linear-algebra

[wintry steppe]

no

[fair wren]

n >= 3

[wintry steppe]

yes

[fair wren]

hahaha

[fair wren]

How did u deduce that fast?

[wintry steppe]

i stared at it

[fair wren]

fax

[fair wren]

can u tell me how does finding the rank of a matrix -the matrix representing an endomorphism -helps in studying its eigenvalues/vectors?

[wintry steppe]

typically you care more about the rank of A - cI as c ranges through your scalars

in this context

[fair wren]

@wintry steppe it doesnt bounds the amount of eigenvalues you have?

[wintry steppe]

well

for one, you're never going to have more than n distinct eigenvalues (n being the dimension)

i don't see how the rank of the matrix says anything about the number of distinct eigenvalues you have, other than simple statements like "if rank \neq n then you have at least one eigenvalue, zero"

[hard drum]

Tbf one fancy way to do this is that A - I is the companion matrix of the polynomial x^3

Or recognise it as a Jordan block

[slender silo]

anyone knows the algorithm? I don't know why.

It's about the computation of the matrix element.

[hoary hull]

Just change the order of the sigmas. You can change the order of finite sums at will.

[slender silo]

Sure, I just took a example and then got the conclusion. Thank you so much.

[dreamy iron]

can you elaborate on this? i always find this confusing.

is there a proof of that fact: You can change the order of finite sums at will. ?

[dusky epoch]

it's just the commutative law of addition

well, commutative and associative

$\sum_{i=1}^n \sum_{j=1}^m a_{ij}$ and $\sum_{j=1}^m \sum_{i=1}^n a_{ij}$ both sum the same $(1:m)\times(1:n)$-indexed grid of numbers

[stoic python]

Ann

[hoary hull]

as Ann says, try to imagine that each cell in column n and row m contains a number, and add them all together
This is good and intuitive

[wintry steppe]

id like to save this

[plush dust]

Hi guys, is this a correct answer?

[native rampart]

Should be

[dusky epoch]

yes

[plush dust]

Thanks

[plush dust]

Is there any fast way to check similarities of these matricies?

[wintry steppe]

Any linear algebra text recommendations

I don’t like axler ladr

[spare widget]

hoffman and kunze, friedberg, halmos, strang

[wintry steppe]

Thank you I’ll check those out

[spare widget]

if you want an all in one calc + lin alg + analytic geometry, then try shifrin

[wintry steppe]

Axler, for me, doesn’t go into enough detail

[spare widget]

then you won't like halmos

[wintry steppe]

Ahhh alright, then I’ll check out the others

[spare widget]

I think Friedberg was extremely detailed

[wintry steppe]

I’ve tried shifrin, but I’d rather delve deeper into Lin algebra

Oooh then I’ll download it rq and check it out

Ty

[spare widget]

you might like the linear algebra problem book from halmos though

it's pretty detailed and has hints + solutions

it's not a textbook - it's a problem book and it's quite nice

[wintry steppe]

👍 definitely will check that out

[wintry steppe]

This is perfect

Looks amazing for self study

[plush dust]

Any guesses, is this the correct one?

[marble pike]

Does it have eigenvectors?

[spare widget]

it has more than that

[spare widget]

you really should write out the matrix explicitly

it's a R^3 to R^3 map

so 3x3 matrix

[plush dust]

I did that, and started to solve eigen values

[spare widget]

then you should have seen that the matrix is symmetric

[marble pike]

In the contents section I only see eigenvalues listed near the end, no sign of eigenvector though.

[plush dust]

But by inner product it is not symmetric

[dusky epoch]

??

2 0 4
0 5 0
4 0 3

this is the matrix

do you claim this isn't symmetric

[plush dust]

I used this definition and there were not equal

[spare widget]

the eigenvalue problems involve eigenvectors

[dusky epoch]

maybe you fucked that up at some point?

[plush dust]

Oops, yeah

[plush dust]

Is this correct answer?

[dawn ocean]

show your work or thought process first

if you want people to check your answers

[slender silo]

anyone ? I wanna figure out why these two entries don't subtract one when constructing the martix to solve the eigenvalue?

[dawn ocean]

what is the matrix I?

[slender silo]

the identity matrix

[dawn ocean]

yes

but like

[slender silo]

oh!

[dawn ocean]

what does it actually look like in matrix form

[slender silo]

I got it!

[dawn ocean]

nice

[slender silo]

Thank you!

[dawn ocean]

np

[radiant coyote]

Can someone spot my mistake ....

[wet stratus]

5-(-2) for that last cross product calculation

as a side note, you need to put brackets around (0,0,0)-(1,-1,0) in your second line

[radiant coyote]

Thx u

[wintry steppe]

I'm having a bit of difficulty with this assignment provided to me by my professor. The idea is that I have a tetrahedron with 4 triangular faces and four vertices.

I have been given components to four vectors normal to the faces opposite to the vertices, pointing outwards.

Now, I need to find out vector 1, vector 2, vector 3, and vector 4 from the vectors I was provided and show that vector 1 + vector 2 + vector 3 + vector 4 = the zeroth vector
How would I approach this? Like conceptually speaking, I'm not exactly sure how to understand this problem.

I can provide a quick drawing of the associated tetrahedron?

[dusky epoch]

probably by solving the system and finding the individual values of x, y and z

or ending up at the impossibility thereof

[radiant coyote]

thx u I'll do that

[sharp idol]

If the tetrahedron is a regular tetrahedron, then every angle is 60 degrees.
That alongside taking the cross product of two vectors that make a face gives the normal vector should be useful.

I'm not sure finding the vectors will be easy, but this seems like a reasonable starting place

[hallow edge]

Question 8, related to 6.

[sharp idol]

What have you tried?

[hallow edge]

Factoring and then simple substitution, but that got nowhere

Tbh I don't even know where to begin with this

[sharp idol]

My only real concern is what does it mean to square a vector here?

[hallow edge]

Me too.

Dot product isn't defined at this stage.

[sharp idol]

Btw, is this an isomorphism under vector addition?

[hallow edge]

It is.

Do we need to consider the inverse image?

[sharp idol]

Okay, I wasnt sure what the operation they were asking about in the isomorphism

Well in that case, $\phi(x,y)^2=2\phi(x,y)$.

[stoic python]

dackid

[hallow edge]

May I ask how?

[wintry steppe]

if $$q(x) = \sum_{k=0}^n a_kx^k$$ is a polynomial, and if $T$ is a linear operator, the notation $q(T)$ stands for the linear operator $$q(T) = \sum_{k=0}^n a_kT^k$$

[stoic python]

TTerra

[wintry steppe]

T^k being the k-fold composition of T. and T^0 is the identity map

[sharp idol]

Hmm, so I'm probably wrong about that then

[sharp idol]

Disregard my claim

[hallow edge]

Well, the text did define the endomorphism of V, so composition is probably it.

[sharp idol]

In groups, raising an element to the n^th power is just doing the operation n times, so that's where I was getting that idea.

But squaring actually has some more significance here

[spare widget]

phi^2(x,y) = phi(phi(x,y)) = phi(x+y, 2x-y) = (x+y+2x-y,2(x+y)-2x+y)

[wintry steppe]

something something squaring in the endomorphism group of R^2

[sharp idol]

I wasnt sure if it was composing phi twice

[spare widget]

I don't see any other obvious interpretation

[wintry steppe]

squaring a linear operator has one and only one meaning

[hallow edge]

Even if the squaring is defined, uhh, how do you substract a scalar from a vector?

[wintry steppe]

T^0 is the identity map. read my message above

[sharp idol]

With Ttera's explanation, composition does make sense here

[spare widget]

a more interesting exercise would have been to show that phi^{-1} = \sum_k (id-phi)^k

[spare widget]

I am guessing the one there would be the identity

[hallow edge]

Huh. I'll try.

[spare widget]

you're not adding vectors, you're adding the linear operators btw

[hallow edge]

Let me get this straight, the squaring is actually composing and the 1 is actually the identity map 1(v) = v.

[wintry steppe]

that's what i wrote here

i am telling you what p(phi) means

[spare widget]

yes 1 is T^0 as Terra said, and the other powers are T^k

furthermore if you have a fractional power, and you have an eigendecomposition, then you can do that too

[hallow edge]

I got what you said, I wrote it again so I myself could 'get' it.

[spare widget]

e.g. T^r = (VDV^{-1})^r = VD^rV^{-1}

[hallow edge]

... So the book I'm reading mentions absolutely none of that.

Nor, to be honest, have I had prior experience with eigendecompositions.

[spare widget]

find a better book I guess

[hallow edge]

The next chapter is bases and dimensions.

[spare widget]

ok, maybe too early, sure

[hallow edge]

Actually, it's abstract linear algebra by Paul Hall.

[wintry steppe]

i can't even find this book on google

[tulip gazelle]

i have a question on laplace transform....would (-e^(-3s))/9 transform into uc(t) or delta(t-c)?

[wintry steppe]

the meaning of p(phi) is on page 19.

you got the author wrong but i figured it was this book anyways

[hallow edge]

..., I'll read that chapter again.

Sorry for the trouble.

[wintry steppe]

it was a small remark so i don't blame you

this is quite the linear algebra book

only one i've seen with a section on normed algebras

hurwitz theorem

[wintry steppe]

It’s not a regular tetrahedron

[wet stratus]

You could maybe first do it in the case of the simple tetrahedron (corners 0, e_1, e_2, e_3, also called standard simplex) and then maybe go form there to the general case (corners 0, x_1, x_2, x_3, note that you can translate any tetrahedron so that one corner is at 0). The transformation sending e_i to x_i then sends the standard simplex to the general case and is linear. Maybe you can argue that the vectors perpendicular to each faces also gets transformed in a linear way (actually not sure if that is correct but I would guess yes) and then from there you are basically done

On second thought, given the corners 0, x_i you can also just explicitly calculate all normal vectors using the cross product and then add them up

[wintry steppe]

Yeah I calculated the normal vectors

I just need to find the individual vector 1-4

Which I’m having difficulty with

@wet stratus

I’m trying to figure out which vectors i have to actually add together

[zinc timber]

$\phi\phi \catthink$

[dawn ocean]

@hallow edge what book is this from?

[halcyon spindle]

Looks like morton curtis abstract linear algebra book.

Not Charles curtis which is a different book.

[subtle gust]

why is this statement false

[wintry steppe]

0 is not an eigenvector but every subspace must contain 0

@subtle gust

[subtle gust]

oh

[wintry steppe]

dumb question i think if you say "or zero" its true

[subtle gust]

i don't get what u mean tbh 💀

how can an eigenspace be a subspace

if it can never contain the zero vector

[wintry steppe]

every eigenspace is a subspace (since it is a kernel), so it must contain 0. but by definition 0 is not an eigenvector

[sharp idol]

If I'm not mistaken the eigenspace is a vector space where the eigenvectors form a basis

[wintry steppe]

but if it said the eigenspace of A corresponding to L is the set of eigenvectors corresponding to L union the zero vector, it would be true (i think)

[subtle gust]

that made me even more confused 💀

it didn't say anything abt subspaces 💀

[wintry steppe]

the idea is that eigenspaces are subspaces, so they contain 0. but zero isnt an eigenvector ever

[subtle gust]

yeah my question is how can eigenspaces be subspaces if they can never contain the zero vector?

i don't even see why eigenspaces being subspaces is relevant to the question

[sharp idol]

Eigenspaces contain the 0 vector

[wintry steppe]

eigenspaces do contain the zero vector, the set of eigenvectors dont contain the zero vector

[subtle gust]

the basis for the eigenspace is the set of eigenvectors right?

[sharp idol]

Yes

[subtle gust]

so the set of eigenvectors of A are basis for the eigenspace

[subtle gust]

isn't this what this statement is ssaying...

[sharp idol]

Linearly independent eigenvectors. Keep that in mind

[subtle gust]

the eigenspace is the set of eigenvectors corresponding to lambda

[subtle gust]

yeah ofc

[sharp idol]

The problem is that the eigenspace contains 0

[wintry steppe]

you want to distinguish two sets of vectors: 1 the eigenspace and 2 the set of eigenvectors

[subtle gust]

the eigenspace is just the span of the eigenvectors eh?

[wintry steppe]

yes

[subtle gust]

oh so i see why zero is in the eigenspace

[wintry steppe]

the eigenspace contains zero and the set of eigenvectors does not

yeah convince yourself of why for each of those

[subtle gust]

yeah got that point

[sharp idol]

And 0 is never an eigenvector

[subtle gust]

yeah

so if this question said

the eigenspace of A corresponding to lambda is the span of the set of eigenvectors corresponding to lambda

it would be correct

[sharp idol]

That would work

[wintry steppe]

yes

[subtle gust]

gr8

tysm y'all

appreciate the help

[wintry steppe]

its enough for the question to have said the eigenspace of A is the set of eigenvectors + the zero vector

thats a bit more subtle

[subtle gust]

yeah i see why now

[sharp idol]

I believe that would exclude linear combinations logan

[subtle gust]

it would still be a subspace i think

[sharp idol]

But not the eigenspace

And no it wouldnt. If linear combinations are excluded, it would not be closed under vector addition

[subtle gust]

oh

fair point

[wintry steppe]

nonzero linear combinations of eigenvectors corresponding to the same eigenvalue are eigenvectors though?

like A(v + w) = Av + Aw = Lv + Lw = L(v+w)

making v + w a eigenvector for A corresponding to L

[sharp idol]

Ah. Fair point

My bad

[wintry steppe]

i think the set of eigenvectors is literally just the eigenspace remove 0

if 0 was considered an eigenvector they would be the same

is the question from Lay?

[subtle gust]

wb this question?

ik that the rank(A)+Nullity(A)=dim(M22)=4

but isn't the nullity the dim of the null space of the transformation matrix A?

which is 1 here

and the rank(A) should be the dim(CS(A)) or the dim(RS(A)) which is 1 here

[wintry steppe]

the matrix is not the matrix of the transformation !

[subtle gust]

it's not ? 💀

[wintry steppe]

so if you have T(x) = Ax where x is a vector, then A is the matrix of the transformation

here you are doing a matrix multiplication instead of a matrix vector multiplication

like your transformation is a transformation on matrices not on vectors

you want to find the matrices such that when you multiply them with [1 3 2 6] you get zero matrix

the dimension of that space is the null space of T

[subtle gust]

ah i see

[subtle gust]

how would we do that tho

[ a b c d]

and set up a system of linear equations?

[wintry steppe]

exactly !!

[subtle gust]

too much work if ask me

😢

[wintry steppe]

it will be 4x4 homogeneous

haha ask wolfram

[subtle gust]

exactly lol

anything harder than a 2x2 goes straight to wolfram alpha

[wintry steppe]

when you solve the system you can get a basis for the null space

which will let you calculate the rank of T

gl

[subtle gust]

nullity*

oh and the rank as well

[wintry steppe]

yeah using rank nullity

[subtle gust]

since it would just be 4-nullity

[subtle gust]

yeah

tysmmmm

'ppreciate the help

[wintry steppe]

ofc np

[tacit pelican]

could someone help me understand how this happened?

im confusd

about

[zinc timber]

what are f, g and h?

[tacit pelican]

they're elements of

wait

my notes say this more concisely

im most confused about these three lines

like

how did they change

that

f_j to f_i?

[dawn ocean]

you still haven't said what f,g,h are

[tacit pelican]

they're all elements of F^\infty

[zinc timber]

do you know that $\left( \sum a_ix^i \right) \left( \sum b_j x^j \right) = \sum_{n=1}^{\infty} \sum_{i=1}^n a_i b_{n-i} x^n$?

[tacit pelican]

uh no i do not

is there a source that explains what that is

[zinc timber]

what is happening here is say you want to find coeff of x^4

then in how many ways you can get x power 4

one is you multiply a_4x^4 with b_0, another is a_3x^3 with b_1x, and so on

[tacit pelican]

yep

[zinc timber]

so you you get x^4 by multiplying coefficients a and b such that their index sum to 4

that's what I have written, but with a summation notation

probably similar thing is happening in your case

it's too long for me to read rn

[tacit pelican]

oh wait

im looking on mse rn

and

apparently that was a typo

but yeah thanks

could you recommend me any

website or something

that

would explain some of these kind of things with summations

this chapter is very heavy on sums and they usually do substitutions and stuff with the indices that always confuse me

[zinc timber]

sadly idk any

[tacit pelican]

ah right

thank you anyway

[dawn ocean]

i think

stuff about tensors

doesn't talk about this exact thing directly

but will help you get used to all sorts of summations things

[zinc timber]

yeah, ask look up tensor to someone who cannot grasp summation notation , good luck with that

[dawn ocean]

i mean

physics people have to learn it all the time

[zinc timber]

$g_{ij;ml}^{kl}$

[dawn ocean]

and i've seen fairly gentle introductions to it

if you can grasp tensors indexes you can grasp pretty much anything else

[slender silo]

hello, any simpler method can solve the inverse of 4D matrix？

[dusky epoch]

,rccw

[dusky epoch]

this matrix in particular is not very hard to invert

it is the product of three elementary row operation matrices

[slender silo]

oh, I remember! Thank you!

[plush dust]

Hi, not sure if I understood task correctly but is this the right answer?

[wintry steppe]

you are correct

[plush dust]

Thanks

And what about this one?

[native rampart]

Yes

That's correct

[plush dust]

Thank you very much

Is option D the right one?

[native rampart]

yea R2=2R3+R4

[plush dust]

Good

Not sure about this one

[native rampart]

False

A is elementary iff A^-1 is elementary

[plush dust]

Oh, okay, in that case, I think its D

[native rampart]

which matrices?

[plush dust]

M1 and M2

[native rampart]

det(M_1),det(M_2)... are all nonzero tho

[plush dust]

But det(M_1 * M_2) is 0

[native rampart]

det(M1 M2)=det(M1) det(M2)

You probably made a calculation mistake somewhere

[plush dust]

Ah, okay, found it

[native rampart]

Sounds like #prealg-and-algebra stuff.(If you mean ratios)

[native rampart]

Should be C

C just means the product is invertible

Also A

#geometry-and-trigonometry

[plush dust]

But besides M_4, all others are elementary no?

[native rampart]

mb,A shouldn't come

so C

[native rampart]

If you don't know why this is true,I can explain

[plush dust]

Would be great if you do 🙂

[native rampart]

ok so you know gaussian elimination?

Each step in gaussian corresponds to multiplying with an elementary matrix(A row operation)

If the matrix were invertible you end up with each column having a pivot,i.e.,your final matrix after applying all the operations is I

so this means (E_1E_2...E_k)A=I(Apply operation corresponding to E_k first,then E_k-1 and so on)

This gives you A=({E_k}^-1 {E_{k-1}^-1)...({E_1}^-1) which is a product of elementary matrices

[plush dust]

Ok, and if im not mistaken, this representation is unique right?

[native rampart]

yes

Actually no,mb

[plush dust]

Alright, thanks 🙂

[native rampart]

You can apply and cancel the same operation multiple times

If you don't do that then yea it's unique

[plush dust]

Ok

[plush dust]

Is B the correct answer?

[dusky epoch]

no

planes P1 and P2 are orthogonal

[plush dust]

Why?

[dusky epoch]

their normals are [1; -2; 1] and [2; 1; 0], whose dot product is zero

[plush dust]

Ah, right, ok

[distant badge]

Do anyone recommend book for linear algebra

[stoic oasis]

Using Euclidean algorithm, find out that the polyname f (x) has an irreducible divisibility of at least 2.

Write the decomposition of the polynomial f into the product of irreducible polynomials (above Z3).

[main lion]

sorry if it sounds dump, but why orthonormal vectors are linearly independent?

[spare widget]

because if you assume they are linearly dependent, then
there exist scalars (a1,...,an) != (0,...,0) such that
\sum_i ai vi = 0. Without loss of generality let a1 !=0, then take the inner product with v1: <v1,\sum_i ai vi> = \sum_i ai <v1,vi> = a1 <v1, v1> = 0 -> v1 is the zero vector, and thus not unit length

[main lion]

thank u

[stoic oasis]

<@&286206848099549185>

[stoic oasis]

Why no one is answering to my question ?

[burnt timber]

Answe on the rhs, can someone explain to me how you compute the matrix in part i? I know that you normally find basis alpha, then apply thi to each element of the basis, and take the coeff of the result to put as the matrix

When I did it I didnt get what they got though

Is their answer incorrect? If the first col of the matrix is meant to instead be 0110 it makes sense to me especially with how they've written it

[native rampart]

They did a typo

0110 is correct

[burnt timber]

Okay thanks for the confirmation im not 100% with the topic atm so thought I may of done something wrong

[zinc timber]

try #groups-rings-fields

[fringe fjord]

I don't think one can exist unless the axiom of choice fails.

There's not necessarily such a thing as "its orthogonal complement" for infinite-dimensional spaces, but given AC you can aways pick a direct complement, which is sufficient.

[distant badge]

Hello do anyone recommend me linear algebra books @fringe fjord

I want it for learning machine learning

[fringe fjord]

No, there are too many of those to be a subspace.

For example, just in two dimensions if you let f(a,b)=(a+b,0), then the image is all vectors of the form (t,0). Now (1,1) and (1,-1) are each linearly independent of that, but their sum is (2,0).

[wise meadow]

If we have two equations:
$$2y=-x+5$$
$$y=x-5$$
If we set them equal to eachother:
$$2y+x-5=y-x+5$$ rewriting this we get a line
$$y=-2x+10$$
Is this valid? What does this line represent?

[stoic python]

Bryan

[hard drum]

Well there's a unique point (x, y) satisfying the two equations above and it lies on that line

But it also lies on the other two, i don't see anything deep about this ig

[fringe fjord]

In particular, if you scale one of the equations by a constant factor before you start, you'd get a different line through the intersection point.

But if you start by scaling them such that normal vectors implied by the two equations have the same length, then your new line is one of the angle bisectors between the two original lines.

[wise meadow]

Now I must ask another question:
Let's say I have a quadric given by $$\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x}+\boldsymbol{B}\boldsymbol{x}+\alpha=0$$
where $$\boldsymbol{x} = \begin{bmatrix} x\y\z\end{bmatrix}$$ and $$ A=\begin{bmatrix}a&0&0\0&b&0\0&0&c\end{bmatrix}$$
I now want to intersect this quadric with a plane of form $$\boldsymbol{Cx}+\beta = 0$$
With $$\boldsymbol{C} = \begin{bmatrix} 1&0&0\0&1&0\0&0&0 \end{bmatrix} $$
My guess would be:
$$(\boldsymbol{Cx})^T\boldsymbol{A}(\boldsymbol{Cx})+\boldsymbol{B}(\boldsymbol{Cx})+\alpha=0$$
But I am not sure what to do with the $\beta$ and I also just realized that doesn't even make sense.....

[stoic python]

Bryan

[fringe fjord]

Hmm, if beta has nonzero third coordinate, then Cx+beta=0 is even impossible.

[wise meadow]

Beta would be a constant

[fringe fjord]

But it needs to be a vector constant for that equation to make sense, since Cx is a (column) vector.

And even in the best case {x | Cx + beta = 0 } is not a plane but just a line.

[wise meadow]

Mmm, I realize I am a monkey

[tribal willow]

what do they mean by “fixed field”?

[slow scroll]

Like fixed as in it won’t change throughout the section of the chapter

[tribal willow]

oh ok

[tribal willow]

i can’t seem to wrap my head around what f~ is exactly

like i know they give the definition, i think an example would suffice

[fringe fjord]

A polynomial is a list of coefficients (together with various other areas of ink that are traditionally included but which don't contribute to the meaning).

[tribal willow]

oh also F is a field*

in case that wasn’t clear

[tribal willow]

mhm

[fringe fjord]

A polynomial function (which is what your image notates with a tilde) is the actual function from the field to itself that you get by evaluating the polynomial.

[tribal willow]

oh

so like f is the polynomial but f(x) is a polynomial function, for example?

[fringe fjord]

A few lines below the image, you can probably find the standard example for why we need to distinguish: If F is the field with p elements (where p is some prime), then x^p and x^1 are different polynomials, but they give rise to the same polynomial function.

[fringe fjord]

I stand corrected. But that is nevertheless the standard example :-)

[tribal willow]

thank you!

[fringe fjord]

So we need to distinguish between whether "these two things are the same polynomial" means that the coefficients are the same for each exponent, or merely that the result of evaluating the polynomials at each field element gives the same result.

[tribal willow]

hmm okay now im a bit confused sorry

oh wait nevermind

oops

[fringe fjord]

By convention the word "polynomial" alone is used when we care about the coefficients, but "polynomial function" when we care only about the black-box behavior of evaluating it.

[tribal willow]

gotcha

[slate hatch]

I am really confused. Why do we use two different formulas for projection?

Here they use this formula

Then this formula is used

Should we or should we not divide by the magnitude squared?

[fringe fjord]

The last image looks like it's part of a description of the Gram-Schmidt process?

In that case, q1 has already been normalized to have length 1, so dividing by |q1|² wouldn't do anything anyway.

We can see on the bottom left that q2 is being normalized. Thus in the next step dividing by |q2|² will not be necessary.

The formula with the division is necessary when you don't know that the vector you're projecting onto has length 1.

[slate hatch]

Thank you

[next dune]

what kind of information do the eigenvectors of an adjacency matrix convey about the corresponding undirected graph

my schooling just taught me how to compute shit like eigenvectors and determinants but i still feel like i understand none of the behind the scenes in LA

and im like months away from finishing my bachelors lmao

[slender silo]

anyone know why? what's the meaning of the multiplication of P1 and P2?

[native rampart]

You interpret P_1 and P_2 as vectors

[spare widget]

Looks like the product from geometric algebra

But with a minus

I guess it's a special case

[odd sparrow]

is it Friedberg?

Does anyone here have an opinion on how does Paul Halmos "Finite dimensional vector spaces" compare to Friedberg's Linear Algebra?

[spare widget]

halmos expects that you' re more mathematically mature

friedberg seemed to be quite detailed

while halmos wasn't

[odd sparrow]

Would you say I'd benefit from Halmos if I already took a course with Friedberg? I want a refresh with a more advanced book

[spare widget]

then use halmos

I would also suggest his problem book

it's great

[tribal willow]

hk

[odd sparrow]

Hong Kong?

[tribal willow]

hoffman kunze

[wintry steppe]

halmos is more advanced imo. something like a "third course" in LA, if a first course is something computational like lay or strang and a second course is more proof oriented (e.g. hoffman and kunze or axler) but still works over subfields of c and doesnt touch too much on very abstract things like dual space, quotient space, tensor products, etc

i think friedberg is similar to those for a second course so probably yes you would benifit

[odd sparrow]

Yes those topics are exactly the ones I want to learn. Although the book is quite old, I find it strange there's not very good recommendations for advanced linear algebra

[wintry steppe]

yeah linear algebra is becoming more computational because of computers

i think halmos is good as a review before taking a course which requires a good understanding of linear algebra -- something like differential geometry or functional analysis

it will change how you think about certain parts of linear algebra, at least did for me

[odd sparrow]

Those are exactly some of the topics I'm interested in learning later on. I'm quite the only person in my class that wants these courses for some reason, they all want to go to abstract algebra because they think it's easy

[wintry steppe]

this is what axler does to your brain

the book by roman is the only one i know of i'd call advanced

curtis-place has some special topics, but i don't know if i'd call it advanced

[odd sparrow]

HK or Halmos?

[tribal willow]

where did this c go

here’s the whole thing for context if it helps

[wintry steppe]

typo

[tribal willow]

ah kk

could someone give me an eli5 on (1) what’s special about the principal ideal and (2) what it means for an ideal to be generated by a polynomial

[zinc timber]

df

[wintry steppe]

eli5 principal ideals: they're very simple

[tribal willow]

oh

then maybe an eli10?

eli10 but with a basic understanding of polynomials

[wintry steppe]

for the second question: the text explains what it means for an ideal to be generated by a polynomial, so do you have a more specific question?

[tribal willow]

hmm i guess a better question is "why is it called generated"

[wintry steppe]

principal ideals are just really simple ideals. in some sense the simplest possible ideals in this ring (all ideals in F[x] are finitely generated, so those with only one generator are "simplest")

[tribal willow]

mm i have not learned about rings yet

[wintry steppe]

then learn about them

[tribal willow]

ok

[wintry steppe]

because every element in the ideal is made from them

[tribal willow]

rings are fields without multiplicative inverse?

[wintry steppe]

in the same way a module is a vector space over a ring, yes (the phrasing is poor, but pretty much)

[fringe fjord]

In general, "the Thingamajig generated by x, y, and z" means "the smallest possible Thingamajig that contains all of x, y, and z".

[slow scroll]

"generated" is like "spanned", but for things that aren't necessarily vector spaces

[tribal willow]

ahh ok that clicks

thank you all

[wintry steppe]

Hi

Pls help @here

[slow scroll]

just ask your question

[dusky epoch]

don't do everyone-pings

[green sage]

LMAOOO

[wintry steppe]

Hi

Here's my question

Question 10

Elimination method

Find x and y

Thank you

[fringe fjord]

That is not linear algebra. Try #❓how-to-get-help, and when you ask in the right place, remember to describe what your trouble is, rather than apparently just expecting someone to do your homework for you.

[dusky epoch]

well ostensibly the systems next to 10 are linear systems (except 10 itself which isn't, due to the y in denom)

[green sage]

Why do people not read the channel description or #rules or #info

[wintry steppe]

Q is nothing more than the Orthogonalization of a....

So ofc it's in the col space of a

[dusky epoch]

you don't need to know anything about QR decomp to answer this

y in Col A means exists z such that y = Az = QRz = Q(Rz)

[wintry steppe]

Ahhh ok

Rs invertable

So let z = r^-1

[dusky epoch]

no and that does not even make any sense

[wintry steppe]

Well if you do AZ = ar(r^-1)x

You fet az=ax

Az=y

[dusky epoch]

......

you are speaking nonsense again

[wintry steppe]

I'll shut up then

Sorry

[dusky epoch]

y in Col A means exists z such that y = Az

Az = QRz

so y = Q(Rz)

that gives you the first part

[wintry steppe]

Thank you. :)

[dusky epoch]

also note math is case sensitive

so you cannot just replace z with Z or R with r

A, Q, R are matrices; y, z are vectors

[wintry steppe]

I'm using a phone..

But that's good to know for the exam

[icy totem]

could you solve it?

[dusky epoch]

we don't do your homework for you here

[icy totem]

it is not homework

it is last year exam question

[native rampart]

Yes we can solve it but you know we can't spoonfed you

[wintry steppe]

Man people are very salty today

I'm guessing he's in a European school that gives problems that don't relate to exams, and hes trying to study... yet being condemed for asking where to start

A + c == a = -c

Or -a = c

Same can be done for b + d

[wintry steppe]

So you can write this as [a b, -a -b]

Then.... a[1 0, -1 0], b[0 1, 0 -1]

Does that help?

And at this time of year who is doing homework problems

[slow scroll]

They didn’t ask where to start. They asked if “you could solve it”

¯_(ツ)_/¯

[wintry steppe]

I think I can

[wintry steppe]

English may not be his first language.... but I see where you're coming from

His explanation made sense to me

[icy totem]

Thanks , my solution was also correct, actually I asked to be sure because unfortunately I don't know the answers.

[wintry steppe]

Np np gl on examen

[icy totem]

Could you also look at options b and c of the question, if possible?

[wintry steppe]

IDK b tbh....

I think c is not isomorphic because it's null dim doesn't equal 0

But I'm not 100% on that

B is 1 I think, but I'm not sure

[icy totem]

shouldn't it be dim(ker(T))+dim(Im(T))=2 ?

[wintry steppe]

It's of a + c

It wants it for the T, not the kernel

The image part is the a + c

[slow scroll]

I agree

As long as T is nonzero (which it is), you can take 1 as a basis for the image

[wintry steppe]

A + c can be any real number. Let it be x. It's just one dim

[wintry steppe]

Is this that rule where U nxm

|| ux | = |x|??

[dusky epoch]

no

$\nrm{Ux} = \nrm{x}$ for \textbf{orthogonal} $U$

[stoic python]

Ann

[wintry steppe]

I'm a bit mistified by that

[stuck tendon]

orthogonal matrices preserve norms (they are isometries)

$\langle Ux, Ux \rangle = \langle U^t Ux, x \rangle = \langle x,x \rangle$

[stoic python]

1345631

[wintry steppe]

Ahhh that makes sense

[zinc timber]

what is E?

then Im t + ker t isn't even defined

you mean t:E->E?

also yes, it is not always true

ex f(x, y) = (y, 0) then ker = (x, 0) and im = (x, 0)

their sum is (x, 0) not (x, y) (I hope you can make sense of the notations)

[wintry steppe]

perhaps what you mean is that if you have T: V -> V there is an n such that Im T^n is intersect Ker T^n = {0}?

so then V is the direct sum of Im T^n and Ker T^n

[zinc timber]

article says isomorphic, not equal

easy argument using dimensions

and yes that's always true

[wintry steppe]

hey

is there a program online that I could use for quick verify if two vectors are orthogonal in R^4?

[winter harbor]

Just use the inner product

[wintry steppe]

oh yeah I could do that as well

[winter harbor]

And there probably is a place online somewhere where you can calculate inner product of vectors.

You can do that with wolfram alpha

[wintry steppe]

yup

[spare summit]

i just join this server where do i get help :o

[fringe fjord]

When you first joined the server you ought to have arrived in the #❓how-to-get-help channel initially. The text there explains it.

[wintry steppe]

let me see if I understood the projection concept, so the projection vector of $b$ over a subspace $E$, is a vector within that space that is orthogonal to $b$ ?

[stoic python]

spatialerror

[zinc timber]

nope

that's the orthogonal projection

let $w$ be the orthogonal projection of $v$ onto $E$ then it means that the difference $v-w$ is orthogonal to $w$ i.e. $w^T(v-w) = 0$

[neon jolt]

Let U = { p in P_4(F) : p(2) = p(5) }. Find a basis of U.

How do I even approach a problem like this? Do I just start picking random polynomials of varying degrees and test that p(2) = p(5)?

Note for my notation since I dunno how to do subscripts: P_4(F) is the set of all polynomials with at most degree 4, and F is just R or C

[dusky epoch]

if you can't type ∈ the next best thing is the word "in" rather than "element-of", and subscripts are typically denoted with an underscore like this: P_4(F)

also could you clarify whether F refers to an arbitrary field whatsoever or if it's just R or C?

[neon jolt]

F is just R or C

[dusky epoch]

right

that's good, because that means there's no nasty edge cases to consider

you don't need to pick "random" polynomials of varying degrees; just having at most one polynomial of each degree ought to do

so get yourself a constant, a linear, a quadratic, a cubic and a quartic each satisfying p(2)=p(5)

(though you will notice that one of these is actually impossible)

[neon jolt]

When you say “get yourself a quartic satisfying p(2) = p(5)” is there a technique for finding this, or is it just relying on me being clever and finding some combo of terms that make it work?

[dusky epoch]

well, a little bit of cleverness can be achieved by requiring that instead of just p(2) = p(5) you have p(2) = p(5) = 0.

which should ring some bells.

[neon jolt]

Bells are not ringing immediately lol, been out of university for a while so I’m very much out of practice. But I’ll think on what you said

[fringe fjord]

This has the advantage that if you find yourself a quadratic with p(2)=p(5)=0, you can get cubics and quartics with the same property just by multiplying more factors of x onto it.

[dusky epoch]

true that. i was thinking of something slightly different.

[fringe fjord]

Alternatively, you can start with a completely arbitrary polynomial of the desired degree, and calculate which multiple of x^1 you need to subtract from it to eliminate the difference between p(2) and p(5).

[winter harbor]

First of all, notice that $\text{dim}(U) < \text{dim}(\mathcal{P}{4}(\mathbb{F})) = 5$ as long as we asume that $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ - or even more generally any field of 0 characteristic.
\
\
This is because not every polynomial $p \in \mathcal{P}{4}(\mathbb{F})$ satisfies $p(2) = p(5)$. For example, think about the polynomial $p(x) = x$ - here the hypothesis that $\mathbb{F}$ has 0 characteristic is important, because over a field of characteristic 3 like $\mathbb{F}_{3}$ we indeeed have that $p(2) = p(5)$ for $p(x) = x$.
\
\
Moreover, notice that:
$$
\mathcal{B} = {(x-2)^{3}(x-5), (x-2)^{2}(x-5), (x-2)(x-5), 1} \subset U
$$
Is a linearly independent set of $U$. And since we know that the dimension of U is stricly less than 5. This implies that $\mathcal{B}$ is a maximal linearly independent set of $U$, thus a basis for $U$.

[stoic python]

MISTERSYSTEM

[neon jolt]

This is helpful thanks. The problem for me is I would have never come up with that list of vectors B pretty much ever

But seeing it now makes me think, oh that’s a pretty clever way to make p(2) = p(5)

[winter harbor]

yeah uhh

The idea to have in mind is that

if you find a polynomial that vanishes simultaneously for 2 and 5.

If you add any constant term

Like c

then you still get a polynomial that satisfies p(2)=p(5)

But now p(2) = p(5) = c

This is the idea Ann was trying to give you

[neon jolt]

I see… thanks all!

[winter harbor]

So you reduce your problem to finding polynomials up to degree 4

that vanish for 2 and 5.

[neon jolt]

I’ll chew on that and practice some more. thinking of it the way Troposphere suggested helped me a ton too

I was hitting a dead end by defining f(x) = ax^4 + bx^3 + cx^2 + dx + e. Then plugging in to find f(2) and f(5) setting them equal to each other but that didn’t really do anything lol

[winter harbor]

Yeah so

If you used the definition that a basis is a linearly independent generating set, then I think you'd have to do some computations along these lines.

To prove that set B I defined is indeed a basis

But there are other ways to prove that a given set is a basis.

Are you familiar with the equivalence:

B is a basis for V <=> B is a linearly independent generating set <=> B is a minimal generating set <=> B is a maximal linearly independent set.

This is really good to have in mind

[fringe fjord]

That does in fact work too. You have 16a + 8b + 4c + 2d + e = 625a + 125b + 25c + 5d + e, which is a system of 1 linear equations in 5 unknowns. Since this is #linear-algebra, you have a general procedure for finding a basis for the solution set of such a system ...

[neon jolt]

Sort of, I think? My book definitely has those statements written but I haven’t internalized it yet. I do understand that for a list of vectors to be a basis you could just prove 2 of the following:

1. Linearly independent
2. Spans the vector space
3. Length equals the dimension of the space

[neon jolt]

Oh! Uh, I do? Okay maybe I haven’t learned that yet

[fringe fjord]

That's usually the first thing you learn. Form the augmented 1×6 matrix [609 117 21 3 0 | 0], do Gaussian elimination to RREF, and read off the solution.

[neon jolt]

This book hasn’t even touched matrices yet lol. Book is “Linear Algebra Done Right” by Sheldon Axler. I’m working through the exercises thru chapter 2. And matrices come in 3 sections into chapter 3. Not a clue about “Gaussian elimination” yet either

[winter harbor]

Axler

[fringe fjord]

Huh. I knew Axler did things in a weird order, but that weird?

[winter harbor]

Anyway, in this case you could just solve for one of the variables.

For instance

Write a as a function of b,c,d,e.

That shouldn't be hard

Notice that this will give you explicitly a basis for the solution set of this particular system of equations.

[tribal willow]

wonder why more people dont use FIS

[neon jolt]

He’s not good? I had a few recommendations for this book when I first bought it

[zinc timber]

axler is weird

[winter harbor]

Idk

I guess the thing is that Axler has a few biases towards what should be taught in an intro linear algebra course.

Which I don't think a lot of people agree with

[winter harbor]

And this is reflected a lot in his book

For example

He really overemphasizes the basis-free approach to linear algebra

He doesn't do pretty much anything with determinants

[odd sparrow]

wtf basis free?

I liked that someone actually focused on basis rather than linear independence first, because the basis is basically what gives you "coordinates", I think it was Linear Algebra Done Wrong but I'm not sure

[wintry steppe]

this is from LADW right?

[neon jolt]

I don’t have much horse in this race other than my time (now I’m wondering if I should try a diff book) but Axler argues for his arrangement in this Preface (which I don’t understand at all haha)

[wintry steppe]

some of his definitions are extremely awkward

[winter harbor]

Is this from linear algebra done right?

I am pretty sure it is not

[wintry steppe]

done wrong im p sure

[winter harbor]

Yeah

Axler does exactly the opposite lmao

He prefers working with linear maps between finite dimensional vector spaces rather than matrices

[neon jolt]

From an uninformed/noob perspective, why is “done wrong” a good thing? Sounds like a bad description?

[odd sparrow]

yes

[stuck tendon]

just the name of the book. it's called that just to be funny

[wintry steppe]

its a joke, he means linear algebra with determinants and elimination and matrices (without emphasis on linear transformations as vector objects)

[neon jolt]

Ohh. Ngl I saw the done wrong version and, joke having been lost on me, I scrolled past it

[odd sparrow]

it's good as a second read but not as an only book IMO, you should be able to identify the concepts even when they are put in a weird manner and demonstrate the equivalence

[winter harbor]

Tbh, if I ever taught an introductory course on linear algebra, I'd probably emphasize a basis-free approach, but not completely eliminating determinants lmao

[wintry steppe]

for instance in LADW he proves rank nullity using the number of columns in the matrices which represent transformations which really lends no intuition to whats going on

but yeah axlers dislike of determinants is kinda extreme

[odd sparrow]

For example, a recursive definition of determinants kills the meaning of determinants for me. When I realized it was a sum of products all from different rows and columns (like filling sudoku) it was so much easier to prove properties IMO

[wintry steppe]

the good definition of determinants (which is given in linear algebra done wrong) is as a multilinear alternating map

props to treil for his determinant chapter its well motivated imo

[odd sparrow]

I want to understand it that way, as of now the sign of the products is effy to me. In Shilov's LA he just says the sign corresponds to the number of out of order rows. I read somewhere that differential geometry gave the perfect understanding of the determinant too

[winter harbor]

I could try giving you a motivation for the determinant as a certain multilinear map.

If you got the time

[wintry steppe]

the idea is the determinant is the unique alternating multilinear map with Det(e_1,...,e_n) = 1

which is pretty cool/ suprising

[odd sparrow]

I kinda remember Friedberg giving that idea

[limber fulcrum]

how do i go about solving this, i dont understand what to do with the singular matrix 1 -3 5

the answer is also given

[odd sparrow]

I think you just choose x2 = 1, x1 = 0 for the first vector, and viceversa for the second. On the equation x . [1 -3 5] = 0

something like that

[limber fulcrum]

wdym by choose tho

and why 1 and 0

@odd sparrow

[odd sparrow]

If $\mathbf{x} = [x_1, x_2, x_3]^t$ (t for transpose), and $\mathbf{x} \in Ker(T)$ then $\mathbf{x}\cdot [1, -3, 5]^t = x_1 -3x_2 +5x_3 = 0$.
You want a set of vectors which are linearly independent and generate the set of solutions for the above equation ($\mathbf{x} = [3,1,0]^t k_1+[-5,0,1]^t k_2$ will get you all $\mathbf{x} \in Ker(T)$ by picking different $k_1,k_2$ values.). So basically pick "whatever works" as long as it has those characteristics (which you have to check). Being in $Ker(T)$ basically just means that it solves $Tx = 0$.

[stoic python]

rain

[wintry steppe]

can anyone explain a simple way to get an adjoint endormorphism?

teacher didn't explain this but gave hw about it and i don't find it in the textbooks, only to check it self-adjoint. smh.

I tried to google but can't find it.

if you're in finite dimensions, pick an orthonormal basis. i leave it to you to find the matrix of the adjoint with respect to the original operator and the basis

I have a linear map, so I know the matrix of the linear map f

I also have a different formula for the inner product (which I must use)

dimension is finite yes

[wintry steppe]

Hmm, basically what you're suggesting is to find an orthonormal basis, change the linear map matrix to that basis and calculate the adjoint of that matrix?

right

in an orthonormal basis, adjoint operators are represented by conjugate transpositions of the original operators

[winter harbor]

There's also a basis free way to get the adjoint. Namely, if you've got a linear endomorphism:
$$
f : V \rightarrow V
$$
Where $V$ is a finite dimensional complex inner product space, then we have an induced linear endomorphism:
$$
f^{\ast} : V^{\ast} \rightarrow \rightarrow V^{\ast}
$$
Given by precomposition, i.e $ \forall \varphi \in V^{\ast}$ we have:
$$
f^{\ast}(\varphi) = \varphi \circ f
$$
Now, since $V$ is a finite dimensional complex vector space endowed with an inner product, we have a canonical isomorphism:
$$
\flat : V \rightarrow V^{\ast{
$$
Where to every $v \in V$ we get the linear functional $v^{\flat} \in V^{\ast}}$ given by:
$$
v^{\flat}(w) = \langle w, v \rangle
$$
Its inverse is denoted by $\sharp : V^{\ast} \rightarrow V$ and to every linear functional $\varphi \in V^{\ast}$ it assigns the unique vector $\varphi^{\sharp} \in V$ for which:
$$
\varphi(w) = \langle w, \varphi^{\sharp} \rangle
$$
for every $w \in V$.
\
\
You can then check, that the map:
$$
f^{\dag} = \sharp \circ f^{\ast} \circ \flat : V \rightarrow V
$$
Is indeed the adjoint operator of the endomorphism $f$, i.e it satisfies $\forall u,v \in V$:
$$
\langle f(u), v \rangle = \langle u, f^{\dag}(v) \rangle
$$

[stoic python]

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

[winter harbor]

You can check the last claim

You can also check that if you choose an orthonormal basis for V

Then f^dagger in the way that we have constructed is indeed given by in a basis manner by the conjugate transpose of f.

As TTerra was saying before.

[winter harbor]

This construction also generalizes to linear maps between (finite dimensional) complex inner product spaces.

[tribal willow]

woah music notation

[winter harbor]

A linear map f : V -> W will induce a linear map f* : W* -> V*

And everything is analogous

You also get that the adjoint of f is given (once chosen an orthonormal basis) by the conjugate transpose of the matrix of f.

[winter harbor]

Yeah, these are known as musical isomorphisms.

[tribal willow]

fascinating

[winter harbor]

I have no idea why they are known like this

Has something to do with raising and lowering indexes of tensors.

But I never remember which is which.

[uneven mural]

j

j

[wintry steppe]

sharp and flat are raising and lowering pitch, respectively

[wintry steppe]

do you have a document/excerpt that explains this in a simple intelligible enough way?

prove it yourself, it shouldn't be hard

if it is, you can probably find it on google or in any good linear algebra textbook

or in MS's message above (i didn't read it but they have a good track record)

precisely: if $T\colon V \to V$ is a linear operator on a finite-dimensional inner product space, and if $B = {v_1,\dots,v_n}$ is an orthonormal basis of $V$, then $$[T^]_B = [T]_B^,$$ where $T^*\colon V \to V$ is the adjoint operator to $T$.

[stoic python]

TTerra

[wintry steppe]

and the star on a matrix is its conjugate transpose

I could just pick the canonical form of R^2 as an orthonormal basis, right?

[uneven mural]

yo

how do you do this

give me hintj

j

[wintry steppe]

definitely

[wintry steppe]

show that every element of the line is an element of the plane

[uneven mural]

how

j

[wintry steppe]

show that (2, -2, 1) + t(2, 5, 9) is in the plane for every t

so check that such a vector satisfies the equation defining the plane

[uneven mural]

how do i get the equation of the plane from given

j

[wintry steppe]

the equation of the plane is (3, -3, 1) dot x = 13

you need to show that x = (2, -2, 1) + t(2, 5, 9) satisfies this

this is what it means to show that the line is contained in the plane

So if T is the linear operator matrix, and P the inner product matrix, I have to conjugate transpose D = PTP^-1 ?

[uneven mural]

what does the dot x mean

j

[wintry steppe]

dot product

[uneven mural]

so i take the dot product

j

[wintry steppe]

yes

j

[uneven mural]

how would i do this sorry

j

[gray dust]

plug in, the same way u show x=2 solves x^2=4

[uneven mural]

this is what i have so far

where do i plug in

j

[wintry steppe]

Yeah what I wrote was confusing, but hear me out, he gave us this inner product to follow, and the f(x_1,x_2)=(4x_1,2x_2), so I thought that T is the linear map matrix T=[4 0;0 2] and I'm trying to find T*

But at this point I think that inner product is worthless

If I'm going to pick the canonical form of R2

you need to be in an orthonormal basis to enjoy "matrix of adjoint is adjoint of matrix"

the standard euclidean basis is no longer orthonormal with respect to this inner product, since <e_1 | e_2> = 1/4

Ohhh okay, let B =<(1,0),(0,1)>

Hmm

So I use Gram Schmidt with the standard Euclidean basis then to get an orthonormal basis

that works

But I'll have to change the basis of T

Right?

yes

no big deal

Alright thanks mate, I'll try it 👍

[hazy rock]

for this set I end up with t = 0.5s and s = -2r when testing for linear independence. Since I wasn't able to isolate any variables does this mean its linearly independent or dependent? (t,s,r are the vector coefficients in that order)

[fleet sun]

(2,1,2) = 2*(1,0,1) + (0,1,0)

[uneven mural]

j

[keen sierra]

or observe that the first and third row are equal

[fleet sun]

Oh yeah even better lol

[slate hatch]

x’(t) =Ax(t)

How do I solve a problem like this?

Our course book doesn’t explain it very well.

[zinc timber]

check some online videos on linear system of ODEs

[winter harbor]

Exponential of matrices

That's the key word

I mean, there also other ways to solve a first linear system of ordinary differential equations.

[winter harbor]

But pretty much anything you want to know about them on a basic level can be obtained in this way.

[runic musk]

I’m not sure how to find the offset “d” when converting from vector to Cartesian using cross product

[quasi vale]

Plug in any point on the plane

[runic musk]

bruh

im such a brick, thankyou

[plain robin]

is this function a linear transformation?

[zinc timber]

why don't you verify?

[plain robin]

working on it :D

[copper crest]

Hey just got a quick question: What is the cross-product of two linearly dependent vectors again? Is it the zero-vector? or is it the scalar 0? or am I completely off track

[fleet sun]

Zero vector

[copper crest]

Thanks

[wise nebula]

hello guys

is it possible to have an application from R^n to R^(n+1) that is bijective?

[wintry steppe]

no

[wise nebula]

ty

any counterproof?

[wintry steppe]

actually

i am not so sure anymore

[native rampart]

https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr?lq=1

If you construct a bijection from R^2->R,you can use that to show there is a bijection between R and R^n for any n

And that implies there is a bijection from R^n to R^(n+1)

https://math.stackexchange.com/a/197743

Injective function from R to R^2 is easy

[wintry steppe]

Guys if I take the psuedoinverse of the svd i multiply V(E^-1)U^T and get a matrix that's the psuedoinverse

How do I use that in a least squares problem To get a solution with A^+

[wintry steppe]

no but its in reverse

Do I just multiply it by x?

[native rampart]

Well, Here's the proper solution.
By CBS,you can show R^k=R for all k and that's a stronger statement than your claim.

There is clearly an injection from R to R^2.

Now,You need to explicitly construct an injection from R^2 to R and we can show this problem is easy to solve once we establish that.

We can reduce this to a problem of finding an injection from (0,1)^2 to (0,1)
To know how this injection is constructed,see the stackexchange links.

So,R^2 is bijective to R.
Now assume R^k is bijective to R for some k,then R^(k+1) can be seen to be bijective to R^2 by
f(x_1,x_2...x_k,x_{k+1})=
f(g(x_1,x_2,...x_k),x_{k+1}) where g is a bijection from R^k to R,now this reduces to a problem of finding a bijection from R^2 to R and we are done

Well This is the induction step and we know R^k is bijective to R for k=2 by explicit construction

And R^1 is bijective to R

So for all k in N-{0} ,R^k is bijective to R

[zinc timber]

do you want linear function or just any function?

[cursive ginkgo]

hi, i have A so that : A^2(A − I𝑛)^2 = 0
knowing that (A − I𝑛)^2 ≠ 0 and A(A − I𝑛) ≠ 0.

I'm looking for the specter of A, I know that spec A ⊆ {0,1} but i don't know what to say else

[wintry steppe]

any old function, yes. see above answers. if you ask it and its inverse to preserve some structure (e.g. vector space structure, topology) then no (violates dimension reasons)

supposedly since you're asking in the linear algebra channel, you care about the case of linear maps. in this case, there is no linear isomorphism from R^n to R^(n+1), for rank nullity would give an instant contradiction to surjectivity

the best you can get is an injective linear map R^n -> R^(n+1) or a surjective linear map R^(n+1) -> R^n, which, up to changes of coordinates, are inclusions and projections, respectively

if you care about other contexts, such as topology or differential topology, no homeomorphism or diffeomorphism R^n -> R^(n+1) exists. the latter case reduces to the linear map case, and the former can be proven using some basic algebraic topology (invariance of domain)

[little crater]

This matrix of a linear system has wouldn't have a solution correct? You have 0x_3 = 1 and there is no such x that could give you that value. Along with this youhave x_3 = -2 but subbing into row 3 would not work as 0(-2) != 1

so I would say this matrix is inconsistent

[lethal merlin]

any youtube reccs for linear algebra?

[slate hatch]

Is the inner product the same as the dot product?

is an inner product space the same as a vector space?

Is the inner product only defined as the dot product when talking about vectors?

[winter harbor]

Well

the usual dot product you know that is defined on euclidean space IS an example of an inner product.

But not every inner product is defined like the dot product.

The concept of an inner product space

Generalizes the notion of a dot product

To other contexts

And I will in fact give you some examples of different inner products than the standard dot product.

But first

I think it is in fact important to clarify your second question.

Is an inner product space the same as a vector space?

Well

Not really

They are related

Yes

And their relationship is very simple

An inner product space is just a vector space, but endowed with more structure, new data. Namely, an inner product on it (a bilinear form that is symmetric and positive define, in the real case, or a sesquilinear form that is symmetric and positive definite form in the complex case)

And it is in fact not always the case that we always have a canonical inner product defined for every vector space (in fact, the notion of an inner product usually makes sense for real or complex vector spaces; but even then, it is not always clear that a real/complex vector space canonically carries an inner product.)

Just to exemplify some of the stuff I have said

I will give some examples.

Consider the set of all $n \times n$ real matrices, which is denoted by $\text{M}{n}(\mathbb{R})$. Then, for any two matrices $A, B \in \text{M}{n}(\mathbb{R})$ we can define their frobenius inner product as:
$$
\langle A, B \rangle_{F} = \text{tr}(AB^{t})
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

You can then show (exercise) that this indeed defines an inner product on the space of n×n real matrices.

Another interesting example

Is to consider the space of real polynomials in one variable, denoted by $\mathbb{R}[x]$, you can then check that:
$$
\langle p, q \rangle = \int_{0}^{1} p(x) q(x) dx
$$
Actually defines an inner product on $\mathbb{R}[x]$.

[stoic python]

MISTERSYSTEM

[winter harbor]

I don't think this question makes a lot of sense.

Could you exemplify?

[novel dirge]

Can someone help me solve this? Tried loads od times and Just can get the right result

Cant*

[stoic python]

Brandon7716

[sacred palm]

if u like matrices : ) #help-24

[fair siren]

Quick question, is this how you write the area of a 3d polygon?

[wintry steppe]

Guys if I take the psuedoinverse of the svd i multiply V(E^-1)U^T and get a matrix that's the psuedoinverse. How do I use that in a least squares problem To get a solution with A^+

Would appreciate any help

[dusky epoch]

what even is the area of a 3-dimensional polygon?

[wintry steppe]

Fuck, well i guess i combine it all together and multiply it by x.

[native rampart]

Area of all the surfaces ig?

[dusky epoch]

what surfaces

let the skew quadrilateral ABCD be given by the coordinates of its vertices: A = (0,0,0), B = (0,0,1), C = (0,1,0), D = (0,0,1).

if you consider its area to be composed of the triangles ABD and BCD, then its area is (1+sqrt(3))/2
but if you instead consider its area to be composed of the triangles ABC and ACD, then its area is 1!

[wintry steppe]

Does anyone know how to use the psuedoinverse of a matrix to solve a least squares problem??

I have the psuedoinverse of VEU^T

What do I need to do next????

[zinc timber]

just multiply

with your y vector

[wintry steppe]

So the inverse of this you multiply it together

And then multiply by b right???

[zinc timber]

ye

$Ax=y$ then $\hat{x} = A^{\dagger}y$

[wintry steppe]

Whats y here???

Change of basis?

[zinc timber]

b in your case

[wintry steppe]

Okay so if I'm Asked to solve least squares with svd psuedoinverse. I get the inverse of svd. Multiply that into single matrix a^+ and then multiply by x

[zinc timber]

yes

[wintry steppe]

Lol, I'm doing exam prep and no problems we're given for this

[wintry steppe]

Tikhonov regularization is relate to this

Is that the same thing? Just a change to the diagonal of E?

Anyways arrigato ryu sama

[zinc timber]

I don't know what that is

[spare widget]

It's just adding lambda * I to a non-invertible matrix so you would make it invertible

https://en.m.wikipedia.org/wiki/Tikhonov_regularization

The svd pseudoinverse gives you the solution with minimum norm

while the Tikhonov regularization imposes some constraint on the norm dependent on lambda

i.e. the difference between the following two problems

$\min_x |Ax-y|^2, , s.t. , \min_x |x|^2$ and $\min_x |Ax-y|^2, , s.t. , |x|^2 = c(\lambda)$

[stoic python]

criver

[spare widget]

The solution of the first one will be given by the svd, while the second would be from Tikhonov with parameter lambda

note that for A^TA singular min_x |Ax - b|^2 has infinitely many solutions. That's what the extra constraints are for - to pick out one solution from these.

I should have probably switched around the two sides of the s.t. for the first one

[wintry steppe]

Yes, but the main difference between this and min norm least squares is.... that min norm has E st 1/sigma(i) while tikhonov has an e that's sigma(i)/sigma(i)^2 + lambda

The notes i took, but I'm trying to understand

[wintry steppe]

I'm just trying to understand how i do this with an actual problem

[novel heath]

how do i find null space of a matrix only not basis for null space

im trying from yesterday but im not quite understanding

[molten ivy]

By null space of a matrix do you mean the kernel?

I don't understand the question here

[wintry steppe]

I think hes refering to row(a) perp

But yeah id have to see the matrix

[novel heath]

this is the question , im talking about . we dont have to find the bases for NULL space of A , we have only to find Null Space of A

[spare widget]

rank(A)+dim(NullA) = 4

https://en.wikipedia.org/wiki/Rank–nullity_theorem

since A is row equivalent to B they have the same rank

[novel heath]

thanks bro

but my question is "how to find the null space of A " and not the bases

[spare widget]

they don't ask you to find the null space, they are asking you about the dimension of the null space

you can derive an explicit expression for the nullspace by finding the shape of vectors x such that Bx = 0

[novel heath]

let me show you soluton

[spare widget]

I had a typo see the update

the solution would go something like this:

x = (x1,x2,x3,x4)

then multiply and solve for x1 and x2

you have x1 + 0 * x2 -1 * x3 + 5 *x4 = 0 -> x1 = x3 - 5 * x4 from the first line
and 0 * x1 -2 * x2 + 5 * x3 -6 * x4 = 0 -> x2 = 5/2 * x3 -3 * x4 from the second line

thus the space is: $(x_3-5x_4, 5/2 x_3 - 3 x_4, x_3, x_4)$ if I didn't make any mistakes

[stoic python]

criver

[spare widget]

you can see that it has two free parameters: x3 and x4, which agrees with dim(NullA) = 4 - rank(A) = 2

[novel heath]

thanks bro , but here look

see this is solution for NUL A but it is "bases for Nul A ". i need to find only "NUL A"

maybe im just too stupid

[spare widget]

you have the exact same thing as me

they just wrote the basis vectors explicitly

they got the two equations by solving Bx = 0

then they solved for x1 and x2

which resulted in what you see

[novel heath]

yeah bro i uderstood what you said it is same . but my teacher said find the NULL SPACE OF A ONLY NOT THE BASIS FOR NULL A

thats is the problem

[spare widget]

the null space of A is just the set of vectors for which Ax = 0

$Null(A) = {x ,:, Ax = 0} = {x ,:, Bx=0}$

[stoic python]

criver

[spare widget]

if they tell you that you don't have to find the basis, then you can basically stop at the point of writing Bx = 0

and not solve the rest

you have to compute dim(NullA) though, which is dim(NullA) = m - rank(A) = 4 - 2 = 2

[novel heath]

thanks bro just one last thing from where i should stop

writing

[wintry steppe]

Is the condition number.... just the highest singular value divided by the smaleste?

[spare widget]

yes

depends which norm you pick though

but typically people mean the euclidean norm, so yes

[novel heath]

@spare widget sorry for disturbing you , can you plz answer my question

[wintry steppe]

He did....

[leaden tide]

Hello, this problem is driving me round the bend right now

Let $p,q\in L(E)$ be two orthogonal projections (that is, projections whose images and kernels are orthogonal, or equivalently, that are autoadjoint)

[stoic python]

Syst3ms

[leaden tide]

Show that if $p\circ q$ is still an orthogonal projection, then $p\circ q = q\circ p$

[stoic python]

Syst3ms

[leaden tide]

This is easy to prove if we take $x \in Im(p\circ q)$, but I haven't managed to prove that $p\circ q(x) = 0 \implies q \circ p(x) = 0$

[stoic python]

Syst3ms

[leaden tide]

oh my fucking god

Found it

It's much simpler than whatever I was searching

If we write u* the adjoint of u, then we can just use the fact that pq is autoadjoint

$p\circ q=(p\circ q)^* = q^\circ p^ = q\circ p$

[stoic python]

Syst3ms

[leaden tide]

sigh

Welp, that's one thing solved

[wintry steppe]

Is that how you can do b???

[stuck tendon]

almost looks like a householder matrix 🤔
"since 1 is an eigenvalue, nullity is at least 1", how? identity matrix has 1 as eigenvalue but its nullity is 0

[zinc timber]

not a householder

projection onto the orthogonal subspace of u

[stuck tendon]

i said "almost" ;o

[zinc timber]

well

why ignore the obvious

[zinc timber]

what are tying to show for b?

[wintry steppe]

I'm saying since theres an eigenvalue of 1 that means there has to be at least one dimension in the null space. I'm prob wrong, IDK.

Whats a householder matrix

[zinc timber]

why null? shouldn't that mean at least one rank for the col space?

null space is the eigen space of 0

[wintry steppe]

Which is what was shown in a

I'm not sure

I thought having eigenvalues means there is a null space

Am I wrong?

[stuck tendon]

Eigenvalue of 0, not just any eigenvalue

[stuck tendon]

It's not householder [householder is for reflections, and it's of the form I - 2cc^t], it's a projection.
Remember for an orthogonal projection P, I - P is also an orthogonal projection
Try to consider P_U(u)

[wintry steppe]

Is that where you setup (I-P)u=u?

And show that p = 0?