linear-algebra

[spare widget]

Say the coefficients of the first poly are c1, and the second c2

You want to find q1,q2 such that cj^T qi =0, and q1^T q2 = 0

[bold sun]

ooh ok i kind of get that logic but the first one and second one already has x term and a constant if you get me?

[spare widget]

You have some freedom of choice for q1

q2 is then fully determined

[zinc timber]

@thorn yacht why did you delete it?

[thorn yacht]

Nvm my question: figured it out.

[bold sun]

we haven't learnt orthogonal vectors yet? whats that?

[zinc timber]

lol I was in the middle of reading

[spare widget]

u^T v =0 -> u and v orthogonal

[thorn yacht]

Ah, sorry, I was just confused about some notation for the double dual vector space

[spare widget]

You probably have something about an orthogonal complement of a subspace in your textbook

[bold sun]

i just checked the notes n didnt see it in chapter 10 we only on 6

havent done any of that yet

[spare widget]

You can think of it as 2 vectors spanning a plane through the origin, and the orthogonal complement being a vector perpendicular to the plane

basically when you have Ax = 0

Then if the rows of A are linearly independent, they can be thought as basis vectors

And the solutions x lie in the orthogonal complement to those

That is if a row is a R^n vectors

[bold sun]

okk slow down lol sorry so here is that what i did initially with the matrix stuff?

[spare widget]

And you have m linearly independent rows

Then you need to find n-m linearly independent vectors orthogonal to the m

And they will dpan you the orthogonal complement

let m<n

[native forum]

look at the polynomial x, its outside of the span of p1 and p2 because you cannot reach it using a linear combination.
so you need to include p3 = x in your basis for P3
same goes for the polynomial p4 = 1
when you include it you will have their span as well and can construct any polynomial in P3

[spare widget]

Then your system is underdetermined and you can write the general solution in terms of n-m unknowns

Plug in something for those unknowns, that's how you get the first vector

Then add it as a row to A, ribse and repeat

Until you find all n-m basis vectors of the orthogonal complement

The plug in something part means you have freedom of choice btw

[proper cradle]

Is there any method to do this one except taking specific vectors?

[bold sun]

lol uno i dont think ive learnt half the stuff uve mentioned here

[spare widget]

You've learned how to solve linear systems?

The above is just an interpretation of where the solutions of Ax = 0 lie

Pick a simple example

in 3d

(1,2,3), (1,1,0)

These two vectors span a plane, and any point of the plane can be written as

p(s,t) = s * (1,2,3) + t * (1,1,0)

Now say you want to find an orthogonal vector to this plane

Then you need a vector v which is orthogonal to both basis vectors

That is (1,2,3) dot v = 0, (1,1,0) dot v = 0

[bold sun]

waitt why are u doing this? the multiplying and coordinates?

[spare widget]

it's just a parametric definition of a plane

To give you some intuition

You can think of 1d spaces as lines through the origin

2d spaces as planes through the origin

[bold sun]

oh havent done that yet

[spare widget]

well the idea is, if you have Ax = 0, and the rows of A are linearly independent, then they form a basis

Any solution of the above lies in the orthogonal complement of the spaces spanned by that basis

So if A is made of 2 3d vectors

Then this basis spans a plane through the origin

And the solution space being the orthogonal complement means that the basis vector for it is perpendicular to this plane

In you problem they essentially ask you to complete the basis, you can do so by finding a basis gor the orthogonal complement

So having a set of vectors a1,...,am

You can find a set of vectors b1,...,b_{n-m}

[bold sun]

wait look this is the only stuff ive learnt so far it dont talk about any of that

[spare widget]

Such that ai^T bj = 0, and bk^T bl = 0

The last guarantees that the b are linearly independent

The former guarantees that they are also linearly independent wrt the as

[spare widget]

Then it's the right time to learn it

Because your problem is essentially asking for that

although it is possible to solve it purely algorithmically, it's probably not a good idea to skip over the understanding

But the purely algorithmic version is what I mentioned: Ax =0 -> find 1 solution, add the solution vector as a new row in the matrix, rinse and repeat

What this process does is it finds a new orthogonal vector to the previous ones at each step

So at the first step it finds an orthogonal vector b1 to all rows of A, and thus it cannot be linearly dependent

[bold sun]

dot here means times right

[spare widget]

Then add b1 as a last row to A to get A1, now A1 x = 0 -> find solution b2 -> it is orthogonal to all rows of A -> cannot be linearly dependent,add it as a last row to A1 to get A2

$(1,2,3)\cdot v = 1v_1 + 2v_2 + 3v_3$

Matrix-vector multiplication can be written as the dot products of the rows with the vector

[stoic python]

criver

[spare widget]

And the meaning of dot(u,v) =0 is that u and v are orthogonal, so u and v cannot be linearly dependent if both of those are non-zero

In geometric terms u and v are perpendicular

I am guessing the problem is meant to prepare you for whst's to come,as in make you think about the above

Or it could be just a problem that's there just for the sake of there being problems in the book 💩

Anyways you know the algorithm now, and I hope you got at least an idea of what thd meaning behind ssid algorithm is

[bold sun]

yeah it kind of makes a little bit of sense now- Thank you so much for everything!!

[spare widget]

https://youtu.be/eeMJg4uI7o0

[bold sun]

Thank you- ill have a look at it 🙂

[spare widget]

Only 13 minutes

[bold sun]

yep- thanks

and thanks 2 every1 else too!!

[bold sun]

and ok yes i watched it and that makes sense so thanks again!

[golden monolith]

Ah yeah I see

[proper cradle]

anyone?

[lavish jewel]

not really, because they are asking you which properties are true. you can spot the false ones by coming up with specific counter examples and plugging them in, but to show which ones are true, you need to show it holds for all vectors and all n

[zinc timber]

I can give you one hint to work on, calculate T^2

also you can write V as $Rv \oplus v^{\perp}$ if that helps

[spare widget]

I highly recommend this too: https://www.engineering.iastate.edu/~julied/classes/CE570/Notes/strangpaper.pdf

it's a fairly short read, but it should give you some motivation regarding what the course you're studying is about

[bold sun]

okk yeah thanks- ill read it in a bit

[proper cradle]

okay

[subtle gust]

Any good advice on doing proofs :)

I understand the questions that involve computing something

But i can't for the love of me understand proofs

Or like come up with them

[spare widget]

try going through some exercises in: Linear Algebra Problem Book by Halmos

he has detailed answers at the end

it takes practice I guess

there are books that are focused on proofs too so maybe that would help

my other advice is to pick several books up on linear algebra and see the different proofs

[twilit flower]

Hello guys, I am a newbie in linear algebra. Can you please give me some advice on what I should focus on or how I should think of linear algebra?

[tacit storm]

Any ideas?

[fringe fjord]

First assume that beta is nonzero.

[tacit storm]

It has to be though

for the last column to be zero

right?

[fringe fjord]

Huh?

[tacit storm]

nevermind

What happens if we assume beta in non-zero

[fringe fjord]

The second row has to be beta times the first.

[tacit storm]

Is this because we are assume we have only one linearly independent vector

so the other is a linear combination of the first

[fringe fjord]

Yes, it is an explicit assumption that the rank of the matrix is 1, unless I'm misreading the image.

[tacit storm]

ok

i understand

thanks

Can i ask one more question?

[crystal plover]

Yo, what should I use to solve this?

My first guess is gaussian elimination but I'm not 100% sure

[lavish jewel]

that will work fine

you seem to have more variables than equations though, so keep in mind how this might affect your solution

[crystal plover]

yeah, i don't remember exact question, i didn't copy it to my paper and now i try to figure it out for a retake xd

it was probably something like "for what w something something"

[icy blade]

where does the 2 * (-2c - b)... come from

[spare widget]

det(A) = dot(r1, cross(r2,r3))

where ri is the ith row of A and it is a 3x3 matrix

It's the part r11 * (r22 * r33 - r23 * r32)

[icy blade]

alr thx g

[humble oak]

if the determinant of a matrix is non-zero, why is it that the matrix is linearly independent?

i've been taught that it's just a fact and i am not too sure why that is

[dusky epoch]

one does not speak of a matrix being linearly independent

a set of vectors may be linearly independent or not

[humble oak]

a set of vectors yes

[dusky epoch]

one must not confuse a set of vectors and a matrix assembled from said set of vectors

but anyway, this all has to do with the myriad different ways to say "this matrix is invertible" in linear algebra which are all equivalent to one another

[humble oak]

hm

i guess i am going to learn more about that next time

[dusky epoch]

you might consider reading up on the formula for the inverse of a matrix that involves determinants

and in particular the formula contains 1/det(A)

which hints at matrices only being invertible when their determinants are invertible

[humble oak]

would this be a seperate property from the "a set of vectors is considered linearly independent if only the trivial solution exists"?

hmmm

nvm i will go read those

[dusky epoch]

there are multiple things wrong with this

[humble oak]

oh no

[dusky epoch]

first is painting linear independence as a matter of consideration

and then the whole "trivial solution" thing needs a lot of added details to make it make sense

[spare widget]

Ax = 0 only if x = 0 there you go

[dusky epoch]

then you have to say that A is a matrix with your vectors as columns

[spare widget]

Equivalent to the orthogonal complement of the space spanned by the row vectors being {0}

[icy blade]

can this be solvable with REF or do i have to do the det(A)....

@spare widget

[spare widget]

det != 0

proceed to find such a,b,c

[icy blade]

yea idk my teacher does not use Ax = b form

[spare widget]

The above is Ax=b form expanded

Whether you use

Matrix notation or the above is the same

[icy blade]

so is this solvable with REF? we have not gone over determinant properties

[spare widget]

idk what ref is

[icy blade]

reduced elechon form

[spare widget]

You'll just get sone denominator there too, so yes

But the easiest is to just compute the determinant

[icy blade]

what we usually do in class is get the bottom row to all equal 0 and then if its unique solution then the equation does not equal 0 like you said

but i cannot seem to get reduced elechon form without crazy fractions

[spare widget]

The point in the above is to make the last row non-zero

just use det!= 0

It's easier

[tacit storm]

Any hints

[spare widget]

Take u from R^n, decompose in u = u_parallel + u_perp

Then negate u_parallel

u_parallel = k * v, k = dot(u,v), u_perp = u - u_parallel

u' = u_perp - u_parallel = u - 2 * dot(u,v) * v

[mortal isle]

What does it mean to say v1, v2, and v3 are linearly independent

[spare widget]

It means that c1 * v1 + c2 * v2 + c3 * v3 = 0 only for (c1,c2,c3) = 0

assume this were not the case, and without loss of generality let c1!= 0

Then c2 * v2 + c3 * v3 = -c1 * v1

And thus v1 = -c2/c1 * v2 - c3/c1 * v3

But this means that v1 is linearly dependent with v2, v3

More precisely it lies in the plane spanned by v2,v3

At coordinates -c2/c1, -c3/c1

[mortal isle]

Wow. Thank you!

[spare widget]

https://youtu.be/eeMJg4uI7o0

I fixed a minus sign that I missed

[spare widget]

To understand the practical importance of this maybe I should mention a corollary. If you have a vector u in the span of v1, v2, v3 and they are linearly independent, then there's a unique set of coordinates (c1,c2,c3) such that u = c1 * v1 + c2 * v2 + c3 * v3. If they were linearly dependent then there could be infinitely many coefficients representing the same vector. This may be undesirable in practice, e.g. when you want a unique representation. It could also be desirable to have many different representation of the same vector, e.g. if some data has noise, then instead of a basis you get an overcomplete basis/frame.

[midnight gust]

I was just speaking to someone and they kept using the term eigenvector to mean basis vector

This person has been working in the field for a really long time so they must understand something I don’t

Is there a relationship between eigenvectors and basis vectors I don’t see?

[fringe fjord]

A basis consisting of eigenvectors (an eigenbasis) is very useful to have if it exists; that's a basis that diagonalizes the matrix (or map) you're looking at.

[subtle gust]

for homogeneous linear systems

the system can only have one solution (the zero vector) or infinitely many solutions

is there any other case?

[dusky epoch]

if your base field is R or C, then no.

[dapper gorge]

The answer to this is to consider vector spaces over non-algebraically closed fields, right?

[zinc timber]

yes, you are in the right direction

[dapper gorge]

well then that's it

[zinc timber]

so what's the example you came up with

[dapper gorge]

I can't think of a nice example tbh, other than constructing artifical examples

i.e. examples where the characteristic polynomial factors, say in the real numbers, as x^mQ(x) where Q(x) is a product of irreducible quadratics

I'm thinking maybe in R^3 define a plane A. If x not in A send it to 0. If x is in A, rotate it in A (with a non trivial angle)

I think that's nicer

well you would define that on the basis. Two vectors on the basis define that plane, the third is sent to zero

[zinc timber]

what about that irred quadratic being x²+1 in R?

say char poly = || x(x²+1) ||

[lilac lichen]

Guys can anyone tell me where people discuss about 8th maths

[dapper gorge]

uhm idk tbh

(responding to ryu sama)

there is a two dimensional space such that T^2=-I, so T^2 acts as some kind of reflection

idk

[lavish jewel]

probably #prealg-and-algebra

[lilac lichen]

Oh thx lemme check

[zinc timber]

what about rotations?

[dapper gorge]

no

[zinc timber]

say rotation of π/2 radians

[dapper gorge]

because rot * rot=rot

oh

[lavish jewel]

[dapper gorge]

that's like the e^{pi i}-1=0 thing

lol

[dapper gorge]

that's a rotation

[zinc timber]

so not nilpotent

[lavish jewel]

you are both saying the same thing but disagreeing

idk why

[zinc timber]

are we?

[lavish jewel]

so far you both said rotate + project

[dapper gorge]

haha lol I thought it was something more misterious. Because I did say about rotations on a subspace and sending to zero on another subspace

[lavish jewel]

ryu's char poly is the result of a block diagonal matrix of rotations and projections

[dapper gorge]

I'm not good at coming up with specific examples anyway

[zinc timber]

oh i didn't read the middle part of the message

lol mb

[lavish jewel]

admittedly only for R^3, but it generalizes

[zinc timber]

[lavish jewel]

so you want some power of x and a product of irreducible quadratics

sounds about right

[dapper gorge]

haha no the thought process was fun

[zinc timber]

yes that should be a general construction

[dapper gorge]

btw

[zinc timber]

ok I'm giving away my example

[lavish jewel]

i think you need the rotated coordinates not to be projected

so maybe the rotations and projections need to be able to commute

at least pairwise?

[dapper gorge]

when solving polynomial equations we talk about field extensions and so on. What is the analog concept to vector spaces?

[zinc timber]

$\dmat{0}{\mqty{0 & -1 \ 1 & 0}$

nice

dammit

[dapper gorge]

[lavish jewel]

what's dmat 0

[zinc timber]

$\m{0 && \ && -1 \ & 1 & }$

[dapper gorge]

yeah

[lavish jewel]

right

[zinc timber]

diagonal matrix with first entry 0

[lavish jewel]

this is the same as $\Pi \circ R = R \circ \Pi$

[stoic python]

19eddy4

[zinc timber]

yes

[dapper gorge]

To determine this kind of maps we need just worry about 2 dimensional spaces when they are over R

[zinc timber]

yes

[dapper gorge]

but I think its not just rotations

[zinc timber]

[dapper gorge]

xd what

[zinc timber]

wym not just rotation?

[dapper gorge]

oh

I mean that the action of the linear operator on these 2 dimensional subspaces is not always a rotation

I'm thinking, but idk

[zinc timber]

yes

they can also be dilation and rotation

[dapper gorge]

because the characteristic polynomial of rotations is x^2-2cost x+1=0, and there are many other forms for irreducible polynomials

true

[lavish jewel]

wait what

how'd you get that poly

(it might just be too early for me)

[dapper gorge]

(cost -x)^2+sin^2 t=0 ?

[lavish jewel]

yeah ok

derp moment

you can try and see if you can rewrite other polys in this form

[dapper gorge]

is it just dilations with rotations?

We would need a+b and ab to be able to take arbitrary independent values. And the condition that the polynomial is irreducible is given by cos^2(t)(a+b)^2<4ab, which can always be satisfied taking arbitrary values for t

I think that's the case, yeah

well not exactly, because a or b could be negative, but we disallow that possibility. I think that's the argument

[stark acorn]

Is addition of two non zero vectors consididered linear cobination

[lavish jewel]

yes

[stark acorn]

What would a Proof of thatlook like

this isnt an assignment

This is just for my own knowing

Let c1,c2...,ck be a scalar

Let v1,v2,.. vk be a vector

the vector in r^n

[dusky epoch]

??

what are you asking again?

"let v, w be two nonzero vectors. is v + w a linear combination of v and w?"

[stark acorn]

Oh im stupid

[dusky epoch]

and the answer is yes, obviously it is. it's 1v + 1w

[stark acorn]

Why is the addition of vectors called a linear combination?

[dusky epoch]

?

'linear combination' is more general

[stark acorn]

I see

So if I add two matrix together

[dusky epoch]

a linear combination of v and w looks like this: av + bw
(where a and b are scalars)

[stark acorn]

would that be a linear combination

[dusky epoch]

sure, the sum is a linear combination of your two matrices

the concept of 'linear combination' makes sense in any vector space

[stark acorn]

Thats exactly how 3Blue1Brown defined it

[zinc timber]

because that's the definition

[stark acorn]

Is my teacher misusing the definition?

Doesnt that simply mean then A + B?

[dusky epoch]

no

[zinc timber]

you supposed to write b as a LC of columns of A

not A+b

[stark acorn]

Can someone help me see where my fault in understanding is so I can work on it more?

[dusky epoch]

your fault seems to be "a+b is a linear combination of a and b therefore any time i see 'linear combination of <blah>' im supposed to just add them"

[zinc timber]

probably u didn't understand the question

[dusky epoch]

you are supposed to write $\bmqty{-9\4}$ as a linear combination of $\bmqty{1\-3}, \bmqty{3\1}$ and $\bmqty{-1\2}$

[stoic python]

Ann

[stark acorn]

I am still confused

Doesnt Linear COmbination just mean add

?

[dusky epoch]

no!

[stark acorn]

I am confused

Can you explain to me what a linear combination is

[dusky epoch]

.

accordingly, a linear combination of more than one vector, or matrix, or whatever, looks like that same sum, except that each term has a scalar coefficient attached to it

[stark acorn]

does that mean just multiple

[dusky epoch]

no

[stark acorn]

im having trouble seeing the conection

[dusky epoch]

it's not "JUST ADD" or "JUST MULTIPLY" stop trying to reduce it to JUST something if you keep trying to do that youll NEVER stop being confused

idk how else to tell you this

[stark acorn]

How is V + B

not a linear combination

[dusky epoch]

what is V

what is B

[stark acorn]

V is a vector

and B is a vector

[dusky epoch]

just some random vectors?

[stark acorn]

in Rn

[dusky epoch]

nothing to do with your problem?

[stark acorn]

yes

[dusky epoch]

and where did i claim "V+B is not a linear combination of V and B"?

are you sure you're not putting words in my mouth rn

[stark acorn]

You did not

I am just asking to understanding

[dusky epoch]

so you are putting words in my mouth rn?

"How is V + B not a linear combination"

as if someone said it was not

[stark acorn]

I was under the impression that it needs a scalar to multiply to be a vector addition

[wintry steppe]

Where do I learn about Lp spaces?

I don't think Axler, for example, will go over them

usually functional analysis

[stark acorn]

Im sorry if i offended you

[proper cradle]

Hey just quick test, If matrix A is symmetric then eigen vector of A is orthogonal?

[zinc timber]

no

@proper cradle

[proper cradle]

Only if eigenvalues are distinct?

[zinc timber]

if the eigen values are distinct then eigen vectors are orthogonal

not the other way around

[proper cradle]

Okay

Unitary diagonalizable is true for normal matrix right and symmetry matrix are normal but here we are getting eigen vectors which are not orthogonal then how can we unitary diagonalize?

[zinc timber]

how are you getting eigen vectors that are not orthogonal?

[proper cradle]

That’s the question, and you just told that it is not true for all symmetric matrix?

[zinc timber]

I didn't say if the eigen values are same then the eigen vectors can't be orthogonal

for example take M = I_2 then (0,1) and (1,0) are EV with ev 1 and they are orthogonal

(1,0) and (1,2) are also EV with ev =1 but they are not orthogonal

[lavish jewel]

they're asking about symmetric matrices, ryu

[zinc timber]

I2 is symmetric

[lavish jewel]

oh i misread your 3rd line

don't mind me

[zinc timber]

[proper cradle]

If suppose A is symmetric and all the eigen vectors are not orthogonal then how we proceed for unitary diagonalizable?

[proper cradle]

Is there exist such matrix?

[zinc timber]

if A is symmetric, you will always find a orthogonal basis

they are orthogonal if corresponding eigen values are different

if ev is the same then you look at the null(A-aI) and find a orthogonal basis of this

[wanton seal]

The answer to the first part of this question is "Yes. Since Col A is a five-dimensional subspace of R5, it coincides with R5.". But we only have four pivot columns, meaning four linearly independent vectors in R5. Hence they aren't a basis in R5 and can't span R5 fully?

[proper cradle]

Then unitary daigonalization is true for any diagonalizable matrix? Due to gram schmith

[zinc timber]

noo, diagonalizable does not mean it's unitarily diagonalizable,, you are mixing up things

if A is symmetric then A is unitarily diagonalizable,

[proper cradle]

Normal matrix are

[zinc timber]

yes normal matrices are

for example take the projection onto x axis along (1,1) then it's diagonalizable but not unitarily diagonalizable

[proper cradle]

Look, we are make U just picking the ev of matirx A if ev are not orthogonal then how we proceed that’s the dout

[proper cradle]

For A=UDU*

[dim epoch]

Gram Schmith gives you an orthogonal/orthonormal basis of a vector space, that's different from finding an orthogonal basis consisting of eigenvectors which you need to diagonalize a matrix

[proper cradle]

So if they are not orthogonal then?

[zinc timber]

because in general case D = P^{-1}AP, if how ever A is happened to be normal, your P is orthonormal (can be made into one)

in that case P^{-1}AP=P*AP

[proper cradle]

You said that, if they are not orthogonal then how we change that vectors in orthogonal and why we can’t do that this process apply for any matrix?

[zinc timber]

i think you may hv mis interpreted my words

as you see eigen vectors are not unique

for example if v is an eigen vector then so is cv (c≠0)

[proper cradle]

And $P=[ev_1 ev_2 … ev_n] $ if p is unitary then $ev_i$ are orthogonal

[stoic python]

BLツ

[zinc timber]

then v and cv is clearly not orthogonal as <v, cv> = c |v|²

[proper cradle]

Yes it is true

[zinc timber]

all I said was if you don't mention the eigen values are distinct then you can't say eigen vectors are orthogonal, it doesn't however mean that you can't find eigen vector corresponding to same eigen values which are orthogonal for example the I_2 matrix

[proper cradle]

Alright got it

If GM>1 then we can always find orthogonal vectors for normal matrix right

[zinc timber]

now ask again what your doubt it

[zinc timber]

yes

[proper cradle]

My doubt is if AM>1 since it is normal so AM =GM, so if we are not able to find all eigen vectors that’s are orthogonal so we need to construct others?

[zinc timber]

your matrix is normal then why won't you be able to find orthonormal eigen basis?

if the matrix is normal then you will always be able to find orthonormal basis wrt which the matrix is diagonal, that's the complex spectral theorem

[proper cradle]

Okay

Got it

Thank you so much

[wanton seal]

just bumping my noob question :^)

[wintry steppe]

Maybe a simple question, but if we want to prove something for inner products of $\langle x, x\rangle > 0$ and $\langle y, y \rangle < 0$, why can we assume that $\langle x, x\rangle = 1$ and $\langle y, y\rangle = -1$ "due to scaling"?

[stoic python]

n/c

[dim epoch]

because the inner product is a bilinear form

[wintry steppe]

Can you elaborate?

[dim epoch]

hence you can always find some scalar a to make it 1 for a* <x,x> and and pull it into the the product

to get new vectors

that make it 1 or -1

i think at least

[lavish jewel]

there is a mistake in what n/c wrote though

inner products are positive definite

<y,y> = -1 cannot be an inner product

[wintry steppe]

Not if we replace the axiom of positivity with <x,x> = 0 iff x = 0

[lavish jewel]

hmm?

[dim epoch]

wait i forgot about it being positive definite lol

[subtle walrus]

you often deal with dot products that are not positive definite

[wintry steppe]

But anyway that's not the point of my question

[subtle walrus]

not sure if you use different terminology in english

(also note that positive definite doesnt make sense in general fields)

[lavish jewel]

the one i often see is positive semidefiniteness, at least

this still excludes getting negative values

[dim epoch]

uuh

but either way if its bilinear

and we wanna prove something about the product itself

and not about the vector in the product

we can scale it

[wintry steppe]

So basically, we can define $\langle x,x \rangle = 1$ because $\langle x,x \rangle = \frac{\langle a, a\rangle}{|\langle a,a \rangle|}$..? Or something?

[lavish jewel]

pretty much, you could multiply the inner product result by some scalar so that the result is normalized

[stoic python]

n/c

[lavish jewel]

and this results in another inner product

[dim epoch]

i mean you're not really defining it but you can assume it without loss of generality in your proof

[wintry steppe]

Sorry that's what I meant yes

[lavish jewel]

are you working with indefinite inner product spaces?

[wintry steppe]

What are indefinite product spaces?

The exercise was just about changing one of the axioms to what I listed above

[lavish jewel]

inner product spaces where indefinite and semi definite inner products are used, kinda generalizations of inner product spaces

because in usual inner product spaces you do need positive definiteness

[wintry steppe]

I don't know what that means sorry

The axioms I'm working with for an inner product is (x,y) = (y,x), (x, y + z) = (x,y) + (x,z), (cx,y) = c(x,y), (x,x) > 0 for nonzero x, and (x,y) = (y,x)* (hermitian symmetry)

[lavish jewel]

you wrote it yourself :x

you can't usually have <y,y> = -1

[wintry steppe]

I know

But there was a particular exercise in this chapter which told us to change (x,x) > 0 to (x,x) = 0 iff x = 0

[lavish jewel]

that's the same as positive definiteness

[wintry steppe]

What is this word, definiteness?

[lavish jewel]

exactly what you're writing

<x,x> = 0 if and only if x = 0

that's the definition

and <x,x> >= 0, and equality is only achieved if x = 0

the result is always positive, unless x = 0

[quaint escarp]

hello, i hope i'm posting this in the correct topic.
What exactly am i doing here?

[zinc timber]

ok that's a long question

hmm maybe not

for the first case, A is diagonalizable means the char eq has 2 distinct root or it's the constant multiple of the identity.

the char eq is given by x^2-(a+d)x+(ad-bc) so it'll have distinct root if (a+d)^2-4(ad-bc) > 0 ( not < 0 as that will give complex roots which we don't want)

and if B^2-4AC = 0 then it means both roots are equal, so it has to be constant multiple of the identity, which means b=c=0 and a=d

follow similar method for ii

@quaint escarp

[quaint escarp]

might take a while for me to understand lol
Thank you for your input 🙂

[zinc timber]

[quaint escarp]

me=

[zinc timber]

[-10 0 \ 0 10] =

[quaint escarp]

wait

ohh

OHHH

[oblique prairie]

[earnest dock]

Looking for help on my logic here
https://i.imgur.com/H3PBbXv.png
I have this transformation and this A and I want to prove that if A != zero matrix then Dim(Im(T_A)) = 2

my idea here is that no matter what matrix A is, you can find an X s.t AX=XA that is also non-zero, which means X is in Ker T_A, so Ker T_a has a dimension of 1 meaning Im(T_A) has a dimension of 2(since the overall dimension is 3 and DimKer + DimIm = DimV)

am I wrong here? X would be the antisymmetrical matrix with a,b,c=1

[zinc timber]

I can give you a different approach to this

try to find the kernel of the transformation and use first isomorphism theorem

see that kernel of T_A is the vs of all matrices that commute with A

be back in a sec

[earnest dock]

I know there are different approaches, but I'm wondering if mine is correct because I wrote this down in the test.

[zinc timber]

FFS, hope no one saw

[earnest dock]

🙂

[zinc timber]

[earnest dock]

😉

[blissful creek]

Hey for |U1-v|² what should i do there ?

I think they asked me to calculate thr firt two to use it there

[proper cradle]

solve (u1-v, u1-v) using inner product axioms

[earnest dock]

can you tell me what you think of my solution?

[zinc timber]

your idea is similar to what I just said

problem is "there is an X s.t. AX=XA" doesn't mean that kerT has dim 1

[earnest dock]

has at least 1?

[zinc timber]

and also your V is M_3x3(R) here not R^3 so dimV = 9 not 3

[earnest dock]

is that the issue?

[zinc timber]

dim is atleast 1 yes but not =1

[earnest dock]

but look at V

or ehm

wait I didn't send it

[zinc timber]

V is not M_3x3(R)?

[earnest dock]

https://i.imgur.com/ihH4Ain.png

[zinc timber]

so set of all skew-symm matrices

yes then the dimV = 3

[earnest dock]

would I need to prove that? 🤔

[zinc timber]

that's a known fact, but if you are not sure then you can try to prove it

ok so X is skew symm matrix and A is given

[earnest dock]

I wrote X explicitly

[zinc timber]

you need the find the dim of the matrices that commutes with A

[earnest dock]

I wrote X explicitly for every A

X is a,b, c = 1

[zinc timber]

you wrote A=[] not X=[]

[earnest dock]

I wrote X

https://i.imgur.com/xtHH7Ja.png

oof

I think it needs to be the opposite

[zinc timber]

that's just one X

[earnest dock]

like

[zinc timber]

you need to find for all X

[earnest dock]

well the kernel is all AX-XA=0

[zinc timber]

ok tell me this, is A = [] what have you written with a,b,c is correct?

[earnest dock]

so AX=XA

[zinc timber]

like that's A right?

and not a typo?

[earnest dock]

wait I don't want to confuse myself

let me just rephrase

[zinc timber]

you surely confused me, so

[earnest dock]

the transformation is AX - XA

so I need to find all X's s.t AX=XA

[zinc timber]

yes and what is A?

[earnest dock]

A is antisymmetrical

[zinc timber]

any?

[earnest dock]

or whatever it's called in English

yeah

in R

not C

[zinc timber]

ok so you don't even know what A is then how are you supposed to calc the dim?

like A could be zero giving you rank = 0

you need to fix A

[earnest dock]

no no the A is non-zero

that is part of the question details

A is non-zero, prove dim(T_A)=2

[zinc timber]

[earnest dock]

now it is known that dimV=3 as we're shown

and so we need to find AX=XA

so X= what I wrote will always give me that

[zinc timber]

huh

[earnest dock]

^

the transformation is defined : T_A(X) = AX - XA

so T_A(X)= 0 => AX=XA

that's the kernel

[zinc timber]

so we only know A is non zero anti symmetric matrix

[earnest dock]

yes, and X is as well

well X can be 0 but it's useless

[zinc timber]

the better approach for this will be to see where A sends the basis then

[earnest dock]

since all vector fields have 0 it doesn't change the dimension

wait, please ryu sama, I'm trying to understand if my solution was correct because I want to appeal...

I got 0 points

(for the question)

[zinc timber]

no, you can't just say the dimension is atleast one so dim is 1

you have to show dim is actually 1

[earnest dock]

well it's exactly 1 because that's the only solution...

but I wrote it in general..

[zinc timber]

then you are literally skipping 5-10 steps of showing that's the only solution

and also you can't assert that there is one X s.t. AX=XA

[earnest dock]

AX=XA, A is non-zero anti-symmetrical, X is non-zero anti-symmetrical
clearly this X is the only linearly independent solution here right?

[zinc timber]

it's like proof by assumption, assume it's true the it's true

[earnest dock]

yeah, I guess.

[zinc timber]

no it's not clear in any way, you have to show it

there's no loose ends in mathematics

[earnest dock]

I should've written that the only solution is X or X^T as I wrote it, but they're linearly dependent by definition(multiple of -1 of each other)

fk.

I guess it does deserve 0 points?

[zinc timber]

heck if you know there's even a non zero X s.t. AX=XA

[zinc timber]

yes

[zinc timber]

like it's not even trivial to me and you are taking an exam

like if you really want a solution, I can guide u to one, but yeah your argument is not valid

[zinc timber]

which problem are you talking about

[blissful creek]

The first question |u1-v|²

Wait let me send a better pic

[zinc timber]

@earnest dock here's an short argument, AX-XA has trace 0 always, so it cuts down one degree of freedom giving you 2

[zinc timber]

so where are you stuck exactly? the instructions are crystal clear

[blissful creek]

No like should i do u1-v and then the ||² or is there a fromule i should follow

[earnest dock]

thanks.

[zinc timber]

like yeah

[dusk basalt]

Given a polynomial, how does one find the companion matrix?. My proffesor just gave the campanion matrix without explanantion

[zinc timber]

that's how

[dusk basalt]

thanks

So there is no way to go from the polynomial to the matrix form without?

Without already knowing the polynomial form.

[zinc timber]

?

[dusk basalt]

So assume that you dont know the matrix A, how do you determine A given a polynomial?

[zinc timber]

^

(polynomial has to be monic tho)

[dusk basalt]

Well, what is give there is that you already know the form of A and then you prove by induction that A is the companion matrix op de polynomial P. (I understand that totally).

But let's assume that you don't know how A looked like, how would you then come up to that?

[zinc timber]

oh let me send again then

@dusk basalt ?

[wintry steppe]

[dusk basalt]

well, I think you don't understand what I really mean. So let me explain once more. Say $p_A = t^n + a_{n-1}t^{n-1} +.....+a-O$. How would you find the companion matrix A?

[stoic python]

Croq

[dusk basalt]

if u never knew that A is this

[zinc timber]

if it's still not clear to you by now, you just reverse the construction

[zinc timber]

given this

[zinc timber]

this is the companion matrix

[dusk basalt]

Yes, I know that the reverse-order construction works (but just considering uniqueness)

I guess I am good, thanks though!

[zinc timber]

what uniqueness? companion matrix is unique

[dusk basalt]

I have been saying "assume". So act like you don't know A, how should A look like if $det(A-t*I) = p_A$

[stoic python]

Croq

[dusk basalt]

But I am good, thxx!

[zinc timber]

if you want a "general A" then will be PSP^{-1} where S is matrix I showed before

[zinc timber]

but that is no longer called the "companion matrix" because companion matrix have a specific form, which looks like this and this only

[dusk basalt]

yess I agree, but how could you know it only looks like that if you didn't know how it looks like?

[zinc timber]

because that's the "definition" of companion matrix

ref: wiki

[stone fern]

Just a quick question:
Is the matrix of a linear map f in the basis B the same as the matrix of the linear map f in the pair of basis (B, B)?

[dusk basalt]

Yes but only if the linear map is an operator

[stone fern]

operator, as in codomain = domain?

[dusk basalt]

yes exactly

If you talk generally about a linear map f, then the basis in the domain and codomain are different

[stone fern]

I meant codomain=domain from the beginning, sorry for poorly formulating the question

thanks

[dusk basalt]

no worries, you are welcome

[brittle gyro]

So, I figured I have to calulate $| xx^T-yy^T |_2$, but I'm not sure how

[stoic python]

Sydd

[brittle gyro]

Something like $\sup_{v \in S^1} | x^T v x - y^T v y |$ felt like the way but I just got stuck

[stoic python]

Sydd

[zinc timber]

what's the def of dist(S1, S2)?

[dapper gorge]

Is this formula of computational interest or nah?

I know there is another formula, but it involves a lot of determinants

[zinc timber]

nah

[brittle gyro]

distance between orthogonal projections onto subspace

[zinc timber]

no that's what I'm asking how are they defining the dist because to me it feels like it should be zero

because 0 is a common element

[brittle gyro]

every subspace S has an unique orthogonal projection, so the way you define dist between subspace is by setting it to be dist between the projections

for a one dimensional subspace like span{x} the projection is xx^T

[wintry steppe]

can you send the source code?

found it.

[trim bear]

Its on my other computer

but uhm

[reef sierra]

hey quick question, does this mean that a can be equal to any value when the rank of the matrix is = 3?

[fringe fjord]

It's definitely not the case that the rank is 3 when a = 0, at least.

[zinc timber]

ye why is it showing rank 3

maybe thinking it as a matrix in R[a]

[zinc timber]

are you sure that this is what you are supposed to do?

[lime zinc]

ant hint?

[zinc timber]

rank A² < rank A³

@lime zinc

[lime zinc]

Rank is increasing

[zinc timber]

so...

[lime zinc]

If somehow we can get its jc form then we can find gm of 0

[dusky epoch]

rank(A^2)<rank(A^3) is literally impossible no matter what A is, lol

[zinc timber]

if it was > you can try to find a possible jcf

[lime zinc]

Yes

[lime zinc]

Lets consider this

[zinc timber]

look at the principal component of x

if B is the submatrix then B⁴=0 but B³=B⁴= 0 means that max size of jordan 0 block is 3

since B² ≠0 gives min size =3

can you construct the jordan form of B now?

[lime zinc]

Ok,I will try

[zinc timber]

don't worry about other eigen values

[lime zinc]

I have no idea about this ( component. From)

[zinc timber]

dont worry about it

just think as if you are working with a matrix with char poly x^4

rest you can ignore

(it comes from the principle decomposition theorem)

[lime zinc]

This is only possible if matrix is a nilpotent matrix

[zinc timber]

yes and how does it look like?

jcf I mean

[lime zinc]

All disgonal entries zero

But how you get here , i have doubt

[zinc timber]

disgonal? off diagonal??

oh diagonal

[lime zinc]

Here our matrix is of order 9

Wait

And B is of order 4

[zinc timber]

since x^3 = 0 and x^2 neq 0 this means that B has a jordan block of size 3 with ev 0

[lime zinc]

So number of block in jc is gm of that eigenvalue ,so 0 has gm =3

[zinc timber]

since 4-3=1 then there's only 1 other jordan block of size 1

[zinc timber]

no 0 has gm 2

why do you think gm=3?

[lime zinc]

First thing is ,how you get this?

[zinc timber]

primary decomposition

(A-kI)^n cannot be singular unless k is an eigen value

you can write $V$ as $V = \mbb{R}[x]/(x^4)\oplus \mbb{R}[x] / ( (x-1)^2) \oplus \mbb{R} / ((x-2)^3)$

[zinc timber]

in other words V has a generalized eigen basis

only the generalized eigen vectors corresponding to eigen value 0 can contribute in reducing the rank of A^k

[lime zinc]

So here i have to focus on x^4 because we want to find gm of 0?

[lime zinc]

This is something new for me

[zinc timber]

have you not heard of primary decomposition

or generalized eigen vectors?

[lime zinc]

Nop

[zinc timber]

ok so don't know if there's an easy way to say this

but say you have any jcf of that matrix

than any jordan block where the eigen value is not zero cannot have rank A^2 < rank A

so the jordan blocks with ev 1 or 2 will never loose rank as you raise the power

only block that will loose rank is the jordan blocks with ev 0

this is very easy to check as say $A = J_2(a) = \m{a & 1 \ 0 & a}$ then $A^2 = \m{a^2 & 2a \ 0 & a^2}$ and $A^3 = \m{a^3 & 3a^2 \ 0 & a^3}$

[zinc timber]

all of them have rank 2

but if a were 0 then see that A^2 == 0

[lime zinc]

This something interesting, is this applicable to any matrix?

[zinc timber]

well yes, if all eigen values of A are non zero then A^n always has full rank

A^n can loose rank if 0 is an eigen value (with am > 2)

i can make this statement more precise but nah

[lime zinc]

Thanyou @zinc timber , i think i should leave this question.

[zinc timber]

for an easy proof, say A^k has rank < rank A means A^k has 0 as an eigen value but all eigen values of A^k are of the form a^k where a is ev of a. so a^k=0 means a=0 so 0 is an eigen value of A

[zinc timber]

your main question is still unanswered, were you able to solve it?

[lime zinc]

Its ok @zinc timber , i will try it myself with my little knowledge

[zinc timber]

okey

[lime zinc]

First i think i should read primary decomposition

[zinc timber]

that's nothing ground breaking really

it's kinda similar to ax+by=1 if (a,b) = 1 from number theory

[lime zinc]

You are making me feel dumb now

Okk i will try🙏 🙏

[brittle gyro]

I must show $| xx^T-yy^T |_2=\sqrt{1-(x^Ty)^2}$ somehow

[stoic python]

Sydd

[lavish jewel]

you can try just expanding the terms first

you could also separate consider the case where x and y are lin dep and lin indep

[brittle gyro]

expand in what sense? like $x=(x_1,x_2), y=(y_1,y_2)$?

[stoic python]

Sydd

[lavish jewel]

i meant expanding the two norm you have there

you can rewrite that as a sum of scaled rank 1 matrices

and then do a change of basis onto an orthonormal one

the expression you have there is a diagonalized matrix of rank at most 2, and this is related to the definition of the 2 norm of a matrix

[zinc timber]

I was thinking of a different approach but the calculation seems long so I'm dropping it

the 2norm of that operator is the largest eigen values as it's symmetric, say we know the angle between x and y then we can project any unit vector in the plane of x, y onto the vectors then they'll have magnitude cos(theta) and cos(theta+phi)

[lavish jewel]

i think the term in the square root is pretty much gonna be the result of doing gram schmidt

[zinc timber]

then we have to maximize the sum of their mag using basic calx

[lavish jewel]

cuz then you get a matrix diagonalized on an orthogonal basis, and the largest singular value is just the 2 norm of one of the basis vectors

[zinc timber]

this might actually work

[lavish jewel]

just a hunch to be fair, but since it's only rank 2, it should be fairly straightforward to do it this way

and you can still consider the rank 1 and rank 2 cases separately, doing the rank 1 one as a sanity check

(at a glance, yes, the rank 1 case checks out)

or you can do it ryu's way

i just immediately go "dat SVD" whenever i see a sum of rank 1 matrices

[zinc timber]

no mine is complicated

I have tried

unless someone points out a better way to calculate eigen values of xx'-yy'

[brittle gyro]

yeah that feels like a good gut feeling to have

gonna try these approaches in a bit

but thanks already, it's so nice to have these ideas thrown around here, it helps a lot!

[stoic python]

In and Out

[lavish jewel]

@wind anvil please check #prealg-and-algebra

[wind anvil]

okay

[wintry steppe]

Let $P$ be the vector space of all real polynomials. Why is $(f,g) = f(1)g(1)$ not an inner product of $P$?

[stoic python]

n/c

[wintry steppe]

Because $f(1)g(1) = g(1)f(1) = (g,f)$, $(f,g+h) = f(1)[g(1) + h(1) = f(1)g(1) + f(1)h(1) = (f+g,f+h), (cf,g) = (cf)(1)g(1) = cf(1)g(1) = c(f,g)$

[stoic python]

n/c

[wintry steppe]

So it should be an inner product

And clearly $(f,f) = f(1)f(1) = f(1)^2 \ge 0$

[stoic python]

n/c

[wintry steppe]

And equality iff f is the zero polynomial.

you sure?

About which part?

[wintry steppe]

this

are you sure?

Hahahah I'm so dumb.

thaaaaanks

[serene tide]

can i get a hint on part 2, it seems really simple and im missing something basic

[indigo stirrup]

Hi, this is a statement that got me confused. "Let vectors u and v be in R^n, vector b be in R^m, and A be an m x n matrix. If there exist scalars c and d such that c(A times vector u) + d(A time vector v) = vector b, then vector b is in the span of the columns of A, but is not necessarily in Span(u, v)." I have to answer either true or false, but I just don't know how to approach it. I first thought that any vector that is in Span(u,v) must be in R^n. If n and m are not equal, b is not in R^n, and therefore, not in the span of {u,v}. Would the above statement be true then? Thanks

[zinc timber]

triangle inequality

|u| = |u -v + v|

[serene tide]

cool, thanks

[wintry steppe]

hey all is this a prediction or inference problem? determining which variables are relevant for predicting an output variable

i got it wrong and id be curious to hear ur thoughts

[brittle gyro]

is that python?

[sleek sundial]

Anyone have a good resource on duals? Only thing I’ve had on it was in axlers LA

[wintry steppe]

is Au * Av = A(u * v)

for dot product

no right?

[tired fossil]

he guys, does anyone know how to approach this problem?

for b)

this is the answer

[wintry steppe]

@tired fossil You just need three polynomials which are linearly independent and of degree 0, 1 and 2

So you're given a list of polynomials, you should check if there are any redundancies

Or, if you can show that some combination of them are linearly independent, and span P_2, then that shows it is a basis too

[tired fossil]

so, if they span P_2 and are independent, it is okay?

[wintry steppe]

@tired fossil That's the definition of a finite basis

First row is x + kz = 0 and last row is -kx + 2y + 3kz = 0. Then {kx + k^2z = 0} + {-kx + 2y + 3kz = 0} gives us 0 + 2y + (k^2 + 3k)z = 0

Then -2 * (y - z) = -2y + 2z = 0, and that with the new last row gives us 0x + 0y + (k^2 + 3k + 2)z = 0

[quaint steppe]

are there any good video lectures about linear algebra on youtube? do you guys have any recommendations?

[zinc timber]

MIT lecture series by gilbert strang

[quaint steppe]

Thanks

[wintry steppe]

how would i describe this in geometrical terms?

like what do 'geometrical terms' even mean?

[lavish jewel]

like a geometric figure

i recommend you do a couple of low dimensional examples and see if you can find the pattern

try R, R^2, R^3

for these cases, you can easily substitute the 2-norm of a difference of vectors for a simple algebraic expression

[wintry steppe]

i guess im a bit lost about what the general form even looks like, do u have any resources on this?

we didnt learn this

@wintry steppe If I gave you the set {x in R : | |x - x_0 | | ≤ 1}

What is this set, given some random x_0?

a hyperplane right?

What is a hyperplane?

[lavish jewel]

you already took geometry

[wintry steppe]

its a subspace where the dimension is one less than n

[lavish jewel]

do you recognize the shape x^2 + y^2 = r^2?

[wintry steppe]

yea circle

[lavish jewel]

ok

now expand the expression you have for a generic pair of vectors x and x0 in R^2

do you know what the || || stands for?

[wintry steppe]

Euclidean length right

/ norm

[lavish jewel]

mhm

now take a the norm of [x,y] - [x0, y0]

[wintry steppe]

its a hypersphere i think

sorry i was working it out

[lavish jewel]

we're in R^2

well i guess hypersphere is fine if you mean a 1-sphere here

BUT

they don't say = 1

it's <= 1

so it's a disk or a 2-ball

[wintry steppe]

ah ok so ur saying since its not = 1 we cant say its a hypersphere

[lavish jewel]

right

[zinc timber]

ball

[lavish jewel]

the sphere is only the boundary

[zinc timber]

[lavish jewel]

hmm?

[zinc timber]

closed ball? it this not what op is asking?

[wintry steppe]

ok awesome tysm

[lavish jewel]

yea

i had put it up there

[wintry steppe]

what shape are matrices typically used to represent?

[lavish jewel]

none

they represent linear transformations

[wintry steppe]

linear subspaces

that's a nice shape

this was the last one i was confused on and its bc ive never seen a matrix in a 'geometrical context' so i didnt know how to describe it

[lavish jewel]

or at best an intersection of flats, if you like

can you show the actual question

[wintry steppe]

its this one^^

[lavish jewel]

the question

there is no question there

[wintry steppe]

just this Describe the following sets in geometrical terms.

[zinc timber]

intersection of flats is a new term to me

[wintry steppe]

and then lists the four sub problems to solve which two were confusing

the first one i posted

and this one

[lavish jewel]

i can't understand what you mean

flat is another name for affine space, ryu

[zinc timber]

i didn't know that, nice

[sleek sundial]

Is axlers treatment of duals good enough for a first class in diff geo

Asking cause that’s all I know

[brittle gyro]

Isn't $P^2=P$ and $| P | > 1$ a contradiction? Cuz if $v$ is an eigenvector of $P$ with $\lambda$ as eigenvalue then $P^2v=Pv \implies \lambda^2 v=\lambda v \implies \lambda \pm 1$, so we'd have $| P |=1$. Where am I wrong here??

[stoic python]

Sydd

[viscid lagoon]

Let $V$ be a vector space over finite-dimensional $\mathbb{K}$ and $U \subset V$ be a subspace of $V$.
For $g \in (V/U)^{}$ define $Ig \in V^$ such that $(Ig)(v) := g(v+U)$ for all $v \in V$. The linear transformation $I: (V/U)^* \to V^$ is referred to as inflation. Show that $\operatorname{im}(I) = U^\perp$, where $U^\perp \subset V:= {f \in V^ \mid f(u) =0) \text{ for all } u \in U$

[stoic python]

lewis

[viscid lagoon]

(Note that V* is the dual space and V/U a vector space V by a subspace U (quotient space))

[stoic python]

lewis

lewis

lewis

[viscid lagoon]

<@&286206848099549185> I'd be glad if anyone could help me

[wintry steppe]

Can someone explain how unitary dilation works? Like how I could embed an operator $V(t) = e^{-iAt} \in C^{n \times n}$ (where A is antihermitian) into another unitary operator $U(t) \in C^{2n \times 2n}$

[stoic python]

AdamF

[wintry steppe]

Sorry if I phrased it wrong, but I've never heard of this

[viral magnet]

maybe also ask in #advanced-analysis . Looks like unitary dilation is an operator theory concept

[serene tide]

how does part b really work?

i believe our professor just used b = <0 0 0>, however that makes nearly no sense to me

[nocturne jewel]

solve Ax=b from part a

via Gaussian Elimination

[serene tide]

right but we don't have specified values for b

[keen sierra]

You can express the solution in terms of b.

If b = 0 then the problem is not very interesting. A is invertible so x would be zero.

[serene tide]

i dont think we're supposed to use gaussian elimination on the problem though

it was discussed today

guess i just dont understand what the problem wants

[keen sierra]

Maybe solve it in a more ad-hoc manner? The first row tells you that x3 = x2 - b1, and the third row tells you that x1 = (x2 - b3)/2. Substituting these into the second row allows you to solve for x2.

[serene tide]

i think thats what i needed

thanks

[serene tide]

need a hint on 6b

given that xu + yv = zw, is it not possible for x = y = z = 0

[indigo stirrup]

It would be helpful if I can get help on this problem. "Let vectors u and v be in R^n, vector b be in R^m, and A be an m x n matrix. If there exist scalars c and d such that c(A times vector u) + d(A time vector v) = vector b, then vector b is in the span of the columns of A, but is not necessarily in Span(u, v)." I have to answer either true or false, but I just don't know how to approach it. I first thought that any vector that is in Span(u,v) must be in R^n. If n and m are not equal, b is not in R^n, and therefore, not in the span of {u,v}. Would the above statement be true then? Thanks

[spice ruin]

So I know how to solve echelon form and all that, but I have a homework problem asking me to determine if a vector represented by a matrix is in the span of S. Im not sure how I should interpret this problem, I put the components of each vector in S on a corresponding row and solved, but I dont know how this implies if it is or isnt in Sp(S)

[halcyon spindle]

From the looks of it you just need to determine if this system is consistent $\x_1+ 2x_2 -2x_3 = -9 \2x_1 -x_2 + x_3 = 12 \ 2x_1 + x_2 + x_3 = 8 \ x_1 + 3x_2 + x_3 = 1$.

[stoic python]

Plegasus

[spice ruin]

ah, that makes sense

[lament loom]

not sure if this would go in linear algebra, but can someone explain to me why

$(\vec{b}-\vec{a})^2 = |\vec{a}|^2 - 2|\vec{a}||\vec{b}| + |\vec{b}|^2$

[stoic python]

Leon

[lament loom]

when we square something is that the same as taking the dot product of itself

I'm very confused on how all this dot produce cross product thing works

I find the $- 2|\vec{a}||\vec{b}|$ especially confusing

[stoic python]

Leon

[halcyon spindle]

Do you mean |b-a|^2?

[lament loom]

whoops

I think

I asked the wrong question

okay

what I meant to be asking is

how does the cosine law work with vectors?

[wintry sphinx]

I don't believe what you wrote is true

[lament loom]

yeah my bad

[wintry sphinx]

the law of cosines with vectors? could be stated as (a dot b) = |a| |b| cos theta

[lament loom]

I'm confused to how you arrive at that

[wintry sphinx]

a bit of rearrangement yields |b-a|^2 = |a|^2 + |b|^2 - 2 |a||b| cos theta

[lament loom]

I think I just took bad notes

okay yeah I see now

I think it's because I wrote (a-b)^2 = |a|^2 + |b|^2 - 2ab cos theta

the a-b should be |a-b| and the ab should be |a||b|

[desert ice]

I dont get it why is this perpendicular

[quaint steppe]

wrong channel but

[lament loom]

I mean

is it tho

[mystic frigate]

it has to do with inverses if im not mistaken

[quaint steppe]

it is because -1/4 (the slope) is negative reciprocal to 4. Thus, 4 is the slope of the perpendicular line

[lament loom]

yes but he's asking

why

[mystic frigate]

not 100% sure but the reasoning is that when you graph the functions, its something to do with how slope works

and then eventually they'll intercept on a point which the lines are perpendicular to each other

[quaint steppe]

Define $T \in \mathcal{L}\left(\mathbf{F}^{n}\right)$ by
$$
T\left(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\right)=\left(x_{1}, 2 x_{2}, 3 x_{3}, \ldots, n x_{n}\right)
$$
(b) Find all invariant subspaces of $T$.

for this question I've found a proof online but there are some parts that I don't understand

[stoic python]

jinichi

[quaint steppe]

someone pls help me walkthrough to this proof

this is from sheldon axler, linear algebra done right book

[zinc timber]

check what happens to the basis e_i

[spiral osprey]

OK So I have started linear alegbra and so far it doesnt seem too bad.
But proffessor is making us come on campus to take exam and we only have an hour to do it.
She also of course wont be grading to solve the aug matrix's using ref rref etc.
My issue is what do you guys do in this situation where you need to calculator the matrix fast to get your variables?
Is there a faster process then just multiple/reducing rows to add or subtract to other rows?

[lavish jewel]

not by hand, no

[quaint steppe]

okayyyy

[zinc timber]

it's an algorithm that you'll implement later, so it's good to know how it works but no, you won't be calculating solutions by hand with it in the higher level

[brittle gyro]

<@&286206848099549185> any insights??

[dusky epoch]

yes, all projectors have norm at most 1

[zealous flame]

Prove that the column space of AB is contained in the column space of A. What
happens with B=transpose(A)?
I know that AB has column space contained in column space of A. Since it can be written as the linear combination of columns of A. But how to prove that column space of AA' = column space of A?

[lavish jewel]

same way

[zinc timber]

column space of AB = ABx, x\in V. take Bx=y

[lavish jewel]

associate the products from the right

[zinc timber]

I can't find an elegant argument for the second one but you can try to show C(AA') \subseteq C(A) and dim(N(AA')) = dim(N(A) and conclude they are same

maybe edd has something better

[lavish jewel]

the what

[zinc timber]

like a better argument to why C(AA') = C(A)

[lavish jewel]

can't you just rewrite the product as linear combinations of the columns of A and gather the coefficients?

[zinc timber]

can you elaborate, it's not obvious to me

[faint dune]

We have a matrix A given and do the lu-decomposition with column pivoting. We try to get the PA = LU structure, where P is a permutationmatrix.

I only show the last part of the solution.

P3 = I btw.

Is the form (P2 L1 P2) always a triangular matrix?

I guess this is always the trick to insert Permutationmatrixes here, since two same permutationsmatrices multiplicated equal unit matrix.

[wintry steppe]

Question. I have an exercise in which I have a 3x3 symmetrical matrix A and it asked me to make A=LDU'=LDL^T and then based on that result to find an R matrix in order to make A=R^TR. How do I use the answer the matrixes L and D to find the R? I hope I make sense

[quartz compass]

you can get a square root of a diagonal matrix by square rooting the diagonal and so $$A = L\sqrt{D} \sqrt{D} L^T = L\sqrt{D} \sqrt{D}^T L^T= L\sqrt{D} (L\sqrt{D})^T = R^TR$$ here I've defined $R=(L\sqrt{D})^T$

[stoic python]

Merosity

[quartz compass]

idk I might be making a mistake, I don't know if the eigenvalues are all positive, just that they're real uhh is it positive definite I'm a bit rusty I'm realizing

[wintry steppe]

well I will try that out. The A matrix is positive definite

[quartz compass]

oh ok then you're fine then

positive definite => positive eigenvalues so sqrt(D) as I've defined it works fine

[wintry steppe]

thanks!

[quartz compass]

yeah, you're welcome 👍

[wintry steppe]

why is beta_hat = arg min_beta RSS(Beta)

[zealous flame]

ya, I was thinking if we show that both have the same null space then by rank nullity theorem we can say they also have the same column space
but then again, how to prove they have the same null space?

[zinc timber]

hint: $\norm{Av}^2 = \ip{Av, Av} = \ip{AA^*v, v} = 0$

[zinc timber]

@zealous flame

[zealous flame]

[wintry steppe]

im a bit confused on this thing

https://cdn.discordapp.com/attachments/495064710278938662/938506103971516456/Screen_Shot_2022-02-02_at_12.47.29_PM.png

so my work for part one is correct i think

https://cdn.discordapp.com/attachments/554278780433334282/938504050335432734/F090F4D5-BBFE-4C16-91C5-2E1DF2F97914.jpg

but i dont understand how to explain the second part

[zinc timber]

what is RSS again?

[wintry steppe]

residual sum of squares

[wintry steppe]

im not even quite sure what the arg min is :/

[lavish jewel]

there seem to be weird issues with what you posted

since the first equation isn't a function of beta hat

[wintry steppe]

wym?

like the problem isnt right?

[lavish jewel]

at least to me, the notation makes no sense

[wintry steppe]

i thought so too but then when i expanded it, it seemed to work out

[lavish jewel]

all right after drinking some water i finally realized that beta_hat is the true solution, not the estimate

kinda weird

your arithmetic for the first part is correct

there's still some mistake in the notation they used, they mixed up beta and beta_hat somewhere

beta_hat being the result of the minimization depends on the rank of X

this overall looks poorly formulated

anyway, to answer your question, argmin_beta means that you look for the argument or parameter beta for which some function of beta is minimized

[wintry steppe]

hmm

i guess my question is

im not quite sure how to get that :/

/ how to find it

[lavish jewel]

that's a convex problem, you can minimize it by simply finding a point where the gradient is 0

[wintry steppe]

well in it, it justs asks why beta hat is the arg min, so it gives us the solution and just asks to explain

[lavish jewel]

well

simply put

the norm is positive definite

the lowest possible value it can have is therefore 0

and for you to get 0, you'd need y = X beta