linear-algebra

[wintry steppe]

So you used the book as the primary resource and then referred to the lectures?

[scarlet quartz]

So it's slightly easier on you

[wintry steppe]

right

[scarlet quartz]

Yes

[wintry steppe]

Mhm okay.

[scarlet quartz]

In the first unit, I went ahead with the lectures and then solved suggested problems, Unit 2 however was a lot better since I read the sub topics first... After which watching the videos on 2x becomes fun rather than a task

@wintry steppe

I'm doing unit 2 rn

Almost done

[wintry steppe]

Nice.

So I'm going to read the chapters of the book first and then refer to the lectures I think. Plus watch 1brown3's (I think that's his name) videos at the same time.

Thanks for the help.

[wintry steppe]

Oh actually, what was your mathematical level going into it?

[scarlet quartz]

Well I knew almost nothing about linear algebra, but I had learnt about matrices and determinants, I had also studied 3-D geometry... I basically knew the fundamentals(but lightly). I was weak at math though since I did not practice it much.

if you want to know about my formal education

Then I just entered my undergrad course

I had studied science up until that time

[wintry steppe]

The only Maths I have is high school Maths plus some BC Calculus. Do you think that'll be enough?

[little cliff]

It is sometimes easier to learn rigorous proofs in a subject if you have already developed some familiarity with the subject through a more computational approach. You might want to supplement it with a book that is just dedicated to matrix algebra. I learned matrix algebra out of the book by Lay.

[scarlet quartz]

yes I do think that is enough if you are diligent with the coursework and the textbook

do the problems well, break your head over the concepts until you understand them and you're good to go

[ionic lion]

What are real life applications of solving systems of linear equations in large number of variables? One use I can think of is polynomial interpolation, which can be used to describe datapoints.

[nocturne jewel]

Kirchoff Laws?

[ionic lion]

@nocturne jewel Ah, thank you

[onyx palm]

when am i allowed to add columns in a matrix

just when calculating det right?

[teal grotto]

in what way were u thinking

[onyx palm]

you know when you calculate a det and you try to make zeroes

you are allowed to do that by adding columns so you can use the laplace thing to get the determinant easier, along this way

[nocturne jewel]

row operations change the determinant, so as long as you keep track of the changes you can get det(A)

[atomic relic]

Linear operations on matrices don't change the rank of the matrix so they are always allowed if you are simply seeking the rank via diagonalization. This is easy to see if you understand the contents of a matrix as a system of vectors. Linear operations on vectors do not change their linear dependence/independence

All matrices are diagonalizable through elementary row operations.

[hollow finch]

column operations are a valid way of simplifying determinants yes. doing so is equivalent to multiplying the original matrix on the right by an elementary matrix.

multiplying a matrix by a column vector on the right is the same as taking a linear combination of the columns. so the columns of AB can be obtained by taking linear combinations of the columns of A

so col 1 of AB is Ab1, col2 is Ab2 etc.

if b1 is the first column of B

[twilit relic]

@hollow finch I also love ooler

So you're telling me that you can operate on the right without changing the determinant?

[hollow finch]

well yeah

det(AB)=det(A)det(B)

same as det(BA)=det(B)det(A)

you adjust the determinant the same doing it from the left and right

another way to look at it is that the transpose does not affect the determinant

and column operations are just row operations on the transpose

[twilit relic]

This is a little confusing to me.

I get that we cannot operate on the rows without changing the determinant, but we can operate on the columns without changing it.

I think of operating on the right as effecting cols and on the left as manipulating rows. Is that incorrect?

[lavish jewel]

both row and col operations have an effect on the determinant

[leaden tide]

It depends which operations though

[lavish jewel]

adding a scaled version of another row or col does not

but simply scaling or swapping multiplies by some factor (the scaling factor, or -1, respectively)

[leaden tide]

by the way, what are the name of each of those operations ? I don't know them in English

[lavish jewel]

i think they all fall under "elementary XXX operation", with XXX as row or col

[leaden tide]

In French they're respectively called "transvection", "dilation" and "exchange"

[lavish jewel]

i think transvection is something else in english

[leaden tide]

i've looked at like 4 different pages, and none of them give any of those names

[lavish jewel]

i had never heard them in this context, yeah

[hollow finch]

I never said it wouldnt change it. I said it would change it identically in the exact same way

Because column operations are still just row operations, just on the transpose which has the same determinant

it's better to think of row reducing to calculate a determinant as applying a linear transformation to make it easier to compute. specifically left multiplying by an elementary matrix. column operations are the same thing but it's right multiplication

[hollow finch]

In English we usually either describe the operation explicitly every time (doing what with which row(s)) or sometimes label them as the first/second/third type of row operation. i like the idea of giving them appropriate names like you describe better tbh.

@twilit relic actually looking back i didn't specify that. I thought when you said "without changing the determinant" you meant that you'd still keep track of how the operations change it like you do with row ops. my bad

[forest quiver]

Alright so I am still new to linear algebra

and i am having trouble with the "free variable" concept

Why do we still keep this equation in the system?

I mean 0=0 yes sure

We don't need to know that

We can just see that the solution can be any one of these points

Why do we put stuff in for z anyway ugh

It says "ignore the third equation; it offers no restriction on the variables"

[hollow garnet]

Read the last sentence

[forest quiver]

Also, I understand that if there are 3 variables, then there need to be 3 equations to find a solution (a single point), but here there are just a wide range of solutions

Yeah I undersatnd that

But why are we allowed to do that ?

0z=0

So z can be anything?

But by that logic 0x=0, x can be anything? 0y=0, y can be anything?

I feel like I'm not making much sense, but I just don't get it

[leaden tide]

precisely that

[forest quiver]

Yes but what about x and y then?

I guess there are other equations for those variables

[leaden tide]

they'll depend on whatever z you decide to pick

[forest quiver]

I see kind of

Because I mean

for there to be a solution to 3 variables

there has to be 3 equations

3 planes on the xyz plane

so a point can be formed

The solution can't be a line

am I thinking correctly?

[leaden tide]

But having three equations of three variables doesn't mean one solution

[forest quiver]

Well if the matrix is consistent

Or equatiosn

[leaden tide]

if they're linearly independent, then yes

[forest quiver]

hmm

i kind of understand linear independence, but only in 2 dimensions

[leaden tide]

Take a line, and imagine three distinct planes passing through it. That's what happens when you have one free variable

[forest quiver]

But if I have 3 variables

then the free variable is a plane

not a line

[leaden tide]

If you have only one coordinate, how are you going to describe anything larger than a line ?

[forest quiver]

I don't follow

But I think I understand now

[leaden tide]

the plane called "x=2" is the set of all triples (x,y,z) satisfying x=2

[forest quiver]

yes

[leaden tide]

You only have one variable being constrained here

[forest quiver]

Right

[leaden tide]

That leaves you with two variables that can take on any value

That makes two free variables

aka a plane

[forest quiver]

yes yes

For a system of 3 equations

there has to be 3 planes intersecting

So for this

Once I solve this into echelon form

[leaden tide]

6 variables ?

[forest quiver]

Augmented matrix

There is bound to be 2 free vars.

right?

[leaden tide]

there is going to be at least three

i think

[forest quiver]

because there are 5 total variables, but only 3 restictions

[leaden tide]

oh, i see

then yes, at least 2 free variables

[forest quiver]

For a system of 5 vars. there has to be 5 planes or whatever these are

[leaden tide]

3 variables -> 2d plane

[forest quiver]

We don't have restrictions on 2 of them because it's small

[leaden tide]

so for 5 variables, you get 4d hyperplanes

[forest quiver]

Wait whaaat

3 variables is xy plane???

[leaden tide]

x+y+z=0 defines a plane

[forest quiver]

Yes

[leaden tide]

x+y+z+w=0 defines a volume (3D) in 4D space

x+y+z+w+t=0 defines a 4D hyperplane in 5D space

[forest quiver]

Oh right

Im not going to worry about 4d intuitively yet XD

But I think I am getting the idea

My next concern is how do I decide which my 2 free variables are?

[leaden tide]

That's on you

[forest quiver]

Do I just isolate every variable I have interms of the other variables, and whichvere variables I can't isolate I will just make them free?

[forest quiver]

Ok got it

[leaden tide]

You can rearrange the equations to make any two variables your parameters anyway

[forest quiver]

Got it

Any way I rearrange them, I will preserve my solution still right? I'm not tinkering with the actual equations]

[leaden tide]

Usually the latest ones are chosen to be free variables, but that's up for choice

[forest quiver]

I have a lot to learn here

BUt thank you for the chat you are very helpful

[leaden tide]

Small geometric interpretation for a second

Wait

[forest quiver]

Yes

[leaden tide]

Actually no, you don't really choose your free variables

[forest quiver]

Ok got it

[leaden tide]

Otherwise that'd mean the x=5 and y=5 planes are the same

Which is... not quite true

[forest quiver]

I'm not there with you on the intuitiion yet, but let me play around with the graph

[urban trail]

my bad on the interruption

[forest quiver]

np

[leaden tide]

When you intersect two planes, you can at least get a line

[forest quiver]

yes

Or you can get a plane

[leaden tide]

You can get another plane if the two are the same, but never a point

[forest quiver]

too

Yeah never a point

[leaden tide]

likewise for volumes, when you intersect them you at least have a plane

[forest quiver]

hmm

Yeah like a cross section

I can see that

[leaden tide]

In general, we call hyperplanes spaces that are one dimension less than the whole space we're working in

A plane is a hyperplane in 3D space

[forest quiver]

That is something

[leaden tide]

A volume is a hyperplane in 4D space, and so on

[forest quiver]

I am not familiar with

[urban trail]

can you guys go through when done with the problem

[leaden tide]

A line is a hyperplane in 2D space

(sure, after this)

[forest quiver]

isn't a line

just a line...

in 2d?

lol

[leaden tide]

Yeah, but look, terminology is hard

[forest quiver]

Right

[leaden tide]

When you intersect a hyperplane with anything else, the dimension of the resulting space can at most have gone down by 1

and it doesn't change iff the hyperplane contained the other space

[forest quiver]

you mean at least?

[leaden tide]

it decreased by at most 1

[forest quiver]

I don't have enough mathematical maturity for this

You shouldn't waste your time on me lol

[leaden tide]

Nahh, we're almost there

[forest quiver]

Let me watch a quick video on hyperplane

1 minute

[leaden tide]

go ahead if you want, i'll wrap this up

An equation with n variables defines a hyperplane in n-D space

e.g an equation with 3 variables defines a plane in 3D space

[forest quiver]

Oh

so it's kind of its natural habitat

for lack of a better phrase

[leaden tide]

yup

[forest quiver]

OK that's what a hyperpl;ane is lol??

what a stupid name

[leaden tide]

so then, a system of k linear equations with n variables is really just the intersection of these k hyperplanes

[forest quiver]

in n-D space

[leaden tide]

And since it decreases by at most one each time, you are left with a space that's of dimension at least n-k

That's it, that's what free variables means

[forest quiver]

if n>k, n-k gives the number of free variables

[leaden tide]

It's the amount of coordinates you need to describe the resulting space

[forest quiver]

coordiantes?

A better way of phrasing it

Is the amount of equations needed to make a point at the intersection

a single point

not a line

not a volume

just one point

is a solution

and that's why we need free variables sometimes because there is just not enough information

and if there is no information, there are no restrictions

[leaden tide]

yeah, i guess ?

[forest quiver]

hence free variable

is my explanation bad?

[leaden tide]

i like the notion of coordinates because people know what that means from physics

[forest quiver]

i haven't taken physics yet

[leaden tide]

the idea is just that if you want to describe the position of something in 3D space, you're going to need 3 different numbers

[forest quiver]

yes

[leaden tide]

that's just what i mean

if the resulting space is a volume, you need three numbers to describe every location within it

those are your free variables

[forest quiver]

hmm

I don't completely follow your last explanation, but I undersatnd this sufficiently now

Thanks for the help

[leaden tide]

this looks correct, the last part could use some better wording but the idea is there

the idea is just "good luck finding an even and an odd number which give the same result"

anyway the injectivity should be obvious

wait

on second thought

[urban trail]

continue

[leaden tide]

oh yeah no my bad

you're shifting every even number to an odd number

and every odd number to an even number

and you're shifting everything by the same amount

there's never gonna be a collision in hell

[urban trail]

so is it correct or nah ?

[leaden tide]

yeah

[urban trail]

even this ?

[leaden tide]

yeah, standard methods here

[urban trail]

aight thanks man

[merry imp]

if V and W are both n-dimensional, and L : V -> W is linear, can we immediately conclude Im(L) = W from dim(Im(L)) = n?

i feel like there's a step or two missing

[north hedge]

think so

suppose (b1...bn) basis for V. then (L(b1)...L(bn)) must be linearly independent else the dimension of image isn't n. then (L(b1)...L(bn)) is a basis for W, so Im(L)=W?

i think something like that

[merry imp]

Ah yeah perfect

[north hedge]

we meet again.

lulz

[merry imp]

O7

[wintry steppe]

you don't even need to pick a basis, just appeal to the result that if a subspace has the same dimension as the larger space then they're equal

[dreamy iron]

is there a "standard canonical" name for the transformation that sends linear maps to their matrix representation?

given $dim(V) = n; ~ dim(W) = m\text{, and a field }\mathbb{F}$

what is the name for :

$$? : \mathcal{L}(V,W) \to \mathcal{M}_{m\times n}(\mathbb{F}) $$

[stoic python]

a rainbow powered ninny

[twilit anvil]

hmm, arent they exactly the same thing

[dreamy iron]

they are isomorphic....so yes.

[wintry steppe]

matrix representation with respect to a base of V and a base of W

idk

you can't get much more than that without forgetting that it depends on choices of bases

[dreamy iron]

there's no name that is common to use? there's no greek letter that is preferred for this?

[wintry steppe]

no

[dreamy iron]

what do you use?

[wintry steppe]

when i studied linear algebra i liked the following notation

[stoic python]

TTerra

[wintry steppe]

wait you asked for name and i gave you notation

lol

idk i just call it by that. it doesn't come up often enough outside of LA to deserve a shortened name

[dreamy iron]

that's fine.....it's kinda what i was looking for anyway.

[native rampart]

Maybe,The "canonical" isomorphism

[dreamy iron]

i've seen $\varphi$

[stoic python]

a rainbow powered ninny

[wintry steppe]

it's a bit of a disappointing answer but "there isn't really one" is all i can think of

i'm hesitant to call it canonical because it depends so heavily on choices

there definitely isn't a standardized notation for this, let alone a standardized "shortname"

[dreamy iron]

i only use the word "canonical" to mean "well regarded as a standard in literature" how like all functions are "f, g, h" linear transformations are "T", etc.

[dreamy iron]

how do you do that $[\cdot] ^{a}{B}$ ?

[stoic python]

a rainbow powered ninny
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry steppe]

[\cdot]_\alpha^\beta

[dreamy iron]

[.]^a_b

[wintry steppe]

maybe it doesn't like the superscript coming first?

unsure

[dreamy iron]

$[\cdot]_{a}^{b}$

[stoic python]

a rainbow powered ninny

[dreamy iron]

okay, nice

[marble lance]

$[\cdot]^a_B$

[dreamy iron]

tyvm

[stoic python]

Lunasong the Supergay

[marble lance]

Superscript first is fine

I don't think they had an underscore

[wintry steppe]

odd

[hardy leaf]

'm looking for details on the following type of generalization of the Legendre symbol:

The function gets a number a and a prime p, and checks whether a^ 1/ 3 exists.

More generally, if exists, whether a^ 1/ q exist for a prime q that divides p-1

[atomic hound]

Is there an example of a linear transformation in V where null T + range T does not equal V (and V is finite-dimensional)?

[hollow finch]

im assuming you mean a linear operator T: V->V right?

[atomic hound]

Mhm that is correct

[hollow finch]

$T:\bR^2\to\bR^2$

$T(v)=\begin{bmatrix}1&-1\1&-1\end{bmatrix}v$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

i think that should work

null(T)=range(T)

not equal to V

[atomic hound]

ahh nice thank you

how were you able to come up with that example?

[hollow finch]

its a rank 1 nilpotent matrix. you can prove that this implies that col(A)=null(A)

since A^2=0

i just chose one where col(A)={(1,1)} since its my favorite vector in R2

[atomic hound]

i see, i haven't gotten to the explanation of nilpotent operators yet in my book (comes later). but i'll take a look.

[hollow finch]

yeah theyre pretty cool

good luck

[atomic hound]

thanks again

[atomic hound]

just reading about it now, did you mean rank 2 here? or does rank mean something different than index?

[hollow finch]

rank is the number of linearly independent columns and index is the smallest power k such that A^k=0

rank can also be thought of as the dimension of the column space

[atomic hound]

ah gotcha, that makes sense

[autumn oar]

for homogenous coordinates, why do we set w to the inverse of the scalar of the length rather than just the scalar of the length

is it literally that much easier to use zero instead of some sort of representation for infinity? so much so that it's worth dividing by w every time instead of multiplying? or is there some other reason?

[woven haven]

It's been 45min since the last question, so I'm going to assume that no one can help that person for now. Pardon me if I'm intruding.

I'm trying to answer this question, and I believe I'm on the right track but I'm not sure how to explain an intermediate step.

I'm trying to show that the matrices will have the same number of linearly independent columns, since this number corresponds to the ranks of the transformation matrices.

But once I expand $[T]^{\gamma}{\beta}$ as $[[T(v_1)]{\gamma}, [T(v_2)]_{\gamma}, . . .]$, I'm not sure how to show that each term in the second matrix is linearly independent.

[stoic python]

Chris24

[woven haven]

Or, more accurately, that the set of transformed basis vectors is linearly independent

Is there a way I can show that to be true? My intuition says that it would really depend on the transformation.

[glacial mango]

What is $[T]^{\gamma}_{\beta}$ ?

[stoic python]

modus ponens

[wintry sphinx]

matrix representation of T wrt bases gamma nad beta

[woven haven]

Yes, just as @wintry sphinx said.

[wintry sphinx]

I mean the cheaty way is to write one matrix in terms of another with change-of-basis matrices and then appeal to the invertibility of those change-of-basis matrices

[woven haven]

So because the change-of-basis matrices are invertible, they will have dimension n, since they will be n x n matrices?

[tall oriole]

simple question: if you have two subspaces U, W of a vector space V, how is it the case for u in U, w in W, that u+w in U+W? in proofs i see this conclusion made in a similar argument but not justified. what is the underlying logic?

[wintry steppe]

by the definition of U + W

it's literally just all elements of the form u + w for u in U and w in W

[tall oriole]

ah, i see. so U+W is all the set of elements of u in U added to w in W so it's true by definition that for u in U, w in W that u+w is in the set U+W?

[wintry steppe]

yes

[tall oriole]

okay awesome, i see it now. thanks

[woven haven]

Is there any guarantee that a linear operator will map a basis to a basis?

I'm still working on my proof from before but I believe I'm very close to a good answer.

I tried what @wintry sphinx suggested beforehand but was unable to make sense of it.

[native rampart]

No, Consider T:R^2->R^2
T(v)=(0,0) for all v

[woven haven]

Rats.

I'm typing up my proof at the moment so I will be able to more clearly communicate my issues.

[wintry steppe]

I'm actually curious regarding a problem I had a long time ago but never really found out how to do.

If I wanted to do a rotation around a custom axis, how would one go about it.

[teal grotto]

if its injective it will

[woven haven]

Unfortunately I have no guarantee that this transformation is injective

[teal grotto]

rats

[woven haven]

Alright this is the section that trips me up:

[native rampart]

cats

[teal grotto]

bats

[woven haven]

What I'm trying to show is that all of the rows of both transformation matrices are linearly independent, so since theyre mapping from an n-dimensional basis to another n-dimensional basis, they will have n linearly independent columns and therefore both have rank n.

But, as is so unpleasantly natural in proofs, there's a problem: I'm not sure how to show that T(v_n) wrt gamma is linearly independent

T being a linear operator, if it was not clear

Sorry, I'm trying to show that all of the $\textbf{columns}$ of both matrices are linearly independent.

[stoic python]

Chris24

[teal grotto]

you can use stars around your text **like this** to make text bold without using the texit bot

[woven haven]

Well said

[teal grotto]

only one star makes it italicized *like this* and you can strikeout text ~~like this~~

[woven haven]

Rats Awesome!

[teal grotto]

okay, anyway

[woven haven]

Oh wait

I think I can get what I need from the definition of linear independence.

Nevermind. @native rampart example also disproves that.

Rats.

[teal grotto]

i was going to suggest trying to show that the images are isomorphic

[woven haven]

I thought of that earlier but I couldnt construct an isomorphism, at least I dont think

I thought $\phi : L(V) \mapsto M_{mxn}(F)$ would be a good idea.

[stoic python]

Chris24

[woven haven]

Since that is an isomorphism, and isomorphisms map sets of linearly independent vectors to linearly independent vectors, I'd have my result.

But I don't think that says what I think it's saying.

[lavish jewel]

doesn't that say that L(V) can be represented with a matrix of elements from F?

[teal grotto]

yes

[lavish jewel]

nothing about the rank of M

dimension is an invariant though

so if you already know it for L(V), you could use that

[woven haven]

I don't know what an invariant is.

I mean, I know what the word means, but we haven't gone over that in class I don't believe so it'd be awkward for me to use it in my homework.

[lavish jewel]

aight

[woven haven]

At this point, I feel like I've committed too hard to a solution which might not be correct.

Because I'm not so sure I can guarantee that the transformations are linearly independent.

[lavish jewel]

what's this theorem 3.3 they say you cannot use

i'd just use a crapton of "change of basis" transformations

[woven haven]

That $rank([T]^{\gamma}_{\beta}) = rank(T)$ where T is a linear transformation and $\beta$ and $\gamma$ are ordered bases.

[stoic python]

Chris24

[woven haven]

Obviously, this would trivialize the proof.

[lavish jewel]

you could change basis to the canonical one and then to the desired one

[woven haven]

canonical?

[lavish jewel]

e_i

[woven haven]

I think that's what I'm supposed to be doing since the L_A maps between F^n and F^m

But I don't see how that's useful beyond guaranteeing that the range of both left-multiplications is the span of the columns

[lavish jewel]

change of basis operations are full rank, so they don't change the rank of the original transformation

assuming you've already shown that

[woven haven]

Someone earlier suggested the same thing and I tried to do so but I hit a roadblock.

I just don't understand how the argument would go.

But I will try again.

[latent flare]

have you seen this

[teal grotto]

yea. seen that a while ago.

[woven haven]

Cupping therapy in general or this delicately balanced gentleman in particular

[teal grotto]

the dapper gentleman in the cups

[wintry sphinx]

what I meant is that you can write $[T]{a,b} = A[T]{c, d} B$, where A and B are suitably chosen change-of-basis matrices; these are invertible and therefore preserve rank

[stoic python]

Saccharine

[wintry sphinx]

In particular, assuming that [T]_{a,b} denotes the matrix repr of T where a is the basis of the input space and b is the basis of the output space, you have A = change of basis matrix from d to b and B = change of basis matrix from a to c

[stoic python]

coycoy

[tropic turret]

Hi, I'm so confused
I have eigenvectors matrix and eigenvalue matrix but the result is not normal. why??? 😩

[lavish jewel]

wdym it's not normal?

[wintry steppe]

Hi

Well

This question is general

What is algebra and its branches and in which order should I study them?

Linear algebra, abstract algebra, modern algebra, commutative algebra and I think there's more

[wintry steppe]

Lol, matrices, I studied them at hs

btw, eigenvectors, eigenvalues, where do you study that?

[wintry steppe]

what is the difference between abstract algebra and modern algebra?

[marble lance]

Linear Algebra is kind of separate from the rest, but is usually studied first. AA and MA I think of as the same thing. And CA is a part of AA, but usually AA refers to the introductory part and then CA is a continuation of that.

[coarse rain]

Usually, linear algebra.

[wintry steppe]

Okay

[wintry steppe]

I don't know, I was searching and I found those branches

Are they the same?

[wintry steppe]

Okay. And do you have more branches there?

[marble lance]

Sure, but there isn't a clear tree for which order you should study things in after that.

[wintry steppe]

Well, so that may be a difficult task

Choosing which order to follow

[wintry steppe]

I've read that Linear Algebra Done Right by Sheldon Axler is one of the best lineal algebra books you can read

[coarse rain]

Most people do LA -> AA though

And then branch off or smth

[silk ether]

Is that possible to have more than one solution for 2x2 matrix linear differential equation involving complex eigenvalues and repeated eigenvalues?

[nocturne jewel]

The only LinAl book I've read is David Poole's Modern Introduction cause that was the course textbook

[warped trench]

hello im trying to understand how to check if a specific matrix is in the span of other matrices

i have those 3 and im trying to see if the bottom M_1 is in the span of the other 3
i found a video that explained it slightly, i put the vectors side by side and did guassian elimination
im just trying to understand if this is what im supposed to do?

[quartz compass]

yeah that sounds good

[warped trench]

i didnt understand how to finish the guassian eilimination

i was using something like this in order to understand how to do the guassian elimination

i dont understand when to solve for x 1 x2 x3 though

[quartz compass]

not sure what you mean by that last comment

idea of gaussian elimination is you try to make the first column have all 0s below the first entry

and then you repeat trying to put all 0s below the next entry

[warped trench]

oh so i keep going until each leading 1

[quartz compass]

until you get it completely to row echelon form

[warped trench]

has 0's under them

kind of like this

[quartz compass]

yeah

[warped trench]

then what do i do for the span because in the video it says to solve for x1 x2 and x3

[quartz compass]

I don't know what 'x1 x2 and x3' means still

[warped trench]

i think im supposed to put them on the side so the third line would be like x3 = 21?

sorry

x1 is the values from the first vector

[quartz compass]

oh I see what you're saying

[warped trench]

x2 is the second colum and x3 is the third column

[nocturne jewel]

they're the scalars

[warped trench]

as far as i understand it

these 3 are supposed to add up to the M vector

[quartz compass]

then you have to go back up and delete things upwards

[warped trench]

but they're supposed to be a multiple of each other

or they can be

[nocturne jewel]

you're saying things we already know..

[warped trench]

yeah i dont understand how to get there

[nocturne jewel]

Gauss Jordan

[warped trench]

after this point in the guassian elimination i dont know how to proof the m is in the span of abc

[nocturne jewel]

well you get a nonsense row

so there's no solution to that system

[warped trench]

so it wouldnt be in the span of abc

how do i know if it would be a nonsense row?

[nocturne jewel]

cause 0 doesnt equal -3

[warped trench]

ah kk thanks

[nocturne jewel]

Im also assuming the gauss jordan was done correctly

[warped trench]

i believe so

i tried to recreate it myself

and didnt see any problem with it

the thing is the first columns are abc

and the 4th is m

[nocturne jewel]

yeah you get 4 equations, one for each entry of the 2x2

[warped trench]

so even though i have 3 leading ones

there is still a problem because 0 != -3

would i only be able to solve it if everything at the bottom was 0?

[nocturne jewel]

if the bottom was a 0 row, then you'd have a subspace of solutions

[hollow finch]

Not in the picture above. 3 pivots and 3 columns (not counting the augmented column) means no free variables

In that case a row of zeros just means the system is consistent

[wraith patio]

how do i prove that if A is a 2x2 matrix with a trace of zero, then A^2 is a scalar matrix? it works if i just square [a b; c -a] but its messy and there must be a better way

[coarse rain]

So consider two cases:

A has distinct eigenvalues a and -a.
A has an eigenvalue of 0 with algebraic multiplicity 2.

In the first case, A² has an eigenvalue of a² with geometric multiplicity 2.
In the second case, A² is 0.

[hollow finch]

Cayley Hamilton makes it pretty simple, as long as you are familiar with the fact that the linear term of the characteristic polynomial of a 2x2 matrix is -tr(A)

[stoic python]

meguuuuu

[drowsy flower]

How do I even begin to proving this?

I introduced basis for U, W and U + W

There is a theorem that says that $dim(V) = dim(W) + dim(W^{\bot})$

[stoic python]

meguuuuu

[drowsy flower]

unsure how to come to this

[twilit anvil]

have you looked at the converse? i had a little time with it and it seems like one can show that the projection of x to U+W is as stated, by playing around with the inner product (while also using some reasoning) to show x minus (proj_U(x) + proj_W(x)) is orthogonal to U+W.

[drowsy flower]

Yeah I have. This was what I had thought of. Suppose U is subset of orthogonal complement of W. Then define orthonormal basis A of U as {a_1, ..., a_k}. Define basis B of W as {b_1, ..., b_l}. Then basis C, of V should be {a_1, ..., a_k, b_1, ..., b_k}. And hence, projection onto V will use vectors from C = projection_{U + W}.

However, I realized later that the basis C I wrote may not be the orthonormal basis of V

so ya still stuck on converse as well rip

[twilit anvil]

one point you might consider is that
C might not even be a basis when U is a proper subset of W^perp (youre missing some vectors from W^perp)

while it is not the approach i would initially use it might be possible to continue like this... maybe you can use gram schmidt on C, and then play around with it?

personally i think the introduction of orthogonal bases might not be necessary, and i would focus on the fact that proj_U(x) is the orthogonal projection of x in U, or that x - proj_U(x) is in the orthogonal complement of U. same for W and U+W.

[drowsy flower]

Okay let me try that approach

[twilit anvil]

although ill say that im still a student, like yourself, learning linear algebra at around this level, my advice might not be the best

im sure a veteran will chime in sometime

[drowsy flower]

thx a lot for your reply though, I am working on that approach right now

let me see if I can figure it out

[stoic python]

coycoy

[drowsy flower]

okay let me give it a try

thanks alot

[stoic python]

coycoy

[teal grotto]

i have no idea the validity of that statement rn tho. i want to say that it is correct

just something that would make this go much quicker, since it would say that U and W intersect trivially.

this would actually just prove the entire biconditional if it’s true

[north anvil]

Given that the dot product of the 0 vector and any other vector is 0, does that mean that the 0 vector is orthogonal/perpendicular to all vectors?

[coarse rain]

Depends whether your definition excludes 0 as a special case.

If it doesn't have a special case, then it is orthogonal

[nocturne jewel]

yes 0 is orthogonal to all vectors

[north anvil]

Geometrically/visually that doesn't make sense

But mathematically it does

[nocturne jewel]

yeah

cause not all vector spaces have geometry

@wintry steppe 🤨

[surreal valley]

Would this be correct?

[wicked marsh]

yeah that looks right

[surreal valley]

Do you mind checking part b,c,and d please

[wicked marsh]

yeah that all looks good

[surreal valley]

okay tyvm

[wintry steppe]

hi guys how can i solve this hw problem?

im very confused

[silk ether]

Is that possible to have more than 2 eigenvectors for a 2x2 matrix?

[wintry steppe]

are you asking me?

@silk ether

[silk ether]

No, my question was ignored and left unanswered yesterday

[wintry steppe]

oh

sorry

[teal grotto]

no

yes im sorry

cannot have more than two eigenvalues

[quartz compass]

depends on if you consider eigenvectors distinct if they're multiples of each other I guess

[wintry steppe]

does anyone know how to go about this

[teal grotto]

i mean, then thats like "cant have more than two equivalence classes of e-vectors"

[silk ether]

What about like [3, - 1] and [-3, 1] 1x2 matrix eigenvector? Are they the same

[teal grotto]

do $4I-A$. try and find the value of $h$ so that the kernel (or equivalently null space) of $4I-A$ is two dimensional

[quartz compass]

I suppose you're supposed to do it by inspection

[stoic python]

coycoy

[wintry steppe]

wym ?

why would u do 4i-A

instead of A-4i

[quartz compass]

since it's clear that [1,0,0,0]^T is a an eigenvector with eigenvalue 4 and then you can look at the 3rd column vector and see what to do with your current eigenvector and pick h etc etc

[teal grotto]

same result

[wintry steppe]

im a bit confused by what u mean here

i have my result from the subtraction

i think i got h = 8

can someone verify this?

[teal grotto]

yup

quick sanity check tho

[wintry steppe]

yup to it being right or being able to verify 😛

[teal grotto]

try and find a different evector with eval of 4 other than something of the form (a,0,0,0)

[wintry steppe]

ah ok

sorry i am new to this stuff

just learning

[teal grotto]

no worries

[wintry steppe]

thank u though! : )

[teal grotto]

do you mind giving more context if there is any?

[twilit relic]

@teal grotto what are you on aboot?

[teal grotto]

wdym? was asking a follow up question

[twilit relic]

What knowledge is it that you seek?

[teal grotto]

language was confusing. couldn't tell if they meant [3,-1] vs [-3,1] as a matrix or as an evector... was asking for clarity... from op. if you know what they are on aboot, then lmk

[last holly]

well, in either case it helps to try out the scenarios and see what make sense. the nebulous areas where you're unsure, review. for example: what does/can the eigenvector represent (rather than how it's calculated)?

[halcyon pollen]

what's a euclidean n -space ?

[coarse rain]

R^n endowed with the euclidean inner product.

[wintry steppe]

When proving a set is a vector space must you prove every condition holds?

[coarse rain]

Yes

[teal grotto]

when proving that a set is a vector subspace however, the laundry list of conditions you need to prove decreases

[wintry steppe]

Yeah I saw

[covert tartan]

How do I calculate the V vectors

[torn hornet]

solve (A-lambda I)x =0 for the given lambda's

[covert tartan]

A - Lambda? they arent the same dimensions

im really new to this sorry

[torn hornet]

right so lambda*I as in the identity

[covert tartan]

Ah that makes sense

[ -2, 1,0]
[0, -2, 1] * x = 0?
[2,-1,0]

is that it @torn hornet

[torn hornet]

for the lambda=2 yes

[covert tartan]

ok is there a quick way to find the inverse of a 3x3 matrix

[torn hornet]

wolframalpha

well this matrix wont be invertible right

thats the point, we want a non-trivial element in the nullspace

[covert tartan]

oh ok, but how do I get V from this

[lavish jewel]

since A - lambda * I is rank deficient, you can find infinitely many vectors v such that (A - lambda*I) = 0

you could introduce a dummy variable t, for example, and find v in terms of that

you can do this by rref-ing the matrix A - lambda*I

you're guaranteed to get a free variable

[covert tartan]

rrefing?

[lavish jewel]

put it in reduced row eschelon form

gauss jordan

[covert tartan]

ah ok

[covert tartan]

okay so how do I solve this

[teal grotto]

put the columns into a three by three matrix, call it A, and find A^-1

then A^-1 [2 0 -1]^t will give you the right answer

[lavish jewel]

either that or make a 3 x 4 augmented matrix and do gauss jordan

[covert tartan]

seemed complicated by hand so

the inverse

the final

im guessing the top result is b1 in C

@lavish jewel @teal grotto

[lavish jewel]

is this a completely separate problem from the other one?

[covert tartan]

same question i used coycoy's method

[lavish jewel]

ah with the inverse

sure, if that's the inverse, that should be right

we can double check

,w rref {{-1, -1, 1, 2},{-2i, 2i, 2, 0},{4,4,4,-1}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Edd#5182

[covert tartan]

woah

I did not know you could do that here

[lavish jewel]

gauss jordan avoids finding the inverse matrix explicitly. it should behave more nicely, especially when you have really large systems

[covert tartan]

okay i will keep that in mind

b3 = 7/8 right?

or is that b1

[lavish jewel]

sounds about right

[covert tartan]

ok

[teal grotto]

edd i was just going to ask in the computing software channel something about implementing a row reduction algorithm in c++. question was kind of ill formed tho, was just asking if there were any tweaks/tips i could make to make the gauss jordan method more accurate or if there are any numerical issues i might run into if i don't look out for something

you seem like you know a lot about this stuff, so

[lavish jewel]

hmm i know there are several issues you can run into if you try to do it "naively"

there are certain reorderings of columns and rows that make it more stable and faster

and tbh i think doing QR decomp might be faster than directly doing gauss jordan

[teal grotto]

hmm. i would probably have to implement a gram-schmidt type of process for that right?

[lavish jewel]

but i think angetenar and anticipation might have more tips for you. try asking in applied computational maths

yeah, one flavor of QR is with gram schmidt

[teal grotto]

alr. ill try there later probably when i have more of a plan laid out. thanks

[lavish jewel]

aight

[covert tartan]

a quick question.
what does "and the first component" part mean. I see the similarities but why are the signs +,-,-

[native rampart]

A^(n-1)B is a column matrix?

[covert tartan]

Here's the whole example

[native rampart]

So,Like first component is the Element in the first row

[covert tartan]

ah of the V vectors

1, -1, -1

hence the signs

right?

anyways i think it's right. Thanks alot for your help guys gonna go submit is like 2:40 am @native rampart @lavish jewel @teal grotto

[bold garden]

Quick question about left multiplication maps:

My proof goes along the lines of $$"Let C = [T]$$, the matrix representation of T. Then $$C = [L_{C}]$$. Then since $$[L_C] = [T]$, we get $$L_C = T$$.

[teal grotto]

nice

[bold garden]

WOOPS

[teal grotto]

lol

[bold garden]

uhh hold up, lemme fix this

[teal grotto]

My proof goes along the lines of "Let $C = [T]$, the matrix representation of $T$ with respect to $\beta$ and $\gamma$. Then $$C = [L_C]$$. Then since $$[L_C] = [T]$$, we get $$L_C = T$$.

[stoic python]

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

[bold garden]

uhh yeah, i think that's as good as i can get it

[stoic python]

coycoy

[teal grotto]

that good?

[bold garden]

yeah

the proof in the book is different, but i don't see what's wrong with this one

[teal grotto]

cool. so im not sure im really following this one.

[bold garden]

The first line follows from a result from before - a matrix is the matrix representation of its associated left multiplication map

The second line follows from our assumption and line 1

[teal grotto]

okay wait nvm i misread

[bold garden]

And the last line follows from another result - if two maps have the same matrix representation, then they are the same map

[teal grotto]

did you show uniqueness?

[bold garden]

uniqueness follows from part (b) of the same question - that A = B iff $$L_A = L_B$$

[stoic python]

kirafa

[teal grotto]

ok cool. ur good then

[bold garden]

hmm, that's strange

alright then, if it works it works lol

tyvm (:

[teal grotto]

i mean, how does the proof in the book go?

typically you get an isomorphism between L(F^n,F^m) and Mat_{m x n}(F) and it would be done in a couple lines.

[bold garden]

the proof in the book goes like this

here is theorem 2.14

[teal grotto]

i guess yea both are fine then

[bold garden]

alright sweet, i appreciate your help (:

[coarse sandal]

To find (v)b, would you just do k1(cos^2(x)) + k2(sin^2(x)) + k3(cosx) + k4sin(x) = 4 +3cos(x)?

and wouldnt that be (4,4,3,0) = (v)b?

[lavish jewel]

looks ok

[coarse sandal]

is that the correct thinking?

[lavish jewel]

yeah

[coarse sandal]

thank you

[nocturne jewel]

yes cause you need to somehow get a constant from the basis

[coarse sandal]

What is the approach to solving this problem?

my thought was to do k1(x^4+x^2) + k2(x^4 - 1) + k3(x^2 + 1) + k4(2x^4 + 3x^2 + 1) = a + bx + cx^2 + dx^3 + ex^4

and put that into a matrix

but im not sure if im doing it correctly

this is the matrix i have in mind

[teal grotto]

yea u just need to find out which vectors from this set are linearly independent. upon inspection, the first one is just the second and third added together, so you can toss that one out

second and third are linearly independent

[coarse sandal]

thats the matrix

oh ok

so basically on inspection

[teal grotto]

the fourth one is twice the second plus three times the third so you can toss the fourth

[coarse sandal]

we notice that the x^4 + x^2 = x^4 - 1 + x^2 + 1

[teal grotto]

you only need the second and third vectors then to span the space

sry keep going

[coarse sandal]

so one of those 3 can be thrown out?

[teal grotto]

yea

[coarse sandal]

vector 4 = vector 3 + 2 * vector 1

so that means vector 4 can be thrown out as well

[teal grotto]

you kind of need to find a maximally linearly independent set, but this is a pretty small set, so everything works out nicely

i believe the method you were going for is more general and will get you a maximally linearly independent family, this set was just so small you could do it by inspection

[coarse sandal]

so the definition of a basis

[teal grotto]

ye

[coarse sandal]

so the new subspace would just be 2 vectors

[teal grotto]

it’s the same subspace

u just got rid of superfluous vectors from the spanning set to make it a basis

[coarse sandal]

oh

[teal grotto]

the subspace only needed to be spanned by two vectors

not four like it was before

[coarse sandal]

so how do you go about finding the basis for a vector space?

[teal grotto]

i mean, you just did for this one

but if you mean, like, in general, that’s a different question

[coarse sandal]

general question

[teal grotto]

there is a theorem that says every vector space has a hamel basis (as opposed to a shcauder basis) and some may not be computable i believe

[coarse sandal]

cause the definition of a basis for a vector space V, is if the basis Spans V and is Linear independent

[teal grotto]

right. there are some vector spaces, like the space C([0,1]), the vector space of continuous functions on the interval [0,1] that do have a basis but no one has ever seen it/it’s not useful anyway/there’s really no good choice of a basis for that space

[coarse sandal]

oh

thank you

[wintry steppe]

people are still exploring the world to this day trying to find a basis for C([0, 1])

much like a proof of the riemann hypothesis its existence eludes all

[zealous junco]

side note is finding basis for this based on stone weiestrass, if it was not periodic

[wintry steppe]

wdym

[zealous junco]

for example polynomial basis, etc.

oh wait they are dense in C([0,1]) but are they considered a basis

[wintry steppe]

Just use Zorn's Lemma

i can't think of a sense in which they do form a basis

[teal grotto]

many lives spent wasted in the homes of parents basements while mathematicians search for a basis for the space of continuous functions on the unit interval

[wintry steppe]

but zorn's lemma is false

[zealous junco]

function basis just means convergence of infinite sum of basis functions right?

[wintry steppe]

then how do you know whether there's a basis?

exactly

[teal grotto]

confused sully

[wintry steppe]

superimpose 🤨

isn't existence of basis for C[0,1] independent from ZF

[teal grotto]

lmao

[wintry steppe]

i actually have no clue, i was just trolling

[teal grotto]

i think the statement that every vector space has a basis is equivalent to axiom of choice/zorns lemma. no idea about independence tho

[wintry steppe]

it is

but C[0,1] isn't every vector space

[teal grotto]

[zealous junco]

ok nvm, ez question: if U is skinny matrix and contains orthonormal column anything can be said about UU^T?

it can be non diagonal though its symmetric pos semi def but apart from that?

[wintry steppe]

what is a skinny matrix?

[zealous junco]

like mxn but n < m

[teal grotto]

contains just one orthornormal column?

or multiple?

[zealous junco]

can be multiple but just less the total number of rows

like i guess my question is very broad but looking for perhaps special structure in UU^T or ways to compute it faster

maybe there isnt anything more to say about it

[teal grotto]

i mean, each of the entries where you multiply an o.n. row vector by an o.n. column vector you’re going to get a 1

[wintry steppe]

op said U has a o.n. column tho so that won't necessarily ever happen

[teal grotto]

it’s transpose will have an o.n. row

[wintry steppe]

yea but were multiplying U by U^T not the other way

[teal grotto]

bruh. *multiply an o.n. column vector by an o.n. row vector

[wintry steppe]

?

[zealous junco]

[wintry steppe]

for example if U was
1 2
0 2

first column is o.n.

but when multiply on the right by transpose nothing interesting seems to happen

[zealous junco]

that is not on, ill give example so U:

1/2 1

1/2 -1

-1/2 0

-1/2 0

[wintry steppe]

you said contains an orthonormal column

1
0 is an orthonormal column

[zealous junco]

wait i mean like

each column is o.n. to each other

sry if it was unclear

[teal grotto]

wait wut

[wintry steppe]

you mean orthogonal to each other?

[zealous junco]

yea

so U^TU is identity but i only know UU^T is just positive semi def

[teal grotto]

oh. i seem to have messed up there

[coarse sandal]

How would i approach this type of problem?

[teal grotto]

M_22 is the space of real 2 by 2 matrices?

[coarse sandal]

i know that a dimension of 2 means that the basis has 2 matrices

[coarse sandal]

yeah

[teal grotto]

do you know a basis for M_22?

[coarse sandal]

the standard matrices of M_22

the 4 matrcies that only have a single 1 in them for a basis for M_22

but that has a dimension of 4

[teal grotto]

so just pick two of them and say look a the space spanned by the two u picked

[coarse sandal]

i believe

[teal grotto]

bam

[coarse sandal]

oh

so pick 2 of the standard matrcies

and then the subspace is just the matrices that can be formed from those two standard matrices

is that the thinking?

[teal grotto]

right on!

[coarse sandal]

👍

This means that c1 *(v1)_B + c2 * (v2)_B = 0 has non trivial solutions

not sure where to proceed from here

[teal grotto]

yes. are (v1)_B and (v2)_B the representations of v1 and v2 in terms of basis vectors from B?

[coarse sandal]

yes

[native rampart]

Do you know c_1(v)_B=(c_1v)_B

[coarse sandal]

wait huh?

[native rampart]

Ok you need to show c_1(v_1)_B+(v_2)_B=(c_1v_1+v_2)_B

[teal grotto]

wait why? the claim is false

[native rampart]

(v_1)_B is writing v_1 in terms of elements of B and then taking the coordinates right?

Like {b_1=(1,-1),b_2=(0,1)} (1,0)=(1,-1)+(0,1)=b_1+b_2

So (1,0)_{b_1,b_2} will be (1,1)

[teal grotto]

just disregard me dude i gtg to sleep fr

[coarse sandal]

yes the coordinate relative to the basis

[native rampart]

So yea this is true

Say $v_1=a_{11} b_1+a_{12} b_2 ,v_2=a_{21} b_1 +a_{22} b_2$

[stoic python]

Buncho Dragons

[native rampart]

So $(v_1)B=(a{11},a_{12}),\newline (v_2)B=(a{21},a_{22})$

[stoic python]

Buncho Dragons

[coarse sandal]

where did you get a11 and a12?

is it because of matrix multiplication?

[native rampart]

Notice that $cv_1+v_2=(c a_{11}+a_{21}) b_1 + (c a_{21}+a_{22}) b_2$

[native rampart]

Change of basis

[coarse sandal]

oh

[native rampart]

If v_1 and v_2 are in span (b_1,b_2) they can be written in that form

By definition of span

[stoic python]

Buncho Dragons

[native rampart]

So $(cv_1+v_2)B=( ca{11}+a_{21}, ca_{21}+a_{22})=c(a_{11},a_{21})+(a_{12},a_{22})$

[stoic python]

Buncho Dragons

[native rampart]

$=c(v_1)_B+(v_2)_B$

[stoic python]

Buncho Dragons

[leaden tide]

what does it mean for two spaces to be linearly dependent anyway?

wait

[nocturne jewel]

the set can span to 0 non-trivially

[native rampart]

Their intersection is not 0

That is you can find a nonzero vector v which is in both spaces

[coarse sandal]

im kinda confused about what happens after this showing "c_1(v_1)_B+(v_2)_B=(c_1v_1+v_2)_B"

[native rampart]

Then just do your normal linear independence shenanigans

[leaden tide]

Wait what, don't we have (v_1)_B = (v_2)_B if v_1 and v_2 are linearly dependent ?

[native rampart]

You then use the fact $c_1(v_1)_B+c_2(v_2)B=(c_1v_1+c_2v_2)_B$

Suppose $(v_1)_B$ and $(v_2)_B$ are linearly dependent. Then there is non zero $c_1 ,c_2$ such that
$(c_1v_1+c_2v_2)_B=0$

[stoic python]

Buncho Dragons

Buncho Dragons

[native rampart]

And that $(v)_B=0$ implies v=0

[stoic python]

Buncho Dragons

[leaden tide]

wait

what is the (v)_B notation

[coarse sandal]

coordinate relative to basis notation

[native rampart]

.

[leaden tide]

ohhh

[coarse sandal]

how does (v)b = 0 implies v = 0?

[native rampart]

You should prove that yourseld

[coarse sandal]

my brain is too small

[leaden tide]

if you have non-zero coordinates for zero it's not a basis

[native rampart]

Feels like you didn't understand this (v)_B thing

[coarse sandal]

oh

oh

[native rampart]

More like 0 co ordinates imply v is 0

v_b gives you a bunch of coordinates

And when I say v_b=0 I mean all the co ordinates are 0

[coarse sandal]

if v1 and v2 are linear independent then (v)_b = a1 * v1 + a2 * v2, and (a1,a2) = (0,0)?

is that the reasoning

[leaden tide]

that being said, i don't understand the point of this question ; vector <-> coordinates is linear

[native rampart]

Yes if b={v_1,v_2}

[native rampart]

I wanted to use this

So one zero implies other is 0 immediately

[lavish jewel]

you could just use linearity twice

and the definition of independence

[coarse sandal]

could you explain this?

[leaden tide]

Ok, unless I'm greatly misunderstanding, wouldn't the following work as proof?
Suppose $\lambda_1(v_1)_B+\lambda_2(v_2)_B=0$ with $(\lambda_1,\lambda_2)\ne (0,0)$. Then $0=\lambda_1(v_1)_B+\lambda_2(v_2)_B=(\lambda_1 v_1+\lambda_2 v_2)_B=0$.
Since $x \mapsto (x)_B$ is an isomorphism, $\lambda_1v_1+\lambda_2v_2=0$ \blacksquare

[stoic python]

Syst3ms
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[native rampart]

Well,He didn't know x -> (x)_B was a isomorphism

This is the same proof

[lavish jewel]

that does work

if you already know that a change of basis is an isomorphism, yeah

but you can build this one up with just a rock and a stick

let's say v1 and v2 are independent

[leaden tide]

That really a change of basis?

[lavish jewel]

then a v1 + b v2 = 0 has only solution a = b = 0

now let's express these two vectors in a basis B containing vectors b1 and b2

then v1 = c b1 + d b2, and v2 = e b1 + f b2

and their sum is something of the form g b1 + h b2

we are given B is a basis, so b1 and b2 are lin indep

this means g = h = 0 is the only solution to g b1 + h b2 = 0

and yes. if you make a matrix M that has as columns the basis elements in B, then x to (x)_B is done via M^-1 x, a change of basis

[leaden tide]

oh right, it's ℝ², not any vector space

[lavish jewel]

yeah

they gave you a lot of info

no need to overextend if you don't have to

[leaden tide]

odd question honestly, seemed obvious that talking about coordinates and a vector was basically the same thing

[coarse sandal]

i just dont think my prof connected how everything is related to each other

[lavish jewel]

it's more to establish that dimensionality is independent of the basis tbh

[native rampart]

Well,you think of Matrices as Linear transforms

But doing that rigorously is a PITA

You have to show matrix multiplication behaves nicely via matrix computations

[leaden tide]

PITA ?

[lavish jewel]

pain in the abutt

[leaden tide]

unrelated, but our prof calls this the "important trivial lemma"

[native rampart]

Indeed

Trivial consequence of vector spaces having a group structure

[zealous junco]

matrix algebra > linear algebra

[lavish jewel]

you can also just straight up show it's a subspace

[leaden tide]

but turns many, many proofs into checking if something equals 0

rather than doing anything else

[lavish jewel]

well

yeah

considering L(v) = w means w is in the image of L, L(v) - w has nice properties

[leaden tide]

I looked back at our material to see how we justify that dimension doesn't depend on the choice of basis

We used the Steinitz lemma, or some version thereof : if some vector space E has a set of n vectors that spans it, then every set of n+1 vectors is linearly dependent.

[nocturne jewel]

yeah, extremal properties of bases

[leaden tide]

Although that's an easy corollary, the proof of the lemma is rather annoying

[coarse sandal]

Thank you everyone for the help

[drowsy flower]

Okay I feel confused about projection more than ever so can someone clarify this to me?
If a vector x belongs to a subspace S, then when we proj_S(x), what is the output?

Isn't it just proj_S(x) = x?

[leaden tide]

Yes

[lavish jewel]

sounds ok

[leaden tide]

If you slap a fly onto a wall, then slap it again

It's still going to be on the wall

[drowsy flower]

oh wow I didnt think of it like that

great thanks alot

[leaden tide]

Yeah, projectors are just flattening things in one direction

(except identity which doesn't flatten much of anything but shush)

[wintry steppe]

Could someone check this for me?

[faint dune]

looks good

[leaden tide]

Statement 1 : if you stretch a vector it doesn't change its orientation
Statement 2 : any orthogonal set of the right size is a basis

[faint dune]

every orthogonalsystem is linear independant. The cardinality of W basis is 3. So dimension is 3, and we have 3 orthogonal vectors which are l.i., so its dimension is also 3, so it must also be a basis of W.

[wintry steppe]

okay thank you

[glacial mango]

What's special about an orthogonal basis?

[quartz compass]

vectors have a nice clean representation and the dual basis is simple as well

[faint dune]

You can easily orthogonal project every vector to determine its linear combination with the orthogonal basis.

[leaden tide]

Does the usual notation for the dot product carry over to vector spaces with potentially different definitions of dot products?

[faint dune]

Im not sure what you mean.

[gray dust]

the definition of dot product on R^n doesn't change. do you mean inner product?

[teal grotto]

depends ig

[gray dust]

if so we use $\ip{x}{y}$

[stoic python]

RokabeJintaro

[leaden tide]

Right, that's the term

In French we call all of them scalar products

[teal grotto]

for example, there is a dot product of matrices that is related to the trace of a matrix. might be confusing if you used the same notation as u did with the dot product in R^n, since we already multiply matrices together

[faint dune]

We use the same dot product notation for L2-Product

[leaden tide]

The usual one is denoted the same

But in the case of inner products and associated spaces, (x|y), <x,y> or the one you showed are common notations

[teal grotto]

lots of tingz. makes computing projections really easy, for one

[gray dust]

<x|y> is used more in physics but i like it bc i'm lazy

[wintry steppe]

simply write xy

[teal grotto]

prefer the <•,•> notation

[wintry steppe]

the functional analysis book on my shelf uses (x|y)

[teal grotto]

booo

[faint dune]

If its a orthonormalsystem then its even more easier to project the vectors to the basis.

[gray dust]

normalizing an orthogonal set makes it orthonormal. if $\brc{e_i}$ is orthonormal we can show the projection of $x$ onto $\Span\brc{e_i}$ is nicely given by $\sum_i\ip{x}{e_i}e_i$

[faint dune]

I actually wanted to write this, but I gave up cus I suck at latex.

[stoic python]

RokabeJintaro

[wintry steppe]

orthonormal bases suck because you can't use einstein notation when representing a vector

[gray dust]

based

[quartz compass]

if you have the concept of dual basis you can work as though your basis is normalized, since your vector $v=v^i e_i$ has components that are just your vector dotted with the dual basis $v \cdot e^i = v^i$

[stoic python]

Merosity

[quartz compass]

if your basis vectors are already orthogonal but not normalized, the dual basis is just the inverse matrix of that matrix of basis vectors, which is just its transpose scaled, so technically you're just putting the normalization step into there lol

that's all to say that people who normalize basis vectors when dealing with tensors are chumps

[gray dust]

you got me good

[faint dune]

xD Will I ever deal with tensors in astro physics 1 ,2 and experimental physics 2?

Sounds like sth one wants to escape from.

[fallow coral]

is an even number modulo an even number always an even number

[coarse rain]

Yes

[north hedge]

^

[wraith patio]

when doing a "two column proof" that cv=cw->c=0 or v=w, i have trouble with what exactly i should write for the justifications. if someone could verify that my proof is good too thatd be great.

Suppose that $v,w\in V$, $c\in F$, and $cv=cw$.
\begin{align}
cv=cw&&\text{Given}\
cv-cw=0&&\text{?}\
c(v-w)=0&&\text{Distributivity}\
\implies c=0 \lor v-w=0&&\text{Some theorem?}\
c=0\lor v=w&&\text{?}
\end{align}

[stoic python]

taxminion

[coarse rain]

you could just write cancellation property

[wraith patio]

for which ones? 2 and 5?

[coarse rain]

Yes

[wraith patio]

alright sure that makes sense

still not sure about line 4. i mean i dont think that property has a specific name, and whatever theorem its a part of is text specific

also would "algebra" work for lines 2 and 5?

[north hedge]

5 requires proof

suppose c is nonzero, then it has an inverse (F a field), so applying it to both sides gives v-w=0

so c=0 and v=w necessary

and sufficiency is p obvious

(im assuming the point is to work from axioms)

[mild current]

if T(a+b) = T(a)+ T(b)
then cant we prove that T(λa)= λT(a) ? (Context: Linear Transformation)

Why create another axiom for that?

Shouldt it be a lemma or a theorem?

[thorn robin]

You can't quite prove it

Look up Cauchy functional equation

There are non-linear solutions to T(a+b) = T(a) + T(b) over the reals. Though they are weird, they exist

[mild current]

Okay, but whats exactly wrong in this "proof"
a+a+a= 3a
then T(3a)= T(a+a+a)= T(a)+T(a)+T(a)
then T(3a)=3 T(a), right?

[thorn robin]

Yeah that's fine

[north hedge]

lambda is a general scalar in ur field

[thorn robin]

But you only proved it for lambda = 3

You will be able to do it for rational lambda if you do some work

But not for all real lambda

[mild current]

So,when considering rational nos, can the abovementioned axiom be considered as theorem?

[mild current]

can u pls explain srry

[coarse rain]

If your field is Q, sure. But usually your field is R or C.

[thorn robin]

You can prove that if T(a + b) = T(a) + T(b) then T(lambda * a) = lambda * T(a) for any rational lambda yes. Here T is a function R -> R. What you did with lambda = 3 was a start in the right direction towards proving this

Well I guess R -> R is not necessary in what I said. It seems like it is a theorem over any Q-vector space

[mild current]

Okay,
Lets take 3.5 a= a+a+a+0.5a
Then T(a+a+a+0.5a)= 3.5T(a) right?
Is it "incomplete" bcuz I cant define 0.5a in a linear combination of "a"s

[thorn robin]

Right that doesn't work