linear-algebra

[wintry steppe]

ah

so what's the best way to find out

row echelon form and check for basic/free variables ?

nice

gg

thank you

[wintry steppe]

Is it sufficient for w to be a multiple of one of the columns of A

or does it have to be a multiple of both ?

to be in Col A

[wintry steppe]

seems fine, just remember you also have to show that T(cv) = cT(v) for any scalar c and polynomial v

also by linearity, to show injectivity, you can just show that T(v) = 0 implies v = 0, but that's basically nitpicking (it does make the notation a bit easier though)

there are a few weird parts ("since x^i = x^i, ..." in the second pic, and "u \neq v" in the third pic), but they don't really take away from your argument so im not going to press on them

[wintry steppe]

@wintry steppe notice that span{Col(A)} = span{-(6,3), (12,6)} = span{-3(2,1), 6(2,1)} = span{(2,1)}

So w is definitely in Col(A). As for null(A), we can see that A is a linear transformation that sends all vectors onto the line in the direction of the vector (2,1). So that means all vectors map to a vector on that line. this means that there is no way to get the original vector back if we tried to invert the transformation. So that means that w is in null(A).

Another way is to just to see if Aw = 0, where 0 is the zero vector

[old flame]

The polynomial could be satisfied by at most m values (because a polynomial of degree m has at most m roots ?) Why is this a contradiction

[dusky epoch]

the contradiction is that (2.3) is true for all z in F, and presumably F is a field with infinitely many elements

[old flame]

Ohhhhh since now z is finite, that's why it doesn't hold ?

[dusky epoch]

"z is finite"? what

[old flame]

At most m solutions now right ?

[dusky epoch]

no i mean
if at least one of the a_i was not zero you'd have a polynomial equation. hence, m roots at most.

but you have more.

[old flame]

So from this list, there is at most m roots for the polynomial, however in P(F) there is infinite of them.

[marble lance]

That is correct

Let a be an arbitrary real number. Then a = a * 1

You do not have to manually check every real number

You just have to show its true for an arbitrary real number

Let (x, y) be an element of R^2. Then (x, y) = x(1,0) + y(0,1)

If you understand what I did, then you should be able to show that the four 2x2 matrices than have one entry of 1 with the rest 0 spans the space of all 2x2 matrices.

You have to show something is true for every element of a set. If you can take any element of the set and show that it's true for that element, then it's true for every element

In my proof (x,y) can be any elemeny of R^2. If it's (5, 10) then (5,10) = 5(1,0) + 10(0,1). So even though I only proved it for one element, that element is arbitrary so the proof holds for every element

Yes

Yes

Just to be clear m = [[a1, a2], [a3, a4]] right? So the coefficients you choose are the entries of the matrix

Okay, perfect

[marble lance]

I will read it again, but your claim is not phrased correctly

Are you saying the set has a basis or the space?

What you said is any spanning set has a basis

any finite spanning*

First paragraph of the proof, you say the set is linearly independent, which it is not

It is only spanning

Oh NVM

I see now you have two cases. Ignore my last comment.

I will edit your paragraphs

If {v1, ..., vk} IS linearly independent, then it is linearly independent and spanning, which means that it is a basis.

First paragraph can be reduced to this. You can write every element of V as a linear combination regardless of whether the set is linearly independent so it doesn't make sense to say you can only do it in this case. Stick to what you have to show. To show its a basis, it must be linearly independent and spanning. We know it's spanning, so if it's also linearly independent then we're done

If {v1, ..., vk} is linearly dependent, then this means that I can extract out a basis out of {v1, ..., vk}. Think for example of {[1 0], [0 1], [1 1]} this set is linearly independent but contains at least 2 basis vectors: {[1 0], [0 1]} or {[1 1], [0 1]} or {[1 0], [1 1]}... This whole paragraph is good for you, but should not be part of a proof.

If {v1, ..., vk} is linearly dependent one of the vectors can be written as a linear combination of the others, say v_i = b1v1 + b2v2 +... + b(i-1)v(i-1) + b(i+1)v(i+1) +... bnvn

Now show the set {v1,..., vk} - {vi} is also spanning:

Let v be a vector in V. Then since the original vector is spanning, we can say that v = a1v1 +... + aivi +... + anvn

[wild pagoda]

I do a lot of machine learning, so I know about tensors in that context. But right now, I feel like I know about tensor a the same way I used to think vectors were just ordered numbers, now they’re so much more. Anyone have any good resources to learn about tensors?

[marble lance]

But we can replace vi with what we just had, and combine the coefficients to get v = (a1 + aib1)v1 +...

So v is also in the span of the vectors without vi. So the set without vi is also spanning.

Now you can say, if the set is still not linearly dependent, repeat the argument until it is. We know it must be eventually since a set of 1 vector is linearly indendent

Np

[wintry steppe]

math isn't a race

go at the pace where you go to the next section once you feel like you understand the first

[wintry steppe]

@wintry steppe lol

For a set to be a basis of R^n it must be linearly independent and also span R^n right ?

yeah, that's a common definition of a basis

the most common basis vectors in R^2 are [0, 1] and [1, 0] - these two are linearly independent and they also span R^2

yeah ?

id call them "usual basis vectors" but yeah

right

This is a base of R^3 because it is linearly independent (only basic variables in ref)
and also it spans R^3 because there is c, u, v that allow us to reach every point in R^3, correct ?

[dusky epoch]

id call them "usual basis vectors" but yeah
standard basis vectors

yes this is a basis of R^3 so long as you've verified it to be LI

[wintry steppe]

usual/standard basis vectors
ive seen both but id assume the latter is more standard

Yeah, standard is the term I was trying to remember lol - from 3blue1brown videos

Why did they provide B here ? They say B is row equivalent to A

Just for convenience ?

Because I can find Nul A and Col A just with A, so why provide B?

[marble lance]

They have already reduced it so you don't have to

[wintry steppe]

usually it's easier to find those spaces using the rref of A

Right, I was confused because they tell us to use rref to find the Nul A but then provide a matrix in ref

in italics too

[marble lance]

They don't want you to have to do long computations, that is the lowest it can be reduced without moving to fractions, presumably that's why it's left with the 2.

[wintry steppe]

ah okay

lucky for us we are allowed a calc with function rref(matrix) lol

[marble lance]

That's nice

[wintry steppe]

thanks btw

[marble lance]

Np

[spice storm]

how do I do this?

nevermind I got it. The wording is just confusing

[wintry steppe]

Damn

That was fast

[spice storm]

I never had a professor who copies questions from the book but rephrase it so you can easily get points taken off

[solar warren]

does anyone have some sort of intuation, or analogy to the real world, regarding SVD?

[spice storm]

What does SVD stand for?

[gray dust]

singular value decomposition

[modest cliff]

Guys, I have a stupid question that I can't find on the internet anywhere...

Can you get the dot product of two matrices?

Or is the dot product only for vectors?

[half storm]

I was about to say, I don't really know how to interpret the inner product of two matrices.

There probably is a definition for that.

I mean when you carry out matrix multiplicaiton of two matrices you are computing the dot product of the ith row with the jth column yea?

[modest cliff]

Yeah, I suppose

[half storm]

It's the same inner product that we define for vectors in euclidean space e.g. R^3 we define a general inner/dot product for two vectors in R^3 as <x,y,z> * <a,b,c> = ax + by + cz yea?

[modest cliff]

Right

[half storm]

When you go to compute the product of two matrices you're literally doing that. If you have two matrices with entries taken from a field you are doing that.

To compute the ijth entry of of the product AB matrix you take the "dot product" of the ith row of A with the jth column of B

[modest cliff]

Oh shoot you're right

i didn't realize that

[half storm]

It does tell you something about the matrices, If you have two matrices and you take their product and you get the zero matrix, then that would tell you that rows of A are orthogonal to columns of B.

Yea.

[wintry steppe]

*if you have two matrices and you take their product

[half storm]

There are probably a variety of ways you would want to define the inner product of two matrices; I'm not privy to them. But I can share with you that sort of relationship of te inner product that you use generally for vectors in R^n.

What did i say wrong?

Nvm I see.

Fixed 👍

There may be some more information there but I don't know enough LA to really tell you anything else besides what I've given you.

[wintry steppe]

while the dot product shows up in the computation of the product of two matrices, it does not make immediate sense to talk about the dot product of two matrices. you can define a "dot product" on matrices; the concept of an inner product generalizes that of the dot product, and you can absolutely define inner products on the space of matrices.

one simple way to define an inner product on the space of matrices is to identify an m x n matrix with a vector in R^(mn), and then take the usual dot product of that. another common inner product is the operator norm. there are probably quite a few more

[modest cliff]

Thanks @half storm you fixed my problem lol

[half storm]

No problem TTerra has a more direct answer.

[modest cliff]

Ah I see, thx @wintry steppe

[stoic python]

TTerra:

[wintry steppe]

if you wanted to define the dot product of matrices in the most straightforward way, this is how you would do it the result isnt a scalar so this is wrong. take the thing above and add the entries, perhaps?

but depending on your level of LA knowledge, you might want to just think of the dot product as an operation on two vectors, and leave any inner product stuff until later

[quartz compass]

why would you say this is a dot product of matrices

if these were both 2x1 matrices you wouldn't get the dot product of the two vectors this way

[wintry steppe]

oops

thanks for calling that out

[half storm]

I don't know enough about inner products yet so I couldn't say anything lol.

[wintry steppe]

an inner product has to take two elements of the vector space and return a scalar

the dot product does this

the thing in my picture does not

[half storm]

yea, I kind of knew that part but wasn't really confident enough to say anything.

[wintry steppe]

quite a silly mistake!

[half storm]

lol

[wintry steppe]

Hey can I dont know how to do this question and would like to know how to set it up

[wintry steppe]

I know that you are taking the vectors v1, v2, v3 and you are making a xyz graph with those vectors. and (w)s is a point on said graph

oo

Nvm i got it

thx btw

[median forum]

inner product is basically a bilinear positive definite and symmetric function from V×V to F

[wintry steppe]

If I say H = Span{v1, v2} - that reads as "H spans v1 and v2" ?

[median forum]

isnt it the opposite?

I mean, the reciprocal

v1 and v2 span H

[wintry steppe]

Possibly

[median forum]

iirc it is

[wintry steppe]

I trust you

[median forum]

dont

lemme check

ye

was correct

H is a span of {v1,v2}

{v1,v2} spans H

[wintry steppe]

There is a linear combination of v1, v2 that reaches every point in H

[median forum]

ye

[wintry steppe]

ok

thanks

H = Span{v1, v2}
B = {v1, v2}

x = [19, -13, 18, 15]

x is in H because c1 = -5/3 and c2 = 8/3.
And now B-coord for x is [-5/3, 8/3]

the new system is the plane created by v1 and v2 ?

or is it a new system where v1 and v2 are the basis vectors instead of [1,0] and [0,1]

[half storm]

Can anyone explain to me why block multiplication of matrices works.

[wintry steppe]

@half storm block?

@half storm depends on the partitioning right?

[half storm]

Yea but I never really get why it works the way it does.

[pallid rampart]

Well you can explicitly write out the sum for each entry

[half storm]

Yea that's where I was thinking I should go with it.

And it will probably reveal it self to me through that.

Like just thinking of normal matrix multiplication of a whole matrix and see that you actually end up performing the multiplication of the block matrices anyways.

It'll probably reveal itself to me if I do that.

Yea, lol it'd be nice if someone wrote a program or something that showed it to convince people of it because it seems very computationally-intensive

For the most part I'm willing to take it as is because in my mind I can probably see how it's supposed to shake out intuitively kind of.

A general proof seems really ... tedious.

[pallid rampart]

Well you can probably see it visually. An entry of the final product is the dot product of some row of the first matrix with some column of the second matrix, then we can write the dot product as the sum of two smaller dot product

The each smaller dot product correspond to an entry of the product of two smaller block matrices, one block matrix from each matrix in the original product

[half storm]

Yea I can see it. I'm not gonna bother with a general proof though lol.

[pallid rampart]

Yeah

[half storm]

I'm sure someone had to prove that though somehwere along the line.. some CS person.

f that though. An uninteresting proof to say the least.

Or maybe interesting but too tedious for such a little reward.

[pallid rampart]

I mean the proof of a lot of facts are pretty tedious but the facts themselves are interesting/useful

[half storm]

True.

[pallid rampart]

Like all vector spaces have a basis

[half storm]

Yea that one is but I find that interesting and the proof of that isn't too bad.

You just need to get acquainted with some terminology such as maximal sets, maximal linearly indepdendent sets and Hausdorff's maximal principle.

[pallid rampart]

Yeah

But that isn't my point lol

[half storm]

and of course the standard LA teriminology and you're off to the races.

[pallid rampart]

But you get what I mean

[half storm]

Yea yea.

I'm just saying that one doesn't seem all that bad for me. But I'm sure there are plenty of examples. Like showing the determinant is unique is kind of tedious one. I'm learning that one now.

I've heard that Fubini's theorem on computing double integrals over reigions is equivalent to iterated integrals, Leibniz Integral rule and Implicit Function Theorem are pretty tedious and hard to prove.

[wintry steppe]

A lot of math follows that kind of pattern. You're dealing with different levels of abstraction. The proofs of the basic theorems of a subject are like the implementation of the standard library for that theory.

And while everyone uses the standard library, it's a much smaller subject who would ever need to or want to dig into how it actually works

beyond the obvious cases where there is pedagogical value.

But even seeing a cool trick or a clever technique in a proof.... if the same trick isn't a common occurrence at the level you're actually interested in working with, it's best left as a curiosity. You see it once in passing as you first learn the subject, or maybe you need to understand it well if you want to teach or write about the foundations of the subject. But otherwise, you just know how the properties of the thing you're working with and call it a day.

[pallid rampart]

Oh boi, implicit function theorem was a pretty tedious proof

Same with the change of variable in multivariable calc

Fubini's theorem isn't that bad

and I haven't seen the proof of the full version of Leibniz integral rule, but the version where the bounds are constant is relatively easy

[wintry steppe]

@dusky epoch

i know the answer is there does not exist one

but i am too bad at this to prove why

[dusky epoch]

if there did exist such a $t$, then both sides of the equation would be equal to $\overrightarrow{OR}$

[stoic python]

Ann:

[wintry steppe]

[dusky epoch]

but considering $\overrightarrow{OR} = \text{LHS}$ gives $t = 2/3$ and $\overrightarrow{OR} = \text{RHS}$ gives $t = 1/4$ and last i checked those two were not the same

[wintry steppe]

why do they to be equal to OR

[stoic python]

Ann:

[dusky epoch]

$(1-t)\overrightarrow{OA} + t\overrightarrow{OB}$ is a vector connecting the origin to a point on line $AB$ (and if $t \in [0,1]$ it's specifically a point between $A$ and $B$)

[stoic python]

Ann:

[wintry steppe]

oh right

so since they intersect at R

[dusky epoch]

ditto for the other side

yes

that's exactly why

[wintry steppe]

thanks

[wintry steppe]

then it's not so trivial

lol

anyway

write on in terms of oa and ob

and om in terms of oa, oc

and ol in terms of ob, oc

yeah

what do you notice

have you done what i said

ok

what is it then

ok

yes

so now

you can notice that for example

c and n are on a concurrent line

l and a are on a concurrent line

m and b are on a concurrent line

and they may intersect at a certain point

see if you can figure out how to show that they must intersect

and then the answer will pop out

nothing is perpendicular here

those are not altitudes.

oh

if you can use part ii

then that makes this trivially easy

just show that they intersect at that point

and that's it

your job is done

lol

w is the centroid

so just use that

and use the definition of a line segment

[spice storm]

what does it mean if $<1,t>$ turns out to be -3 and not orthogonal? Should I still contiune with the two other polynimals? So the next will be $ <1, t^2-\frac{19}{6t} +\frac{29}{19}> $ and the other will be $ <t, t^2 -\frac{19}{6t} +\frac{29}{19}> $

[stoic python]

KillerWhale2498:

[wintry steppe]

what is the equation for a line

... in vectors..

it's hard to tell without latex

write it as a parametric equation in t

where the coefficients will add up to 1

[zealous widget]

what is the component of a matrix defined as?

$a_{ij}$ for some matrix $A$?

[stoic python]

catfood:

[zealous widget]

or is it the columns/rows of matrix $A$?

[stoic python]

catfood:

[half storm]

first index is the rows second index is the columns

So the it's the entry in the ith row and the jth column of the matrix A.

[dusky epoch]

a matrix is a grid of numbers

those numbers are called its components

[wintry steppe]

Where does x2[1, 1] come from ?

[half storm]

From the rref of the matrix right?

You've got a row of a zeroes and free variable

That top row corresponds to the equation $x_1 - x_2 = 0 \implies x_1 = x_2$

[stoic python]

JohntheDon:

[half storm]

So the solution set is given by the vectors $(x_1, x_2) = (x_2,x_2) = x_2 (1,1)$ make sense?

[stoic python]

JohntheDon:

[wintry steppe]

Could it also be x1(1,1) ?

[half storm]

Yup

[wintry steppe]

gotcha

thanks

[limber sierra]

Seems good

[wintry steppe]

So if F was R or C we could imagine it as the 3-d space

And the subspaces of a such a space are the planes and lines passing through the origin

However here we have an arbitrary field

So I think the field isn't relevant in the question?

And all that matters is that the "plane" or "line" should pass through the 0 of that field?

The channel was occupied but go on

Sure

Okay so I actually think since fields are only concerned with the scalar domain

(I hope)

The only difference is in how you image a field's Cartesian product(s)

If it were R it's easy to digest

So now then

@wintry steppe id be careful trying to visualize C^3, and would recommend sticking to base field R for visualization purposes in this case.

the subspaces of R^3 are the zero vector, the lines and planes through the origin, and the entire space

not sure what you mean by *the field isn't relevant". the problem asks you to show that U_a is a subspace considered as a subset of F^3, where F is an arbitrary field. since it's over any field, it certainly doesn't matter which one you choose; R is just a convenient choice for visualization

Yes yes

Exactly

So it only needs to have to 0 of the field to be a sub space of the vector space F^3

And if x_1+x_2+x_3 is not equal 0 it means that it doesn't have the zero of the vector space

And hence isn't a vector subspace

So it must have the 0

you seem to have the right idea, but id highly recommend writing out a more formal proof

you already know what you have to do for that, so most of the hard work is done

Yes working on it

The idea is what matters

I can formalise it

@wintry steppe done

How did they find the basis ?

[dusky epoch]

scale [1/2; 1; 0] by 2 for no reason beyond the convenience of not having fractions

[wintry steppe]

ok thx

@wintry steppe will this arguement work that in order to be a sub space a scalar times an element of the set U_alpha has to be in U_alpha

And if alpha is any constant

We can just keep multiplying it by some scalar until it's not in U_alpha

there should be an easy "subspace test" for vector spaces that you can apply here. unfortunately im a bit preoccupied so i can't directly answer your question

Oh idk about that

Anyways thank you for the help

[pallid rampart]

These are good intuitive justifications of why should the statements be false/true, but it's probably better if you proof or find a counterexample explicitly. For example, for a) I would say
If S:={v1, v2, ..., vk} where v1=0, then 0 = v1*1 + 0*v2 + ... + 0*vk is a linear combination of vectors in S where not all coefficients are 0, so S cannot be linearly independent

Yeah my point is that you should prove the statements from the definitions

He didn't say {0} is a basis for the null space

He said {0} spans the 0 vector space, which is true

[pallid rampart]

You don't need to prove d iteratively, you can just notice that any linear combination of a subset of the vectors is automatically a linear combination of the original set of vectors

Also the reason why the empty set is linearly independent is that, for a set of vectors to be linearly dependent, there must exist a nontrivial linear combination that is equal to 0, but no linear combination exist so the empty set is vacuously linearly independent

[wintry steppe]

So I've pinned it down to this

U1 is a subset of U2 or the other way around or they're equal

Since they're already subspaces

They share the same 0

And addition and multiplication are closed

After that we have been given that their union is V

Well if They're equal and the union is equal to V then it's trivial

If not then one is the subset of other

Oh shit hold on

Hmm well it's clear from the Venn diagram how do I prove it analytically

[pallid rampart]

Suppose neither U1 and U2 is a subset of each other, meaning there is a u in U1\U2, and v in U2\U1. Consider u+v, since the union of U1 and U2 is V, u+v must be in U1 or U2

And arrive at a contradiction.

[wintry steppe]

Oh nice that's very neat

Thanks whoever

But if I were to go down my path how could I complete the proof?

Ah ok ok

If u is in U1 and v in U2 then u and v must be in the same set

[pallid rampart]

U1 is a subset of U2 or the other way around or they're equal

How do you know this

[wintry steppe]

They are both sub spaces of V

And hence share the same 0 and are closed under the operations

And if X is in u then so is -X

[pallid rampart]

Yeah but does that mean U1 is a subset of U2 or the other way around?

[wintry steppe]

It can be any

It doesn't affect the question

[pallid rampart]

The whole point of this exercise is to prove that U1 is a subset of U2 or the other way around

If you know that already, let's say $U_1\subseteq U_2$, then $V=U_1\cup U_2=U_2$

[stoic python]

Whoever:

[wintry steppe]

Isn't the point to just prove any one of them equal to V

Then the other one automatically becomes a subset

Or equal to it

[pallid rampart]

Yes, but if you know that one is a subset of the other then it becomes trivial

[wintry steppe]

Ok hear me out

Oh wait

So that's the crux of the problem?

[pallid rampart]

There might be other ways of proving it but this is what I would do

[wintry steppe]

I mean do you agree if I can prove that one is the subset of the other then the question is solved?

[pallid rampart]

Yes

[wintry steppe]

I really don't see why that's the tough part

I probably lack the rigour

But they're both subspaces of V

One has to be bigger than the other

Or be equal

[pallid rampart]

You can't just say that

[wintry steppe]

Hmm ok

[pallid rampart]

What about the subspaces y=0 and x=0 of R2

They are both subspaces

But neither is a subset of the other

[wintry steppe]

But the zero of a vector space is unique

[pallid rampart]

So?

[wintry steppe]

They're the same

[pallid rampart]

Yes any subspace contains the unique zero vector

[wintry steppe]

Yeah so one is either the subset of the other or they're equal

That's my claim

Is that wrong?

[pallid rampart]

The claim is not wrong

But you haven't proved it

[wintry steppe]

But I need to prove it

Okay okay

Got it

[pallid rampart]

That was my whole point bro

The claim is correct but you didn't say why

[wintry steppe]

Sorry I got mixed up in all the things

My bad

I'll try to prove it

I suppose a Venn diagram isn't worth much here,is it lol?

[pallid rampart]

No

I don't think it helps

[wintry steppe]

Your proof is exactly what mine would become

Or what I would hope it becomes

Thank you very much

Meh yeah you're right

Sorry bad internet

[pallid rampart]

ok

[zealous widget]

With the $A=QR$ factorisation ($Q$ is a matrix with orthonormal columns $q_1$ to $q_n$, $A$ is a matrix with independent columns $a_1$ to $a_n$. $Q$ is created from the columns of $A$ with Gram-Schmidt), is the following statement true?
$r_{jj}=\rVert a_j \lVert$

[stoic python]

catfood:

[zealous widget]

@half storm maybe?

i don't have much time lmao

[zealous widget]

nvm#

[inland gull]

COPY AND PASTE BROTHA

With the $A=QR$ factorisation ($Q$ is a matrix with orthonormal columns $q_1$ to $q_n$, $A$ is a matrix with independent columns $a_1$ to $an$. $Q$ is created from the columns of $A$ with Gram-Schmidt), is the following statement true?
$r{jj}=\rVert a_j \lVert$

[stoic python]

Corgie:

[wintry steppe]

[delicate zealot]

If you have a change of basis matrix that changes from base alpha to base beta and have a change of base matrix that goes the opposite way and you multiply them together- shouldn't you get a matrix whose columns are the basis vectors of alpha because that's what you started with?

That might have been EXTREMELY terribly worded so lmk if I explained it poorly

[marble lance]

They should be inverses and should give the identity matrix

If you change to a different basis, and then change back, you are changing nothing so you need to be multiplying by the identity to have the effect of changing nothing

[delicate zealot]

Ok thx

[wintry steppe]

I'm trying to find the eigenvector associated to the eigenvalue 2

For the general solution I get S = {[-r-w, r, w]}

Does that mean that there are more than one eigenvectors ?

[marble lance]

Lol, are you going to give us the matrix?

[wintry steppe]

Sure, I didn't want to overcomplicate the question

one sec

[subtle walrus]

the eigenvector?

nonzero multiples of an eigenvector will always also be eigenvectors to an eigenvalue

[marble lance]

But possibly there are independent eigenvectors for the same eigenvalue

[wintry steppe]

Right, so in S = {[-r-w, r, w]} I would have r * [-1, 1, 0] and w * [-1, 0, 1]

so there are many eigenvectors

But the thing is that I'm trying to diagonalize the matrix, so which basis do I pick of the two ?

Because I need 3 linearly independent vectors

idk if I make sense lol

[marble lance]

If you have two linearly independent eigenvectors, then the multiplicity of your eigenvalue is two

So you use both eigenvectors

And the diagonal matrix will have a 2 in both of those columns

[wintry steppe]

Hm okay, I have not yet written a diagonal matrix it's my first exercise in this

When you diagonalize, you need 3 linearly indep. vectors

The eigenvalue of 1 gave me one vector.

Now the eigen value of -2 gives me 2 vectors.

And the last eigenvalue wil give me another vector.

So I have 4 vectors to choose from

[marble lance]

Do you have three eigenvalues?

[wintry steppe]

Yes

[marble lance]

Lol, let me do this from scratch to make sure I'm not talking shit again

[wintry steppe]

Here you go

I think if I find three eigenvalues, and the first eigenvalue gives me one vector, and the second gives me 2 vectors

I don't have to evaluate the third eigenvalue then

I already have 3 linearly independent vectors

[marble lance]

Absolutely not

[wintry steppe]

Alright, that was my confusion

thank you

[zealous widget]

With the $A=QR$ factorisation ($Q$ is a matrix with orthonormal columns $q_1$ to $q_n$, $A$ is a matrix with independent columns $a_1$ to $an$. $Q$ is created from the columns of $A$ with Gram-Schmidt), is the following statement true?
$r{jj}=\rVert A_j \lVert$, where $A_j$ is the column vector produced by Gram-Schmidt before it is divided by its length to produce $q_j$ of length 1

[stoic python]

catfood:

[marble lance]

@wintry steppe Btw what is your third eigenvalues?

eigenvalue?

[wintry steppe]

ahh

lol I saw {1, -2, -2}

and didnt notice they are the same

[marble lance]

Omg dude

That's why you get 2

[zealous widget]

any help?

[marble lance]

If the multiplicity is 2,then you get 2 linearly independent vevtors

[wintry steppe]

ok, I wasn't aware, that's good to know

[marble lance]

Sorry, catfood, idk

[zealous widget]

tis ok

lmao

[marble lance]

I can barely remember how to find eigenvalues

[wintry steppe]

The book is like, here is how for a 2x2 matrix

use a computer for a 3x3

[zinc copper]

Eigenvalues arent hard to find

You can rederive the method pretty easily

For a matrix A and eigenvector v you have $Av = \lambda \cdot v$

Which is $Av = I\lambda \cdot v$

[stoic python]

Little Narwhal:

[zinc copper]

Which means $(A - I\lambda)v = 0$

[stoic python]

Little Narwhal:

[zinc copper]

So $det(A - I\lambda)=0$

[stoic python]

Little Narwhal:

[zinc copper]

Just solve for lambda then

[stoic python]

Little Narwhal:

[marble lance]

I remembered, but just barely :P

[zinc copper]

Sorry was correcting mistake

I always get eigenvalues first and then use them to get eigenvectors

[marble lance]

Is there any other way? Haha

[zinc copper]

Ah i thought there was and you did it that way

Didnt read up

[wintry steppe]

How did they get D ?

Oh it's the eigenvalues

[zinc copper]

Matching eigenvalues of eigenvectors yeah

[zealous widget]

@zinc copper you got any ideas for my question?

i'll paste it here again

With the $A=QR$ factorisation ($Q$ is a matrix with orthonormal columns $q_1$ to $q_n$, $A$ is a matrix with independent columns $a_1$ to $an$. $Q$ is created from the columns of $A$ with Gram-Schmidt), how would I prove?
$r{jj}=\rVert A_j \lVert$, where $A_j$ is a column vector produced by Gram-Schmidt before it is divided by its length to produce $q_j$ of length 1

[stoic python]

catfood:

[zinc copper]

Haha im not at all qualified enough in linear algebra for this

[zealous widget]

awww

my brother refuses to help me

[zinc copper]

Idk what Gram-Schmidt is or what orthonormal means sorry

[zealous widget]

:(((

alright then

[wintry steppe]

is this a graduate course ?

[zealous widget]

no

its from strangs textbook

[zinc copper]

Ive only done basic linear algebra

Up to eigenvectors and eigenvalues but that's it

[marble lance]

Ugh, you've posted this question so many times, I feel bad. If no one else helps you today, I will go relearn Gram-Schmidt lmao

[wintry steppe]

There is Gram-Schmit process in David Lay's book

[zealous widget]

yes

[wintry steppe]

I can send it to you if you want

[zealous widget]

ik how gram-schmidt works

my question is kinda related

to it

but it isn't really focused on it

i know that A=QR isn't unique

but if one always uses gram-schmidt to factorise A=QR

then A=QR will always be the same

hence why i think this is provable

all my attempts end in impossible equations of unequal vectors supposedly being equal to each other

[wintry steppe]

maybe try mathematics stack exchange

[zealous widget]

hmmm

does that support latex?

[wintry steppe]

yeah

[zealous widget]

huh

i will try it

thanks

[marble lance]

Let me know if someone answers your question there

[zealous widget]

@marble lance sure

[wintry steppe]

How can I describe vector [-3/2, 1, 0] geometrically?

It's a line on z = 0, but what else

it spans the xy plane in R^3 ?

i dont think it spans the xy plane though

so its just a line on the xy plane in R^3

[zealous widget]

no, it doesn't

it spans the line 2x-3y=0, with domain limited to z = 0

@wintry steppe

[wintry steppe]

The vector isn't following 2x-3y = 0 tho ?

[frosty vapor]

why do you have it as -3,2,0

the first field in a vector is the x component, then the second field is the y component

[wintry steppe]

well catfood said 2x-3y = 0 so I tried that

I also tried -3/2x + y = 0

didn't match up either

[marble lance]

The vector (-3/2, 1, 0) can be seen as lying on any line with gradient 1/(-3/2) = - 2/3, depending on where you start drawing it, but if you say it starts at the origin, then it is on the line y = - 2/3 x

[zealous widget]

@wintry steppe my bad, should be 2x+3y=0

@marble lance my post was answered here: https://math.stackexchange.com/questions/3778749/linear-algebra-gram-schmidt-a-qr-factorisation-and-diagonal-components-of-r/3778772#3778772

[marble lance]

Excellent, now I don't have to relearn Gram Schmidt

[zealous widget]

lmao

haha

i found my problem was that I kept on trying to derive equations

what I should have done was attempted to derive a single expression to a point that I can clearly see that it was equal to my second expression
i suppose you would have to do this same process twice to attempt to prove an if and only if statement, except that for the second time you would have to swap the expressions around

because finding b = b or 0 = 0 does me no good, does it? lmao

@wintry steppe

[wintry steppe]

how'd you find 2x + 3y

lol u 1 TB RAM xd

[zealous widget]

@wintry steppe 16gb lmao

10gb used with csgo, discord, steam and my chrome tabs open

[wintry steppe]

xD i said it cuz ur 20k tabs opened

[zealous widget]

@wintry steppe it's really easy - [-1.5, 1, 0] has it's componenets satisfy 2x+3y

@wintry steppe lmao

i don't have that much money

[wintry steppe]

hahahahahaha

[zealous widget]

i built my pc in 2018

[wintry steppe]

I don't get it lol

[zealous widget]

shitty time to build it

[wintry steppe]

im jking

[neon zephyr]

@zealous widget Regarding your question about QR factorization: Can't you argue like that:
I write $q_j= \frac{1}{\rVert \tilde{q}j \lVert}\tilde{q}j$. First notice, that when $Q$ is orthogonal, you have $R=Q^TA$, so $r{ij} =<q_i,a_j>$.
The Gram-Schmidt-Algorithm tells you: $\tilde{q}j = a_j -\sum{i=1}^{j-1}r{ij}q_j$. So
$$r_{jj}=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,a_j>=\frac{1}{\rVert \tilde{q}_j \lVert}(<\tilde{q}j,\tilde{q}j>+ \sum{i=1}^{j-1}r{ij}< \tilde{q}_j, \tilde{q}_i>)=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,\tilde{q}_j>=\rVert \tilde{q}_j \lVert$$

[stoic python]

Ceana:

[zealous widget]

@wintry steppe $2 \times (-1.5)+3 \times 1=0$

[stoic python]

catfood:

[zealous widget]

@wintry steppe my memory costs 50% now than when I bought in 2018

[wintry steppe]

lol

[zealous widget]

fuck me :((((

another thing

[wintry steppe]

i built in 2019

[zealous widget]

i have a b350 motherboard

[wintry steppe]

also 16gb

[zealous widget]

b450 came out 6 days after i built my pc

with b350

[wintry steppe]

me b450 xd

[zealous widget]

i also was a victim of the dunning-kruger effect

poor me

i thought choosing the right parts would be simple

wrong

i got a shitty $30 case

with an inconsistent as fuck powerbutton

anyway

@neon zephyr i'll have a look at what you wrote now

[neon zephyr]

somethings missing in the last equation. It goes on like that $$\frac{1}{\rVert \tilde{q}j \lVert}(<\tilde{q}j,\tilde{q}j>+ \sum{i=1}^{j-1}r{ij}< \tilde{q}_j, \tilde{q}_i>)=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,\tilde{q}_j>=\rVert \tilde{q}_j \lVert$$

[stoic python]

Ceana:

[zealous widget]

pasting your message here for my clarity's sake

Regarding your question about QR factorization: Can't you argue like that:
I write $q_j= \frac{1}{\rVert \tilde{q}_j \lVert}\tilde{q}j$. First notice, that when $Q$ is orthogonal, you have $R=Q^TA$, so $r{ij} =<q_i,a_j>$.
The Gram-Schmidt-Algorithm tells you: $\tilde{q}j = aj -\sum{i=1}^{j-1}r{ij}qj$. So
$$r{jj}=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,a_j>=\frac{1}{\rVert \tilde{q}_j \lVert}(<\tilde{q}j,\tilde{q}j>+ \sum{i=1}^{j-1}r{ij}< \tilde{q}_j, \tilde{q}_i>)=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,\tilde{q}_j>=\rVert \tilde{q}_j \lVert$$

[stoic python]

catfood:

[neon zephyr]

Not sure why it messes with my latex code when i copypast something from overleaf to discord

[zealous widget]

you seem to mess up a lot of subscripts

not sure why

[neon zephyr]

well its correct on overlaf. Give me a second

[zealous widget]

lmao

would you mind rewriting it completely

kind of but hard to assimilate for me rn

@neon zephyr take your time lmao

no rush here

[neon zephyr]

I write $q_j= \frac{1}{\rVert \tilde{q}j \lVert}\tilde{q}j$. First notice, that when $Q$ is orthogonal, you have $R=Q^TA$, so $r{ij} =<q_i,a_j>$.
The Gram-Schmidt-Algorithm tells you: $\tilde{q}j = a_j -\sum{i=1}^{j-1}r{ij}q_j$. So
$$r_{jj}=\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,a_j>=\frac{1}{\rVert \tilde{q}_j \lVert}(<\tilde{q}j,\tilde{q}j>+ \sum{i=1}^{j-1}r{ij}< \tilde{q}_j, \tilde{q}_i>)$$
But the sum vanishes since the $\tilde{q}j$ are orthogonal. So
$$r{jj} =
\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,\tilde{q}_j>=\rVert \tilde{q}_j \lVert$$

should be fine now ? i guess...

[stoic python]

Ceana:

[gray dust]

not super needed but physics package has inner product cmd

$\ip{x}{y}$

[stoic python]

RokettoJanpu:

[neon zephyr]

oh good to know

[zealous widget]

@neon zephyr I don't understand how the dot product of $q_j$ and $a_j$ produces the dot product of $q_j$ with itself. Would you mind explaining?

[stoic python]

catfood:

[neon zephyr]

Do you mean this step? $$\frac{1}{\rVert \tilde{q}_j \lVert}<\tilde{q}_j,a_j>=\frac{1}{\rVert \tilde{q}_j \lVert}(<\tilde{q}j,\tilde{q}j>+ \sum{i=1}^{j-1}r{ij}< \tilde{q}_j, \tilde{q}_i>)$$

[stoic python]

Ceana:

[zealous widget]

Yes.

I understand how the 2nd term on the right in the sum comes to be.

but not $q_j \cdot q_j$

[stoic python]

catfood:

[neon zephyr]

I hope i understood your problem. You solve this $\tilde{q}j = a_j -\sum{i=1}^{j-1}r_{ij}q_j$ for $a_j=\tilde{q}j +\sum{i=1}^{j-1}r_{ij}q_j$ and then use the linearity of the inner product.

[stoic python]

Ceana:

[zealous widget]

Yes.

I don't understand what linearity is, but we can come back to that later

[gray dust]

respects vector addition & scalar multiplication

[zealous widget]

@gray dust I understand your statement, but how does that apply to inner product for example?

[neon zephyr]

You have $<x+y,z>=<x,z>+<y,z>$ And $<ax,z>=a<x,z>$ for a scalar $a$.

[stoic python]

Ceana:

[gray dust]

$\ip{cx+y}{z}=c\ip{x}{z}+\ip{y}{z}$ for all vectors $x,y,z$ and scalar $c$

[stoic python]

RokettoJanpu:

[zealous widget]

ah

so like a vector space

fully understood, thanks!

[paper ether]

$\langle \rangle$

[stoic python]

DuChat:

[zealous widget]

@neon zephyr continue with your original proof

and my questions about the first term

[neon zephyr]

If you plug $a_j=\tilde{q}j +\sum{i=1}^{j-1}r_{ij}q_j$ in the equation $r_{jj} = \ip{\tilde{q}_j }{a_j}$ and then use the properties RokabeJintarou meantioned , you get the right hand side

[stoic python]

Ceana:

[zealous widget]

@neon zephyr ah i see, i'm an idiot

when i was trying to understand your proof

[neon zephyr]

$r{jj} = \ip{\tilde{q}j }{\tilde{q}j +\sum{i=1}^{j-1}r{ij}q_j} = \ip{\tilde{q}_j}{\tilde{q}j}+\ip{\tilde{q}j}{\sum{i=1}^{j-1}r{ij}q_j}$ and so on.

[stoic python]

Ceana:

[zealous widget]

for that certain step

i tooke $a_j$ to be the same as $\tilde{q}_j$

[stoic python]

catfood:

[zealous widget]

yep

[neon zephyr]

oh alright

[zealous widget]

i fully understand your proof now lmao

flashbacks to losing so many unnecessary marks in my school maths tests

due to silly errors like this

idk why i'm so prone to those

i'm not that overconfident like i used to

@neon zephyr thanks for the alternative proof :D

[neon zephyr]

sure, glad i could help=)

[zealous widget]

:D

💝

[wintry steppe]

do u have any info on tensor product space construction?

i saw the tensor product like conceptually very applied, and not like the logical construction

[ocean sequoia]

can anyone help me walk through this proof

im a bit lost as to where $u1 \in U_1,...u_m \in U_m$ is coming in as in why the capitalization is shifting

[stoic python]

brzig:

[pallid rampart]

Well by definition $U_1+\cdots+U_m=\brc{u_1+\cdots+u_m:u_i\in U_i\text{ for }1\leq i\leq m}$

[stoic python]

Whoever:

[ocean sequoia]

yea im with that

[pallid rampart]

So any vector u in U_1 + ... + U_m must be in the form u_1 + ... + u_m

[ocean sequoia]

why is the capitalization switched then is it just stylstic?

[pallid rampart]

?

U_i is the vector space

[ocean sequoia]

no not for oyu

in the answer

sorry

[pallid rampart]

I’m not sure what you mean by capitalization shifting

[ocean sequoia]

am i misreading this

[pallid rampart]

I think you are

[ocean sequoia]

im reading this as u1 in element of U_1 ... u_m

[pallid rampart]

That part just means $u_1\in U_1$ and $u_2\in U_2$ and $u_3\in U_3$ and ... and $u_m\in U_m$

[stoic python]

Whoever:

[ocean sequoia]

oh

lmao duh thanks

[pallid rampart]

Alright

[ocean sequoia]

wow thank you sorry sometimes i get lost in notation and forget to ask myself why something im trying to work on

[pallid rampart]

That’s ok

[ocean sequoia]

[wintry steppe]

Is this arguement valid that

Let's assume the set is linearly independent

Then (v1,....,vn) is a basis

But then it means that the dimension is r and not dimV

Which isn't possible

So the set has to be linearly dependent

[marble lance]

Not every linearly independent set is a basis, so what are you using to say it is a basis in the second line

[wintry steppe]

Oh shit you're right

I forgot it doesn't span V

Yeah so exchange lemma is the way

Thanks

[cursive narwhal]

You don't really need the exchange lemma for this. Just use the basis extension theorem directly (though, the exchange lemma is just a corollary of that result).

Let $V$ be a finite dimensional vector space with $dim(V) = n$. Suppose that $(v_1,\ldots,v_r)$ is a linearly independent list, with $r > n$. This vector space has a basis $(u_1,\ldots,u_n)$ Now, by the basis extension theorem, we can add in vectors from the given basis into the linearly independent list to turn it into a basis for $V$. In particular, if we add $k$ vectors, then it is the case that $r+k = n \implies r = n-k \leq n$ and that is a contradiction.

@wintry steppe

[stoic python]

Abhijeet Vats:

[spice storm]

Oh shit you're right
@wintry steppe is the "shit" really necessary??

[wintry steppe]

sorry, no profanity allowed

time for trillium to get permabanned

<@&268886789983436800> how can you let this be?

wtf

[pale orchid]

what

[plain violet]

uh

[jagged pendant]

I'm guessing you didn't intend to ping mods

[wintry steppe]

i tried to do backslash ping mods for a joke yeah

[jagged pendant]

but it still worked

[wintry steppe]

my bad for the ping

i thought that backslash @ name would not ping (tested in another server) but here we are

[wintry steppe]

Yeah abhi basis extension+exchange lemma is what I did

Oh yeah I don't need the exchange lemma sorry

Lol

I didn't actually use it in my notes

[brittle lark]

like how do I plug that into i correctly

I feel like u shouldn't be x and v shouldn't be p(x), because then idk how that'd work with the second one

[dusky epoch]

check whether or not for any polynomials $p, q \in P_1$ and any scalar $c$ you have $T_1(p+q) = T_1(p) + T_1(q)$ and $T_1(cp) = cT_1(p)$

[stoic python]

Ann:

[half storm]

You can do this actually in one step if you want by modifying what Ann said and showing that $T(cp + q) = cT(p) + T(q)$

[stoic python]

JohntheDon:

[half storm]

That is equivalent to showing linearity of a function. But until you prove this fact, it might be better to just do what Ann said.

[brittle lark]

i need to do it the way my professor does or he'll assume i cheated on my hw in some way he's kinda a dick about doing things his way

[half storm]

But this is just a headsup for future reference.

[brittle lark]

thats good to know though

thank you

[half storm]

👍

[brittle lark]

So I’d be right saying T_1 is linear?

[dusky epoch]

no

[half storm]

Ya kind of assumed what you were trying to prove.

[dusky epoch]

T_1(f(x)+g(x)) is not x + f(x) + x + g(x)

it's x + (f(x)+g(x))

[brittle lark]

ohhh

[dusky epoch]

and in fact for this map a stronger property is true: it's not just not linear, it never preserves addition.

[brittle lark]

So would T_2(f(x)+g(x)) = x^2 + 2(f(x) + g(x))

[dusky epoch]

no

[brittle lark]

x^2 part is okay though? I assume I have a misunderstanding of what to do with the 2?

[dusky epoch]

you are misreading the formula that defines T_2

$T_2(f(x)) = x^2 \cdot f(2x)$

[stoic python]

Ann:

[brittle lark]

oh

okay

I see

[dusky epoch]

$T_2(f(x)+g(x)) = x^2 (f(2x)+g(2x))$

[stoic python]

Ann:

[brittle lark]

yeah makes more sense

I rewrote it wrong on my paper

wait so would this be the next one:
$T_2(cp(x)) = x^2(cp(2x))$

[stoic python]

hypnos:

[dusky epoch]

yes

[brittle lark]

woohoo i think i get it now lol

[brittle lark]

Another question now, where did i go wrong on T(sinx). I can’t seem to find a linear combination that works...

I'm trying to find [T]_B',B

All the given info in the question is written on the top

[half storm]

I'm a little tired so my explanations will probably be a litting sparing; @ me if you need clarification. I haven't check your work but you've got the basic procedure down. If you want to find a change of basis matrix for a linear transformation - in your case from B to B' - then you take each fo the basis vectors in B, find their image under T and then find their coordinate vector relative to the basis B'.

It looks like you're trying to do that. Good job. Let's check to see how you found the first one.

$T(cos(x)) = 3cos(x) - 2sin(x) + cos(x) = 4cos(x) - 2sin(x)$ Now, we need to find the set of scalars / constants so that we can express $4cos(x) - 2sin(x)$ as a linear combination of beta; This seems to be the part where you are struggling.

[stoic python]

JohntheDon:

[brittle lark]

i figured out the cosine one

the Sine one is the issue

[half storm]

Are you sure it's correct?

I haven't checked

so I was going to through and make sure it's right.

[brittle lark]

i got to where you are

[marble lance]

It's right. I checked.

[half storm]

👍

cool

So all we have to deal with is sine

[marble lance]

Is f(x) just cos(x)? Because you wrote something else there

Wait wait

[brittle lark]

look at the top of the paper

[marble lance]

What is f(x)?

It says f(x) = a cos x + b cos x?

[half storm]

That doesn't really make much sense.

Oh I see.

[marble lance]

Actually nevermind, I'm dumb. f(x) is just the input of T

[half storm]

Yea

It's just a representation of a arbitrary element of the vector space; the set of all linear combinations of cos and sine

[marble lance]

In general if you have an element v you want to write as a linear combination of b1, b2,..., bn. Say v = a1b1 + a2b2 +... + anbn. Then solve for the an's.

In the first one you got a clever way to rewrite it to find it immediately, but if you can't. You resort to finding it manually.

[brittle lark]

yeah in the second one i cant seem to get it to work out to a linear combination = to 4sinx+2cosx so im wondering if i messed up or just cant think of the combination

[marble lance]

Find it manually like I said

[half storm]

So we do what you did in the last problem yea? We take $T(sin(x)) = 3 sin(x) + 2cos(x) +sin(x) = 4sin(x) + 2cos(x)$. We want to find scalars $a$ and $b$ such that $$4sin(x) + 2cos(x) = a(cos(x) + sin(x)) + b(cos(x) - sin(x))$$

Solve that equation

The values that you obtain for a and b will be your coordinate vector - which will be the second column of your matrix.

[brittle lark]

yep i just cant figure out the a and b ill just continue brute forcing it lol

[marble lance]

What do you mean by brute forcing? Don't just try numbers

[half storm]

Try grouping the cosines and sines together.

[marble lance]

Solve that equation.

[brittle lark]

so like ill get to 2(sinx+cosx) then i need two more sinx so i try -2(cosx-sinx) which cancels the 2 cosx's from the first part

[stoic python]

JohntheDon:

[marble lance]

You are not listening, lol

Solve a and b in John's equation

[brittle lark]

idk what that is lol

[marble lance]

$4sin(x) + 2cos(x) = a(cos(x) + sin(x)) + b(cos(x) - sin(x)) = (a+b) cos(x) + (a-b) sin(x) $

[half storm]

You're trying a little too hard. We don't want to just plug in numbers. If you want to do that, there's a whole field of study where you do that called numerical analysis.

[stoic python]

Lunasong:

[half storm]

We can deduce what a and b are in this instance.

[marble lance]

Then the only way that is true is if a+b = 2 and a-b = 4

Then you solve those simulataneously

[brittle lark]

oh

that makes it so much easier

to solve

[marble lance]

You do this always when you try to find coordinates in terms of a basis

[half storm]

This is what Luna meant when they said "solve manually".

[brittle lark]

i havent been good at that throughout the entire course because the professor never explained how to solve manually as you say for a and b

[half storm]

lol, then this is an exceptionally bad professor.

[brittle lark]

yes

!

[half storm]

Because I have no idea how else you would do this

And cannot concieve what he may have been making you do before this.

[brittle lark]

literally his words

Brute force it

[marble lance]

It's not always the same thing, but it's pretty similar. But because we know a and b have to exist (because it has to be a linear combination of basis vectors) you can set up the equation. Also the coefficients have to be unique, so somehow you will be able to force out a unique solution

[half storm]

Brute Force computation is saved for numerical analysis. We have direct ways of finding a and b in this instance - and probably anything else you encounte in LA.

[brittle lark]

a=3 b=-1 woohoo

[half storm]

So what does your matrix look like?

You can type it out.

[brittle lark]

idk how to do special formatting so I'll do this {(1, 3), (3, -1)}

[half storm]

Noice.

[brittle lark]

if you cant tell i know a fair amount about the topic but the professor is trash when it comes to application

I've learned more from sitting in this discord than i have from him as to application of concepts

luckily this is my 2nd to last assignment for him

in fact im not really sure how what he did solved for his example because he never explained it

luckily he did explain the rest of this assignments questions well

after this

[half storm]

O.K. to do what the question asks of you. You need to understand what the matrix you constructed actually tells you / what it does.

The matrix that we just constructed basically is an "equivalent" way of representing the linear transformation that that you've been given. But the difference is is that it takes coordinate vectors right?

For it's input, the matrix that you just found takes the coordinate vector relative to the basis $\beta$ of $2cos(x) - 3sin(x)$ and gives you the cooridnate vector of $T(2cos(x) - 3sin(x))$ relative to $\beta '$.

[stoic python]

JohntheDon:

[half storm]

So I'm going to ask you what is the coordinate vector relateive to $\beta$ of $2cos(x)) - 3sin(x)$

[stoic python]

JohntheDon:

[brittle lark]

So are you saying i also need [T]_B?

[half storm]

I'm not sure what you're notation is indicating here. What I am asking you, what are the scalars so that you can express $2cos(x) - 3sin(x)$ as a linear combination of $\beta = { cos(x), sin(x) }$

[stoic python]

JohntheDon:

[brittle lark]

2, -3

[half storm]

Yup.

So we take that vector $ \langle 2 , 3 \rangle$ and multiply it by our matrix that we found.

[stoic python]

JohntheDon:

[brittle lark]

so that's (-7, 9)

[half storm]

It might be. Let me check your work quickly.

[brittle lark]

i can use a calculator for that (which i did)

[half storm]

Oh o.k.

Good

So now, what do those coordinates tell you? Can you remember?

[brittle lark]

Coordinates of 𝑇(2cos𝑥−3sin𝑥) relative to B'

[half storm]

Right so $\beta ' = { cos(x) + sin(x), cos(x) - sin(x) }$ so that vector tells you that $$T(2cos(x) - 3sin(x)) = -7(cos(x) + sin(x)) + 9(cos(x) - sin(x))$$

[stoic python]

JohntheDon:

[half storm]

You're pretty much done. But you still have one final step right?

It's the easiet part

[brittle lark]

i think ik what to do lemme catch up in my writing

[half storm]

O.k.

[brittle lark]

so i distribute and boom there's my answer in the format he requested

right?

[half storm]

I can't see it.

[brittle lark]

i mean I take the linear combination there

and i distribute the -7 and 9

[half storm]

Yup

So you should get $2cos(x) - 16sin(x)$

[stoic python]

JohntheDon:

[brittle lark]

yep

thats what i got

[half storm]

Ur done 🙂

[brittle lark]

🙂

theres a B.3

"Is 𝑇 invertible?" Seems simple, but i dont have the definition of that in my notes(i write everything he says, he doesnt say much, jesus i wont miss this class) How do I know if a Linear Transformation is invertible

[half storm]

There's a lot of ways to describe invertibility.

Here's one that is quick that you can use if you remember him mentioning something in passing in class about it. A linear transformation T is inveritible if and only if it's matrix representation is invertible.

Do you remember him mentioning this at all?

[brittle lark]

no lol but ill work with it

[half storm]

O.k. do you know how to show whether a matrix is invertible or not?

It involves the determinant of the matrix?

[brittle lark]

ik what a determinant is but not how it relates to this

[half storm]

Did he ever mention to you that if the determinant of a matrix is nonzero then it is invertible?

[brittle lark]

actually i think i recall that

so in this case, the matrix representation of T would be the {(1, 3), (3, -1)} right

[half storm]

yes.

Calculate it's determinant. If it's nonzero it is invertible. If it is zero, then it is not invertible. If the matrix representation is / isn't invertible, then the linear transformation is / isn't invertible.

There is a particular definition of what it means for a function / linear transformation to be invertible. But again, this is a shortcut way of proving it.

[brittle lark]

okay, yeah the det is -10

so it is invertible

thanks again!

[half storm]

np

[marble lance]

Only the first one is correct

The second basis does not consist of symmetric matrices

Yes, that's correct

But if you just have [0 1][0 0] then it's not symmetrical

Sorry, I'm not going to fix the spacing, the 0 1 is first row and 0 0 is second row

Okay, are you sure?

Oh but I'm lying

You need three basis matrices

Not just 2

It's actually very easy to find

Do you know what the general form of a symmetric 2x2 matrix is?

Yes

So take a general 2x2 matrix [a b] [c d]. What is its transpose?

If you equate those, what do you get?

Yes

So the diagonal can be anything it wants, but the off diagonal elements have to be the same

So the general form is [a b] [b d]

Yes

Exactly

This also shows you why the other matrices were not symmetrical, because b was not the same as c

Np

[gray dust]

#prealg-and-algebra

[tiny wren]

aight

[native river]

Hi, i have know that we can use linear map to represent matrix and also know that the matrix multiplication of S and T is equivalent to $ S \circ T$. My question is how can i compute $S \circ T(e_{1})$ and $S \circ T(e_{2}) $ without initially finding S×T?

[wintry steppe]

put dollar signs around latex to make it render

[stoic python]

joseph2531:

[native river]

The problem in reference

[wintry steppe]

$S(T(e_1)) = S(2e_1 + 3e_2) = \cdots$

[stoic python]

TTerra:

[wintry steppe]

there is a certain property of S you can use

it's kind of implicit in the statement of the exercise that S has this property

[covert nest]

does anyone have any youtubers/resources they would recommend for linear algebra?

[native river]

Is the property that S is associative

[wintry steppe]

no

linearity

"S is associative" doesn't even make sense

[native river]

I understand, thank you very much

[wintry steppe]

$S(T(e_1)) = S(2e_1 + 3e_2) = 2 S(e_1) + 3 S(e_2) = \cdots$

[stoic python]

TTerra:

[wintry steppe]

and so on

you know S(e_1) and S(e_2), so you can simplify this into something in terms of e_1 and e_2

then do the exact same thing for S(T(e_2)), and you're done

[supple hemlock]

Hey, I need help with the following question:

How would I apply the subspace test on this?

[native river]

You need to show that the set S:
Contains the zero vector, it is closed under vector addition and scalar multiplication

[supple hemlock]

I know that

but how would I put that in symbols

I have a hard time visualizing this subspace

[wintry steppe]

try writing down the general element of S

if q(x) is a polynomial of degree 2 (ax^2 + bx + c), what is xq(x)?

if you really wanted to "visualize" this space

that's what you'd do

[supple hemlock]

Ohhh

so ax^3 + bx^2 + cx is all polynomials?

[wintry steppe]

the set S is the set of polynomials of that form

[supple hemlock]

Ahhh, ok

Thank you very much!

[wintry steppe]

now that you know that, the problem should be easier

[supple hemlock]

I need help with closed under addition

do I prove that a(x1+x2)^3 + b(x1+x2)^2 + c(x1+x2) = P(x1) + P(x2)?

@wintry steppe

[wintry steppe]

that makes no sense to me

to show that the set S is closed under addition, you have to show that if p_1(x) and p_2(x) are two elements of S, then their sum p_1(x) + p_2(x) is also in S

so suppose p_1(x), p_2(x) are in S. what does this mean?

[floral thistle]

@supple hemlock How are you going with the problem?

@supple hemlock Complementing what @wintry steppe said:

1. Closure under addition: Show that if $$p_{1}(x), p_{2}(x) \in S \rightarrow p_{1}(x) + p_{2}(x) \in S$$
2. Closure under scalar multiplicacion: Show that if $$p_{1}(X) \in S ; & ; r \in \mathbb{R} \rightarrow r \cdot p_1(x) \in S$$
3. Zero vector: Show that there is a $$p_0(x)$$ such that $$p(x) + p_{0}(x) = p(x)$$
4. Identity operation: Show that there is a $$1 \cdot p(x) = p(x)$$

[stoic python]

Max Hetfield:

[supple hemlock]

I got that

[floral thistle]

I omitted other subspace properties...

[supple hemlock]

Same approach! Thank you!

I just need the 3 basic ones

I realized you need to play around w/the coefficients

[floral thistle]

Yup!

[supple hemlock]

Now, I need help with part b of that question 😦

[wintry steppe]

you only need to show that S is closed under addition and scalar multiplication, because

1. as a subset of a vector space, it already satisfies a bunch of the properties of a vector space (commutativity & associativity of addition, identity scalar multiplication, scalar multiplication distributes over addition, etc.)
2. if it's closed under addition and scalar multiplication then it necessarily has a zero element, since v + (-1)v = 0 must then be in the subset
   i say this because of the ominous sounding

I omitted other subspace properties...

you might already know this but it's good to keep in mind

[floral thistle]

Now, I need help with part b of that question 😦
@supple hemlock
Start from the standard basis for M_2x2, adapting it to the problem. (Check for linear independence if you wish to do so, although a basis is linearly independent). You will get the dimension once you arrive to the basis.

[median forum]

hint ; one of the inclusions already comes from the question

[slow scroll]

you have by assumption that $\ker(P) \subseteq W^\perp$. You just need to show the other inclusion

[stoic python]

kxrider:

[slow scroll]

um, jinx

[median forum]

true

[median forum]

@quick sparrow by def QQ^T = Q^TQ = I

you can do it explicitly since the matrix is smol

[quick sparrow]

ty!

[marble lance]

It's not a formula

[wintry steppe]

^

[marble lance]

Just plug x1 + x2 into the LHS of the equation and see if you get b

[wintry steppe]

@brittle lark if $Ax_{1} = b_{1}$ and $Ax_{2} = b_{2}$ is a solution

[stoic python]

robo™:

[wintry steppe]

then what does that say about $A(x_{1} + x_{2})$?

[stoic python]

robo™:

[brittle lark]

i dont have b just that b isnt 0

[wintry steppe]

after all, A is a linear transformation

well it doesnt matter

[marble lance]

A(x1+x2) = Ax1 +Ax2, and you know what Ax1 and Ax2 are. Write it in terms of b

You don't need to know what b is, only if that is equal to b

[wintry steppe]

the only reason why $b \neq 0$ is because we dont want $x1$ and $x2$ to be in the nullspace

[stoic python]

robo™:

[brittle lark]

gotcha

[marble lance]

So what is A(x1 + x2) in terms of b?

[brittle lark]

uhh im still kinda lost

[marble lance]

Pretend you don't even know what x1 and x2 are

[wintry steppe]

What is 2(x + y) equal to in regular algebra?

[brittle lark]

2x+2y

[wintry steppe]

What is A(x + y) in regular algebra

[brittle lark]

Ax + Ay

[marble lance]

So what can you do with A(x1 +x2)?

[wintry steppe]

😛

[marble lance]

I actually already said this part

[brittle lark]

Ax1+Ax2

[marble lance]

Now what Ax1?

And what is Ax2?

[brittle lark]

b and b

so

2b

ohhh

i see

[marble lance]

So A(x1 + x2) is 2b