linear-algebra

[marble lance]

Is 2b the same as b? If not, why not

[wintry steppe]

well 2b still lies in the space that b lives in

[brittle lark]

because its a multiple of b? but can't you scale the vector and have it still =

so

[marble lance]

Because b is not 0

[brittle lark]

that's still a solution

[marble lance]

If b was zero, they would be the same

No it's not? We just said it's equal to 2b and not b

So A(x1 + x2) ≠ b

[brittle lark]

okay so its not what i was thinking of

i was thinking of like scaling a basis or w/e

[wintry steppe]

@marble lance i dont think the question necessarily says that they map to the same vector

[marble lance]

You're right you can't scale a vector and have it be the same, except if the vector is zero

[wintry steppe]

i think its just a general statement

like x1 -> b1 and x2 -> b2 not x1 -> b and x2 -> b

[marble lance]

No, I think A and b are fixed

x is a variable to that equation

x1 and x2 are solutions to the equation

[wintry steppe]

oh i see

i misinterpreted my bad

[marble lance]

But because A(x1 + x2) = 2b ≠ b, x1 + x2 is not a solution

[wintry steppe]

yep

[brittle lark]

So it was simple algebra and i over thought it

[marble lance]

Simple is relative, but yes

[brittle lark]

(A.2) Obtain a non-trivial solution of the associated homogeneous equation 𝐴𝐱=𝟎.

Not sure how i'd solve this

[half storm]

Linear algebra is one of those classes that is sort of an introduction to math-thinking i.e. proof based thinking

[wintry steppe]

@brittle lark non-trivial solution here means a vector that isnt the zero vector

[brittle lark]

Linear algebra is one of those classes that is sort of an introduction to math-thinking i.e. proof based thinking
@half storm Yeah I get what you mean

[wintry steppe]

@brittle lark is this related to the previous problem

[brittle lark]

yes

[wintry steppe]

like is $x = x_{1} + x_{2}$?

[brittle lark]

part 2

[stoic python]

robo™:

[brittle lark]

we just proved that is not the case

[wintry steppe]

?

we proved it for $b \neq 0$

[stoic python]

robo™:

[wintry steppe]

but $b = 0$ here

[stoic python]

robo™:

[brittle lark]

i see

[wintry steppe]

so if $Ax = 0$, then $A(x_{1} + x_{2}) = 0$

[stoic python]

robo™:

[marble lance]

No

b is still not 0

[brittle lark]

it is

[wintry steppe]

$Ax = 0$

[stoic python]

robo™:

[marble lance]

Yes, b is still not 0

[wintry steppe]

well ok i see ur point

yes technically speaking its not 0

[marble lance]

This is a separate equation

[wintry steppe]

because we defined $b \neq 0$

yes

[stoic python]

robo™:

[marble lance]

And you need to use the solutions of Ax = b to find solutions for this equation

[brittle lark]

yea

[marble lance]

Any guesses hypnos?

[brittle lark]

x1 + x2?

x1 * x2

[marble lance]

Wait

[brittle lark]

im leaning towards the first one

[marble lance]

What is x1 * x2?

[brittle lark]

dot product

idk how to use the bot

[half storm]

Matrices are functions on coordinate vectors.

[marble lance]

If it's a dot product you get a number not a vector

This is a vector equation

[brittle lark]

oh

then

i mean multiplication

[marble lance]

How do you multiply vectors? It's not defined

[brittle lark]

oh

idk my professor really is bad at teaching LMAO

[wintry steppe]

if you know $A(x_{1}+x_{2}) = 2b$ then $A(x_{1}+x_{2}) - 2b = 0$

[median forum]

Ill create my own pointwise algebra ok

jk, somebody prolly already done that

[marble lance]

And it's not x1 + x2 because we already said A(x1 + x2) = 2b which is not 0 because b is not 0

[brittle lark]

is it what robo just said?

[marble lance]

You want to put a vector in here A() and get 0

I don't know where robo is going with that

[stoic python]

robo™:

[marble lance]

You have an equation in terms of b for which you have solutions. So can you write 0 in terms of b

please note 0 and b are vectors not numbers. So only use vector operations

[brittle lark]

so would what robo just said work? that makes sense? unless i should be using just x1 or x2

[marble lance]

I have no clue what they are doing

Ax1 is just b and Ax2 is also b. So that's not 0

You need to combine them in a way to get 0

[median forum]

$A(x_1+x_2) = Ax_1+Ax_2$

[stoic python]

Fractal:

[marble lance]

What can you do with b and b to get 0?

[brittle lark]

b - b

[marble lance]

Yes

So Ax1 - Ax2 = b - b = 0

Now how can you write Ax1 - Ax2 as A times a single input

[dire thunder]

(you may consider that when you miltiply coefficients by unknowns you get a_1(x_1+x_2)+..+a_n(z_1+z_2)

[median forum]

did the q change?

[brittle lark]

so

A(x1-x2) = 0

[marble lance]

Yes!

So x1 - x2 is a nontrivial solution

[brittle lark]

so in this case

that ='s

[marble lance]

Does that make sense?

[brittle lark]

yeah

(-3, -3, -3)

[marble lance]

You know Ax1 and Ax2 are both b, so if you subtract them you get 0

[brittle lark]

im doing the - vector operation right getting (-3, -3, -3) right?

[marble lance]

Yes

[brittle lark]

just verifying my understanding b/c the professor literally just talks w/o explaining

and only teaches 1/3 of the stuff we need to know

[wintry steppe]

Welcome to your education 🙂

[brittle lark]

i took this at a community college over the summer

my advisor at my university almost lost it when she heard this

"If that was here they'd be fired"

[wintry steppe]

The reality is with any sufficiently difficult math subject, you can't realistically expect to 'be taught' everything you need to know in class. It really is up to the students to learn. The lectures are just part of that.

(Although it can be really frustrating to be stuck in a bad class, no doubt)

[brittle lark]

no but this guy really doesnt teach much

like maybe 3 hours a week of pre-recorded videos

in a short summer class

[marble lance]

3 hours a week seems fine?

[brittle lark]

thats on a good week

[wintry steppe]

Luckily, with Linear Algebra, if you don't like the lectures, there's so much material out online for free, and the subject is so routine that the cirricula will be compatible.

[brittle lark]

sometimes its 30min

[median forum]

I had 30 min a week of pde analysis and it was fine o.o

[brittle lark]

idk maybe its just me

[wintry steppe]

(I'd argue linear algebra is also the last "easy" math class anyone ever takes, for some value of "easy")

[median forum]

Id argue thats prolly ODEs

[wintry steppe]

I never took ODEs in school. I imagine the curriculum is slightly less standard compared to LA, but otherwise, I would agree.

[marble lance]

I know some people struggle to self study, but a good student should be able to self study the material and only use lectures to clarify points you're uncertain about. Obviously this might be unrealistic if you've never done proofs before or something, but there are a lot of good linear algebra books for you to learn from if you don't like your lecturer

[brittle lark]

yeah... I'll probably have to look into that...

So how would I find another solution to 𝐴𝐱=𝐛, that haven't been given in the problem yet

[half storm]

Yea prob ODEs lol. You just figure out what the equations look like and follow the algorithms

[median forum]

depends on your b

[marble lance]

You can't for a general b

[median forum]

ODEs is p much another calculus. complex variables is also another "easy" last class

[marble lance]

Oh, you can, I lie

All you have to work with is x1 and x2

[brittle lark]

yep

[median forum]

you could try some linear combinations of them

[half storm]

[marble lance]

So you have Ax1 and Ax2 = b. Just create a linear combination of x1 and x2 that will give you b again

[brittle lark]

so like ax1+bx2

[marble lance]

Yes

But you have to choose a and b(scalar) in such a way that you get b(vector) again

[median forum]

you can even generalize that. given 2 solutions, any linear combinations of a certain kind will give you b

[brittle lark]

we proved 1 doesnt work

for a and b

[marble lance]

Yes

[median forum]

(avoid using b again)

[brittle lark]

so like 0 for b

and 2 for a

[marble lance]

No

Let's use ax1 + cx2 (to not use b twice)

[brittle lark]

okay

[marble lance]

Then what is A(ax1 +cx2) in terms of b?

[brittle lark]

any chance its ab+cb

[marble lance]

It's a(Ax1) + c(Ax2) = ab + cb = (a+c)b

Exactly

So you need to choose a and c to make a + c = 1

And there are infinitely many ways of doing it

[brittle lark]

do i just pick 2 random numbers

oh

[marble lance]

That will give you 1

[brittle lark]

that = 1

i see i see

[floral thistle]

An affine combination, right?

[marble lance]

Since the general form is (a+c)b and you want b, you need a+c = 1

[brittle lark]

let's say a=-2 and b=1 then

then i scale the vectors by those amounts

and add the vectors

[marble lance]

If you mean c = 1

[brittle lark]

yes

[marble lance]

That gives you - 1, lol, not 1

[brittle lark]

oh

2 -1

[marble lance]

Yes

[brittle lark]

my phone is dead and i cant take a picture but i got it from there

[marble lance]

Great

[brittle lark]

would i be correct in saying this system would have 3 equations in it

[marble lance]

Good luck. Study from your course notes, or find a book to study from. You need to rely on yourself if a professor is bad

What system?

Ax = b?

[brittle lark]

yeah

[marble lance]

If you mean turn it from a vector equation into a system of linear equations, then yeah it has 3

[brittle lark]

like A i'm pretty sure

A would have 3?

[marble lance]

A would have 3 what?

[brittle lark]

rows/equations

[marble lance]

Actually I lied

A has 3 columns because x is 3x1

But the number of equations depends on b's size

b can be any mx1 vector

And A will then be mx3

with m equations

[brittle lark]

so it cannot be determined

w/o knowing b

[marble lance]

Yes, I think so

Was it a question?

[brittle lark]

How many equations are in the system of linear equations represented by 𝐴𝐱=𝐛, or is there not enough information to determine this?

[marble lance]

Oh, then the latter

There are 3 variables because of x's size, but b's size determines the number of equations

[brittle lark]

I should be good for the rest of my hw after this one "How many unknowns are in the system of linear equations represented by 𝐴𝐱=𝐛, or is there not enough information to determine this?" I'm not sure what he means by unknowns

[marble lance]

How many variables

[brittle lark]

oh so that's 3

right

[marble lance]

Yes

Yes

columns in a matrix*

[brittle lark]

perfect

[wintry steppe]

@brittle lark Here are some notes I wrote out for when I was teaching linear algebra:
https://imgur.com/gallery/kCBVi

probably isn't exactly how your class teaches things. But it's basically what I find useful to keep in mind when thinking about the subject 🙂

[brittle lark]

Thank you Im going to save that

Can I augment [T] with the zero vector , rref that then get a Basis which would be my kernel?

[wintry steppe]

To do what?

[brittle lark]

i want to find the Kernel of a Linear transformation

is that the right way to do it

[wintry steppe]

That sounds like it might work. I don't know that you need to augment the matrix at all.

[brittle lark]

how would you go about it then?

[wintry steppe]

I think row reduction sounds right. It's the kind of thing I would always work out as I needed to.

So what, you have Tx = 0 right?

and you want to find a basis for ker T?

yeah. ok.

So when you do row reduction, you know that each "elementary row operation" (or whatever the heck your book calls them) is just multiplication by a special matrix

So E1 might be the matrix which "swaps rows 1 and 3" and row E2 might be the matrix which "multiplies row 2 by the scalar 5", etc

If M is your original matrix, then E1 M would be "M with rows 1 and 3 swapped". And E2 E1 M would be "M after you swap rows 1 and 3 and then you scale row 2 by the scalar 5".

Does that sound like crazy talk to you? Or does that make any sense?

[brittle lark]

yes

you are describe what rref does

[wintry steppe]

Yep.

And they are invertible matrices, since you can always "unswap" rows or "unscale" rows

From that, you can see that E2 E1 M x = 0 has the same solutions as E1 M x = 0 has the same solutions as M x = 0

So yeah. I'm just proving it to myself on the fly because I forget. But the rref of M would have the same kernel as M itself.

There's no need to augment with 0, I don't think

(The reason you ever augment is that when you have Mx = b, you can keep track of E1 M x = E1 b and then E2 E1 M x = E2 E1 b etc in a single, tidy matrix).

But when b = 0, that augmented column never changes from 0, so you don't need it.

[brittle lark]

okay so rref gives me

the example i was referencing if it isnt augmented idk what to do next

so im curious as to what you'd do

[wintry steppe]

Play around with it a bit. I don't know what I'd do either 🙂

but imagine what vectors you might multiply against that thing.

you want the kernel of that thing. What vectors can you right-multiply to get the zero vector?

alternatively, take a generic vector [x, y, z] and multiply it to see what system of equations you get.

[edgy anvil]

Let $\beta = {1, 1 +x, 1 + x + x^2}$. Show that $\beta$ is a basis for $ℙ_2$. Make sure to use words to explain what you have shown.

How do I go about this?

[cursive narwhal]

Use the definition of "basis"

What are the two things you need to check?

[dusky epoch]

\{ ... \} by the way

for braces in tex

[edgy anvil]

thx

[stoic python]

Domino:

[cursive narwhal]

So, what are the two things you need to check?

[edgy anvil]

not sure ermmm

[dusky epoch]

do you know what a basis is?

[edgy anvil]

yeah kinda

[dusky epoch]

no

there is no kinda

either you do or you don't

[cursive narwhal]

You need to check that the set you've been given is:

1. Linearly Independent
2. Spans the given space

[edgy anvil]

i guess i dont

[cursive narwhal]

Are you familiar with these concepts?

[edgy anvil]

linear independence yes

[cursive narwhal]

What about the second point?

[edgy anvil]

not really

[cursive narwhal]

If you're not familiar with it, then there is no sense in doing the question

You should go back to your notes and learn these things before doing more problems.

[edgy anvil]

is there another term commonly used other than spans?

[cursive narwhal]

I've seen linear hull being used for the span of a set

or linear span

[edgy anvil]

I think i know now from thinking back. Spans ℙ_2 because none of the polynomials have a power greater than 2

right?

[cursive narwhal]

?

If i have a set of vectors S, do you know what span(S) refers to?

Yes or no?

[edgy anvil]

no where in my notes is the word span

let me put it that way 😭

[cursive narwhal]

You should go and learn what it is. It's a very simple concept but it's important in talking about a basis

[edgy anvil]

watching a youtube video about it 😉

[cursive narwhal]

ping me when you're done and still need help with the question

But in any case, since you're familiar with the concept of linear independence

What do you need to check if your given set is linearly independent?

[edgy anvil]

if none of the vectors can be defined from a linear combination of the others

okay so lets pretend for a moment it said R_2(just to line up with the video I watched)

and that the basis was made up of real numbers

A basis spans R_2 if it is made up of 2 real numbers

@cursive narwhal is this an accurate understanding

[cursive narwhal]

? what's R_2?

You mean R^2?

[edgy anvil]

yea sorry

that's what I meant

I was looking at P and sorry the subscript it was a typo

[cursive narwhal]

$\bR^2$ is a different object from $\bR$. The elements of $\bR^2$ are ordered pairs of real numbers while the elements of $\bR$ are the real numbers. Two different things entirely.

[stoic python]

Abhijeet Vats:

[edgy anvil]

R^2 is made up of all tuples of 2 real numbers

right?

then R is a singular real number

[median forum]

ordered tuples

[cursive narwhal]

Well, I'd rather just say it's made up of all of the ordered pairs of two real numbers

[edgy anvil]

okay so thats the more accurate way to say it

rather the accurate way

[cursive narwhal]

You can say it in a way that's more accurate than that but it's not really necessary to go down that path

[edgy anvil]

span basically means that basis B fits the criteria of whatever you are saying it spans right?

[cursive narwhal]

Now, the set {(0,1),(1,0)} is a basis of R^2 because it's a linearly independent set of vectors and it spans the whole of R^2. That is, every vector in R^2 can be written as a linear combination of the vectors in the abovementioned set.

[edgy anvil]

okay

[median forum]

span means every possible linear combination
as a verb, spans means every vector in the vector space you want can be expressed as a linear combination of the basis

[cursive narwhal]

span basically means that basis B fits the criteria of whatever you are saying it spans right?
This is the equivalent of saying that the span of a set of vectors is defined as the thing that behaves like the span of the set

(Or were you using it as a verb lol)

[edgy anvil]

Now, the set {(0,1),(1,0)} is a basis of R^2 because it's a linearly independent set of vectors and it spans the whole of R^2. That is, every vector in R^2 can be written as a linear combination of the vectors in the abovementioned set.
idk but i get what you mean after this

[median forum]

domino, everytime you use spans you need a subject and an object(in the synthatic grammar sense)
B spans V means that every vector in V can be expressed as a linear combination of vectors from B

[edgy anvil]

gotcha

[median forum]

so its always something spans another something

[cursive narwhal]

where the "something" has to be particularly specific

[edgy anvil]

so how does this definition differ when speaking about something spans P_2 rather than something spans R^2.

[median forum]

the vectors in P_2 arent the vectors in R^2

[edgy anvil]

ik

its completely unrelated

but the question i in the end need to answer is about P_2, I watched a video that talked about span in regards to R^2

[median forum]

you need to show every vector in P_2 can be expressed as a linear combination of vectors of B

[edgy anvil]

ik im just confused as to what "fits" in P_2

[median forum]

sure

P_2 is all quadratic polynomials(I assume in R, but they didnt specify)

[edgy anvil]

like P_2 vs P_3 vs P_4 etc

[median forum]

so consider their general form ax²+bx+c for some a,b,c constant

[edgy anvil]

okay so it is what i thought, P_n n being the highest power

[cursive narwhal]

P_2 is the set of all polynomials up to degree 2

[median forum]

so you need to write polynomials of that form as linear combinations of the basis polynomials

(then show they are LI)

[edgy anvil]

so since I have $\beta = { 1, 1 + x, 1 + x + x^2 }$

[stoic python]

Domino:

[edgy anvil]

I would take and get for the 1, a=0 b=0 c=1

I could then take and make like a matrix of this and prove linear independence?

[median forum]

youre probably better off writing the equations you wanna solve for that explicitly

ah

well, you could, but that would be kinda unjustified

ideally, youd show by definition

[gray aspen]

I'm trying to study linear algebra and came across this. I have no idea how to start solving this problem. I'd be very thankful of any help.

[edgy anvil]

Where did I go wrong here?

Let 𝑇: ℙ2→ℙ2 be the linear transformation given by, 𝑇[𝑝(𝑥)]=𝑝′(𝑥)−𝑝(𝑥).
Obtain the matrix of 𝑇 relative to ℬ Obtain the matrix of 𝑇 relative to ℬ.

My Work:

T(1) = (1)' - 1 = 0 - 1 = -1 \
T(1+x) = (1 + x)' - (1 + x) = 1 -1 -x = -x \
T(1 + x + x^2) = (1 + x + x^2)' - (1 + x + x^2) = 1 + 2x -1 -x -x^2 = x - x^2

[T(1)] = $\bmqty{-1 \ 0 \ 0} \
[T(1+x)] = \bmqty{0 \ -1 \ 0} \
[T(1+x + x^2)] = \bmqty{0 \ 1 \ -1}$

$[T]_\beta = \bmqty{-1 & 0 & 0 \ 0 & -1 & 1 \ 0 & 0 & -1}$

I assume something is wrong because on the next part which asks this:

Use the matrix obtained to demonstrate that 𝑇[1+3𝑥+𝑥^2]=2−𝑥−𝑥^2.

I don't get "2-x-x^2" from doing the following:

$\bmqty{-1 & 0 & 0 \ 0 & -1 & 1 \ 0 & 0 & -1}\bmqty{1 \ 3 \ 1}$

If someone knows how to force the bot to make a new line thatd be great to know lol

[torpid portal]

type 2 backslashes

$a \ b$

[stoic python]

Domino:
Compile Error! Click the reaction for details. (You may edit your message)

[neon zephyr]

What is your ℬ? From your calculation it seems like your ℬ is the basis given by ${1,1+x,1+x+x^2}$. Then your $T$ would be the matrix representing a base change from ${1,1+x,1+x+x^2}$ to ${1,x,x^2}$. When calculation $T[1+3x+x^2]$ you have to write $1+3x+x^2$ with respect to the basis $1,1+x,1+x+x^2$ and use these coordinates to calculate to image of $1+3x+x^2$ under $T$.

[stoic python]

Ceana:

[edgy anvil]

you are right about ℬ

how do you know it's to {1,x,x^2}

[neon zephyr]

Because you write the image of each element in your basis with respect to the basis {1,x,x^2}

For example $T(1+x) =-x$. With respect to the basis ℬ your coordinates would be $T(1+x) =1\cdot 1 -1\cdot(1+x)+0\cdot (1+x+x^2)$. Then your coordinate vector would be $(1,-1,0)$

[stoic python]

Ceana:

[neon zephyr]

What you did was $T(1+x)=-x=0\cdot 1 -1\cdot x + 0\cdot x^2$. So you wrote the image of $1+x$ under $T$ with respect to the basis $1,x,x^2$ and not your Basis ℬ

[edgy anvil]

oh i see

i think i get what to do now

[stoic python]

Ceana:

[neon zephyr]

perfect

[edgy anvil]

still not getting the right thing

do those numbers look right

@neon zephyr

[marble lance]

Yes

Oh, nvm, I just multiplied the matrices

Didn't look at context, so idk

[neon zephyr]

Okay. So your coordinates for $1+3x+x^2$ are again with respect to the basis $1,x,x^2$. You need the coordinates with respect to your ℬ. I will explain it this way:
Your matrix $T$ represents your linear transformation 𝑇: ℙ2→ℙ2 with respect to the basis ℬ. So if you want to calculate the image of let's say $v \in ℙ2$ under 𝑇: ℙ2→ℙ2 using your calculated matrix $T$, you have to use the coordinates of $v$ with respect to the ℬ. So in our case
$$1+3x+x^2=-2 \cdot 1+ 2\cdot(1+x)+1\cdot(1+x+x^2)$$. So your the coordinates of $1+3x+x^2$ relative to ℬ are given by $(-2,2,1)$.

[stoic python]

Ceana:

[edgy anvil]

OHHH

i see

[edgy anvil]

idk what happened but my internet went out

right as that happened i tried to say that when I multiply that against my matrix for [T]_B I do not get the answer i am supposed to get still

I essentially replaced (1,3,1) with (-2, 2, 1)

[neon zephyr]

exactly and you should get the vector (3,0,-1)

[edgy anvil]

yeah

thats what i get

oh

i have a feeling ik what youre about to say

thats a linear combination

of my Basis

[neon zephyr]

bingo

[edgy anvil]

bahhh i just realized it

then how would I prove if T is one-to-one in this scenario? Its the last part of the question and im not really sure where to go with it

[neon zephyr]

Do you know what a determinant is?

[edgy anvil]

yes

[neon zephyr]

whats the determinant of $T$?

[stoic python]

Ceana:

[edgy anvil]

-1

[neon zephyr]

what does it tell you about your matrix?

[edgy anvil]

uhhh it's not 0 so it's invertible?

idk thats the first thing i can think of

[neon zephyr]

okay and what does it mean that a function is one-to-one

[edgy anvil]

it will always have a unique output from the transformation. no 2 unique vectors will output the same output

[neon zephyr]

Okay lets's say $p(x),q(x) \in ℙ2$. Let's say that $v$ is the coordinate vector of $p(x)$ with respect to your basis and $w$ is the coordinate vector of $q(x)$. So $T(p(x))=T(q(x))$ is the same as saying $Tv=Tw$, correct?

[stoic python]

Ceana:

[edgy anvil]

ok

[neon zephyr]

now use that T is invertible

[edgy anvil]

so invert T

[neon zephyr]

$T^{-1}Tv =T^{-1}Tw$

[stoic python]

Ceana:

[edgy anvil]

so this cancels out the Ts?

[neon zephyr]

correct

so $v=w$

[stoic python]

Ceana:

[neon zephyr]

which means p(x)=q(x)

[edgy anvil]

thus T is one-to-one

[neon zephyr]

correct

[gilded bobcat]

what does it(the marked dim expression) mean? I've never seen it before.

[marble lance]

It probably just indicates the field

I don't think it's particularly meaningful. Still means the dimension of V is 3

[gilded bobcat]

ok. thank you.

[marble lance]

Perhaps the dimension could be different over a different field? So over emphasis

[gilded bobcat]

might be the case.

[half ice]

Indeed seems to not really indicate anything

[weary isle]

For this toom-cook algorithm at https://arxiv.org/pdf/1803.10986v1.pdf#page=6 , how do I get the value 4/2 in the matrix G ?

[edgy anvil]

Explain what the row reduction tells us and why it shows that the set is linearly independent (based on the definition).

Wouldn't that be dependent though?

because of the last row

Im confused

[median forum]

those are LD ye

[edgy anvil]

supposedly they are independent based on the question

[median forum]

can you show the question?

[edgy anvil]

"Explain what the row reduction tells us and why it shows that the set is linearly independent (based on the definition)."

[half storm]

They are linearly independent because this is an augmented matrix

[median forum]

maybe theyre trying to show the first 3 are

needs context ig

the null vector is always LD

[edgy anvil]

im gonna assume that ius what they mean

[median forum]

what is an augmented matrix

[edgy anvil]

well in this case the matrix is augmented with the zero vector

[half storm]

That last column is a column os zeroes right, if you want to show that the columns of a matrix are linearly indepdent or linearly dependent you solve the matrix equation Ax = 0. Right The matrix in this instance because A is a 4 x 3 matrix yea? $L_A: \mathbb{R}^{3} \to \mathbb{R}^4$. But when you find the RREF what you're finding is $(x_1, x_2, x_3) = (0, 0, 0)$ i.e. the nullspace is the set only containing the zero vector.

[median forum]

is A 4x3?

[stoic python]

JohntheDon:

[half storm]

That's what it looks like to me 4 rows and 3 columns

[edgy anvil]

yeah

[median forum]

there is a zero vector there

[half storm]

That's because it's matrix equation $Ax = 0$

[stoic python]

JohntheDon:

[half storm]

But A is a 4x3 matrix.

[median forum]

am I missing something

[half storm]

So the new matrix is 4x4.

[median forum]

ok, but you are saying that, because of context, right?

[half storm]

No, I guess I'm assuming

[median forum]

well yes

assuming

youre probably correct

[half storm]

because it's saying it's linearly independent.

So I'm kind of making an inference.

It says they are linearly indepednet and the only way that's going to be the case if it's an augmented matrix.

[edgy anvil]

i am assuming you are right

[half storm]

But yea they would certainly be LD if it weren't.

[median forum]

thats why I hate linalg computations with coordinates

[edgy anvil]

so back to what you said before

[ocean sequoia]

but even then isnt it still linearly dependent if its a 4x3?

isnt the fourth row still the zero vector?

[half storm]

No because the last vector is not a column of the original matrix

[median forum]

ye

4x3, not 3x4

[ocean sequoia]

ok but then the fourth row is just 3 zeros

[half storm]

The only way it's going to make sense that the columns are linearly independent if you're interpreting it as an augmented matrix.

[edgy anvil]

so how should i go about answering this from all that...

[median forum]

well, the rows are LD. the columns are LI

row space has the same dimension as column space

if it is 4x3, it has at most 3 dimensions

which it does

but, by convention, we generally assume the columns to be vectors

in most contexts

[ocean sequoia]

wait if the rows are LD then dont the columns have to be LD?

[half storm]

No.

[median forum]

rows are vectors in R3, columns are vectors in R4

[half storm]

[ocean sequoia]

but the rank is the same?

row rank = column rank

[median forum]

yes

3

[edgy anvil]

i only care if the column vectors are linearly independent

pretty sure

[median forum]

ye, they are

no linear combination of the third and first will give the second

[edgy anvil]

but how do i explain that, i still dont totally get

oh

[median forum]

even easier to see that on the right

[ocean sequoia]

show that there is no non trivial combination to get zero

[edgy anvil]

and why does the row reduction not impact linear independence

[median forum]

there isnt

every combination has to be trivial

that is

[half storm]

To say that the rank of the span of a vectors is the same doesn't mean they span the same space.

it just means you need the same number of vectors to span both spaces

[ocean sequoia]

typoed that lmao

[half storm]

But the vectors can be completely different.

[median forum]

$a c_1 + b c_2 + d c_3 = 0 \iff a,b,d=0$ with each c being your columns

[stoic python]

Fractal:

[median forum]

you could give them any name tho. these are terrible names

[half storm]

If you weren't interpreting this as an augmented matrix, then you could say that the columns are LD though. So you probably need to specify that if you interpret this as an augmented matrix, then they are linearly independent yea.

If you are just asking "are the columns of this matrix linearly independent" then they are LD because you've got the zero column vector. Whoever gave you this question needs to be more specific in their wording.

@median forum I require help

[median forum]

@half storm cant guarantee any help, but go ahead

[edgy anvil]

The next part is
Determine whether or not each of the following vectors are in the subspace of ℝ4 spanned by 𝑆. Show work to justify your answers and clearly indicate your answer.
𝐯𝟏=(0,0,0,0), 𝐯𝟐=(1,2,3,4), 𝐯𝟑=(−5,11,5,19)

wouldn't that just be

[median forum]

better send pic

[edgy anvil]

Yes for each

[median forum]

half of these symbols dont load

[edgy anvil]

oh they did for me

[half storm]

I know how to show this is a subset of the solution space and I also know how to show the base case for the strong induction. It's the next part that's pretty hard.
Hold on I'm posting theorem statement. One sec.

[edgy anvil]

...

[median forum]

what S?

the one from prev?

[edgy anvil]

yeah

ignore the zero vector

[half storm]

I'll let you finish this first.

[median forum]

you need to show if there is any linear combination that gives those vectors

[edgy anvil]

v_1 is easy then 0 for each vector in the combination right?

[median forum]

(but notice that the last coordinate is always zero)

ye

0 is always spanned

[edgy anvil]

Express each of the vectors in part 2 that was in the span of 𝑆 as a linear combination of the vectors is the next part. is there some easier way to answer this that im missing that doesnt involve writing out linear combinations

[median forum]

uhmm

which are which?

[edgy anvil]

part 3

image above is part 2

[median forum]

right

you sure those are the only ones?

cause otherwise you just need a)

[edgy anvil]

only ones of what

[median forum]

or maybe the vectors are in the matrix on the left?

not the one in the right

[edgy anvil]

they are the ones on the left

yeah

[median forum]

(also, you need to brute force linear comb. no other way)

by brute force I mean solve systems

assign a variable to each vector representing their constant in the linear combination and sum them to solve a system

so for example

say I had the vectors

(1,2,3,4) and (4,3,2,1)

and want to see if (3,3,3,3) is in there

then

a(1,2,3,4)+b(4,3,2,1)=(3,3,3,3) needs to have a solution

meaning

a+4b=3
2a+3b=3
3a+2b=3
4a+b=3

if there is no solution then (3,3,3,3) is not in their span

[wintry steppe]

It's like getting around in Chicago. Chicago's streets form a grid (which is like your standard [1, 0] [0, 1] basis). But there are also diagonal streets that point towards downtown (which would be like [1, -1] if you're on the north side). The question is, can you get from downtown to anywhere on the north side of Chicago without using the streets that run east-west?

[ocean sequoia]

imagine using chicago and not NYC for that example smh

[median forum]

what an odd analogy

[wintry steppe]

In other words, do the North-South and north-northwest diagonal streets span the north side of Chicago 🙂

[edgy anvil]

i live near nyc lmao nyc > chicago

[wintry steppe]

Yeah, but that's a nonlinear coordinate system.

[ocean sequoia]

@half storm did you need to ask a question or do you mind if i fire one away?

[half storm]

Go ahead because mine may take a sec.

[ocean sequoia]

if U is invariant under T does that mean that T must have eigenvectors?

[wintry steppe]

@ocean sequoia Can you think of any examples in 2 dimensions where U is all of R^2?

[ocean sequoia]

i think im overthinking what you are asking

are you asking if U can be all of R^2?

[wintry steppe]

I'm saying, you should try out "trivial" or "degenerate" cases to see if things hold there, where they are easy to calculate or explore.

With every operator T : U → U, you have at least two trivial invariant spaces: all of U and {0}

[median forum]

well
U contains{0} so o,o

[ocean sequoia]

if U is invariant under all L(V) then U is either V or 0

or is that the same thing?

[median forum]

ah

thats totally different

[wintry steppe]

I'm saying if T : V → V is an operator, and you let U be an invariant subspace, consider the case where U = V.

(Sorry, I was screwy with my wording above I think)

[median forum]

U cant be V, otherwise anything different than the identity would not do that

find the general form of linear maps that only preserve {0}

[wintry steppe]

Truefact

[median forum]

if you can show they exist, then U={0}

[ocean sequoia]

honestly i think i need to reread the chapter on eigenvectors/values

thanks anyway

[half storm]

I mean I don't know much about eigenvectors but you can probably do something about thinking about the restriction of T onto U and $T_u: U \to U$. Then consider a matrix representation for $T_U$ with respect to a basis $\beta$ for U

[stoic python]

JohntheDon:

[median forum]

looks hard

[half storm]

Yea I was thinking that

might be worth not thinking about what I said.

[ocean sequoia]

honestly my question might be too complex for what i was even going for

it was just off the cuff

[half storm]

I was gonna say then look at $det({[T_u]}_{\beta} - \lambda )$

[ocean sequoia]

its not even an exercise

[median forum]

there is certainly a difference about asking which U is invariant under a certain linear map and under all of them

[half storm]

but that might be a bit too much and not worth your time.

[median forum]

eigenvectors dont refer to the vector space
theyre specific to your linear transformation

unless you mean eigenvectors of L(V)

[ocean sequoia]

i know that if U is invariant under all of them it must be V or {0} right

LADR had that question and i had to look up the proof i couldnt figure it out

[median forum]

how would it be V?

[stoic python]

JohntheDon:

[median forum]

if youcan find a single linear transformation that doesnt make V invariant wouldnt that automatically exclude V?

[ocean sequoia]

sorry thats why i said V

[half storm]

This is different than what I thought you were asking.

[median forum]

T=0

[ocean sequoia]

its completely different

[median forum]

U is not invariant under everyT anymore

[ocean sequoia]

honestly my question more came from my reading of upper triangle maticies but i didnt expect this to happen

its not a well fomulated question tbh

@half storm you can ask your question i think i need to read a bit more

[median forum]

wait

when you said every T, you meant endomorphisms?

[ocean sequoia]

honestly i dont even know what endomorphism means

[wintry steppe]

Morphism just means "linear map". Endo means "from itself to itself"

[half storm]

I map from a mathematical object into itself. So the identity map is the obvious first thing to think of.

[ocean sequoia]

oh

[wintry steppe]

So endomorphism just means linear operator T : V → V 🙂

[ocean sequoia]

neat

like isomorphism is a map to a space with the same dimension?

[wintry steppe]

Almost

Isomorphism just means a morphism with an inverse.

[ocean sequoia]

ahhh

[median forum]

in any case, T=0 should be a counter example to everyhing

homomorphism*

[ocean sequoia]

so endomorphism = operator?

[median forum]

also, not always
thats only true for vector spaces

[wintry steppe]

"between spaces of the same dimension" means "square matrix", while "isomorphism" means "invertible (square) matrix"

[median forum]

homeomorphisms are not continuous functions with inverses

[wintry steppe]

For what you're learning right now, @ocean sequoia, yes, they are always the same.

[half storm]

Do you mean homomorphism?

[median forum]

ye, tbey the same

[ocean sequoia]

[wintry steppe]

But in math, you can always generalize things to invalidate any truth you find 🙂

[median forum]

nah, I mean homeo
counter example to isomorphism being a morphism with an inverse @half storm

[wintry steppe]

isomorphism just means "a morphism with a two-sided inverse morphism which combine to give the identity morphism". That definition holds always.

you just swap out what kind of "morphism" you like.

[median forum]

oh ye, but it can be invertible and not have an inverse morphism

[wintry steppe]

whether it's linear, continuous, measure-preserving, whatever you like

yeah.

inverse-as-a-morphism, not inverse-as-a-function

[median forum]

right

[wintry steppe]

I love how arguments between math people always end up with agreement.

🐸

[median forum]

unless its about foundationalism

[half storm]

Because there is almost always only one answer.

[wintry steppe]

well, that's not math. that's religion, isn't it?

[median forum]

what is

[wintry steppe]

clings to his univalent type theory

[ocean sequoia]

wait so the whole point of upper triangle matrices is that they are easy to use as operators and have the eigenvalues on the diagonal

honestly i feel like im missing something in this section

[wintry steppe]

I forget exactly how to think of them, but don't upper-triangular matrices give you a nested chain of subspaces, each invariant under the transformation?

[ocean sequoia]

that seems like what you aare saying

is c yea

[wintry steppe]

Oh, is that LADR?

[ocean sequoia]

yea

[wintry steppe]

Yeah. I like that one. (although the proofs always seem much less clear than they could be)

[ocean sequoia]

i agree

sometimes i get lost in them

[wintry steppe]

Yeah, like, you get an ordered-bassis, v1, v2, v3, ..., vn of the vectorspace

and you can pick as many as you like, as long as you pick them in order, and the span is an invariant subspace.

eigvals on the diag

So in particular, you can quickly see if it's invertible

[ocean sequoia]

so its invariant for every v_j right in V?

[wintry steppe]

Not quite

err

well, depends what you mean

span{v1} is invariant

span{v1, v2} is invariant

span{v1, v2, v3} is invariant

etc

[ocean sequoia]

but span{v2} isnt invariant?

[wintry steppe]

correct. You can't expect v2 to be invariant

[ocean sequoia]

ok

so it has to be ordered

wait thats kinda weird so would

span {v2,v1} be invariant?

it should right

[wintry steppe]

Well, span {v2, v1} is the same as span {v1, v2}

[ocean sequoia]

yea

[wintry steppe]

You can also think of it like this:

define U_i = span{v1, v2, ..., vi}

[stoic python]

OrangeTacTics:

[ocean sequoia]

and here U1 = span{v1}, U2 = span{V1,V2}... etc?

[wintry steppe]

yep

[ocean sequoia]

thanks!

[wintry steppe]

And the cool thing is every matrix can be made upper-triangular. Even without leaving the Real numbers.

... I think... let me check that

no nevermind

if you have a complex eigenvector, you have a complex number on the diag

but I think all complex matrices can be upper-triangled

yeah. the statement is, in fact, that as long as all of the eigenvalues are valid scalars (so, if they are all real), you can upper-triangle a matrix.

[edgy anvil]

is it possible for a 5x7 matrix to form linearly independent rows and/or columns? Wouldnt this be true for both?

[wintry steppe]

what does 'and/or' mean here?

[edgy anvil]

i mean like it needs to be answered for both

but they dont necessarily have to have the same answer

[wintry steppe]

if you have a 7 columns, treating each column as a vector in R^5, then you have 7 vectors in R^5, right?

[edgy anvil]

yeah

[wintry steppe]

Work with smaller numbers for intuition

Imagine the question was a 3x2 matrix

Then you would have 3 vectors in R^2

[edgy anvil]

columns can be LI but not rows

right

[wintry steppe]

I'd think so.

[edgy anvil]

I have a Trapezoid with the points (-2, 0), (1, 0), (1, 1), (0, 1). I need to find the standard matrix for orthogonally projecting onto the line y = x.

So would I find that T(x, y) = (y, x)
Then find T(1, 0) and T(0, 1) and use that to make a matrix?

[half storm]

You mean a triangle?

[edgy anvil]

i missed a point

fixed

is that the correct approach to the problem

or am i misunderstanding the question

[marble lance]

Why is T(x,y) = T(y, x)?

[edgy anvil]

So like y = x? idk that was my best guess as to what id do

[marble lance]

Then T(1,0) is (0,1) which is not on y = x

[edgy anvil]

yeah...

thats why i asked

im not sure how to project to anything but the axes

[marble lance]

Orthogonal projection, means you have to project from your point to y = x so that the line from the old point to the projected point is perpendicular to y = x

[half storm]

I'm kind of confused like what points are you projecting? The set of all points that lie on the edges of the trapezoid?

[marble lance]

I don't understand the purpose of the trapezoid either

[edgy anvil]

the points are vertices

after I find [T]

[edgy anvil]

I have to graph it

[marble lance]

It's dark so my lighting is bad

Consider the point (1,0)

You need to find a point on y=x, so of the form (x,x), so that the line from (1,0) to (x, x) is perpendicular to y = x

Does that make sense?

[edgy anvil]

okay

[marble lance]

So when are two lines perpendicular? When the product of their gradients is -1

So what is the gradient of the line from (1,0) to (x, x)?

[edgy anvil]

-1?

[marble lance]

Yes, you need it to be -1

[half storm]

When he says gradients he means the slopes of the lines.

That might be easier terminology.

[marble lance]

So now find it using the two points (x, x) and (1,0) and the gradient formula: m = (y2 - y1)/(x2 - x1)

And then set what you get equal to -1

And solve for x

Then that gives you the orthogonal projection of (1, 0) onto y = x

[edgy anvil]

I get what you're doing but to get full credit

he wants me to make a matrix

lemme copy paste the whole thing

one sec

[marble lance]

NO

I understand

How do you make the matrix?

[edgy anvil]

okay lol

[marble lance]

You first transform the basis vectors (1, 0) and (0, 1)

[edgy anvil]

ohhh

[marble lance]

And use that to make the matrix

[edgy anvil]

okay

i see

[marble lance]

So I did (1,0), can you do (0, 1) while I go sleep?

[edgy anvil]

i think i can figure it out

[marble lance]

It's basically the same thing

Okay, good luck

[delicate zealot]

Does a matrix representing a linear transformation in $\R^n$ have n eigenvalues? Or is that not necessarily true?

[wintry steppe]

it will have n (not necessarily distinct) complex eigenvalues

the matrix does not have to have real eigenvalues, e.g.

0   -1
1   0

also you should say linear operator, if you want to discuss eigenvalues

the reason is that the term operator is reserved for linear transformations from a vector space to itself, in which case the notions of eigenvalues and eigenvectors make sense

@delicate zealot It's an interesting topic when it comes to counting eigenvalues, by the way.

elaborate

[prisma pier]

^

[wintry steppe]

The 2x2 identity matrix, for instance. We know "1" is an eigenvalue. But how many eigenvalues does it have?

[delicate zealot]

Oh dang yeah cause characteristic polynomial could return complex results yup yup thanks for the catch

[prisma pier]

hmm so is it debatable whether you count it once or twice? @wintry steppe

[wintry steppe]

Yes.

There are at least three meaningful ways to "count" eigenvalues.

The two you mentioned being two of them.

(The other having to do with jordan blocks)

[prisma pier]

isn't the number of jordan blocks the same as the number of distinct eigenvalues? (I don't know much about them so I might be wrong)

or do you mean counting them in a different way

[pallid rampart]

No that's definitely not true, consider the identity matrix

[wintry steppe]

Each jordan block has an eigenvalue associated to it. So you can turn that around and ask, looking at all the jordan blocks associated to a fixed eigenvalue λ, how many blocks are there?

[prisma pier]

oh wait

[wintry steppe]

$\begin{bmatrix}
1 & 1 & 0\
0 & 1 & 0\
0 & 0 & 1
\end{bmatrix}$

[stoic python]

OrangeTacTics:

[wintry steppe]

Depending on how you count, this matrix might have 1, 2, or 3 eigenvalues.

well, not quite

I forget how it works out exactly. But you end up with two notions of "multiplicity" when you do it properly.

[prisma pier]

that's really interesting

[wintry steppe]

when you have vocabulary such as "distinct eigenvalues", "algebraic multiplicity", "geometric multiplicity", this feels like it boils down to a semantics issue

[prisma pier]

though is the general convention to say 3 eigenvalues here?

[wintry steppe]

where can i read about this @wintry steppe ?

What you said is the more appropriate phrasing, @wintry steppe

I wish I knew a place it was written about concisely. If I did, maybe my own thoughts wouldn't be so hazy on it 😅

i've always seen "n eigenvalues" to mean "n not necessarily distinct eigenvalues", and then if you want to specify how many times each one shows up you use algebraic multiplicity

[prisma pier]

is geometric multiplicity then the number of jordan blocks, or is it different?

[wintry steppe]

Right, but then what does "n" mean 🙂
(I think it's usually meant counting with their algebraic multiplicities)

yeah

honestly to give a precise meaning to "n eigenvalues" is probably unnecessary when you can specify all you want using extra words

Yeah.

is geometric multiplicity then the number of jordan blocks, or is it different?
yeah i think so

number of jordan blocks corresponding to that eigenvalue

[prisma pier]

oh i see thank you

[wintry steppe]

i might be wrong since it's been a long time since i studied jordan forms, so you should look it up and see if it's right

something something minimum polynomail

and "eventual kernels", where you take the kernel of the matrix exponentiated by a large number...

those are called generalized eigenspaces in some books

i believe

ker (A - lambda I)^N?

I think I made up the word "eventual kernel" because I find the word "generalized" offensively imprecise.

but yeah, that's what LADR calls them

haha that's a good reason

[edgy anvil]

This was a bit back, but I have T(1, 0) = (1/2, 1/2). would that be accurate?

When I try to calculate T(0, 1) I get 0 = -1 so im doing smth wrong

for T(1, 0) I did $-1 = (x_2 - 0)/(x_2 - 1)$ and solved for $x_2$

[stoic python]

Domino:

[edgy anvil]

when i do the same for (0, 1) x_2 gets canceled

[median forum]

you can build a system with the points of the polynomial

and that system will have (a0,a1,a2,a3)

meaning there will be a matrix A^-1 which you can multiply by it

to get (b0 b1 b2 b3)

[ocean sequoia]

hey quick kinda a dumb question but for Ax=b is A the linear transformation or is x?

[pallid rampart]

Ax=b is an equation

But a linear transformation can be represented as a matrix

[ocean sequoia]

honestly i am embarrassed to admit Ive forgotten it

But a linear transformation can be represented as a matrix
so its literally just a system of equations

[pallid rampart]

No

[ocean sequoia]

not a matrix

[pallid rampart]

Yes Ax=b is a system of linear equations

[ocean sequoia]

ok so i was just over thinking it

[pallid rampart]

Linear transformations are functions

[ocean sequoia]

yea

[pallid rampart]

Ax=b is an equation

[ocean sequoia]

there isnt necessarily like a basis vector or anything involved here

this isnt a change of basis

[pallid rampart]

??

[wintry steppe]

"Transform" or "Transformation" or "map" or "mapping" or "function"... all of these words mean the same thing.

So with Ax = b , A is the function acting on x. So A is a transform.

[ocean sequoia]

yea so Ax=b is just a linear mapping of x to b?

[wintry steppe]

no

Ax = b is an equation

(I see @pallid rampart wrote this above lol)

[ocean sequoia]

so then wouldnt A have to be the linear transformation from x to b?

[wintry steppe]

You can say: "A maps x to b"

When you say a linear map is "from something to something else", you're usually talking about the domain and codomain

like maybe: "A maps R^3 to R^2"

which is written as A : R^3 -> R^2

[ocean sequoia]

yea ok sorry thanks kinda a dumb question i appreciate it

felt like i shouldve known that

[wintry steppe]

Don't worry about it. The language you use is something you have to pick up over time.

[ocean sequoia]

[wintry steppe]

If you use the wrong language, it makes things ambiguous. So it's good to spend some time learning the right way to word things in math. (Because a slightly different wording change completely change the meaning)

[lilac stratus]

you said "change" twice :p

[ocean sequoia]

yea ive noticed that i try to use specific language when asking questions so someone can point out the errors

[wintry steppe]

An example from earlier today, I forget who, but someone was slurring their words slightly with eigenvalues. I forget the exact wording, but it was something like "The eigenvalues of the vectorspace"

but vectorspaces don't have eigenvalues. Neither do vectors 🙂

[ocean sequoia]

thanks i appreciate the kind words 🙂

[pallid rampart]

eigenvalues of a vectorspace

[wintry steppe]

Yes. Eigenvalues are the vectors which don't change direction when they are changed @pallid rampart 😮

[pallid rampart]

[wintry steppe]

hehe

[pallid rampart]

Eigenvalues are the vectors

Eigenvalues are vectors

[bitter hornet]

hey guys I forgot what this notation means, can someone help me out?

[pallid rampart]

The matrix of the linear transformation with respect to the basis B and B?

That'll be my guess but it's probably defined in the book

Or you can check the end of the book for a definition

Usually all the notations are at the end of the book with the page number where it is first defined

[gray dust]

yes the matrix rep of a linear operator wrt B

[wintry steppe]

If it's LADR, then I think that's correct.

most likely correct regardless

[minor obsidian]

For the T in part b is that some kind of operator

Or does that just mean it’s symmetric

[bitter hornet]

T means the transpose of the matrix

[minor obsidian]

is that hard

[wintry steppe]

I'd say it's pretty hard. You have to do double-replacement of indices.

[minor obsidian]

crap lol

[wintry steppe]

it just means "turn the matrix sideways" 😛

Rows become columns and columns become rows.

[minor obsidian]

i have an hour and 6 minutes to teach myself this stuff and then take a quiz lol

lets hope i can do it

[wintry steppe]

best of luck

[bitter hornet]

gl

[minor obsidian]

thanks lol, one other thing, is the Gauss Jordan reduction hard?

[bitter hornet]

takes a bit of practise but not too bad overall

[pallid rampart]

What's Gauss Jordan reduction?

Is it just row reduction?

I've never heard it being called as that

[bitter hornet]

i think he meant guass jordan elimination

[median forum]

for every linear transformation you have a matrix and vice versa

[jaunty grove]

does simultaneous diagonalization for two commuting hermitian matrices A and B always work with the unit eigenvalue matrix of AB?

If I have U unitary whose columns are the eigenvectors of AB

Can I always have $D_A=U^\dagger AU$ and $D_B=U^\dagger BU$ ?

[gray dust]

@jaunty grove that's precisely what A & B simultaneously diagonalizable means, you can build a unitary transition matrix for A & B using the same orthonormal eigenbasis

[jaunty grove]

yes, but does this method work for computing the unitary transition matrix? or does it have to be something else

[gray dust]

@jaunty grove can be a nice little exercise actually

if you already showed a pair of hermitian operators commute iff they share an orthonormal eigenbasis, then show the same for a set of ANY number of pairwise commuting hermitian operators

& additionally show that, assuming A*=A & B*=B & A,B commute, that {A,B,AB,BA} can be simultaneously diagonalized

[weary isle]

in https://www.12000.org/my_notes/circular_convolution/index.htm , could anyone explain how to go from step 5 to step 6 ?

[potent wraith]

Is the scalar the 1x1 matrix?

[dire thunder]

no

scalar is a number

if scalar would be a matrix aA would be allowed only for 1xn matrices A

[potent wraith]

@dire thunder I don't understand your last line

OK, you talk about mutliplication

I see now.

What confused me a bit is that when you multiply nx1 (matrix multiplication) 1xn you basically get 1x1 matrix with the value of the scalar product.

So I thought, OK, a vector scalar product is just a case of matrix multiplication. But this is not correct.

Right?

[dire thunder]

recall that if u want to dot product of vectors u multiply V and V^T and get scalar

[potent wraith]

@weary isle In circular convolution you wrap your bottom sequence around to the right side. So all the elements that go beyond index 0 (that is less than 0) are wrapped to the other side. So if the size of the convolution is 5 (from 0-4) the element on -1 is wrapped to the position 4, the one at -2 to the position 3, and so on.

recall that if u want to dot product of vectors u multiply V and V^T and get scalar
@dire thunder Yes I know that. I thought maybe this can be made a special case of matrix multiplication. But this is incorrect. It is not a special case of matrix mutliplication. Matrix multiplication has nothing to do with scalar product of vectors. (except that the value of element (i,j) is the scalar product of vectors on position i from frist matrix and position j from second matrix)

[weary isle]

@potent wraith What do you mean by -2 ?

[potent wraith]

-2 is the element from the second array that (when shifted) moved back to position -2. The element 9 on the above picture.

[potent wraith]

@weary isle

[weary isle]

@potent wraith okay

[potent wraith]

@weary isle Actually when reading again, I don't see what is unclear. You just take the left hanging part (the part left of the vertical line) and you put it in the gap on the right.

[half storm]

Can we can consider a differential operator for a homogenous linear differential equation with constant coefficients as a linear functional on the solution space of the DE?

Since such an operator takes elements of the solution space and maps them to 0.

[weary isle]

Could anyone help with https://www.reddit.com/r/algorithms/comments/i4k6er/modified_toomcook_algorithm/ ?

[real stirrup]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[spice storm]

is there a short way to find a 4x4 determiant to show is invertible?

[limber sierra]

for "nice enough" matrices

ie if theres a convenient row/column

in the general case, not really

row reduction to a diagonal matrix might be your best bet

[spice storm]

I can do that? so have the first two columns be on the right side?

[limber sierra]

not sure what you mean

row reduction changes the determinant but in predictable ways

so as long as you "remember" how you changed the determinant

you can "undo" those changes

to get the determinant of the original matrix

[spice storm]

so like row reduce?

[half storm]

I mean the row reduction of a matrix does affect the determinant yea?

[spice storm]

SO the last row should be zero and that will make it invertiale?

[limber sierra]

yes john but not in a meaningful way here

[spice storm]

because i did the row reduct and I have 0 on the last rows

[limber sierra]

we just need to determine whether the determinant is 0 or not

[half storm]

But none of the transformations caused by the row reduction change it's invertibility

yea

I remember now

If you express the reduction as the product of elementary matrices then they just change it by scaling factors ; non of which are zero.

[limber sierra]

anyway, if you can row reduce it to have a 0 row, that means the matrix is noninvertible

this fact doesnt require a knowledge of determinants though

[spice storm]

This is what I have when I did the row RREF to the 4 x 4 matrix

[half storm]

oh you listed the rules.

I see them now

lol

[half storm]

That is noninvertible then.

row of zeroes.

[spice storm]

So the matrix I have is not invertile?

[half storm]

Yup

You can see this by doing a cofactor expansion on the 4th row.

So you're original matrix is noninvertible by the remarks of @limber sierra

[spice storm]

so this matrix is saying is noninvertible but the practice problem says show its invertib;e

I will get this. Thank you

[half storm]

You might have done your row reduction incorrectly

[spice storm]

yea, that is what I am checking

[half storm]

What's the original matrix that you are trying to determine whether or not it's invertible?

[wintry steppe]

sounds right to me @wintry steppe

Yes

It might be kind of difficult

unless you copy and paste your gif above 😛

oh

for nxn

you have to use a verbal description

or if you're feeling saucy, a formula

That's the exercise, isn't it?

Even though each matrix is 3x3 in your example above, you can describe it much more succinetly by giving the (i, j) row and column index of the "1" entries.

Or.... for a first pass... just try and explain in words how you would have someone write down the basis.

Good idea. Try to find a handy notation suitable for the problem.

[marble lance]

@wintry steppe You don't have to write out the matrices explicitly, just define their entries

So let $A_k (1 \leq k \leq n)$ be defined by $(A_k) _{ij}$ be 1 if i = j and 0 if i ≠ j

[stoic python]

Lunasong:

[marble lance]

This would be the n matrices that take care of the diagonal elements

Try to think of a way to represent the basis elements that take care of the off diagonal elements

I mean, it's sufficient for me because I understand what you mean

It's sufficient as long as the person reading it can understand what you mean, whether it will be sufficient for your professor (assuming you are in a class), idk

[prisma pier]

the diagonal needs to have all 0s and the elements on opposite sides of the diagonal need to be the negatives of each other

yes

the diagonal elements need to be 0 for a matrix to be antisymmetric because their position remains fixed under the transpose, so if you have a diagonal element a, you get the equation a=-a, which implies a=0

oh haha I misread your question but I'm glad it all worked out

[prisma pier]

Is that correct?:
@wintry steppe Not quite. Test the first matrix for example and see that it is not antisymmetric

looking good

(assuming that this is for all non-diagonal elements and not just the ones directly above and below the diagonal)

[wintry steppe]

Does anyone have a good resource that explains how to find the number of solutions to a linear system

Specifically how to check if it has no solutions, one solution, or infinite solutions.

[wintry steppe]

@wintry steppe thanks, that seems like a good explanation, I will read it in more detail

I need to show that v1, v2, and v3 form a basis of R^3

I combined all three and put that matrix in rref, which gave me the 3x3 identity matrix, therefore they are linearly independent.

Now I need to show that they span R^3, or is it sufficient to show that they are linearly independent ?

[spice storm]

well if you know is a basis of r^3 then there is a theroem that says the span is R^3

[wintry steppe]

I don't know it's a basis, tough.

I want to show that it is.

But I'm missing the span step.

[half storm]

@wintry steppe you’re good to go because there is a theorem that states any linearly independent subset of a vector space that has a cardinality I.e number of vectors equal to the dimension of the space is a basis for it

So you can say it also spans the space as well.

Since R^3 is a 3 dimensional vector space and you found a set in R^3 of exactly three linearly independent vectors. It’s a basis.

[wintry steppe]

I gotcha, thank you!

[wintry steppe]

T(x,y,z) = (x + e^y + 3z , 2x + y , y + 10z)

For this transformation, is it enough to say that it is not linear because e^y is not a factor of x, y, or z?

[half ice]

What's the definition of a linear function?

It's not linear because T(u + v) ≠ T(u) + T(v) @wintry steppe

However, good call - the e^y is the big reason why it's not linear

[wintry steppe]

What's the definition of a linear function?
@half ice
a linear function is one such that $f(c\mathbf{v}) = cf(\mathbf{v})$ and $f(\mathbf{v} + \mathbf{w}) = f(\mathbf{v}) + f(\mathbf{w})$

[stoic python]

polynomial:

[wintry steppe]

no

how can i know that

please don't insult me

[half ice]

Agreed it's not necessary to berate, but yes I wasn't looking for the definition

Not everyone is English speaking

[cursive narwhal]

can we not have this discussion here

[wintry steppe]

i actually do not insult people

[dusky epoch]

@warm briar poly gets a kick out of answering questions not directed at them

[median forum]

I think theyre trolling and trying to be funny
which isnt inherently bad, but they do it with people that clearly dont like their jokes

which is the easiest way to get mod slapped in any server :v

[tribal nebula]

Yup

[wintry steppe]

how am i trolling

i answer questions

i never have insulted anyone, even though whoever has insulted me like 20 times