linear-algebra

[hushed hedge]

im just gonna read through what you said earlier one sec

ok so it would be

xn + c(n*n) = d as you said earlier

so in this case it would be

(-1,-2,0)(1,-1,1)+c((1,-1,1)(1,-1,1))= 0 (since d is 0 ?) so we just solve for c here

and if c = 0 then it would mean that it goes through the origin if i understood that correctly? 🙂

[hard drum]

Wdym by 'it goes through origin'?

like by 'it' I mean

[hushed hedge]

intersection i think it is called?

[hard drum]

btw here is a terrible diagram but yeah lol

The idea is you take ur point and find the point on the line through x parallel to the normal and see where it hits the plane

[hushed hedge]

oh it just says in my book that you have to use that formula since the plane goes through the origin

[hard drum]

Well, so the orthogonal projection of x onto pi is just x + cn

So if c = 0, that just means x is already on the plane

[hushed hedge]

ah yeah

[hard drum]

Ah so I think maybe what you mean is if d = 0 then the plane goes through origin?

[hushed hedge]

yeah excatly

[hard drum]

There's a nice interpretation of the d actually

[hushed hedge]

im gonna try it on another problem and see if i understand

[hard drum]

Sure

but yeah so for the d, essentially the interesting thing is uh

Well do you understand the geometric significance of d? (distance from origin)

[hushed hedge]

i dont think so

i have not studied the geometric part of it

[hard drum]

So to calculate the distance of a plane r.n= d from origin, what you need to do is find the line segment between the origin and the plane of least length, and this ends up being the line perpendicular to the plane

[hushed hedge]

im guessing it has something to do with "add x+y+z= d" where d is the sum of points?

or like the point it like reaches

[hard drum]

Not quite that

[hushed hedge]

ah so the d is the perpendicular line?

[hard drum]

d is just a real number

So the significance really comes from the fact that if i want to find the distance from 0 to a plane, I need to find the least possible length of a line segment from 0 to the plane

But that comes from considering a line perpendicular to the plane

does that make sense?

[hushed hedge]

trying to understand it haha

[hard drum]

It's this sort of diagram I have in mind

[hushed hedge]

honestly nothing in la makes sense to me

[hard drum]

Aw well it may be a good idea to kinda think about how you get the equation of the plane etc to demystify it a lil

For these sort of geometrically-motivated things I find it helpful to consider 2D diagrams and all

(Because ultimately, it's geometry, and by translating and rotating we can use the same intuition as we would if, say, the plane were always z = 0 or something, although that can be a crutch)

[hushed hedge]

yeah i think i know what you mean

i think...

there was explanation for it in the book, trying to look that up

[hard drum]

Sure.

[hushed hedge]

im gonna read through what you said earlier and let it sink in, thank you for the help!

[hard drum]

Np

[spare widget]

$\vec{n} \cdot (\vec{p}-\vec{c}) = 0 \implies \vec{n} \cdot \vec{p} = \vec{n}\cdot \vec{c} = d$

[stoic python]

criver

[spare widget]

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

c being a point on the plane, n being the normal, p being another point on the plane

so d is the signed distance from 0 to the plane multiplied by the length of the normal

sibce dot(c,n) = cos(c,n) * |c| * |n| = length to plane * |n|

[hushed hedge]

and this means that the plane is going through the origin?

[spare widget]

No

Only if d = 0 does it go through the origin

What does u dot v = 0 mean?

It means that u,v are orthogonal or one of those is 0

Noe let u = n

and we pick a non-zero normal

And let v = p-c

If p and c are both vectors on the plane then p-c is perpendicular to n

That's what n dot (p-c) = 0 means

Using the linearity of the dot product we can rewrite this as

n dot p - n dot c = 0 or n dot p = n dot c

So it turns out that d = n dot c

But n dot c is the signed length of the projection of c onto n multiplied by the length of n

If |n| = 1

Then this is exactly the signed distance from the origin to the plane

i.e. if you set the normal n at the origin, and you scale it by d you end up on the plane

If the plane passes through the origin then d = 0

For a non-normalized normal d = distance to plane * length(n)

[hushed hedge]

oh yeah i'm with you on that

[spare widget]

See the wiki page I linked

[hushed hedge]

that's what we were discussing earlier, i forgot to bring my question here

[spare widget]

They have an illustration

[hushed hedge]

will do 🙂

[vocal isle]

Hey guys I'm trying to play around with the proof for eogenvectors and eigenvalues

Can I use this result to somehow find lambda and X ?

[spare widget]

What you wrote can be used to find an eigenvalue provided you feed its corresponding eigenvector

[vocal isle]

i agree

is there an alternative way i can find the eigenvector (rather thant det(A-lambda I) = 0)

using this method?

[lavish jewel]

through minimization and maximization, for special kinds of matrices A, you can find the largest and smallest eigenvalues of A, and their corresponding eigenvector this way

[prime thicket]

why is this true

nvm

i got it

[vocal isle]

Hi Edd, I'm really curious about this. Can you elaborate a bit?

[lavish jewel]

you can wiki up the rayleigh quotient

[vocal isle]

Thanks Edd. I had no idea this theory exists. Thanks 🙂

can I ask you another question?
Do you think this video is a solid proof for why det(A-λI) = 0?
https://www.youtube.com/watch?v=hpE9Iom55N0&lc=Ugwmn4mbrTPGjmmrro14AaABAg.9_eDxlGeZX09_eKkcALjSV

To me, the logic seems a bit off because he assumes that that other solutions that aren't v= 0 exist to motivate why the inverse can't exist. Once you know other solutions that aren't v=0 exist, then you can use your logic to show that det(A) =0. What are you thoughts?

[old raft]

Can we consider a vector space to be a vector?

[halcyon spindle]

wdym?

what does vector mean in this setting.

[viral olive]

Wait maybe i am missing something what logic is bothering you exactly?

[wintry steppe]

sure, if k is a field and S is the set of all vector spaces then take the free k-vector space on S

[wintry steppe]

how exactly can you get det(A) = 0 here?

the logic in the video is fine

[hard drum]

that isn't a set tterra right smh

smh

[wintry steppe]

i don't know

do i look like i know set theory

[halcyon spindle]

yes. 🙂

[wintry steppe]

i don't.

google says it's not a set

whatever

if V is a vector space take the free k-vector space on the set {V}

[zinc timber]

{V}?

[hard drum]

free vector space on a singleton

[zinc timber]

I think I've read too much module theory for my own good

[hard drum]

impossible

[zinc timber]

why not?

[hard drum]

modules r cool

[zinc timber]

until rings are non unitial and non commutative

[old raft]

So can we consider a vector space to be a vector?

[zinc timber]

no

not in the sense you are thinking rn

[old raft]

So consequently an inner product cannot be considered as an inner product space?

[old raft]

What sense do you think I'm thinking about?

[elfin agate]

so in general, a vector space is defined over a field K with two operators. "+" : VxV-->V and "*":KxV-->V, with V being just any set you want it to be, yet K being a field has an algebraic structure. also take care that for the outer prodcut you take elements out of the field K

so a vector would be an element of V (not K), so you see directly that the vector space isnt just an ordinary space so to speak

[zinc timber]

inner product is a function, inner product space in other is a vector space equipped with an inner product. Definitely not the same thing

obviously

[wintry steppe]

as i wrote above, the answer is yes. but it's bound to not be useful for you, since free vector spaces are way too big to be any useful

consider it a tongue-in-cheek "yes"

[vital monolith]

Guys I am arguing with my friend

But the columns span R3 since there are three linearly independent vectors right?

So the answer is yes?

[gray dust]

yes but its a bit roundabout

u can write any vector in R^3 as a combo of its cols (the easiest set is ||cols 1,3,5||)

try to see what such a combo is for any vector, say (x,y,z)

[old raft]

How can you say inner product is a function?

[zinc timber]

[old raft]

Ok, so can you tell me what are the problems I'll face if I considered a vector space as a vector?

[spare widget]

$\langle \cdot , \cdot \rangle : V \times V \rightarrow \mathbb{F}$

[stoic python]

criver

[wintry steppe]

the worst problem you'll face is that your time will be wasted

[gray dust]

an inner product maps a pair of vectors to a scalar

[modest cliff]

Can you have a rotation, then shear, and reflection all happen within the same linear transformation? Or do you need to compute separate transformations?

Is all I have to do matrix multiply all three matrices?

[spare widget]

yes

note that the order matters

[modest cliff]

Oh nice, thanks

[wintry steppe]

how do i work out the direction of 1.18 i + (-0.857) k

img is answer but i dont the derivation

[old raft]

How is it read?

[spring pasture]

Wdym?

[sinful bane]

how do you prove that this matrix is its own inverse?

[tacit pelican]

is that the row swap elementary matrix

good question

hmm

[sinful bane]

yeah it is

the row swap explanation makes perfect sense

but it doesn't feel too rigorous

[dusky epoch]

just multiply it by itself and verify you get the identity lol

@sinful bane

[sinful bane]

But if the dimension is arbitrary, how do you carry out the multiplication?

or are you talking about the RHS

[lavish jewel]

the rhs indeed

[dusky epoch]

i am talking about the RHS, yes.

[lavish jewel]

the product has a bunch of terms, but they're pretty simple

[dusky epoch]

the product has two dozen and one terms

[sinful bane]

Ah jeez that is pain

[dusky epoch]

but it's rigorous.

[sinful bane]

Price of rigor

I imagine things cancel nicely after the I^2 term

[lavish jewel]

that's pretty much it. you expect many products to yield something that is already there, I, or 0

a good place to start is checkimg which things are idempotent or nilpotent

[spare widget]

https://en.m.wikipedia.org/wiki/Function_(mathematics)

[spare widget]

V being the vector space, F its associated field, <.,.> being the inner product (e.g. (u,v) -> <u,v>)

[stone sand]

Hello

please how to prove that a norm is symmetric

[dusky epoch]

what would that even mean?

[zealous pond]

what is this eij eji

[tacit pelican]

eij is prob the matrix with the same dimensions as the left side with 0 entries everywhere except the entry 1 at the intersection of the i row and j column

same thing for eji but j th row and i th column

[zealous pond]

too ambiguous. question needs to be defined properly

[dusky epoch]

how is "e_ij is the matrix with entry 1 at position (i,j) and zeros elsewhere" too ambiguous? @zealous pond

[tacit pelican]

hey guys is this proof fine or am i just saying bullshit?

im pretty sure the first part about linear independence is fine

im proving that

im a bit unsure if i did the sums correctly

indexes always confuse me

[zealous pond]

is there a structure to (i,j) pair.

[dusky epoch]

what do you mean "structure"?

i and j are just natural numbers between 1 and n

distinct natural numbers between 1 and n.

[zealous pond]

(i,j) can be anything from (0,0) to (n,n). There will be clearly some pattern which (i,j) turns to 1 maybe u can convey that to me.

[dusky epoch]

??

"(i,j) turns to 1"?

what are you talking about?

[zealous pond]

show me a example of 6x6 matrix defined with the conditions in the question.

[dusky epoch]

ok, sure, let's take $n=6$, then $e_{43}$ will refer to the matrix $\bmqty{0&0&0&0&0&0\0&0&0&0&0&0\0&0&0&0&0&0\0&0&1&0&0&0\0&0&0&0&0&0\0&0&0&0&0&0}$

[stoic python]

Ann

[zealous pond]

thanks was unaware of this notation.

[dusky epoch]

e_ij is the matrix with entry 1 at position (i,j) and zeros elsewhere

you were aware of it the moment i wrote out the definition of e_ij to you

[zealous pond]

Na i had some different visualisation in mind

now the problem become very trivial it just switches the 2 basis-axis. like basis i(x-axis) beccomes j (y-axis) and vice-versa. if you do the same operation it will flip the basis again returning everything to original state

[sinful bane]

Ah, well let’s say you have e_ii * e_ii. I know the product is e_ii (idempotent, right?), but how would we prove this?

[manic nest]

If I'm given 3 vectors, how would I find the vector in the middle of all of them

I'm thinking about getting the midpoint several times

[spare widget]

What do you mean by middle

As in having the same distance geodesic to all 3 on a sphere?

[manic nest]

Let a,b,c be 3 vectors, then m_a = mid(a,b), m_b = mid(b,c), m_c = mid(c,a) then middle might be mid(mid(m_a, m_b), m_c)

Im not sure what a distance geodesic is

[spare widget]

what do you mean by middle?

[manic nest]

but intuitively, given orthogonal vectors, it would be the one poking from the center

[spare widget]

Same angle between this one and the other 3?

[manic nest]

e.g for 2x2 case, with standard basis, the middle vector would be c(1,1) for any c in R

oh wait

im being an idiot

[spare widget]

So are the vectors always the axes?

[manic nest]

would just adding them give it to me?

yeah so for orthogonal it would be the axes of rotation

[spare widget]

Are the vectors always the canonical basis vectors?

If yes then sure sum them up

[manic nest]

No, not necessarily

[spare widget]

then no

[manic nest]

Well for orthogonal, adding them up should work

For non orthogonal it seems more difficult

But it would be that geodesic thing for non orthogonal

[spare widget]

Then: https://dl.acm.org/doi/10.1145/502122.502124

[manic nest]

Like the sphere between them, the middlemost part

[spare widget]

There's no closed form solution, it's an iterative algorithm

[manic nest]

Massive rip

Was hoping this would be trivial

[spare widget]

In the above they use projected gradient descent and projected newton

[manic nest]

This intuitively feels like it should be trivial

[spare widget]

It's not unfortunately

For 2 vectors it's slerp

Well actually

I shouldn't be that certain

Because you asked for the midpoint, and the above is for arbitrary weights

[manic nest]

I think assuming they're normal at least should be fine

[spare widget]

So maybe for the mid-point there's a closed form solution

You mean orthonormal?

[manic nest]

No not orthogonal

So slerp in 3d, at 0.5

[spare widget]

Eitherway, they provide an algorithm in the above paper

Just use that imo

[lavish jewel]

it really depends on your definition of middle

[spare widget]

I think he wants middle as in equal angles

at least that's what I understood

[manic nest]

Ok for 2x2 case, it's where the angle is half. (Which should be easy)

[spare widget]

Just use the paper

[sinful bane]

Middle as in the center of the circle intersecting all three points? @manic nest

[manic nest]

Given two normal 2d vectors u & v, and the angle between u & v θ

Let middle vector be, the vector between u & v, which is either u or v rotated θ/2

It's not the best definition, but I hope that expresses the idea

[spare widget]

Yes it's exactly what the paper handles

See the algorithm inside

[sinful bane]

Really? That sounds so easy though

[manic nest]

I mean shouldn't it be finding the middle point of 3 points on a sphere?

[spare widget]

For 2 vectors yes

[lavish jewel]

yeah this doesn't seem trivial

[sinful bane]

Can’t you convert to (r, theta) form and then convert back

Oh oh 3d

I see

[spare widget]

for 2 vectors it's just slerp

For 3+ it's what they do there

[manic nest]

Well for 3D, it seems like we're just finding the middle point of 3 points on a sphere which I thought would be trivial in spherical coordinates.

[spare widget]

They perform a minimisation

With projections using the exp and log map for the sphere

[lavish jewel]

i wouldn't even know if the vector will necessarily be in the middle anymore, it could point in the opposite direction

you'd have to optimize over the sphere

[spare widget]

One algo is gradient descent based, the other is newton - relying on the hessian

[lavish jewel]

what criver mentions would probably rely on trust region methods

[spare widget]

https://dl.acm.org/doi/10.1145/502122.502124

[lavish jewel]

take small gradient or newton steps, project onto the sphere, check whether some condition is satisfied

and continue iteratively

[manic nest]

So even allowing opposite directions, its not trivial?

[lavish jewel]

not just allowing, you kinda need it

[manic nest]

Oh sorry, meant changing it so we don't care about opposite direction

Basically turning it into just finding the 'axis' between three vectors

Yeah for 2D, it's angle bisector, so in other words, 3D angle bisector is non trivial

[spare widget]

It's not even bidector at that point

You have potentially 3 angles lying in different planes

And you want to make those equal

[manic nest]

Well the intuitive generalization of bisector to 3D

[spare widget]

I can't vouch there's no closed form solution specifically for the midpoint in 3d, but for arbitrary weights in n-d the paper claims there isn't

[spare widget]

Bisector in 3d would still bisect an angle lying in a plane

[manic nest]

Yeah, but you'd imagine it'd do it for 3 vectors.

[spare widget]

What happens instead is that you have 3 angles potentially lying in different planes

[manic nest]

Yeah, so it becomes a approximation problem

For the algorithm in this case

Well I guess that's good to know, thanks

[sinful bane]

@lavish jewel ?

[lavish jewel]

hmm? that it's idempotent?

write e_ii * e_ii either as sums or entriwise as dot products

then note that almost everything is zero except for a specific row and column

then you need only focus on one product

well, this works in general

more easily though, note that e_ii is diagonal

so e_ii^2 is diagonal, and the diagonal entries are equal to the entry-wise squared elements of e_ii

[zinc timber]

In general $e_{ij}e_{kl}=e_{il}\delta_{jk}$

[zinc timber]

it's handy to know

[lavish jewel]

just to make sure, what are delta and e here

[zinc copper]

been struggling with this question for a bit now. I want to show that a symmetric real matrix A is positive semi-definite iff all its symmetric minors are >=0. Ive shown that => direction, and for the opposite one my idea is to proceed by induction. Then for the sake of contradiction suppose A is not positive semi-definite and hopefully find a vector with one of its coords 0 that gets taken to a negative real by the quadratic form. This would be the same as taking a quadratic form on a matrix formed by removing a row and column symetrically from A which by induction must be positive semi-definite and id get a contradiction

the "coord 0" part is making me struggle though

[sinful bane]

Ah, i guess my question is then how do you prove anything about multiplication with arbitrary-sized matrices? Is there a way without using ... and only using strict logic?

[lavish jewel]

using sigma notation is one way

[sinful bane]

can you even define a matrix using sets

or prove things about them using set theory

[lavish jewel]

idk what you mean

[sinful bane]

like

I can prove something about functions easily because functions have very strict definitions in set theory

do matrices have a set-theoretic definition

[lavish jewel]

you can take basic properties as true at face value, but those can also be shown using sigma notation

i would just say that the columns (or rows) of a matrix are a set of vectors v_i with some property

for diagonal matrices, you have that splitting a matrix into rows and columns yields the same vectors, and you can use this to show that D = D^T

then write D*D = D^T D and note a generic element of the product as u_ij

and that the v_i are mutually orthogonal

something like that

or do it all with sums if you prefer

[sinful bane]

do you mean a literal set of vectors? if they're the same vectors then the matrix would collapse into one vector then right

not sure if I'm making sense lol

[lavish jewel]

i have no idea what you mean

[sinful bane]

ok you know how functions are defined as like

[lavish jewel]

i would describe the columns of a matrix as a set of vectors v_i with a special property

[sinful bane]

A function is an ordered triplet (X, Y, f) where f is a subset of X x Y such that for all x, blah blah blah

[lavish jewel]

i mean, that's exactly what you want to do here

but there's no unique f

you can let f be the inner product and take X = Y, the set of columns of the matrix

[sinful bane]

I'm imagining a matrix is some bijection from [1, n] in N to a set of vectors or something, idk

[lavish jewel]

or you can let X = Y be scalars in R, and f is just summation

[sinful bane]

oh

[lavish jewel]

and both of these will work

idk why you're confusing yourself

you don't even need to think of this as matrices if you don't want

[sinful bane]

I'm past the diagonal eii stuff now, I'm just asking about matrices and set theory

[lavish jewel]

you can do whatever you find useful

[sinful bane]

wait what, inner product

[lavish jewel]

i like to think of matrices as what they are, a linear transformation in a given basis (fin dim)

then you can either choose to treat the action of matrices as taking linear combinations of their columns, or taking inner products with the rows

whatever you like best

[sinful bane]

yeah to be clear there's nothing I'm confused about on the conceptual level, i'm more asking how matrices are defined in set theory

[lavish jewel]

it depends on what you want to do with them

[sinful bane]

ah

[lavish jewel]

you can see them themselves as functions

or not

whatever you find useful at the moment

[wintry steppe]

an m by n matrix with entries in R is a function {1, ..., m} x {1, ..., n} -> R

[lavish jewel]

a set of functions with some property, perhaps

[sinful bane]

ah that makes sense ok

[lavish jewel]

or vectors of some kind, so a set with some extra structure

[wintry steppe]

this point of view is useless but it works if you really want a rigorous definition of "matrix"

[sinful bane]

yeah it doesn't really seem like linear algebra would need definitions like that

like it's just kinda pointless

but I was just curious

[lavish jewel]

whatever you prefer, you could come up with arbitrary set theoretic ways to describe them

[sinful bane]

Ye

It’s pretty fascinating how the multiplication process seems really weird at first

But then has all these extremely nice properties

It looks exactly like a group or ring or something

[lavish jewel]

they're there by construction, since they are meant to represent linear transformations in a basis

[sinful bane]

Oh

Wait so like

They had the idea first and then came up with the definition?

[lavish jewel]

that's why i keep telling you to look at the rows and/or columns as vectors

[sinful bane]

Ohh I see

[lavish jewel]

a matrix vector product is a linear combination of the vectors that make up the matrix columns

and equivalently, a vector whose entries are inner products between rows of the matrix and the vector it is multiplying

[sinful bane]

Yeah I noticed when I have a matrix like
[a c]
[b d]
Then left-multiplication would take <1,0> to <a, b> and <0, 1> to <c, d>

[lavish jewel]

matrices arise naturally when dealing with linear systems of equations as a way to manipulate things more easily, and then it was further abstracted

[sinful bane]

Ah

Yeah when I was reading the textbook they presented the definition first, and then all the properties and examples

Makes a more sense now with context

[lavish jewel]

they are there by design, yeah

notice similar things happen in several places, then look for a nice, general way to describe those properties

[sinful bane]

Inner product? Is dot product a type of inner product

[lavish jewel]

yes

[dusky epoch]

yes, inner product is a more general notion

[sinful bane]

If you have a row vector times a column vector, you’re basically dot producting the two vectors

[lavish jewel]

indeed

[sinful bane]

Which extends to matrices if each column/row is considered as one “object”

[dusky epoch]

what are we discussing right now anyway

[lavish jewel]

abs discovering that matrices don't just happen to have nice properties, they were made like that by design

[sinful bane]

I asked about how matrices are defined in set theory and now we’re just talking about them in general I guess

Ye

[dusky epoch]

well i mean like

how do you define something like a sequence rigorously

it's a function from N to R (or N to whatever else for sequences of things other than real numbers)

[sinful bane]

Function from N to a set

[dusky epoch]

right

informally a matrix is a grid of numbers (or, in general, ring elements)

[sinful bane]

(wait don’t you want multiplicative inverses)

[dusky epoch]

no

not for the purposes of defining matrices

[sinful bane]

ah

[dusky epoch]

it's enough that the entries of a matrix be something that you can add and multiply

anyway

a grid of size m by n is naturally identified with the set 1:m × 1:n

as is hopefully clear

[sinful bane]

Yep

So it would be a function from that to a ring

?

[dusky epoch]

yes, a matrix is a function from 1:m × 1:n to R

where R is the ring your matrices are over

[zinc timber]

has it been answered?

[zinc copper]

Nope

[zinc timber]

have you heard of Sylvester theorem?

though u don't need it, your method is fine

A is symm then take EVD of A

then try something with that

[dark ruin]

Can someone help walk me thru this problem? Im confused on how to row reduce since i cant have fractions as answers. At least i assume fractions cant be part of the answer.

[dusky epoch]

can you show what system of equations you are starting with? @dark ruin

[dark ruin]

im starting with P+N+D=13 &1P+5N+10D=83

[dusky epoch]

okay

that checks out

it sounds like this is not something that can be solved with simple row-reduction alone, given that (a) there should be only 2 equations despite 3 variables and (b) there is the extra constraint that all variables are natural-valued

...you aren't being faced with any stupid constraints imposed by your teacher like "any solution that uses anything other than row-reduction will be rejected outright and zeroed", are you?

[dark ruin]

I don't think he would. However we have only been practicing row reduction so im not sure of another method.

[dusky epoch]

this isnt the kind of problem row reduction alone can solve, as i said above

i mean, ok, we can do one step of it (really the only thing we could do) and end up with our second row as 4N + 9D = 70

and then treat this as a diophantine equation (which is what it is, let's face it)

look at what natural solutions it has

identify at least one, such as (N, D) = (13, 2) and use it to construct all integer solutions, then filter to only those where N and D are both positive

but this of course relies on knowing how to solve linear diophantine equations

which im not sure if i personally can say anything meaningful about at the moment

[dark ruin]

I assumed 3 pennies that way I only had 2
and I got an answer that way.

[spare widget]

Express N and P as a function of D for instance and start plugging numbers from 0 to 8. It's pretty slow but it's a finite number of cases.

Alternatively you can express N and D through P, or P and D through N in order to check for the others

If you haven't studied diophantine eq. then the goal was probably for you to "guess" a solution in such a way

[dark ruin]

oooh I see what you're saying. Does that mean that there may be multiples answers?? I go one answer using 3 P and didnt plug in numbers for N or D.

[spare widget]

You cannot not plug in numbers for n and d

[spare widget]

As for getting any potential other solutions: https://en.m.wikipedia.org/wiki/Diophantine_equation#One_equation

But I think they will violate nonegativity if your problem was set up to have a unique solution

[wintry steppe]

If B* (dual basis) is given, is B unique?

[wintry steppe]

Can someone help me understand how k/k^2 + k simplifies to that

I was understanding that both ks would cancel

And you’d have like 1/k^2

[hard drum]

numerator k, denominator k(k+1) you mean?

[wintry steppe]

yes

[hard drum]

You just can divide numerator and denominator by k to get 1/(k+1)

[wintry steppe]

hmm okay I was thinking that you’d cross out the k on the top and the k on the bottom

and be left with 1/k^3

2*

My understanding of fractions is bad

[hard drum]

really you're dividing by k

[wintry steppe]

Okay so it’s just like you do to the top what you do to the bottom

and u can divide thru because like terms ?

[hard drum]

for example if you were to cross out k on the denominator by replacing it with 0, by consistency you ought to be replacing the numerator with 0 too

[hard drum]

Not quite

[wintry steppe]

Oh i thought it was replaced by 1

Okay so like

[hard drum]

You can multiply/divide the denominator by the same factor

[wintry steppe]

If you had k + 2 numerator and k^2 + k denominator you would end up with 2/2k?

[hard drum]

It wouldn't simplify in this case

I'm not sure how you got 2 or 2k

[wintry steppe]

It’s just an example

I was thinking you’d cross out the ks

[hard drum]

(Also this isn't linear algebra)

[wintry steppe]

Yeah i dont know where to go

Lol

[hard drum]

But no, yoy don't just cross out ks

[wintry steppe]

Im trying to figure out discrete

and don’t understand basic fraction rules

Is there ever a situation where you cross out a variable

[hard drum]

Afaik #prealg-and-algebra would fit this

[wintry steppe]

or is that simply incorrect

[hard drum]

Well cross out is vague

[hard drum]

I would recommend thinking in terms of multiplication/division instead for the moment

Yeah so that's not permissible

[wintry steppe]

stuff like this is what’s left in my memory after doing hs math

[hard drum]

But yeah that doesn't simplify

[wintry steppe]

so all terms must have a common factor to be able to multiply or divide by something

[hard drum]

You can always multiply/divide both the numerator and denominator by something nonzero

[wintry steppe]

ok ya ik that

[hard drum]

And if there's a common factor that's a special case

[wintry steppe]

I mean more with variable simplification

I think?

I also didn’t learn how to factor so discrete is just more difficult than it needs to be

Zzz

[hard drum]

I thin #prealg-and-algebra would be better suited for this btw

[wintry steppe]

okay

Im just going to move on for now since I understand the specific example

Thank you

[robust stratus]

Is it just me or are quotient spaces unbelievably unintuitive?

[sinful bane]

I agree

Very weird

Well it’s just quotient anything really, quotient sets, quotient groups, quotient rings, etc

[hard drum]

They're ultimately quite nice to work with and give pretty proofs of e.g. Jordan normal form existence but just take a while to get used to

[viral olive]

What are quotient spaces all i know is that there elements are equivalence class

[robust stratus]

You have a vector space V and a subspace of it called U. The quotient space V/U is the set of all affine subsets of V parallel to U.

[sinful bane]

I think it’s easier to learn quotient sets first, and then extend the idea to a vector space later

[robust stratus]

The dumb thing is that those affine subsets of V aren't subspaces themselves

[sinful bane]

What’s an affine subset? Is that the same as an affine space

[viral olive]

Wait what?

So affine sunsets dont have to necessarily be subspaces ?

[hard drum]

Affine subspaces are generalisations of linear subepaces which are basically what you get when you translate a subspace

So of the form a+ U where U is a (linear) subspace of V and a is some vector in V

It needn't be a subspace (e.g. the plane z=2 in R^3)

[robust stratus]

They're the set with all vectors of the form v+u for a fixed v in V and all u in U

They're not usually a space since they don't necessarily contain the additive identity aka 0 vector

[viral olive]

Wait we never got introduces to linesr subspaces are they just normal subspaces. To be accurate per my learned definition. A Subspace is a subset of Vectorspace which also fullfills the rules for vectorspaces and so is a vectorspace.

[robust stratus]

Yes, all subspaces are vector spaces

[hard drum]

Those are sometimes called linear subspaces to avoid ambiguity

But yes

[robust stratus]

It's infuriating how much harder it is to understand quotient spaces from product spaces, but axler's linear algebra done right just treats it like no big deal

[hard drum]

Have you done quotient groups before?

[robust stratus]

No, the only thing I've seen about them was gluing sides together of a manifold in topology

It wasn't for a class, I was just watching some lecture Tadashi Tokieda gave on it for fun

[hard drum]

Ah ok

[robust stratus]

If I learned group theory would make it easier to understand?

[hard drum]

Well i reckon it might be common to cover quotient groups before quotient spaces at some unis (that's what my uni did anyway) and so it may be that Axler assumes more familiarity because of that or something

Oh well if you just want to understand linear algebra I'd not worry

Although groups are cool anyway

[robust stratus]

I'm just doing math for fun, I have a group theory book by Herstein that I haven't started

[hard drum]

Ah epic

Well looking at groups can be cool, the only thing is that the group underlying the vector space is abelian whereas groups aren't in general

[subtle gust]

how can i find the rank of the matrix lamda I - A if i know the eigenvalues of A

is there a way besides using gaussian elimination

like is there a relation between rank and eigenvalues

[paper fractal]

Hey guys what does the modulus of a direction vector mean

[sinful bane]

Depends what u mean by direction vector @paper fractal

[fleet sun]

They're strange at first sight but they're not that bad

an element of V/W looks like (v + W)

How would you scalar multiply it?

c * (v + W) := (cv + W)

How would you add them?

(v + W) + (u + W) := (v+u + W)

with those definitions it can be shown to satisfy the definition of a vector space

and the dimension of V/W turns out to be dim(V) - dim(W)

so like, R^3 / R^2 is isomorphic to R^1

[native ore]

can someone confirm a something for me

[fleet sun]

what's that?

[native ore]

the gram schmidt process allows us to convert a basis of a subspace to an orthogonal basis of that subspace right

not an orthonormal one

[fleet sun]

well, maybe, but if you have an orthogonal one, you can always normalize each vector and get an orthonormal one

[native ore]

but the process itself doesn't normalize it right

[fleet sun]

correct

[native ore]

ok cool thanks

[meager harness]

<@&286206848099549185>

[hardy inlet]

So I haven't started on this homework yet; but my friend in the class was asking about if this was possible, and I've got that <Av,v> = ax²+2xy+2xz, which im not sure when that would be positive. But then we were trying to google what a positive thing is and she found something about the upper 2x2 corner needs to have a positive determinant for positive definite. Is this possible?

Also idk if its asking for a positive matrix or a positive transformation (or if those are the same)

our Axler definition was a positive operator <Tv,v> is nonnegative for all v

A positive matrix is a matrix in which all the elements are strictly greater than zero. The set of positive matrices is a subset of all non-negative matrices. While such matrices are commonly found, the term is only occasionally used due to the possible confusion with positive-definite matrices, which are different.
from wikipedia; so all alpha >= 0?

[fleet sun]

I doubt they mean positive in the same sense as wikipedia

otherwise it'd be too easy

alpha > 0, done

they might mean positive-definite

[hardy inlet]

but then it cant be because of this thing

idk we'll prob just have to ask the professor tomorrow

[fleet sun]

yeah, the determinant of that matrix is 0

[robust stratus]

The thing is I have trouble grasping exactly what (v+W) is. For instance there is a statement that a nonempty subset A of V is an affine subset of V iff cv+(1-c)w is in A for all v,w in A and c in C. I can't mentally connect the definition of affine subsets with why this statement would be true.

[fleet sun]

an affine subset is a translation of a subspace by a vector v

if v happens to lie in that subspace, then it hasn't actually been moved

like consider the xy-plane in R^3, with z = 0

so all points (x, y, 0)

take v = (1, 2, 0)

then translating all points of your xy-plane by v, it's still just the same xy-plane

because (1, 2, 0) + (x, y, 0) = (1+x, 2+y, 0) and you get all the same vectors as you already had

but now take w = (0,0,1)

w + {xy-plane} is a horizontal plane through z = 1

because (0,0,1) + (x,y,0) = (x,y,1)

[fleet sun]

that statement is saying all points linearly in between v and w are also in A

that's a property shared by subspaces

it means they're flat/linear looking

it just probably doesn't pass through the origin

[robust stratus]

Yeah, the cv+(1-c)w looks kinda like the closed under addition and scalar multiplication property of subspaces, but the (1-c) part trips me up

[fleet sun]

look at what it becomes with c = 0 and c = 1

you can rearrange cv + (1-c)w to c(v - w) + w

when c = 0, it's w

for c > 0, it's a vector starting from w and in the direction of v - w

so it's on its way to v

since v - w points from w to v

when c = 1, it's v

[robust stratus]

Oh so (v-w) is like the vector representing the original subspace and w the vector added to it?

[fleet sun]

well, v and w are any two elements of A

to be an affine subset, everything linearly "in between" v and w also has to be in A

those are of the form c(v - w) + w

[robust stratus]

c can be greater than 1 and less than 0 and those still count as in between v and w?

[fleet sun]

they'll be on the line connecting v and w

so not necessarily in between

but in line with

you should think of an affine set as some (hyper)plane, not necessarily passing through the origin

a subspace is a hyperplane that does pass through the origin

[robust stratus]

Yeah the properties of a subspace making cv+cw in it makes so much more sense to me but I think I'm starting to have a better feel for an affine subset

[fleet sun]

are you working with complex vector spaces? your statement calls the scalar c in C

those are harder to picture for me

[robust stratus]

The book I use says F for any field, so C or R works

Apparently complex vector spaces have nicer properties than real ones, but I haven't gotten that far yet

[wintry steppe]

I'm very confused on how to do this questin

question*

For reference^

[wintry steppe]

try to use the diagram to compare the kernel of T and the kernel of L_A

same for their images

[zinc copper]

you mean SVD?

or spectral decomposition

[teal grotto]

eigen value decomposition

[zinc copper]

yrsh just realized

yeah i thought of that but the problem is the eigenvalues of A arent realted to eigenvalues of its minors in any obvious way

i mean it's the way you go about a proof for positive definite matrices

because you can show that a even number of eigenvalues must be negative

and get two linearly independent vectors that you can combine to get a vector with last coord 0 and finish the proof

in our case though, the product of all eigenvalues could well be 0

in which case i dont really know how to conclude anything about which eigenvalues are positive or negative

[zinc timber]

signs of minors = signs of the product of the Eigen values

[zinc copper]

how do you see this 🤔

[zinc timber]

that's Sylvester law

though you aren't allowed to use jt

btw only holds for symmetric matrices

[zinc copper]

can you clarify exactly what you mean here?

are you saying that the sign of a symmetric minor is the sign of the product of the eigenvalues of our original matrix or some subset of those eigenvalues?

[zinc timber]

https://en.m.wikipedia.org/wiki/Sylvester's_law_of_inertia

[zinc copper]

yeah ill read that thanks

[zinc copper]

i mean this is sylvester's theorem

but i dont see how it relates to the minors

[zinc timber]

diagonalize and it'll make sense

[zinc copper]

so if i understand, you're suggesting that taking the EVD of A and looking at a minor matrix of A, i can look at how that minor matrix gets expressed as a product of submatrices of our decomposition?

isee how we sort of implictly talked about sylvester's theorem in the proof for positive -definite matrices yeah

it just doesnt seem adaptable to semi-definite

[lusty swallow]

my friend solves this problem like this, is that work or just coincidence?

[dusky epoch]

there are some things your friend is leaving unspecified

namely, taking the formula at the top as the definition of a linear map $T : P_2 \to \bR^{2 \times 2}$, your friend's matrix is the matrix of $T$ with respect to the ``standard'' bases of each space: the monomial basis for $P_2$ and the matrix-unit basis for $\bR^{2 \times 2}$

[stoic python]

Ann

[lusty swallow]

If I solve this problem I need to put the standard basis of a polynomial into T(X) and then find the coefficient from the combination...
but he just vectorized the matrix of the right-hand side and get the answer, is he doing the right things...? I have no idea why he can solve it like that.

[dusky epoch]

i mean

if you know how matrix-vector multiplication works

you can recognize this sort of stuff

[languid sphinx]

I think what you are asking is how did someone come up with this method

[timid sage]

I've got a question on an assignment that I don't quite understand. We're given a matrix A, a vector p and a vector v_p, and are told some properties about them.
We're then asked to state the rank of the matrix in the attached image.

I understand how to calculate the rank of a matrix of numbers, but am confused on what exactly I'm doing when the two elements of the matrix are a matrix and a vector

[spare widget]

The rank is the # of lin indep rows/columns

So if the rows of A and (v-vp)^T are lin indep then it is full rank

If the rows of A and (v-vp)^T are lin dep then it's the rank of A

Maybe I should clarify

[lusty swallow]

thank you, that is what I want to ask

[timid sage]

So I'm checking if the rows of A are all linearly independent of (v-vp)^T ?

[spare widget]

well you would first try to find a basis

For the span of A^T

And then check whether the vectors from said basis and (v-vp) are lin indep

[timid sage]

alright yeah

So I'm checking if all vectors in the basis for A are independent of (v-vp)^T

[spare widget]

That's a strange way to write it

A set if vectors can be lin indep

idk why you say all vectors are independent of ...

They must be lin indep as a set

There's no "of"

you are not checking every 2 or something of that sort

[lusty swallow]

thank you answer me, but I think there have some misunderstanding, I know how to split coefficient and variable

[timid sage]

So a set of the basis of A and (v-vp)^T are linearly independent?

[fleet sun]

a vector is independent of some set of other vectors if it's not in their span

[spare widget]

Basis for the span of A^T

[timid sage]

Yeah that makes sense

thanks

[zinc copper]

@zinc timber https://math.stackexchange.com/questions/1831988/how-to-prove-that-a-is-positive-semi-definite-if-all-principal-minors-are-non so this is one way to adapt the positive definite case

you add a bit to the matrix to make it positive definite and then take the limit as th rror tends to 0

i dont think i wouldve thought of that

considering i dont have jacobi's formula anyways this isnt really usable

[subtle gust]

if the diagonal of matrix A is all ones

can we say anything about its rank?

[zinc copper]

Well first note you can multiply any matrix with no zeroes on its diagonal by an invertible matrix such that it only has ones on the diagonal, so that should hint to the fact that you cant

[subtle gust]

Hmmm makes sense

[zinc copper]

it's not too hard to construct a matrix of arbitrary rank with ones on the diagonal

[subtle gust]

yeah... tysmmmm

truly appreciate ur help

[wintry steppe]

How can I use Remainder Theorem?

[wintry steppe]

Please ping me, back to reading lol

[lavish jewel]

i and j are indices

you are told there are n linearly independent eigenvectors

i and j are indices between 1 and r, where r is the rank of the matrix A

[wintry steppe]

I'm being told lots of things here. This is far from easy to comprhend lol.
Tyty

[glad pewter]

did you get it?

If i above r, they are not L.I anymore, if it's 0, well... You won't go any further

[wintry steppe]

I get the idea now. Read the chapter for hours lol

[native ore]

Can someone help me understand the proof for 4.5.2a

I dont understand what exactly they mean by A = (a_ij)
and how <Ae_i, e_j> = a_ij

<,> is the standard inner product

[spare widget]

<Ae_i, e_j> = (e_j)^T A e_i

right mult extracts column i, then the left mult extracts row j

[native ore]

@spare widget Ok I think I get what you mean by that but

but I am confused about how you get <Ae_i, e_j> = (e_j)^T A e_i

is this a property of the inner product?

[native ore]

Ok nvm I figured it out

[subtle gust]

Ummmm guys...

I'm doing this report on LU decomposition

And i kinda need help with a few questions

Cuz i'm getting mixed answers on google

1. is LU decomposition ALWAYS possible?

2. is LU decomposition unique, and if not is there any situation in which it is?

Yeah i think those aee the points i'm mixing up for some reason

Also, regarding the three main algorithms for finding the L and U matrices (choloskey doolittle and crout)

Is it safe to say that in choloskey A=LL^T

And in doolittle the diagonal of matrix L is all ones

And in crout the diagonal of matrix U is all ones

[limber sierra]

no, but it is subject to some constraints

e.g. requiring the diagonal entries of L to be 1

[subtle gust]

Yeah that's what i found but i wanted to make sure i understood correctly

So

If diagonal(L)=1s

The L and U matrices are unique?

[limber sierra]

there is only one LU factorization of a matrix A such that all the diagonal entries of L are 1.

there are other similar constraints that achieve this

[subtle gust]

Is it the same thing for U

[limber sierra]

yes

and you can change 1 with another number in [1, n] and it works as well

[subtle gust]

So if diagonal(U)=1s L and U are unique

[limber sierra]

yes.

[subtle gust]

If i want to generalise this

I would say

[limber sierra]

you can see this by expanding the equation A = LU

[subtle gust]

If the diagonal of matrix L or matrix U is all a digit n, L and U are unique

[limber sierra]

youll notice that theres n² equations in the entries of LU, but n(n+1) possible values of entries in L and U that satisfy the criteria (why?) A = LU

so you wont have uniqueness unless you "get rid of" n possible options

and you do this by fixing the diagonal to 1 of the n possible values

i'm being vague, you can make my reasoning more clear by just trying it on paper

[subtle gust]

Yeah i couldn't quite understand

[subtle gust]

Is fhis right tho?

[limber sierra]

take this equation and expand the right hand side via the rules of matrix multiplication

[subtle gust]

Oh

[limber sierra]

you'll get a system of equations

[subtle gust]

Yeah

[limber sierra]

n² equations, specifically

[subtle gust]

9×9 i think

No wait

[limber sierra]

but there's n(n+1) possible options

since this is twice the triangular number of n

[subtle gust]

Yeah

[limber sierra]

(since theres two triangle matrices)

so your problem is underdetermined and hence your factorization is not unique unless you make choices to "remove" n options from consideration

and fixing the values of 1 diagonal works (again, try doing this and expanding out)

[limber sierra]

not quite

[subtle gust]

I kind of still don't understand why it's not

[limber sierra]

its a phrasing thing

as phrased, it sounds like you get the same unique values for L and U if L's diagonal or U's diagonal is chosen

but they could be different unique valuies

(in fact, in general they will be)

[subtle gust]

I meant that they're both unique

Didn't mean that they have the same value

[limber sierra]

then yeah, that's fine

[subtle gust]

Great then

Tysmmmm

[limber sierra]

wait scratch that last point that i just deleted

its not necessary

[subtle gust]

So it works for all values of n

[limber sierra]

yeah

itd also work for another constraint

really all you need to do is define the diagonal with nonzero values

if you fix the diagonal of L or the diagonal of U, that leads to a unique factorization

[subtle gust]

They have to be equal tho eh?

The entries i mean

[limber sierra]

not even!

they could differ as long as theyre nonzero

its just that you need to fix them

[subtle gust]

Oh

Yeah yeah got ur point

[limber sierra]

look back at this image

imagine we're choosing values for each a_ij

there are n² (in this case, 3² = 9) different values we need to pick

but on the right hand side, youll notice there's more than n² different entries

indeed, we have 2 triangles of "height" n

[subtle gust]

Yeah

[limber sierra]

so the total number is twice the n'th triangular number

the n'th triangular number is n(n+1)/2, so since there's two matrices, there are n(n+1) values to choose

that is to say, n² + n

so we have n² choices for the LHS but n² + n choices for the RHS

so if we want the LHS to be uniquely factorized, we need to "get rid of" n choices

but you'll notice that the diagonal has n entries on it

so by simply restraining a diagonal, we "lower" the number of choices on the RHS to just n²

since n² + n - n = n²

[subtle gust]

Aha

So 2 constraints

The entries on the main diagonal are all n

Or

We fix the main diagonal entries

[subtle gust]

The green line are the fixed values right?

[limber sierra]

YEP

sorry caps

you dont need to fix them to the same value

we often choose to fix them all to 1 just as a convention

since thats simple to do and makes for easy computations

but it could be whatever

[subtle gust]

Ah

[limber sierra]

you could fix them to 69, 420, 69, 420, ...

[subtle gust]

Lmao

[limber sierra]

you just need them to be nonzero, for reasons that become clear if you expand out the RHS by hand

(if they're 0 a lot of terms cancel out and you're not always able to determine the LHS)

[subtle gust]

Yeah

I think i got your point actually

That's pretty much what i need

Wb the other questions tho?

[limber sierra]

existence? LU decomposition does not always exist

theres some classification criteria but i forget it off-hand admittedly

[subtle gust]

Only if the leading minors are all non zero right?

[subtle gust]

Too small to read

[limber sierra]

wikipedia specifically says the leading principle minors, and the "only if" condition only holds if it's invertible

otherwise it's only one-directional

i do not recall enough details to be able to produce a counterexample right now.

[subtle gust]

Idk what one directional means but ok 💀

[limber sierra]

as in like

if A is invertible:

• if it has an LU factorization, then its leading principal minors are nonzero
• if its leading principal minors are nonzero, then it has an LU factorization

if A is not invertible (with rank k):

• if its first k leading principal minors are nonzero, then it has an LU factorization

[subtle gust]

Oh

[limber sierra]

but the final direction (if it has an LU factorization, then its first k leading principal minors are nonzero) does not necessarily hold in this case

[subtle gust]

Yeah yeah got u

[limber sierra]

again though idk a counterexample

[subtle gust]

Wb the algorithms' questions tho

[limber sierra]

im just trusting wikipedia here

i know nothing about algorithms for this stuff so i'll ask you to repost the question and hope someone else comes along.

[subtle gust]

Is what i said accurate

[subtle gust]

Sure

Tysmm

You've helped me a lot

Appreciate it

[native ore]

Given two matrices A, B, and a vector x.

(A+B)x = Ax + Bx ?

[fleet sun]

it's true

[limber sierra]

matrix multiplication is linear

[clever prairie]

Can anyone clairfy how I did part A of this wrong,
This is my work for it

|-4A| = (-4)^3|A| = -64|A| ~ -64(14.5) = -928

[limber sierra]

that's just... not how determinants work

$k\abs{A} \neq \abs{kA}$ usually

[stoic python]

Namington

[limber sierra]

instead, review how matrix operations affect the determinant

i'll give you a hint: if $I$ denotes the identity matrix, then $\abs{-4A} = \abs{-4 \cdot I \cdot A} = \abs{-4 \cdot I} \cdot \abs{A} = \abs{\begin{matrix}-4&0&0&\dots\0&-4&0&\dots\0&0&-4&\dots\\vdots&\vdots&\vdots&\ddots\end{matrix}}\cdot \abs{A}$

[stoic python]

Namington

[limber sierra]

does this make sense?

now with this info, can you compute $\abs{-4A}$?

[stoic python]

Namington

[limber sierra]

as for inverses, note that $\abs{I} = 1$ (again $I$ is the identity matrix) and $AA^{-1} = I$, so $\abs{A}\abs{A^{-1}} = \abs{AA^{-1}} = \abs{I} = 1$. That is to say, $\abs{A}\abs{A^{-1}} = 1$ so $\abs{A^{-1}} = 1/\abs{A}$.

oops

[stoic python]

Namington

[limber sierra]

now can you compute $\abs{A^{-1}}$?

[stoic python]

Namington

[fleet sun]

but they raised -4 to the power 3 when they pulled it out. That seems right

[limber sierra]

oh fuck

i misread that

huh

sorry @clever prairie your work for |-4A| is correct

ignore me there

ask your grader i guess

because that's definitely right

[clever prairie]

Yeah I sent them an email about it give me my points back 😛

Thank you though!

[oblique prairie]

can someone tell me how $\langle u_2,v\rangle=\langle u_2,u_2\rangle$?

[stoic python]

quantum

[oblique prairie]

i don’t see it

wait i think i might see it now lol wow

[fleet sun]

why is it? I don't see it either

oh because <u1, u2> = 0?

[oblique prairie]

yeah

[fleet sun]

and v = u1 + u2

ok

[meager harness]

how to do 2b)

[quartz compass]

I'd use Gram-Schmidt

[meager harness]

how tho

wouldnt that give me 3 vectors

how do i get the 4th

nvm its a subspace

[quartz compass]

😎 👍

[timid sage]

I'm given a square matrix A and know that the basis for the nullspace of A is two dimensional. If I need to find the orthogonal projection of a vector v onto the nullspace of A, what exactly am I actually doing there?

As in, I understand how to find the orthogonal project of one vector onto another, but I'm not sure what I need to do when projecting onto a two+ dimensional vector space

[fleet sun]

if the nullspace is 2 dimensional, it has a basis of two vectors

and if you want to orthogonal project v onto that subspace, you can find its projection onto those two basis vectors

[timid sage]

I found a page on it, so I'm finding the projection onto each of the two basis vectors and then adding them?

[fleet sun]

yeah

[timid sage]

oh alright, thats pretty easy then

thanks

[fleet sun]

sure

[rugged hamlet]

Hi! My first question here so lmk if I'm asking this wrong. I'm not sure how the hint provided in the question is helpful... does it suggest proof by contradiction? If so, how can I evaluate the expression on the bottom? Any additional hints are appreciated :) (notation: V* represents the dual space of V)

[fleet sun]

it intends you to show each a_i = 0

which would let you conclude {f_0, ..., f_n} is linearly independent

[rugged hamlet]

ok, so my hunch about this being a proof by contradiction was correct. sorry if i'm missing something here (still new to dual spaces), but any tips on going about evaluating the inner product w/ summation expression on the bottom?

[fleet sun]

so, I assume you've already shown V* is a vector space

so you can split up the linear functional <p, sigma a_i f_i>

basically the sigma pulls out front

a_i can pull out as well by linearity

so you end up with a bunch of <p, f_i> guys adding up

then use the definition of <p, f_i> and your given expression for p

[rugged hamlet]

haha ok I'll work to pull out those <p, f_i> guys :)

thanks for the tip Manifold!

[fleet sun]

sure thing

[wintry steppe]

Is lay just applying the change of coordinate matrix there????

Ahhhh I think I see it. Its exactly that. It's just changing [ab1]b to the change of coordinates matrix.

But I don't understand why p^-1 is being used here. I thought that was used to convert an x in the standard basis to another basis (p)

Is this saying d is a change of basis to the eigen basis and p^-1 is that change of coordinates matrix, right????

It's essentially a way to map one linear transformation with two different bases right????

[spare widget]

It's change of basis for matrices

The intuition is that if you have a basis E and basis F

And if you are given a vector v wrt the basis F, but you are given A wrt the basis E, then you can't apply A as is. If you have a change of basis matrix P which maps vectors expressed wrt E to vectors wrt F, then you can do A[v]_E =A P^{-1} [v]_F

And since you want the result in F then you have to map from E to F also

P [A]_E P^{-1} [v]_F = [A]_F [v]_F

Then [A]_F = P [A]_E P^{-1}

[wintry steppe]

I understand change of basis

It's just the diagonalization

[spare widget]

Which step do you not understand?

[wintry steppe]

I'll get back to you. I'm gonna re read it.

[spare widget]

What's the book btw?

[wintry steppe]

linear algebra and its applications, lay, sixth edition

I can send it to you if you want it.

[spare widget]

P^{-1} comes from P [x]_B = x -> [x]_B = P^{-1} x

It's right above the multiline derivation

In this specific case [Abi]_B = P^{-1} Abi

[wintry steppe]

Bro I think it's clicking :)

Tyty for the help

[spare widget]

It would have been probably clearer if they wrote P [x]_B = [x]_E instead

I think that by x they mean x wrt the canonical basis e1,...,en

[stuck tendon]

Make everything one
Yes.
How did the 3 get here?
A factor of 1/3 was removed from each element of the matrix.
How did 1/27 get here?
det(cA) = c^n det A for an n x n matrix

[wintry steppe]

But shouldn't it be 1/3 lambda?

[stuck tendon]

1/3 - lambda = 1/3(1-3lambda)

[wintry steppe]

Shit I see

Thank you lol

[wintry steppe]

The 1/3 is raised to the n.... what's n here 3???

What does that come from even??? The n

Is it the col space of a?

[stuck tendon]

Yes, here n is 3. The matrix is 3x3, so det(1/3A) = 1/3^3 det A

[stuck tendon]

n is just the number of rows/columns

[spare widget]

det[a * A] = a^n * det[A]

[wintry steppe]

This is only for nxn matrix this det rule?

[spare widget]

as opposed to what?

[stuck tendon]

determinant is only defined for square matrices

[wintry steppe]

Lol, I should know that

Thank you so much :)

[spare widget]

if you need a generalization, they usually do det[A^TA] or det [AA^T] for non-square

e.g. for integration on manifolds embedded in higher dimensional spaces you get non-square jacobians

then you do sqrt(det[J^TJ])

[wintry steppe]

Oh I'm not that advanced. I'm taking a linear algebra course that's the second one people usually take here.

My first one was over 10 tears ago. So I've forgotten basic things. Hehe

This course seems heavily focused on orthoganility

[quartz compass]

I wouldn't think of it as a rule, think of it as putting in an identity matrix

[spare widget]

that's an unrelated remark yes, just in case someday you need to compute "a determinant of a non-square matrix", sometimes det[A^TA] makes sense

[quartz compass]

det(aA) = det(aI * A) = det(aI) det(A) = a^n det(A)

since det(aI) is just a along the diagonal

[spare widget]

just for clarity det[AB] = det[A]det[B]

Merosity is using the above in the special case with A = aI

[wintry steppe]

I see. I get it now thank you for the help :)

[placid willow]

Hello, I've been trying to figure out how to go about this problem for a little while now without making any progress. I already have the eigenvalues and eigenvectors, but I'm unsure where to go from there. Any help would be greatly appreciated!

[lavish jewel]

is this an exam?

[placid willow]

no, just some practice from last years homework. I'm just curious how to go about it just in case something similar to it is on the exam.

The prof gives us previous semester's HW but no answer key for some reason

[lavish jewel]

i would first note that multiplying two matrices BD, where B is some generic matrix and D is a diagonal matrix, has the effect of scaling the columns of B by the values on the corresponding diagonal elements of D

then we note that for a matrix A with eigenvalues c_i and eigenvectors v_i, we have that A v_i = c_i v_i = v_i c_i

these two things should let go push forward

[placid willow]

okay, that's on the lines of what I was thinking but I wasn't positive, I appreciate the help!

[lavish jewel]

i guess to give a further hint, i would note that if we have a matrix W and multiply it by some other matrix U, we can think of U as being made up of column vectors, so that U = [u_1 u_2 ... u_m]

and then the product WU can be expressed as WU = [Wu_1 Wu_2 ... Wu_m]

these should be all the hints you need, play around with it for a while

[placid willow]

alright, i'll give it a shot. Thanks for the tips!

[plush dust]

Hi guys, will have an midterm the day after tomorrow and I wonder, maybe you know some online quizes/exams to prepare to it. Mostly, it should be about these topics.

Part 4. AbstractVector Spaces;
Part 5. Matrix Computations in Spaces;
Part 6. Determinants and their Applications;
Part 7. Linear Transformations;

Thanks

[wintry steppe]

open up a linear algebra textbook and go to the "exercises" section

[plush dust]

Any textbook suggestions?

[open locust]

If I have an invertible matrix, how can I check whether or not it is the matrix representation of the identity transformation?

[native rampart]

If it's the identity transformation,the matrix has to be the indentity matrix

If not it's definitely not identity transformation

[open locust]

@native rampart That's definitely not true.

[native rampart]

Example?

It's definitely true , because T(v)=v

So whatever basis you pick,you get identity matrix

[open locust]

Just take any basis to another basis, you won't get an identity matrix in general.

[native rampart]

I mean wrt identity transformation

[open locust]

It's not an identity matrix in general, I don't want to explain that right now because I don't want to confuse myself more than I already am.

[native rampart]

An identity transformation corresponds to identity matrix in any basis

Because well A matrix corresponding to operator T when you pick a basis {e_1,e_2...e_n} has columns {T(e_1),T(e_2)...T(e_n)}

For T=id,T(e_1)=e_1 T(e_2)=e_2 and so on no matter what basis you choose

[open locust]

If you take the representation of a vector u in one basis, and take the representation of u in another basis, the coefficients are generally different.

So the matrix that takes one to the other is not the identity matrix

[native rampart]

Oh you mean change of basis matrix?

[fleet sun]

In other words, if B1 and B2 are different bases of V, and I: V -> V is the identity transformation, then [I]_B1 is not the same as [I]_B2

[open locust]

@fleet sun Exactly.

@fleet sun Do you have any thoughts on my question?

[fleet sun]

I think what you're asking is equivalent to how to tell M is similar to the identity matrix

[open locust]

@fleet sun I'm not sure about that

[fleet sun]

so M = Q^-1 I_n Q

If M represents the identity transformation, then in the right basis its matrix really is the identity matrix

It's a matter of finding that basis

[open locust]

@fleet sun My understanding is that if an invertible matrix is really a representation of the identity transform, then if we do change of basis transforms on the left and the right we should still have a representation of the identity transform. Your point is that in one particular to and one particular from basis, we'll get the identity matrix?

[stoic python]

joesmith1042

[open locust]

Oh my fault, the identity transform has to be from $U$ to itself. Never mind my deleted question

[stoic python]

joesmith1042

[fleet sun]

right

[open locust]

Why do we have to "find" a basis at all? Can't we just say the to and from bases are identical?

[fleet sun]

I'm not totally sure about the similarity thing I said

[open locust]

@fleet sun Well I think you pointed us in the right direction. We could just do a change of basis on the right to change the from basis to be the same as the to basis.

[fleet sun]

well, you can consider the identity from a space to itself, but with the standard basis on the domain, and some stupid basis on the codomain

[open locust]

Or do one on the left to change the from basis to the to basis.

[fleet sun]

And then its matrix won't look like the identity

[open locust]

Okay, so an invertible matrix represents the identity transform if and only if it is similar to the identity matrix.

That's pretty interesting.

[fleet sun]

I think another way to say it is it can be diagonalized to the identity matrix

[open locust]

I haven't really gotten to diagonalization yet. Thank you for your help figuring this out, I really appreciate it.