7 times now i've got totally different weights c

[lavish jewel]

gimme a few min to get to it

[pseudo thicket]

sure, thank you

[lavish jewel]

see

same thang

method 1 is your "theorem 10"

method 2 is the ortho decomp. theorem

the weights should be these

[pseudo thicket]

oh wait.... the weights should be different for these two formulas

right?

since one uses the unit vector, the other is not

hence the weights differ

[lavish jewel]

they should be identical

in the ortho decomp, you divide by the norm squared of u_i

that normalizes the vector

those are the weights

you can rewrite the ortho decomp. theorem so that the weight is the scalar projection of y on a unit vector in the direction of u_i

[pseudo thicket]

as compared to using the non-unit vector the weights differ,no?

[lavish jewel]

and then that weight multiplies another unit vector in the direction of u_i

well but using a non-unit vector would be wrong

both methods are using unit vectors

as i just wrote

[pseudo thicket]

this methods gives me 4/3 for C1 and -1/3 for C2

while the other method gave me and 10.xx for C1 and -0.81xxx for C2

and i still got the yhat of -9 1 6 respectively

[lavish jewel]

don't forget the norm squared in the denominator

you can group that cleverly to show that it is equivalent to having used unit vectors u_i

the two methods say exactly the same thing

just with different parentheses to associate different terms

[pseudo thicket]

ah

sorry i was referring to this

this is the weight right? for u1

assuming U1 is not unit vector

however when U1 is unit vector, the weight is different

that is my confusion

and it should be different

[lavish jewel]

aha, that is not a unit vector u_1

because you divided by the norm squared

if you realize that u1 dot u1 = norm(u1)^2, you can divide each of the u1s on the numerator by one norm(u1)

it's equivalent because this is a unit u1 and you divide by an extra norm(u1), but the other u1 is not normalized

i.e., again, it is equivalent

just throw some parentheses around

it is the same equation

[pseudo thicket]

ah

thank you so much for taking for the detailed explanations, really appreciate it man @lavish jewel

[lavish jewel]

no prob

[pseudo thicket]

i got it now, thank you!

[ornate loom]

does anyone know at least where i may be able to find help with this/ good places to look etc?

[lavish jewel]

is that game theory?

i have no idea

[wintry steppe]

Hello, I have a question for myself. Given a matrix A = vv^t where v is an nx1 column vector
I have shown A is idempotent, symmetric and of rank one - it represents the orthogonal projection of R^3 onto Lin{v}
Given B = I - 2A, I have shown B is symmetric, orthogonal and B^2=I
How can I get the determinant of B? and eventually figure out the transformation it represents

[ornate loom]

yes it is 🙂

[lavish jewel]

B has determinant 0

[wintry steppe]

Can't be, its orthogonal

[lavish jewel]

ah 2A, sorry, thought it was just A

[wintry steppe]

So it's obviously 1 or -1

[lavish jewel]

you could get the determinant from its diagonalization

shouldn't be that difficult to find out what the singular values are

[wintry steppe]

But I wanna know if its orientation preservering / reversing

[broken oxide]

Would someone be able to help me with this?

I manipulated the A^2 = -I to A+A^-1 = 0

But idk how to continue

[lavish jewel]

A has as singular values 1, and n-1 0s, yes?

2A will have sing val 2, and n-1 0s

you should be able to do something with that. i get the impression it will reverse the orientation, but i can't check rn

(assuming v is unit norm, anyway)

otherwise, a simultaneous diagonalization/generalized eigenvalue approach might help

[quartz compass]

you could write A=[a,b;c,d] and square it then solve for the system of equations

[wintry steppe]

Nvm, I got this solved

I have another question I need a hand with

how can I find such a matrix?

[wintry steppe]

Please help

Given h(x)=2x−3,find, h(6)

#prealg-and-algebra

[rose umbra]

it T linear map?

[nocturne jewel]

are you asked to see if T is linear in the question?

[rose umbra]

@nocturne jewel it asking if it is a linear mapping

if its linear or not

[nocturne jewel]

ok so you need to check if 0 maps to 0 and if a linear combination maps to the linear combination of transformations

namely does:
T(0)=0
T(cu+dv)=cT(u)+dT(v)

[rose umbra]

well f(0) + g(0) = (f+g)(0)

[nocturne jewel]

$T[cf+dg]=(cf+dg)(0)=cf(0)+dg(0)=cT[f]+dT[g]$

[stoic python]

moshill1

[rose umbra]

@nocturne jewel what is the c and d?

[nocturne jewel]

scalars in the field

[rose umbra]

is it necessary?

[nocturne jewel]

Yes, cause you want to show any linear combination of vectors

Or you test scaling and addition seperately

[rose umbra]

yes i do it seperately

like T(kf) = kT(f)

[nocturne jewel]

yeah that works too

[rose umbra]

is this true?

[nocturne jewel]

yes

[rose umbra]

like how can i proof it?

or its known sentence like f+g(0) = f(0) + g(0)

[nocturne jewel]

T(kf)=(kf)(0)=kf(0)=kT(f)

[rose umbra]

gottcha thanks

@nocturne jewel btw does the [-pi,pi] change anything?

[nocturne jewel]

Well if the interval was [1,2] then T wouldn't make sense

[rose umbra]

why [1,2] ?

[nocturne jewel]

cause 0 isnt on [1,2]

so it doesnt make sense to ask about f(0)

[rose umbra]

ohh yea oops

right ty

[wintry steppe]

I’m listening to a lecture on youtube where the lecturer states that the vector u - v points from the head of v to the head of u.
I’m a bit confused if he means that the tail of vector u - v actually starts at v, or that it’s just the same length. If the former, how can you know when a vector starts out at the span or somewhere else?

[wary lily]

you know that the position of vectors is irrelevant, yes?

only their direction and magnitude matters

[wintry steppe]

I’m not sure if the position of vectors is irrelevant, that’s what I’m wondering

[wary lily]

it is irrelevant

to better visualize this problem maybe it helps to look at it this way:

[wintry steppe]

He says it starts at one head and goes to the other, and I’m here wondering if it matters that it does so, or if it just matters that it’s length is equivalent

[wary lily]

you can describe it this way or another way

you can start from the origin and go to the head of u

then from there draw -v (v in the opposite direction) and travel along it

the vector that you create from the origin to the head of -v is the same

now if you want, you can move -v so that it looks like the teacher is saying in the video

[wintry steppe]

Right, that makes sense. Thank you.

[wary lily]

np

[reef prism]

why does multiplying a vector by a matrix over and over again converge to the eigenvector span

[lavish jewel]

it might become easy to see if you consider the diagonalization of the matrix

say, A = QDQ^-1

multiplying Q^-1 by a vector v will do a change of basis that tells you how much of each of the vectors in Q was contained in v (you can treat the elements of the resulting vector w as dot products to see why this is)

then these coordinates are multiplied by D, which will scale them by the eigenvalues of the matrix

the result is a weighted set of coordinates that will tell you how to combine the vectors in Q to get the result of Av

already, this means that the result has to be in the span of the eigenvectors of A

but then, if the eigenvalues in D are not all equal, repeating this procedure will make the coordinate of v corresponding to the largest eigenvalue of A grow faster than the others as you multiply by A over and over

until the largest eigenvalue dominates, and you get a result almost parallel to the corresponding eigenvector

(parallel in the limit)

[reef prism]

i understand it kinda

but not really

when i multiply v in the basis of Q^-1 a bunch of times with the eigenvalues it must converge to something but its not intuitive for me rn

[wintry steppe]

how can i show that $f'(\lambda) \geq 0$ for $\lambda \geq 0$ where $f(\lambda) = X(\lambda I + X^TX)^{-1}X^T y)$, $X \in R^{d\times n}$, $y \in R^n$?

[stoic python]

tramell

[wintry steppe]

im not sure how to work with all these matrices and inverses

[fervent elm]

if a set is linearly independent, does it mean that a set of vectors which consist of additions of the vectors from the original set is also linearly independent

[fleet sun]

not necessarily.. in kind of a dumb way

[limber sierra]

no

suppose {u, v} is lin indp

then {u, v, u + v} is not

[fleet sun]

yeah

[limber sierra]

since u + v = u + v

[fleet sun]

or even {u, 2u}

[limber sierra]

...that came out sounding dumb

but

yeah

if you mean "excluding the original vectors", also not necessarily

consider {u_1, u_2, u_3, u_4} as a subset of R^5 and count the number of possible sums

youll note it certainly exceeds 5

the dimension of the space

hence cant be linearly independent

[fervent elm]

hmmm

ok thanks, i get it now

[marble lance]

Take that long equation and group the coefficients for each vi so you can use the fact that the original set is linearly independent

[fervent elm]

if there are 3 distinct basis for R3, does it mean all 3 basis are linearly independent from each other since they are distinct

[lavish jewel]

what do you mean by basis here?

did you mean basis vectors, perhaps?

[marble lance]

And if you mean if you put them all together in one set, that set is linearly independent, then the answer is no.

Adding any fourth vector to the basis, nevermind 6 more vectors, would make it linearly dependent

[fervent elm]

@lavish jewel yeah i meant basis vectors

[brazen venture]

how do i do part 2 for this qn

[lavish jewel]

i have to make sure. you mean 3 vectors in a set, or putting together 9 vectors?

[fervent elm]

so if they are distinct,does that mean they are linearly independent

[brazen venture]

im stuck

[lavish jewel]

in one basis, the vectors are linearly independent

[fervent elm]

oh wait

is bases

[lavish jewel]

if you look at 2 bases for the same subspace, those are linearly dependent

[brazen venture]

yes

i reduce it to rref and i got this

[fervent elm]

so assuming the bases only have 1 vector each?

do they all have to be linearly dependent

[lavish jewel]

a basis for R3 contains 3 linearly independent vectors

[marble lance]

If a basis has one element, then all vectors are multiples of each other and any set with more than one vector is linearly dependent

[lavish jewel]

to make it more clear, say the basis is Sv = {v1, v2, v3}, with basis vectors v_i

if you have another basis Sw = {w1, w2, w3} for the same subspace, then w1,w2,w3 are linearly dependent on v1,v,2,v3

[brazen venture]

so whats the ans

edd

[lavish jewel]

the answer to what

i haven't read your problem

was talking to razerac

[brazen venture]

sry i tot u were solving mine

i disrupted my bad

its ok i solved it alr

[fervent elm]

@lavish jewel so does that mean for the different bases, the respective vectors in them are linearly dependent on the same fixed vectors in other bases

[lavish jewel]

just think of it this way

the basis vectors are in the subspace they span

so any other basis is made up of vectors also in the span, so a set of basis vectors is spanned by another set of basis vectors for the same subspace

[fervent elm]

ohh

okay i kinda understand

also sorry if im disturbing you but if i have two subspaces U and V where U is a subspace of R5 and V has a dimension of 2, will the dimension of the dot product of the two subspaces be 2 as well cause its the lower number

[lavish jewel]

what is the dimension of U?

wait, dot product?

[fervent elm]

yeah

dot product of them is 0

[lavish jewel]

you mean dot product of vectors in the subspaces

[fervent elm]

err no

i think i misread

i mean U is a subspace and V is a set

[lavish jewel]

can you post the exact question?

that's a lot clearer haha

[fervent elm]

im thinking cause the dimension of S is 2 so since the dot product is 0 , dimension of V should be 2 as well?

[lavish jewel]

no

say you make a matrix S = [u1^T: u2^T] (2 rows, 5 columns)

what you want is Sv = 0

[fervent elm]

yeah

[lavish jewel]

what's the dimension of the null space of S?

[fervent elm]

3?

[lavish jewel]

yep

[fervent elm]

cause 5-2 is 3

[lavish jewel]

that's the dim of V

[fervent elm]

wait why?

is it cause V is a subspace of R5

[lavish jewel]

yes

and so is S

and S is orthogonal to V

[fervent elm]

oh so thats why V has dim 3

so the basis has 3 vectors

thank you for the help lol, bases is so confusing for me

[wary lily]

I found the general solution for this.

To find positive integers, I can try and test, but is there a methodical way to do this?

[lavish jewel]

find the kernel/null space of the corresponding matrix

[dusky epoch]

presumably az already found the general sol without the positive integer constraint

im not sure about anything methodical but you might notice that x must be congruent to 4 mod 5 since x + 5(y+2z) = 44

if you wanna do some research on equations like these, they're called diophantine equations

[wary lily]

OK, thank you. I'm going to read up on that.

Also, Edd, thanks. I haven't learned kernel/null space yet. But good to know for the future.

[dusky epoch]

it turns out you can find the value of x from here

||x must be 4, since it's congruent to 4 mod 5 and the next lowest value (9) guarantees that the first equation will be violated||

[lavish jewel]

oh that's really nice

that really made short work of the problem haha

[tulip glacier]

@lavish jewel for any subspace the basis is not unique, but could i use the null space of one basis to find null space of the other basis?

[lavish jewel]

should be the same null space

yea

[tulip glacier]

wat do u mean by same

like is there any particular relationship between the null spaces

[lavish jewel]

it is the same null space

[tulip glacier]

oh wait

so the null space is the same one

even if i use other basis?

[lavish jewel]

yes

[tulip glacier]

oh thks but any particular explanation

like if the basis is not unique shouldnt the null space not be unique

the basis for any subspace isnt unique, wouldnt that suggest that we could get variety of null space too

[lavish jewel]

sure you can get several basis for the null space

it is the same space though

[tame mural]

If the nullspace is the set of inputs which get mapped to the zero vector, then no matter which basis you choose for the space of inputs...

[lavish jewel]

you already started from the premise that you have a subspace with several bases

the same is true for its orthogonal complement

(the null space)

you could use exactly the same basis if you wanted

[tulip glacier]

um any possible videos online where they could visualize it

[lavish jewel]

use a 3x3 matrix if you want

you can't really "visualize" it in more than 2 or 3d

[tulip glacier]

how would starting from the premise that i have a subspace with several bases affect this?

[marble lance]

What exactly do you mean by null space of a basis? Because the null space of a linear map/matrix has nothing to do with a basis.

[lavish jewel]

it doesn't, i 'm just saying exactly the same thing you said

[marble lance]

But you say null space of a basis as if it's something to do with the basis

[dire thunder]

take identity mapping

it has the same null space for all bases

[tulip glacier]

oh i see i think the more appropriate way to say it is the null space is shared among all basis

[lavish jewel]

no

[tulip glacier]

huh

[lavish jewel]

the orthogonal complement of the subspace

you have 1 subspace and its orthogonal complement

each of these have infinitely many bases

[tulip glacier]

wait orthogonal complement means projection right

[dire thunder]

no

it means set of all vectors orthogonal to vectors in subspace

[tulip glacier]

oh so the normal

and since normal is a direction im guessing its the same for all

[lavish jewel]

take a matrix

it has a row space and a null space

[tulip glacier]

yes

[lavish jewel]

those two are orthogonal to each other

[tulip glacier]

yeap

[lavish jewel]

any vector in the null space is not in the row space

[tulip glacier]

yes

[lavish jewel]

the null space is the orthogonal complement to the row space

and they both have infinitely many bases

but the null space is related to the row space

not to the basis you chose for the row space

[tulip glacier]

Ax=0 x is the complement is what ur saying

[lavish jewel]

yeah, the vectors x such that Ax = 0 are a subspace

and you can pick whatever basis you like for that subspace

[tulip glacier]

oh wait so basis for any other basis that i pick it gives the same null space

but the main reason is becos the other basis are row equivalent to the row space

[tame mural]

hmm

I think it's far more clear to say that

[tulip glacier]

and this null space is related to th row space

so basis would only have one kind of null space

[tame mural]

The nullspace is the set of all inputs mapped to the 0 vector

And that basis discusses how you wish to generate a space

But regardless of how you wish to generate the space

[lavish jewel]

the null space is a property of an operator

[tame mural]

It's still the same space

[lavish jewel]

if you talk only of a subspace in general, then you mean its orthogonal complement

not the null space

they just happen to be the same when looking at a matrix

but if you look only at a basis and the subspace it spans, then you have ortho complement

[tulip glacier]

but wouldnt that imply that orthogonal complement is also unique

[lavish jewel]

it is

[dusky epoch]

whats the original question

[tame mural]

The confusion was over why the nullspace was basis-independent

[lavish jewel]

pretty much

[tulip glacier]

yea like i think i sort of understand

[lavish jewel]

why is the null space the same regardless of which basis you pick for the row space

[dusky epoch]

why wouldnt it be basis-independent

the zero vector is the zero vector no matter how you look at it lol

Null(A) = {v ∈ dom(A) | Av = 0}

[lavish jewel]

that's why i'm pushing for thinking of the subspaces, not the basis you arbitrarily pick

[tame mural]

Imo, bringing in orthogonal complements, row-space, even matrices, is quite extra

it's not minimal in its conceptual dependencies

[tulip glacier]

but if the null space was two column would it still be unique or is there now a parameter by which it can change

[lavish jewel]

the thing is that if you say "subspace" and "basis" only, there is no "null space"

[tame mural]

Gilbert Strang teaches this perspective only after he goes over the basics first

[tulip glacier]

um guys did gilbert strang explain this in the basis video

[lavish jewel]

vectors and bases can exist separately from matrices that do linear transformations based on them is what i mean

[dusky epoch]

matrices are a scourge upon linear algebra

a necessary evil

[tame mural]

He explains this in "The big picture of linear algebra"

that's what he calls it

[tulip glacier]

id it under mit course material video on utube

is

[tame mural]

If you google "the picture of lienar algebra, gilbert strang"

you'll find his discussion

he is very passionate about this lecture

[lavish jewel]

when he discusses the 4 fundamental subspaces of a matrix, he addresses that

[tame mural]

he says it's his very favorite

he has many variations of this same lecture

[lavish jewel]

but as i said, you can speak of subspaces and bases without there being a matrix

[tulip glacier]

lol i watched that video but i didnt get the revelation that the null space is unique

[lavish jewel]

the 4 spaces are unique

not just the null space

[tulip glacier]

maybe cos i had the perception that basis isnt unique

[lavish jewel]

a lot of stuff is being mixed up right now

[tulip glacier]

lol my knowledge is literally fumbled up

if only i had a clear button in my brain

[tame mural]

Let's say I have the vector space generated by [0, 1] and [1, 0].

We can then have all the values of [x, y]

Let's say I have a function which takes some of these values to [0, 0]

[tulip glacier]

is that 2x2 matrix

[tame mural]

It doesn't matter how I generated that space

It's still a fact that I map some values to [0, 0]

[tulip glacier]

but could the null space have linear combinations of each other

assuming they are 2 or more columns

[tame mural]

You mean, does the nullspace have a basis?

[tulip glacier]

and technically couldnt i say the null space of a "null space" be the basis

or row space

since they are perpendicular

[tame mural]

ahhh

~_~

[marble lance]

What is a null space of a null space?

[tame mural]

my explanation didn't work

[tame mural]

not as simple as I thought

[lavish jewel]

what is even going on anymore

[tulip glacier]

the row space is perpendicular to null space

[lavish jewel]

yes

[tulip glacier]

similarly null space is perpendicular to row space

if i were on the null space or row space

[lavish jewel]

this is why i told you not to use null space

[tulip glacier]

i wouldnt be able to differentiate which is which no?

[lavish jewel]

you mean orthogonal complement

[marble lance]

@tulip glacier do you understand that a vector space does not have a null space? It does not make sense to talk of a null space of a vector space. You can talk about a null space of a matrix/linear map.

[lavish jewel]

that is why i used a different word for it

the row and null space are orthogonal to each other

and if you are just given the two spaces with no reference, you cannot tell which is which

only that they are orthogonal to each other

so you have a subspace and its orthogonal complement

and that is all

it doesn't matter what you call each

they are still a subspace and it's ortho comp

[tulip glacier]

@steep sprucenaspace um vector space isnt a matrix?

[lavish jewel]

no

NO

[tulip glacier]

oh yea the span thing

[tame mural]

But you can have subspaces and orthogonal complement without talking about linear maps

[lavish jewel]

they are NOT the same

[marble lance]

Idek what this means, but that question is very worrying

[lavish jewel]

holy crap, i'm out. i already explained that like 1000 lines ago

[tulip glacier]

span(set) is a subspace

[lavish jewel]

subspaces exist without the scourge of matrices

[tulip glacier]

set is not a subspace

[tame mural]

I'm afraid some of your concepts might be a bit muddied

And you have to clean them up

[marble lance]

How are you learning linear algebra? Watching videos, reading a book, in a course?

[tulip glacier]

yea no joke my school messed it up real bad, i see no link between every chapter with how my teachers teach it

[tulip glacier]

like through the notes the school provides

[native rampart]

{0} is a subspace

[lavish jewel]

can you show the definitions they gave you

[tulip glacier]

and recently i came across projection vector which is largely different from how gilbert strang taught

[tulip glacier]

yes iknow that

[tulip glacier]

um for basis or null space

[lavish jewel]

for subspace

[marble lance]

It's possible the notes are really bad, but they can't be so bad that they have caused this confusion. I suggest you forget about everything you think you know, and go through the notes again. Don't make assumptions. Just read the notes as they are. The question "Isn't a vector space à matrix ?" is genuinely shocking. It seems you don't know what a vector space is at all

[lavish jewel]

i second that

you will have to reread everything and start over

even if the lectures are bad, your notes should have the correct definitions

[tulip glacier]

um if i were to use mit material where should i start?

[lavish jewel]

from 0

right from the beginning

[tulip glacier]

i think im gonna dump my school nots

[lavish jewel]

strang's courses are already too advanced

restart linear algebra from 0

[tulip glacier]

like right from eRO?

[lavish jewel]

yes. you can't talk about matrices before first doing vector spaces and subspaces all over again

so from there, i guess

[tulip glacier]

ok i will see what i can do about it, anyway thanks guys

@marble lance , @lavish jewel @tame mural

[marble lance]

Np

[lavish jewel]

godspeed

@wintry steppe

what are you bringing onto this cursed land

[solid flower]

when finding my P and D matrix when i solve a differential system of eqns do i have to arrange the D vector so that it goes from largest to smallest eigenvalue diagonally top to bottom and similarly align my P matrix so that each corresponding eigenvector is in the correct column?

[dusky epoch]

you dont have to arrange the eigenvalues within D in any way but the eigenvectors do need to be in the same order as the eigenvalues

[solid flower]

oh

so even if i have $D= \begin{pmatrix}1&0&0\ ::0&2&0\ ::0&0&3\end{pmatrix}$ thats fine as long as my $\begin{pmatrix}x_1&x_2&x_3\end{pmatrix}$ corresponds with the eigenvalues in D?

[stoic python]

ahmad_11

[dusky epoch]

yes

[solid flower]

interesting

am i just making this greatest to smallest arrangement up or does this come up somewhere?

[lavish jewel]

if you pay attention to the diagonalization, you're essentially saying that the original matrix is a sum of rank 1 matrices

and sum is commutative, so the order doesn't matter

it is by convention that in the SVD, the singular values are arranged from largest to smallest

it's also not necessary

similarly with the EVD: it's nice, but not necessary, as ann said

[solid flower]

sorry i just start diffy eqs yesterday, what are rank 1 matrices and SVD?

[lavish jewel]

oh, ignore everything i said then

[solid flower]

xD

[lavish jewel]

it is nice of you if you order it that way, it is often done by convention

but not needed

[solid flower]

ah

so lets say i ordered it as $D= \begin{pmatrix}3&0&0\ ::0&2&0\ ::0&0&1\end{pmatrix}$ thats fine too?

[stoic python]

ahmad_11

[solid flower]

oh

[lavish jewel]

as long as you make sure to reorder the eigenvectors accordingly, as was stated earlier, yeah

[solid flower]

ok that makes a lot more sense

thanks a lot 😄

[dusky epoch]

bad tex?

[zealous junco]

whats best way to get good at applied lin alg?

i took a proof based course a while ago and forgot alot of things esp on innor products and actually i havent learn alot of decompositions and stuff

[dusky epoch]

sounds like itd be heavily application-dependent

[solid flower]

is there any easier way to do matrices in LaTeX lol

i just end up copying from symbolab

[dusky epoch]

well you could get rid of those weird \:'s peppered all throughout

[zealous junco]

hmm i think stuff related to ML optimization and numanalysis is good

[dusky epoch]

& separates entries within a row, \\ separates rows

and for this bot theres even a handy command to abbreviate \begin{bmatrix} ... \end{bmatrix}

[stoic python]

Ann
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusky epoch]

bruh lmao

i mean ok

[solid flower]

lol

matlab is so much easier for writing matrices

[zealous junco]

matlab

[solid flower]

if only latex could implement that

[zealous junco]

i use matlab tho so no criticism there

[solid flower]

i mean just notation wise its less heavy

like this is all you gotta do A=[11/4 -13/10 -4/5 -3/20; 3/4 7/10 -4/5 -3/20; 3/4 -3/10 1/5 -3/20; 63/4 -93/10 -24/5 -23/20]

[zealous junco]

yea that true

[solid flower]

; where ur row ends

boom done

[zealous junco]

easy computation too

[solid flower]

yup

[dusky epoch]

yeah matlab syntax is good

[zealous junco]

ig latex has to be like that cuz some people need alot of

ldots vdots and whatever

but they could add a command or u could create a macro maybe

[zealous junco]

i was also thinking abt spectral radius

suppose if rho(A)<1, then let B = diagonal w/ diagonal entries either 1 or 0

can we ensure rho(AB)<1 always, similarly rho(BA)?

[wary lily]

How can I solve this?

can't use inverse matrices bc they aren't square.

Tried to reason about it case by case, like what matrix do we need to left multiply with B to give C, AK must give that matrix, but that was too much.

[native rampart]

Just write a few linear equations and solve?

[wary lily]

hmm

[native rampart]

It's kinda horrible

But it works

[wary lily]

so I need to have a matrix K with proper dimensions

and use variable for all of its entries

multiply out

then I have a system of equations?

like that, right?

or is there an easier way?

[wintry steppe]

guys i got a question

how can i prove this subspace is inf-dimensional?

https://gyazo.com/6b761266f18f893d0638cbd8b1124236

the orthogonal vector subspace to p(t) = 1 + t with that to product

i ended with that the orthogonal subspace is

3/2 * a0 + 5/6 * a1 + 7/12 * a2 + 9/20 * a3 + ... = 0

maybe because this is 1 equation with infinite params and according to the fundamental theorem of algebra?

i mean, i have inf - 1 possible solutions

[zealous junco]

ah iirc i got a proof

i mean our TA proved it but i forget XD

[spare pier]

just assume the matrix K with a lot of variable according to its order and do normal matrix multiplication and you should be able to arrive at an answer

[zealous junco]

i think this just legendre polynomials right

[reef prism]

yesterday i asked myself why a vector converges to the span of the eigenvector of a matrix when you multiply this vector by the same matrix over and over again

to understand it intuitively im posing a question whether the eigenvector is the vector that gets stretched the most in a single direction

and if every other vector gets stretched towards that eigenvector

and if you do the matrix multiplication with a random vector enough times the dimension collapses onto the eigenvector with the bigger value and bigger impact

can someone tell me if my understanding is correct

[reef prism]

answer: no, the eigenvectors are not the vectors that get stretched out the most

and again

i still dont have a clue what the fuck this all means

[zealous junco]

i havent got to jordan decomp but at least for

[zealous junco]

matrices that are diagonalizable, any vectors can be a lin combo of eigenvectors so the idea is that when u apply A iteratively the largest eigenvalue will dominate

tbh im not sure as well but i think this make sense

[reef prism]

yeah i know its a lin combination of the eigenvectors

but i cant for the love of god visualize it and ive found no visualization program

[zealous junco]

i guess if u want visualizing then its hard, but if k->infty then pretty much lambda_max^k v_max is the only vector that matters

[reef prism]

tbh ill copy a python code to visualize

what is the v_k?

[zealous junco]

corresponding eigenvector

shoud wrote v_max oop