linear-algebra

[lavish jewel]

(UNLESS: T = 0 and lambda = 0)

you might need to think about that degenerate case

but i guess that means 0 is an eigenvalue, so it falls under the previous scenario

so nvm

[quiet wren]

Dunno where I shall put this but can someone help me see the connection between the two problems ?

They define the same optimisation problem

I assume we replace w by w x M no?

[wintry steppe]

because it's an injective operator on a finite dimensional space

[lavish jewel]

ei can't read what it says to the left of <aj, w>

[quiet wren]

b_j

[lavish jewel]

is that scalar? vector?

[quiet wren]

sorry scalar

so like a_j are the feature vectors, b_j is a scalar (1 or -1) in a classification problem

I suppose there was a mistake in that w is set to w/M instead i think it is set to wM

if w is set to wM then norm(Mw) = 1 translates into norm(w)M = 1 ie M = 1/norm(w) so that maximising M is equivalent to minimising norm(w)^2

[lavish jewel]

division seems right

if you have <aj, w> >= M, you can divide both sides by M if M > 0

so that <aj, w/M> >= 1

then since you divided w by M

maximizing M minimizes the norm of w

i dont see why norm(Mw) = 1

you have that norm(w) = 1

so norm(Mw) = M

[hardy inlet]

So people (including me ) did bad on the exam proofs, so the teacher is giving us 20 minutes to redo it when we have class. Can I receive help on 'studying'/proving them so Im ready to redo them?

[lavish jewel]

does the prof allow outside help

[hardy inlet]

I'd assume so. He never explicity said we had to do it all on our own, ik some people in the class worked together to do it; just to "come prepared" to re-prove them (i.e. my preparation isn't getting graded)

[lavish jewel]

did he say to prepare the questions ahead of time? will the questions be the same?

[hardy inlet]

yeah we had the exam and he fully graded them; gave it back with marks and whatnot. Is letting us redo the proof section with the same exact proofs. So if u had them right the first time ur fine, otherwize u can correct the ones u did wrong for partial credit back

so we have the physical copy of our exam

[lavish jewel]

i have to say no

this treads too close to the border with academic dishonesty

[hardy inlet]

Ok

[hardy inlet]

but just to clarify on this the redo is happening in-person without any notes/textbook (we just know whats going to be on it again)

[quiet wren]

if f(x) = 1/2 || x ||^2 then the gradient with respect to x is x right?

[gray dust]

@quiet wren yes

[subtle gust]

i asked this question here yesterday and i think @lavish jewel answered it

but still i don't think i quite understand it

if XX^T is skew symmetric prove that X=O3x3

[fair plinth]

Does there exist a matrix which can encode incrementing a binary-encoded number of length N? I mean, assume we have natural number

x = a_n * 2^n + a_{n-1} * 2^{n-1} + ... + a_0

So, let

v(x) = [ a_n, ..., a_0 ]

Does there exist A s. t.

forall x in { 0, 1, ..., 2^(n-1)-1 }
A * v(x) = v(x + 1)

?

[hardy inlet]

idk but if u find one i'd be interested to know the solution

[fair plinth]

Actually inc of 0 is 1, but no matter the matrix, matrix multiplied by zero vector should be zero vectorActually inc of 0 is 1, but no matter the matrix, matrix multiplied by zero vector should be zero vector

So seems like no 😦

[hardy inlet]

thats true but maybe u could write a piecewise function for when x = 0 to output 1

Then A(0) = 0 doesn't matter, but still holds

[fair plinth]

Maybe

[hardy inlet]

wait im dumb

T(x) = x+1 is non linear

thats like a fundemental definition

cause like T(x+y) != T(x) + T(y) doesn't hold; as well as some other issues

T(2+2) = 4+1 =5
T(2) + T(2) = 2+1 + 2+1 = 6

[wintry steppe]

T(0) =/= 0 and that's enough

[hardy inlet]

fair enough

back to the HW, im not sure what these have to do with each other

$\sqrt{\langle Tv, Tv \rangle} \leq \sqrt{\langle v, v \rangle}$?

[stoic python]

MattDog_222

[hardy inlet]

does that do anything

im still trying to figure out why (T - cI) is invertible when c is not an eigenvalue

is it because (T-λI) != 0 (if it equaled zero then its an eigenvalue). Then this means (T-λI)v != 0 for all v in V. so nullspace is empty and thus injective implies invertible? Not sure if that logic works out. (don't think it requires showing such just trying to believe it)

[wintry steppe]

if you take the determinant of (T - c Id) it becomes clearer that T - c Id is invertible when c is not a root of the chararcteristic polynomial

[hardy inlet]

ok so then this proof holds?

[wintry steppe]

just a typo I assume

[hardy inlet]

yeah = \sqrt{2}u

thanks for pointing that out

[wintry steppe]

other than that, it's correct 👍

[wintry steppe]

and your conclusion is made possible because "lambda is an eigenvalue <=> T-lambda Id is not invertible"

so the contrapositive is "lambda is not an eigenvalue <=> T-lambda Id is invertible"

[hardy inlet]

oh so its just the contrapostive; okay... cause the not contrapositibe is in the book

i seriously have no clue how to approach these

[subtle gust]

Anyone guys?

[wintry steppe]

what's O3x3

[night wren]

Given any two distinct non-zero vectors, is there always a linear functional that sends both of them to something non-zero?

[teal grotto]

the group of 3 by 3 orthogonal matrices

[subtle gust]

0

[gray dust]

use the fact linear maps are uniquely determined on a basis

[subtle gust]

The zero matrix

[teal grotto]

hard to tell without more context lol

[subtle gust]

That's not what i meant but sure lol .. i put O instead of 0 cuz 03x3 looks weird

[subtle gust]

Ummm that's the whole question lol

X is a 3×3 matrix. XX^T is skew symmetric. Prove that X is a zero 3×3 matrix

[teal grotto]

oh my bad, though Entelechy asked an isolated question

i just didn’t read above

[subtle gust]

Oh

Can you answer this question tho 💀

[night wren]

I don't think I know what you mean. Are you sayin its false ??

[subtle gust]

@teal grotto only using thing we covered .... (matrix multiplication and the def of skew sym matrices)

I could only prove that the product XX^T is zero

LET C=XX^T
C^T=-C => C^T+C=O3x3

[teal grotto]

okay that’s half the battle. think about what that means for each of the entires in XX^T. how are they defined?

[subtle gust]

Huh

[gray dust]

a linear map can be defined by specifying how to map each basis vector, use this to think whether the given statement is true

[subtle gust]

(XX^T)^T+(XX^T)=0 matrix

[night wren]

i kind of get you but I dont think its sufficient to send the basis vectors to something nonzero because maybe they cancel out to 0

[subtle gust]

X(X^T+X^T)= 0 matrix

X(2X^T)=0

2XX^T=0

so XX^T=0 matrix

That's where i stopped ....

[fringe fjord]

If X is not the zero matrix, then there is a row vector v such that vX != 0, and therefore vX·vX = vXX^Tv^T is nonzero too. This means that XX^T must be nonzero.
By contraposition, XX^T = 0 implies X = 0.

[gray dust]

i dont follow the bit about canceling

[subtle gust]

Confused tbh

[wintry steppe]

ahah I thought it was O(3) too but it was weird in the context

[night wren]

what if you send e1 to -1 and e2 to 1 and v= e1+e2? then v goes to 0

[gray dust]

nothing wrong with that

[night wren]

but I want v to not be 0 and same for some other w nonzero

[subtle gust]

Maybe you can do an easier proof...? 💀

[teal grotto]

let the rows of X be x_i. each entry on the diagonal of XX^T is given by x_ix_i^T and is equal to zero. expand out x_ix_i^T @subtle gust

[fringe fjord]

That looked pretty easy to me ...

[subtle gust]

One that a student who only did 4 weeks of linear algebra would understand

[gray dust]

no, we only want e1,e2 to have nonzero outputs

[night wren]

oh so make my two different vectors the basis elements? then its true?

[gray dust]

even if e1,e2 are independent theyre not necessarily a basis

[night wren]

i mean just like make them part of one

[subtle gust]

Still don't understand.... 🥲

Can we explain it with an example 3×3 matrix?

One that has like a b c d e f g h i as entries

[night wren]

i was asking about two arbitrary distinct vectors, not the basis btw

[teal grotto]

let X = x_{ij} be a square matrix

we know that XX^T = 0

[subtle gust]

Mmhm

True

[gray dust]

lemme clarify

given nonzero distinct vectors x,y does there exist a functional f where f(x),f(y) are nonzero?

is that ur q?

[night wren]

yes

[fringe fjord]

You can also avoid the detour via contraposition:
Assume XX^T = 0. Then for every row vector v we have vXX^Tv^T = 0. But vXX^Tv^T = (vX)(vX)^T = vX·vX, and the dot product of a vector with itself is only 0 if the vector is 0. So vX = 0, and since this is true for all v, it must be that X = 0.

[gray dust]

thats my interpretation all along

[night wren]

ok srry, i'm dumbo. How do you do it??

[teal grotto]

on the diagonal of XX^T, the entries are of the form x_{i1}^2 + x_{i2}^2 + … + x_{in}^2 = 0. you have the sum of a bunch of non-negative numbers being zero. what’s the only possible choice for all of those number to be?

[subtle gust]

Yeah we didn't cover dot products..... got last halfway through ur proof

[subtle gust]

Oh

See that's what i thought abt ..

[teal grotto]

…

[subtle gust]

But i was like "hmm maybe there could be 3 numbers squared that add up to 0"

I was stressed af lol

[teal grotto]

if ur working over a field with characteristic 2, there can be

[fringe fjord]

The dot product is simply (a1,a2,...,an)(a1,a2,...,an)^T = a1²+a2²+...+an², and that can obviously only be 0 if each of the ai's are 0.

[subtle gust]

Ok but like how does this prove that X is zero

This only proves that the diagonal elements are zero

[teal grotto]

because this shows that x_{ij} = 0 for all i,j

[subtle gust]

I still can't see how.... only the elements on teh main diagonal are of the form a^2 + b^2 + c^2

[subtle gust]

Well that doesn't seem hard

Still would prob learn more abt it when we cover it in class

[teal grotto]

just try writing it out with a three by three matrix

[fringe fjord]

You're not proving that the diagonal elements of XX^T are zero --- you already know that, since you have proved XX^T = 0 in general. In c²'s argument, you're using the known fact that the diagonal elements of XX^T are all zero.

[gray dust]

if x,y are independent, extend to a basis. if x,y are dependent then pick out one, extend to a basis. define a functional by mapping each basis vector to 1

[subtle gust]

Oh

Yeah yeah you're right

Idk why but i still can't understand it fully 🥲

XX^T is the zero matrix

That i get

[night wren]

tysm

[teal grotto]

sry to bat in, but as an interesting comment, if you don’t assume the axiom of choice, there exists an infinite dimensional vector space whose dual is zero dimensional, so the only linear functional is the zero functional

[gray dust]

ur welcome

we're definitely using aoc coy

[subtle gust]

And the entries on the main diagonal of XX^T are of the form a^2+b^2+c^2 and are all equal to 0

Oh

Ohhhhhh

[fringe fjord]

And then I and c-squared have each presented an argument that XX^T = 0 implies X = 0.

[teal grotto]

do ik u lol?

[subtle gust]

I think i understood it omg

Omgggg

Ty guys smmmmmm

Appreciate the help !!!!!

Well now i'm mad i missed such an easy question

[gray dust]

idk but i gave u trusted

[teal grotto]

wait how did u know my name tho

[gray dust]

ive seen u here since last august

its also on ur profile

[teal grotto]

oh lmao. dude that freaked me out

[gray dust]

https://i.gyazo.com/6a96c5be47c21bf02598ff5137515e84.png

[shut tree]

Ok this is a dumb question but why can’t we define linear transforms or inner product/norm on an abstract vector space

[gray dust]

who says u cant?

[teal grotto]

it’s hard if your field isn’t R or C for an inner product

[shut tree]

Without a basis being specified I mean

[teal grotto]

typically you don’t need to specify a basis when defining a norm or inner product

[shut tree]

Oh ok

[teal grotto]

however, linear transformations are uniquely determined by how they act on a basis. you don’t need a basis to define some of them

[shut tree]

I thought an inner product was a mapping from the space to a scalar and we need a basis to define it

[teal grotto]

the standard inner product on R^n is just x^Tx

no basis needed to define this

[shut tree]

I mean in abstract space

Not R^n

R^n already has structure on it I think

So not very abstract

[gray dust]

do u know any other 'common' spaces

[shut tree]

Examples?

There was space of functions on an interval, C^n, no idea tbh was just working with V, with nothing assumed on it

[gray dust]

sure continuous functions [0,1]->R

one inner product on it is $\lbr{f,g}=\int_0^1fg$

[stoic python]

RokabeJintaro

[shut tree]

Yea I got that as a canonical example but like you need something more than an abstract space to define it though right

Like if all you had was an abstract space would that still work as an inner product

[gray dust]

no, integration isnt a thing in an abstract VS

[teal grotto]

i don’t think there is a canonical way to define an inner product on an abstract vector space without more structure.

[shut tree]

Ok great. So say there are two defined bases of this space V. Inner products defined won’t change with the basis right?

Since you should be able to represent it as matrices and all that when taking the inner product

[teal grotto]

i’m confused on how you are defining your inner product from the basis

[shut tree]

I mean I’m not defining it from that

I have inner products as positive definite symmetric Nikon ear forms

Bilinear forms

So now that we have structure on V because of the basis, all inner products can be defined without worrying about basis right

[teal grotto]

if you have an inner product, <•,•> on V (somehow), then <a,•> is a linear map for each fixed a in V. assuming that V is finite dimensional vector space, then the matrix representation of <a,•> will be different depending on the basis. <•,•> isn’t linear, so i’m not sure what you mean by matrix representation. i’m doing the best interpretation i can lol

[shut tree]

It’s ok I am slowly getting it thanks

But as long as it has a basis /matrixrepresentation I guess it works

What is the motivation behind studying self adjoint and normal operators? I see the point of inner product, but what is the point of this?

[fringe fjord]

It is standard to represent an inner product (or in general bilinear form) on R^n with the matrix A such that <v,w> = v^TAw for all v and w. This also gives a canonical way to define matrix representations of inner products on a finite-dimensional abstract vector space once you've chosen a basis.

[fringe fjord]

Self-adjoint operators are wildly important in quantum mechanics, for one thing.

Normal operators are important because they can be diagonalized -- i.e. one can pick a basis where they have a particularly simple expression.

[shut tree]

oh

yea I do recall reading about hilbert spaces and operators in my qm class (but the class was mostly a survey class anyways)

[blissful vault]

Does my explanation make sense? The value of cos(theta) must be between -1 and 1. Since cos(theta) is equal to the inner product of x,y over norm(x)*norm(y), we must check that this value is between -1 and 1. It is straightforward using the Cauchy-Schwarz inequality.

[shut tree]

I think you can derive Cauchy shcwarz from inner product and work from there if it’s a positive definite symmetric bilinear form if you can’t assume that Cauchy Schwarz is known already

[blissful vault]

I believe we assume Cauchy-Schwarz is true for this exercise since it is proven in the section above the exercise.

[spare widget]

The angle between two vectors in R^n is pretty clear geometrically, since if those are linearly independent they span a 2d plane, and the angle in that plane is taken

That is, it's the same as in 2d or 3d

If they sre not linearly ibdependent then the angle is either 0 or pi

As far as cauchy-schwarz goes, note that the range of cos is [-1,1]

So you're not supposed to feed acos anything outside of [-1,1]

[spare widget]

Then again this exercise is poorly motivated as noted above

[winged nebula]

can anyone help me with this?

<@&286206848099549185>

i just need like one person to walk me through some basic lin algebra

explain somethings

[teal grotto]

what parts do you need help with?

[winged nebula]

i dont know how to give a parametric description of two individual vectors

[teal grotto]

parametric description of the plane*

[winged nebula]

i dont know how to do most of this

mhm

i just dont know what to even start writing down

i dont know what the plane is because its all variables no?

[wintry steppe]

wdym ?

[torn stag]

"parametric description" probably means to write the plane as the image of a smooth function

[teal grotto]

given two vectors v and w in R^n, the surface (plane, line, or point) spanned by v and w is the span of v and w, i.e., the set {sv + tw : s,t in R} (this is the parametric description of the surface)

if x = (x1, x2,..., xn) is perpendicular to both v = (v1,...,vn) and w = (w1,...,wn), then
x1v1 + x2v2 + ... + xnvn = 0 = x1w1 + x2w2 + ... + xnwn.
The set of all vectors x such that x1v1 + x2v2 + ... + xnvn = 0 = x1w1 + x2w2 + ... + xnwn is the parametric description of C for part b (in this case, you can write x3 as a function of x1 and x2 to get a nice formula).

for part c, any vector in P is a linear combination of the vectors you started out with. use the properties of the dot product and the fact that any vector in C is perpendicular to (1,0,1) and (2,1,0)

for part d, C = {x in R^n : <x,v> = 0 for all v in P}. show that this set is a vector subspace by showing that it is closed under addition and scalar multiplication

[misty bear]

Hi i'm having trouble with determining whether vectors are subspaces of R^3 - I know we check through closed through addition and closed through scaling but im not too sure where to start - to understand what I am doing to confirm this

if someone could help with these problems much appreciated!

[shut tree]

1. check zero vector
2. closed under scalar multiplication
3. vector addition @misty bear

[misty bear]

could I get an example of how to come to those conclusions? I dont really know how to check for those 3

[hardy inlet]

This is an example of the exact 3 things he listed:

[shut tree]

thanks I was getting annoyed at typing out vectors in latex lol

[hardy inlet]

and heres W2 for another example

1.) show that the identity is of the form
2.) show that an arbitrary vector v_1 of that form, and another vector v_2 of that form, added together still have that form
3.) show that scalar multiplication on an arbitrary vector v_1 preserves the form

so to get u started on 2a, is 0 of the form [x, y, (y-4x)]? (recall 0 = (0,0,0) in R3)

[misty bear]

yes(?)
since x = 0, y =0, 0=(0-4(0)) ???

[hardy inlet]

yup

0=(0-4(0)) is sort of confuzing, but
z = y-4x = 0 - 4(0) = 0

i guess thats what u had but without the x and y

[misty bear]

I see! So my question for b then - would b not be a subspace because y + z + 1 would not have (0,0,0) in R3? and it's vector does not contain the origin?

finishing a)
it would both be closed under addition and scalar multiplication right
since any arbitrary number would still preserve the forms as you stated?

[hardy inlet]

im quite confident that a.) is a subspace

and your explanation on b seems partially correct

we want to show that [0,0,0] is in the form (y+z + 1, y, z) so the last 2 coordinates must be 0 to fulfill this, and then we'd have [y+z+1, 0, 0] but y=z=0 so y+z+1 = 1 but we need it to equal zero; hence a contradiction

basically if (y+z+1, y, z) = (0,0,0) {a condition for a subspace} ur solving the system of equations

y+z+1 = 0
y = 0
z = 0

by equations 2 and 3 you can substitute into equation 1 to get 1 = 0 which is obviously false

[misty bear]

oh I see so plugging in (0,0,0) into the system of equations and seeing if it makes sense. is what the first condition is?

[hardy inlet]

not necesarily, you need the first component to equal zero, the second = 0, the third = 0

so you take the form, in this case (y+z+1, y, z) and set each component equal to 0, not substituting with 0

[misty bear]

Ooohh I see
setting to 0 not substituting. It's just in this case since they are just one variable kinda looks like it's substitution but that's not I'm doing

[hardy inlet]

well u use substitution in that example to solve the system

but u first equate them to 0

[misty bear]

yes

so checking for each set I equate them to 0 and see if it makes sense

[hardy inlet]

I believe so. someone can @me/you if i'm wrong {im quite tired} but that should be a correct way of checking

[misty bear]

in the case of b
y + z + 1 = 0 does not make sense since both y = 0, and z = 0

but in the case that maybe like
x = y + z + 1
y = z - 1
z = 0;

this would work?

lol it's understandable thanks for helping out though! I think I kinda get it

[hardy inlet]

so, z = 0, therefore y = (z - 1) = (0 - 1) = -1 false

[misty bear]

Oh lol I see why that wouldnt work

[hardy inlet]

and as such x = y + z + 1 = (-1) + 0 + 1 = 0 holds; only the middle coord fails

[misty bear]

yeah

[hardy inlet]

idk if theres a way to get zero from adding +c and stuff; but even if there is it will fail the closed under addition/scalar check

[misty bear]

oh I see!
So the closed under addition and scalar stuff how do I check those? :')

[hardy inlet]

to prove its true is a little more work (i.e. those snippets i showed earlier), to prove its false you only need 1 counter example

so if we have

lets chooze the vector (3,1,1) since its of that form

and (2,1,0) which is of that form too

doesn't really matter, just wanted 2 different ones (the same would be scalar multiplication which also fails)

so (3,1,1) + (2,1,0) = (5,2,1)

is (5,2,1) of the form?

y + z + 1 = _____

[misty bear]

it would be 4 right

so no

[hardy inlet]

correct. 2 + 1 + 1 = 4 and 4 =/= 5

so its no longer in the set of vectors of the form; hence it is not "closed"

we take 2 things from this set, and we get something outside of the set

[misty bear]

ah I see! that makes sense - in the case of scalar multiplication how would that work?
take a constant like 2 x (3,1,1) i would get (6,2,2) but that would also not be closed right?

since 2 + 2 + 1 =/= 6

[hardy inlet]

yup

proving something is true is a bit harder

[misty bear]

Nice okay thank you so much for explaining this to me! Makes a lot more sense now

Yeahhh, checking random vectors to see if they are false. Is that a good way to check?

[hardy inlet]

Showing any of them is false with any example is enough to show it is not a subspace

but to show it is a subspace, you can't just give an example that works

because there are infinite combinations, you would have to check them all

take the example (x,2x), is this closed under addition?
let v be of the form (a, 2a) and w be of the form (b, 2b)
v + w = ___
it equals
= (a,2a) + (b,2b) {replace v and w with their arbitrary forms}
= (a + b, 2a + 2b) {add components}
= (a+b, 2(a+b)) {factor out the 2}
so its of the form (x,2x) {where x = (a+b)}

[misty bear]

the answer is yes since it's still in the same form (x, 2x)

closed under addition

[hardy inlet]

yes

[misty bear]

cool cool

[hardy inlet]

lastly you need closed under scalar mult

in this example, you'd also need to show

λ(x,2x) is in the subset of R2

for any v = (a, 2a) and scalar λ
λv
= λ(x,2x)
= (λx, 2λx)
which is obviously of the form (x,2x)

as for a counterexample {where scalar doesnt hold}, if we had (x, x+1) as the form
λ(x, x+1)
= (λx, λ(x+1))
= (λx, λx + λ)
which is only of the form if λ=1

[misty bear]

hmmm I see
similar to b) in which we had y + z + 1 which also didnt work out scalar mult wise

[hardy inlet]

yeah

if u wanna show a counterexample showing 0 isn't in it is probably easiest, seconded by scalar

[misty bear]

okay omg this makes so much more sense

thank you so much!

[hardy inlet]

np

[snow totem]

what type of connections do there exist between diagonal entries of a (may be restricted to symmetric) matrices and the eigenvalues of the matrix, I know that the trace of a matrix is equal to the sum of the eigenvalues, but are there other connections/identities or is this the only one?

[tranquil steeple]

depending on the structure, you can define spectral symbols that describe the eigenvalue distribution

also you have https://en.wikipedia.org/wiki/Gershgorin_circle_theorem which is a crude approximation

[snow totem]

that is a bit vague, but does that link describe it in detail?

[tranquil steeple]

For the univariate case https://libgen.is/book/index.php?md5=36AA2B2781737BD3AA1D72097B10FC41

[snow totem]

ok thx a lot

[tranquil steeple]

so-called GLT sequences can be very general (for example PDE/FDE discretizations are of this type)

but for an arbitrary matrix Gershgorin is useful

[junior ravine]

how do yo find the value of k for infinitely many solutions?

[zinc timber]

have you performed the row operation yet?

/row elimination whatever

[subtle gust]

Guys regarding the question i asked yesterday, (if X is a 3×3 matrix and XX^T is skew symmetric prove that X=0 3×3)

I thought of a much

Much

Easier approach

X= [a,b,c
d,e,f
g,h,i]

We know XX^T is 0 3×3 and the entries on the main diagonal of XX^T are
a^2+b^2+c^2
d^2+e^2+f^2
g^2+h^2+i^2

So

a^2+b^2+c^2=0
d^2+e^2+f^2=0
g^2+h^2+i^2=0

Which is a homogeneous system of equations

If we prove that the augmented matrix of this system is invertible (by proving that the determinant is not zero) we can say that this system only has the trivial solution

Or
a=b=c=d=e=f=g=h=i=0

or in other words X= 0 3×3

I can't believe i couldn't think of that during my test 🤦🏻‍♂️

Nvm i just realised it's not a linear system of equations

[lavish jewel]

that's super convoluted

you can just see you have a sum of squares = 0

real numbers squared are >= 0

and that's all

[grave kettle]

suppose $U, W$ are subspaces and $S = { \bd{u} + \bd{w} \vert \bd{u}, \bd{w} \in U \cup W }$, how to show $B_U \cup B_W$ is a basis for $S$?

$B_U, B_W$ are bases for $U, W$

[stoic python]

Chromium

[zinc timber]

can't you conclude from XX^T that XX^T = 0 then since X=0 directly?

or you aren't allowed to do that?

[lavish jewel]

you are given that XX^T is skew sym, ryu. you have to prove X=0

[zinc timber]

ya but it's symmetric that we can check, so it's both symm and anti symm so 0

[stoic python]

Chromium

[lavish jewel]

right, so XX^T = 0 from that

[zinc timber]

now N(XX^T) = N(X) = entire space so X=0

[lavish jewel]

they have to prove that

i mean, they would have to prove what you just said

[zinc timber]

$\ip{X^Tv, X^Tv} = \ip{XX^Tv, v} = 0 \implies \norm{X^Tv} = 0 \implies X\equiv 0$ but ig they are not allowed to use it yet?

[lavish jewel]

that's essentially the same thing everyone suggested already

[zinc timber]

oh

[lavish jewel]

but yes, they have to prove all of this

[zinc timber]

yes true

[lavish jewel]

so it's the same thing

[zinc timber]

@grave kettle did you mean to write $S = { u+w | u\in U \qq{and} w \in W }$?

[grave kettle]

not much difference ig

[zinc timber]

ya but it'll be easier to work with so

also B_U \cup B_W may not form the basis of U+W

for example U = (0, y, z) and V = (x, 0, z),, x,y,z are reals

[grave kettle]

i can't see why

[zinc timber]

one easy way it suppose both U and W have dim 2 in R3 then B_U \cup B_W will have 4 vectors but then it must be LD since we are in R3

so can't be a basis

although it is a generating set, we don't have "linear independence"

[wintry steppe]

Suppose i have a 2×2 matrix whose entries are matrix then how am i going to calculate determinant of original matrix?

https://en.wikipedia.org/wiki/Determinant#Block_matrices

Ty !

[grave kettle]

what makes linear transformations easier to work with than other tranaformations

[wintry steppe]

basically, being able to say T(ax+by) = aT(x) + bT(y) is very useful to determine a lot of properties of transformations and spaces, and even for computation it's a lot simpler

[grave kettle]

wdym

[wintry steppe]

It's hard to give a specific answer to your question cause it's quite vague; now you know that linear transformations are maps that satisfy the following condition: T(ax+by) = aT(x) + bT(y), (for a,b scalars and x,y vectors). This property is absolutely crucial to prove a lot of propositions involving bases, dimensions of spaces, isomorphisms, determinants and so on... Just like you can get the intuition that studying a linear function in the plane is a lot easier than studying a logarithmic or exponential function, because the two latters are not linear (although in vector spaces it's different).
I don't exactly know where you are in your course, but if you want a specific example of a proof that involves linearity, I can give you one

[grave kettle]

yea sure

i’m actually self learning and not following a very concrete outline

[wintry steppe]

Oh, that's nice 😄

[wintry steppe]

Do you know about linearly independent vectors ?

[grave kettle]

yea

[wintry steppe]

It's not really a short proof, but I think it's a good exercise to try to understand it (which I'm sure you will easily !)

And also, notice how linear independence only makes sense because of linearity 😉. Now to give a final answer to your question, Linear algebra is made possible because it studies the properties of... linear objects. The study of non-linear objects, non-linear maps etc. is far less common and requires more advanced mathematics knowledge. The reason behind linear algebra is because it's all nice and smooth to work with maps and spaces that satisfy good properties

[grave kettle]

is linear algebra the study of just linear equations

idk where i’m heading with subspaces, transformations and stuff

[nocturne jewel]

No

It's the study of vectors and linearity

[grave kettle]

properties such as?

(useful properties that build up for future things?)

[wintry steppe]

wdym by "future things"

[grave kettle]

future concepts in lin alg

that build on linearity

or properties of linearity

i can’t convince myself of its importance and specificity of the definitions of linearity

[wintry steppe]

oh thanks God I thought for a moment you were talking about real life application

well all of linear algebra is based on linearity of the maps

you probably know already about determinant

[grave kettle]

no

i haven’t studied matrices

[grave kettle]

can i still get names, just to see if i’ve heard of them

[wintry steppe]

names of properties, theorems, etc. ?

[grave kettle]

concepts

that require linearity to make sense

T(a + b) = T(a) + T(b) and T(cv) = cT(v) seem such specific requirements

i can’t see the motivation

[zinc timber]

it's more than just linear maps

just don't hv the energy to type

[grave kettle]

how?

[wintry steppe]

"T(a + b) = T(a) + T(b) and T(cv) = cT(v)" is the formal and mathematical way of describing transformations that keep the "grid lines" parallel and evenly spaced, if you look at your space in terms of graded grid lines

[grave kettle]

(that’s the 3b1b description? still seems a bit goofy)

[fringe fjord]

You know how it's a common error for beginning students to attempt rewritings of the form $\sqrt{x+3}=\sqrt{x}+\sqrt{3}$ or $\sin(2t)=2\sin t$? Linear maps are the ones where you \emph{can} get away with such things, which makes everything much easier.

[wintry steppe]

goofy ?

[stoic python]

Troposphere

[grave kettle]

keeping lines ‘parallel and evenly spaced’ sounds weird

[wintry steppe]

How so ? I find it very easy to visualise

[grave kettle]

for you ig

[wintry steppe]

and that helped me a lot when I started linear algebra

[grave kettle]

i don’t get the importance of keeping lines ‘parallel and evenly spaced’

[zinc timber]

you don't have to have linearity, but if u do, it's linear algebra, if u don't, it's not linear algebra

[lavish jewel]

to be honest trying to think about it geometrically is also limiting

[grave kettle]

what’s a good way one should interpret linear transformations

[grave kettle]

i know the definitions but rn they seem a bit random and out-of-nowhere

[lavish jewel]

axiomatically

one always looks out for objects with nice properties

playing around with these turned out to be nice

transformations with these properties are "nice" in that they are relatively simple and easy to study

[grave kettle]

please elaborate

[lavish jewel]

and one can often approximate other transformations that are nonlinear with linear approximations

well, as tropo wrote above

they "split" nicely

no matter how complicated the input is, you can break it up into parts and study them separately

you cannot do this with sqrt(3 + 2) = sqrt(3) + sqrt(2)

but you can with a linear transformation

[grave kettle]

so mathematicians just came up with these in a flash of inspiration and decided ‘wow they’re such nice things to work with !’?

[lavish jewel]

the concept of vector spaces, basis, and linear combinations all go in this direction

well, that's how most stuff goes, isn't it?

you see something, you wonder if this property is shared by other objects, you try to generalize it

[grave kettle]

i want to understand the thought process of how one might invent these T(a + b) and T(cv) things

[lavish jewel]

sometimes it works, sometimes it doesn't

and further, sometimes it's actually useful

[wintry steppe]

it wouldn't have gone that far if it wasn't nice to work with

[lavish jewel]

but you don't know a priori

[grave kettle]

at least for this linear transformation thing

[zinc timber]

there's also one rigorous way, using group actions

[lavish jewel]

sometimes you have a problem that you know could be solved if you were working with objects that had extra structure

and the motivation comes from there

but i think linalg developed recently along with abstract alg

[zinc timber]

but meh, if you aren't satisfied with the geometric one, this one won't change yr opinion

[wintry steppe]

something like 19th century ?

[lavish jewel]

more recent, i think

a passing wikipedia survey says around 1900, yeah

[zinc timber]

LA is that recent?

[lavish jewel]

the current flavor, at least

[grave kettle]

so in a nutshell

[lavish jewel]

i'd be hard pressed to say linear systems of equations are recent 😛

and i guess those were the main motivation

how to deal with large systems of equations

which in general are super nasty

[wintry steppe]

Descartes my man

[lavish jewel]

but when they are linear, they can be solved in a simple fashion

tried and true

and then these properties abstract and generalize to the more abstract linalg

[grave kettle]

linear transformations are special cases of transformations in general?

and uh

we study them specifically because they’re nicer to work with

and in investigating transformations, mathematicians have discovered they boil down to this thing called linear tramsformations

[lavish jewel]

idk what you mean with that

[grave kettle]

geometrically, this is implied

preserving linear entities is cool????

[zinc timber]

LT arise naturally in the context of module homomorphisms

[fringe fjord]

"Transformation" is mostly just a nonsense word here. It doesn't mean anything appreciably distinct from "map" or "function".

[grave kettle]

does further investigation mean abstract algebra

ok so geometrically, why would ‘keeping lines parallel and equidistant’ make things easier to work with?

[lavish jewel]

checking wikipedia says solving systems of linear equations has been an interesting problem for a long time. makes sense, since this is related to resource allocation

[grave kettle]

hm ok

why are linear transformations considered the ‘simplest to deal with’ and also retaining a sensible degree of generality?

[fringe fjord]

Do you have a proposal for something simpler?

[plush canyon]

I've never seen a hw question more complex

[lavish jewel]

these properties, chromium, make it possible to completely characterize a function simply by its effect on a basis. and in the finite dimensional case, really by looking only at a small set of vectors that make up a basis

[plush canyon]

U talking to me?

[lavish jewel]

no

[plush canyon]

Ok

[zinc timber]

8 variable linear equation

[plush canyon]

Yeah

Can someone help

Wtf

It's only a 10mark question. For a hw

An 8x8 matrix WTF

[grave kettle]

why are transformations that you can’t describe using change of bases nastier/less simpler to work with?

[fringe fjord]

Do you have a proposal for something that would be simpler to work with?

[grave kettle]

the way i see it (as of now) if the change can be described with change of bases, it can be put in some matrix

i don’t see anything else

[fringe fjord]

That's linear algebra!

[plush canyon]

Can someone help with my question.i have to put it here cause its complex

[fringe fjord]

Huh? It looked real to me.

And you haven't actually asked any question that it makes sense to try to answer.

[grave kettle]

does putting things in a rectangular array make transformations objectively easier to analyse?

[plush canyon]

Who u talking to m

M

Me?

[fringe fjord]

For the third time: What is it you keep implying you think would be easier?

You can just keep claiming that you have something that would be easier to deal with than linear algebra without revealing what your proposed alternative IS.

[lavish jewel]

at face value, anything you can put at in that form can be solved with a fixed recipe without worrying about anything else, but i can't tell if that's the explanation you want

what are you fishing for

[grave kettle]

now just wanting to know the uses of putting things in a matrix

and how it makes stuff simpler to calculate

[fringe fjord]

For the fourth time: which alternative method is it you think would be easier?

[lavish jewel]

also, modern linalg is not about matrices

they just happen to be handy representations in some cases

[plush canyon]

Why is my case handy then?

8x8 is not ok

[lavish jewel]

your problem is just back substitution

what's your question

[plush canyon]

This

8x8 is scary

[fringe fjord]

Why?

[plush canyon]

I did the other stuff easily and then I see T's

[fringe fjord]

This is not a do-my-homework-for-me service. If you have something to ask about that problem, ask it and people can try to give you a helpful answer. If you just want to complain, though ...

[plush canyon]

I don't know how to do it

And I would like a guide

To explain if possible

[fringe fjord]

Ar you implying that you would know how to do a similiar task if it was 3×3 instead of 8×8? What's the relevant difference?

[plush canyon]

Well you see

I have multiple T terms

[lavish jewel]

replace them

[plush canyon]

T3, T5 etc

[lavish jewel]

T1 = a

[plush canyon]

With 1?

[lavish jewel]

T2 = b

etc

so that it looks like something you recognize

[plush canyon]

Oooooo

[grave kettle]

can i get some ideas as to how matrices may help beyond only representations?

[plush canyon]

Thanks

[fringe fjord]

No. You have shown that you reject every answer you're given, why should we keep repeating ourselves?
You have also rejected repeated calls for you to reveal what the heck it is you think we should use instead of linear algebra.

[lavish jewel]

they don't, anyway

it's the same thing with different notation

[grave kettle]

i’m getting that representing basis change with arrays can help somehow? idk if that’s justified (in terms of computations or whatever) i’ll accept linear transformations the way it is now

[fringe fjord]

It's pretty clear that you're not going to accept anything.

And that, really, is fine with us.

[grave kettle]

i’m just asking

not sure why you’re against that

[fringe fjord]

Go ahead and continue thinking linear algebra is useless,

[grave kettle]

i’m not against your answers

[fringe fjord]

You have rejected each and every answer people have given you.

You have repeatedly refused to reveal what it is you want to be using instead of linear algebra.

Very well, you don't have to.

Go ahead and do your own thing instead without telling us about what it is.

[wintry steppe]

[fringe fjord]

But then there's not much point in having a conversation.

[lavish jewel]

i think they're looking for some philosophical "why" that doesn't exist

not trying to be dismissive

[grave kettle]

so i’ll continue

[lavish jewel]

no, don't

[grave kettle]

?

[lavish jewel]

the discussion doesn't make sense either way because all of tropo's points were valid regardless

[spare widget]

matrices are representations and you have efficient tools designed to work with said representations regardless where said array of numbers arose from

does that satisfy you?

[grave kettle]

no, not really, sorry

[spare widget]

it makes your life easier

[lavish jewel]

you give us an example then

can you answer these questions for anything else in math?

what answers satisfy you

[spare widget]

See 1.3 in Hoffman and Kunze

they motivate it as a shorthand for writing linear systems of equations

[zinc timber]

https://tenor.com/view/frodo-keep-your-secrets-alright-then-lotr-gif-12907121

[spare widget]

I have also seen determinants motivated in the same way through Cramer's rule, should be fairly straightforward for students out of highschool

[grave kettle]

i’m just asking things i can’t wrap my head around and sometimes in having some resolved new questions pop up and i’m looking to understand them

not sure if i can describe it, either

[lavish jewel]

then we can't help you

i suggest you pick up a couple different linalg books and read the preface and first chapter or two and see what you come up with

[grave kettle]

i did

[lavish jewel]

because you have "questions", but don't know what they are

so naturally they don't have answers

[grave kettle]

on a grand scale no but i can ask smaller questions that hopefully shape a big one

[lavish jewel]

idk what "why" you want though

and you don't know either

look up practical and abstract motivations

https://en.wikipedia.org/wiki/Abstract_algebra

[zinc timber]

thing is what's to point of having extra sctucture on your space if you are not going to use it anyway, what's the point of defining homomorphism if you aren't going to use the group structure of the inderlying set and use set map anyway

[lavish jewel]

read wikipedia

idk

as a part of abstract algebra, my main take is just "what are the consequences of having additional structure?"

[zinc timber]

this convo is pointlessly being dragged

[spare widget]

is the question "how would one come up with linear spaces"?

[lavish jewel]

we already presented ideas, ranging from practical to abstract

so the rest is on you

[grave kettle]

so far
‘why linear transformations? because it’s the simplest vector transformation worth studying.’
‘why is that? because it’s representable by basis changes that can be arranged into matrices.’
‘why matrices?’

[spare widget]

try it in the opposite direction

say you want to solve

a11 * x1 + a12 * x2 = b1
a21 * x1 + a22 * x2 = b2

write the solutions for x1 and x2

you'll see that the numerator and denominator are functions of the coefficients, notably the denominator depends only on the as, and the numerator is a mix of the as and bs

studying larger systems you can try and generalize the rule

and you'll end up with determinants (to be precise you'll end up with Cramer's rule)

then you'll realize only the coefficients matter, and for brevity you may want to order them in tables, i.e. matrices

I think this process should be understandable even for people that have never seen a matrix

studying the properties of matrix multiplication and addition you'll naturally arrive at A * (x+y) = A * x + A * y and A * (c * x) = c * (A * x)

if you start drawing some examples in 2D and 3D you'll also see that said transformations preserve parallel lines and you'll be able to interpret the columns of A as a frame

sutdying x * A instead would do the same for the rows

[grave kettle]

scratch the matrices thing

so i’m seeing that being able to study transformations of vectors in general as only their bases is a main reason linear transformations are nice (easy to work with)

[spare widget]

you may be given a frame instead of a basis, but linearity is still nice to have

e.g. after studying equations, you may decide to study polynomials, and then realize that you can again take the coefficients and work with those using the same tools

eventually if you try to generalize this you'll reach something like the definition of a linear space at some point

it's a standard procedure to take some example (e.g. R^3) and then see what properties generalize for similar examples (e.g. R^n, polynomials of degree <=n, etc.), and what do not

eventually you'll end up with a set of assumptions, such that if they hold, you have a number of results that are valid

regardless from what field the initial problem arose

e.g. it will abstract away whether your problem arose from a PDE, from some geometric intersection problem, or something else - you'll be able to use the exact same results and tools provided the axioms hold

[icy blade]

how can i build a matrix using the info i have, the p'(1)= 7 stumps me

i understand from p(2)=0 we get; 4a2 + 2a1 + a0 = 0
from p(-2)=-32 we get; 4a2 - 2a1 + a0 = 32

but what do i do with p inverse of 1

[nocturne jewel]

It's the derivative

Not the inverse

[icy blade]

so how can i build the matrix with the derivative what does 2a2(1) + (a1)*1 = -7 give me

[spare widget]

use the p(2) = 0, p(-2) = 32 and p'(1) = -7 for your 3 equations

a0, a1, a2 are the unknowns

the vector [x^0, x^1, x^2] is set accordingly based on the argument of p

[icy blade]

[spare widget]

$p(x_1) = a_0 + a_1x_1 + a_2x_1^2 = y_1 \
p(x_2) = a_0 + a_1x_2 + a_2x_2^2 = y_2 \
p'(x_3) = 0a_0 + a_1 + 2a_2x_3 = y_3 \implies
\
\begin{bmatrix} 1 & x_1 & x_1^2 \ 1 & x_2 & x_2^2 \ 0 & 1 & 2x_3 \end{bmatrix} \begin{bmatrix} a_0 \ a_1 \ a_2 \end{bmatrix} = \begin{bmatrix} y_1 \ y_2 \ y_3 \end{bmatrix}
$

[stoic python]

criver

[wintry steppe]

How do i find the kernel of a linear transformation

or

my bad

i know how

but im getting it wrong

according to the dimension theorem

i have the transformation T(f(x))=xf(x)+f'(x)

from P_2(R)->P_2(R)

[nocturne jewel]

ok

[wintry steppe]

I put in a arbitrary function from P_2(R)

[nocturne jewel]

you're solving for all vectors st $xf(x)+f'(x)=0$

[stoic python]

Mosh

[wintry steppe]

all functions

but yes

I arrive at $a(x^3+2x)+b(x^2+2)+cx=0$

after plugging in an arbitrary function from P_2(R)

[stoic python]

jswatj

[wintry steppe]

but i'm not sure where to go from here

also it follows that {x^3+2x, x^2+2, x} is a basis for the range of T

right?

since its linearly independent

wait no its not

{x^2+2, x}

there

[lusty pumice]

Not sure what this one is saying

and this one

it's studying questions btw not tests or quiz

[fallen karma]

True or false I take it?

[lusty pumice]

yeah

but not sure what it's asking

or how to picture that

I'm studying older exams and writing why in my answer

[nocturne jewel]

Can you find a non-invertible upper-triangular matrix?

That's all this amounts to

[lusty pumice]

What does Ax = 0 have to do with an upper-triangular matrix being invertible?

[nocturne jewel]

Ax=b has a unique solution for all b iff A is invertible.

Thus suppose A is both upper triangular and non-invertible, then Ax=0 won't have exactly 1 solution

[lusty pumice]

woah that's pretty cool

[nocturne jewel]

well yeah, it's one of the TFAE for invertibility from fundamental theorem

[fallen karma]

Yeah, linear is such a nice property

[fallen karma]

I think if it's consistent and more equations than unknowns, then there has to be linear dependence

[compact tartan]

hello hello! im not entirely sure im grasping the process for finding basis of null/row/column space. If someone wouldnt mind checking what i have here for the given matrix, id appreciate it

[wintry steppe]

The vector space given in the que is vector space of smooth functions??

[slow scroll]

yes

[wintry steppe]

Okay ty !

[slow scroll]

np

[fallen karma]

@compact tartan looks good

[compact tartan]

you are wonderful

thank you very much

[fallen karma]

Np

[dire copper]

If the determinant of a 2x2 matrix is negative, does that mean we can't find the square root of the matrix?

[fallen karma]

Yes. Because det(A^2)=det(A)det(A)>0 so if a matrix has a square root its determinant is positive

[wild hollow]

can someone help me out with my understanding on span of a set of vectors

ive been taught that the span of some arbitrary set T is the smallest subspace containing T.

im a bit confused what it means by smallest subspace, smallest set of vectors that satisfies the rules of being a subspace, which has the least elements in it ???

im a bit confused what it means by smallest set, how can a set be bigger than another one, is it the amount of elements within the set ?

[fallen karma]

T is a set of vectors, so span T is the set of all linear combinations of the vectors in T

So T is not a vector space but span T is

And by smallest it is meant that...

If U is a subspace containing all the vectors of T, then spanT is a subspace of U

[dire copper]

Alright thanks

[calm flower]

Can anyone help me with this?

I am confused

[wintry steppe]

Which one in particular ? Did you do the first one ? It just comes down to a definition that should be in your course

[calm flower]

I solved it thanks.

[tacit dagger]

the questions is determine which value a makes the system have infinite solutions. Doesn't it already have infinite solution just by the fact there's more variables than equations? after some gauss-elimination i got the last row to be -a-2 = -3 (the -a-2 takes one spot in the matrix). It will definitely have infinite solutions if the last row becomes 0 = 0, but for which value a should z(-a-2) = -3. If the last row becomes 0, it will just equal with the matrix having no solutions.

nvm

figured it out after sending the question

[fringe fjord]

If it has any solution at all, there are infinitely many of them, but you might be unlucky and it has none.

[hearty breach]

In Python/other language, I might get the last three elements of a vector like so x = v[-3:]. How can I denote this using conventional math notation?

[spare widget]

$x = \begin{bmatrix} v_{n-2} \ v_{n-1} \ v_n \end{bmatrix}$

[stoic python]

criver

[hearty breach]

fair enough. There's no more concise way to do it though? What if I wanted the last 100 values?

[spare widget]

$x = \begin{bmatrix} v_{n-99} \ v_{n-98} \ \ldots \ v_n \end{bmatrix}$

[stoic python]

criver

[hearty breach]

haha, that's fair. I just thought maybe I could just do something like $x = v_{-3:}$ But I didn't know if this notation is actually used outside programming

[spare widget]

I haven't seen this notation, but if you define it you can use it. Albeit I would recommend against

[hearty breach]

Thanks for your input! It's not a rigorous math paper, but I'd prefer to use whatever's most understandable. I will go with your notation

[spare widget]

It always depends on the audience that the material is meant for. If the intended readers are python programmers then defining the python notation and using it will probably be more convenient. If the readers are arbitrary then it makes more sense sticking to standard notation.

[wintry steppe]

any idea how to show that A B = det(A) * det(B) in commutative rings

[wintry steppe]

what

[wintry steppe]

I have a matrix [A 0 0 B] and I need to show that the det([A 0 0 B]]=det(A) * det(B)

A 0
0 B?

you could write out the summation formula for determinant and it becomes apparent

Thats the way I would justify it

[wintry steppe]

There might be another way however

[wintry steppe]

with summation formula would be AB-0?

something similar to that

i would need to work out myself

another way is to notice where the matrix would have linear independence

I could probably say more sorry for short help

[fringe fjord]

I think A and B are matrices here. The ad-bc formula for 2x2 determinants doesn't generalize to block matrices.

Toysem was speaking about the Leibniz expansion, I believe.

[tranquil steeple]

The eigenvalues of [A 0; 0 B] are the same as the eigenvalues of A and B separately, so it should be clear that det([A 0; 0 B])=d(A)*det(B)

since it is block diagonal

[plush canyon]

can someone pop to help 14

I didn't know if to put question in here or there

[stoic python]

@feral mountain

[feral mountain]

some shit like that

doesn't seem too massive

[stoic python]

@feral mountain

[spare widget]

Is there a proof somewhere that A^TA is non-singular if A arises from an inconsistent system Ax = b, with row rank >= dim(x)?

[lavish jewel]

are you sure you mean dim(x) there?

[spare widget]

A^TA being the matrix from the normal equations that one can get by taking the derivative of |Ax - b|^2

dim(x) = # of rows of x = # of columns of A

[lavish jewel]

that's not what dim means

[spare widget]

consider it abuse of notation

[lavish jewel]

but anyway, A^TA is nonsingular if A has full column rank

the lazy proof is using an SVD

let A = USV^T

then A^TA = VS^TU^TUSV^T = VS^2V^T

S is diagonal, and this result is also in the form of an SVD, i.e. VS^2V^T is the SVD of A^TA

so that A^TA has the same rank as A

so A needs to have linearly independent columns in order for A^TA to be invertible

[spare widget]

how would you understand dim here other than the dimensionality of the vector?

[lavish jewel]

dim is dimension of a subspace

so dim(x) = 1

or rather dim(span(x))

[spare widget]

Ok, I got it

And if A has row rank >= # of columns -> it has full column rank?

[lavish jewel]

the row rank cannot be > # of columns

[spare widget]

I have a rectangular matrix

More rows than cols

Oh, yeah

It cannot

[lavish jewel]

yes, the rank cannot exceed the smaller of # rows, # cols

[spare widget]

they'll be linearly dependent

So at most it is row rank = col rank

[lavish jewel]

for a tall matrix, the rows are lin dep. yep

[spare widget]

Great. Thank you

[wild hollow]

im a bit confused about the upside down V notation

is this the line created between these two planes or something

[nocturne jewel]

And

[wild hollow]

i dont understand why they use and

[nocturne jewel]

cause the line is the solution space to -x-y+1=0 and x-z-10=0

[wild hollow]

so the intersection between the two planes?

[nocturne jewel]

yes

all points st both equations are satisfied means the point is on the line l

[wild hollow]

would it be better worded to so that the line l is the solution to the system of equations

or is it implied

[nocturne jewel]

Implied

"given by the system of equations"

[wild hollow]

idk why im getting this wrong

[spare widget]

You have to solve the system

$\begin{bmatrix} 0 & 1 & -1 \ -1 & -1 & 0 \ 1 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ -1 \ 10 \end{bmatrix}$

[stoic python]

criver

[spare widget]

I am trying to figure out how to complete a matrix to full rank. I have a matrix of the form:

$M = \begin{bmatrix} L & A \ A & B \end{bmatrix}$

[stoic python]

criver

[spare widget]

where L is not full rank, but [L A] is full row-rank and [L; A] is full column rank

I am not sure whether it's relevant but A is symmetric

I want to pick B such that M is full-rank

Then I believe I have to choose B such that [A B] and [A; B] are respectively full row and column rank

I figured that B's kernel must be the pre-image of A's range and the pre-image of B's range must be A's kernel

Is there an (algorithmically) simple way of constructing such B?

If A has the eigendecomposition

$A = Q\Lambda Q^T$

[stoic python]

criver

[spare widget]

And let Q be chosen such that the first k eigenvalues are non-zero and the remaining are zero

then would B be given as:

$B = QDQ^T$

[stoic python]

criver

[spare widget]

where the first k eigenvalues are 0, and I have free (non-zero) choice for the remaining n-k

[drifting ravine]

Been stuck on this task for a while

Now this is the solution i could find, i just don't understand the first part marked in red:

[spare widget]

What is E_lambda?

[drifting ravine]

Just the eigenvectors for the corresponding eigenvalues

[wintry steppe]

This relates to row independence but its not an obvious truth in my opinion

column independence*

[spare widget]

They're showing that the intersection between E_lambda1 and W is {0}

technically the lhs says sum_i E_lambda_i, while the rhs says directsum_i E_lambda_i

So if you show that the intersections are only {0} then the equality holds

[brave cliff]

I'm confused about how they're doing what they do after finding what numbers to use for the cofactor

[nocturne jewel]

You remove the column and row that has 3, then check the position to determine the +-

Then compute determinant of the resulting matrix

[dull rose]

The answer to this question is Re(z) = 2 and Im(z) = 0, and i was wondering how to get those values, could anyone help explain? thanks.

[glad acorn]

z is a real number since it's an absolute value of a complex number

[dull rose]

oooh i see, thank you!

[wintry steppe]

Hi I have a linear algebra question

Could you have a dot product of a matrix? Or would it only work if that matrix was considered as a vector?

Let me give you an example to make my question clear to you.

So if I try to find the product of a matrix with dimension (2x1) with another matrix with the same dimension (2x1), it wouldn't be valid because the number of columns of the first matrix does not equal the row of the second matrix.

However, if I were to define the 2 matrix to be a vector, I would get a scalar value by dot product.

If that's the case, you could consider any matrix as a vector and make every matrix dot product valid.

I'm just so lost. Help me plz 😦

[dusky epoch]

it is possible to equip the vector space of all m×n matrices with an inner product

the ``standard'' inner product on $\bR^{m \times n}$, to my knowledge, is given by $$\ang{A,B} = \tr(A^TB).$$ under this dot product, the basis consisting of all matrix units (i.e. matrices with one entry 1 and the rest 0) is orthonormal.

[stoic python]

Ann

[lavish jewel]

the one ann mentions is indeed the standard one. the one you wrote, in contrast, is called frobenius inner product

[wintry steppe]

I need to prove that the eigenvalues of the Bessel functions are real. How can I do that?

[wintry steppe]

mind if i ask what course is that part of

[wintry steppe]

Bessel Equation. Mathematical Methods for Physics

batchelors level ?

Yes

help i posted a question in #help-13

[lavish jewel]

do you mean fixed points of bessel functions?

or eigenvalues of the bessel differential equation?

[wintry steppe]

Eigenvalues of the Bessel DE

[spare widget]

If I have A symmetric not full-rank, then would the following matrix have full row-rank?

$\begin{bmatrix} Q\Lambda Q^T & QDQ^T \end{bmatrix}, , A = Q\Lambda Q^T, , D_{ii} = \begin{cases} 1 & \Lambda_{ii} = 0 \ 0 & \Lambda_{ii} \ne 0 \end{cases}$

[stoic python]

criver

[spare widget]

And at the same time the following will have full column rank:

$\begin{bmatrix} Q\Lambda Q^T \ QDQ^T \end{bmatrix}$

[stoic python]

criver

[spare widget]

I think this is correct, but I am not sure how I would prove that

[fringe fjord]

Is Lambda diagonal? Is Q invertible?

[spare widget]

Yes Lambda is diagonal Q is orthogonal

A is symmetric, this is its eigendecomposition

D is also diagonal

My logic was that the kernel of A is nontrivial

And its subspace is associated with the 0 eigenvalues of A

I can associate to the 0 eigenvalues some eigenvectors in Q

[lavish jewel]

both will have full rank, then

[spare widget]

And then to the same vectors I can associate non-zero eigenvalues in QDQ^T

[spare widget]

Yes but how would I go about proving that

[lavish jewel]

factor out the matrices in a way that makes it evident

[spare widget]

And is this an iff scenario - i.e. are the only completions of the form QDQ^T

[lavish jewel]

should be able to diagonalize the overall matrix

though there might be easier ways

lemme see if i can come up with something before i start eating

[spare widget]

The overall matrix is rectangular, so are you suggesting SVD?

[lavish jewel]

one easy way in the first case is to notice you can equivalently write Q[L D]Q^T

now multiply in, for example, the Q on the left

[QL QD]D^T

since L and D are diagonal, QL and QD are simply Q with scaled columns

and conveniently, L and D have complementary entries

so the products QL and QD each have 0 columns where the other has a nonzero column

their images are therefore linearly independent

or more strongly, orthogonal complements of each other

after that you can show that QL has the same rank as QLQ^T

and similarly for QDQ^T

and you're done

you can do something similar when you stack them vertically, but the eigenvectors need to be handled more carefully

[spare widget]

Ok so change of basis for [L D], I tried to get something like that but couldn't get the matrices right, I didn't realise it was as simple as Q[L D]Q^T

[lavish jewel]

hmm?

what change of basis?

you mean Q Q^T, i guess

[spare widget]

yes

[lavish jewel]

it's a consequence of how matrix mult is defined

[spare widget]

I was thinking of it as a tensor product

Where each element gets transformed

[lavish jewel]

if you have AB, then each column in the result is Ab_i

[spare widget]

wait

[L D]Q^T won't work

That's where I got stuck last time too

[L D] has column # 2n, while Q^T has n rows

[lavish jewel]

then the product is not defined in the first place

let's see

the factoring needs a little mclovin

you'd stack two Q^Ts on top of each other

that works

[spare widget]

[L D] [Q Q]^T

[lavish jewel]

or if you prefer use kronecker products with identity mats

yeah

hmmm wait

[spare widget]

Won't this sum them up

[lavish jewel]

i think you HAVE to use the kron

[spare widget]

LQ^T +DQ^T

[lavish jewel]

[Q^T 0
0 Q^T]

this one works

which as you way is equivalent to a tensor product

$I \otimes Q^T$

[spare widget]

yes, that shoild wprk, and a similar thing for Q

[stoic python]

Edd

[lavish jewel]

hmm

Q doesn't need it

[spare widget]

Let me think it through

yeah, that should be ok

[lavish jewel]

the arguments work pretty much the same way

just had to be more careful

and something similar should work when you stack the columns

maybe easier

[spare widget]

would the matrix needed to make [A ?] full row rank always be of the above kind?

[lavish jewel]

no

[spare widget]

so what other options are there

[lavish jewel]

what exactly do you need?

this is a very restrictive case where the images are orthogonal

[spare widget]

sure, but A is symmetric

Assuming A is symmetric, my guess was that ? is always of the form QDQ^T

If [A ?] is to be fulk row-rank

Basically does the converse hold

[A B] is full rank -> B = QDQ^T

[lavish jewel]

no

[spare widget]

where A = QLQ^T

[lavish jewel]

making B an identity matrix works

you mean B not being symmetric?

i'm pretty sure any full rank A works

sorry, B

any full rank B

[spare widget]

no I mean if I know that [A B] is full row-rank and A = QLQ^T, then must B be of the form B = QDQ^T, but I guess it doesn't have to be

[lavish jewel]

yes, any full rank B works

and it doesn't have to be full rank either

and B with rank at least the size of the nullity of A

and with image complementing that of A

so if A is n x n and rank r

[spare widget]

well QDQ^T is not full rank, it's rank(B) = n - rank(A)

[lavish jewel]

B can be rank anything between r-n and n

[spare widget]

But I guess a higher rank is fine too

[lavish jewel]

and can have any structure

as long as its image complements that of A

and it doesn't have to be an orthogonal complement

[spare widget]

Ok, that makes sense. But if I want it to not have any linear dependence with A would I have to set it QDQ^T?

[lavish jewel]

no

compute the orthogonal complement of im(A)

make up whatever basis you want for that

and that's enough

[spare widget]

but isn't that exactly QDQ^T?

[lavish jewel]

no?

look

say Q is size 10 x 10

and it's the 1st n columns that we keep due to L

[spare widget]

Well it's made up of non-zero eigenvalues where A has 0 such

[lavish jewel]

take the remaining columns, 10-n of them

put them as columns of B

let the remaining columns be 0

and you're done

this matrix B is not of the form QDQ^T

[spare widget]

D_ii != 0 if L_ii = 0 and D_ii = 0 if L_ii !=0

[lavish jewel]

no

i'm gonna go eat, re-read what i wrote

the answer is no

[spare widget]

I think I get what you're saying

I failed to mention that [A; B] also has to be full column-rank, idk whether that changes anything, but I think it enforces that B is symmetric (namely requiring that [A B] and [A; B] are full rank)

[lavish jewel]

it can't be full column rank if A and B are square

it can only have col rank = row rank

so n, at most

not 2n

[spare widget]

[A; B] means A stacked on top of B, I agree that the row-rank can be at most n

[lavish jewel]

oh oops

anyway, the same analysis applies to that case

just this time with the rows, for simplicity

[spare widget]

I understand that, the question is whether these two constraints make it so that B = QDQ^T

[lavish jewel]

no

oh, the 2 together?

still no though

B can be any full rank matrix

for example

[spare widget]

Because I believe B has to be symmetric if [A B] and [A; B] are to be max possible rank

[lavish jewel]

it could be any random full rank mat

[spare widget]

what if I require that its image is the orthogonal complement of As image?

[lavish jewel]

then yes, by construction

[fringe fjord]

Isn't the vertically-stacked case just "transpose everything throughout" of the horizontally stacked?

[lavish jewel]

yep

hmm

maybe not

look at what tropo wrote

while i go eat

[spare widget]

yes that's correct

So it implicitly requires B^T = B

A is already symmetric

[fringe fjord]

Is B still QDQ^T?

[spare widget]

If I take the argument for [A B] then it follows for [A^T; B^T]

[lavish jewel]

i guess that's what they wanna know

whether B has to be symmetric

[spare widget]

If B is forced to be symmetric yes

[lavish jewel]

it doesn't in the individual cases

[spare widget]

But I think that the requirements force it to

That is

[fringe fjord]

I scrolled past a bunch of stuff and haven't found where the letter B was introduced, sorry.

[spare widget]

[A B] full row-rank, [A; B] full column-rank, and im(B) perp im(A)

B is any nxn matrix (A is nxn)

The question evolved to must B be of the form QDQ^T if A=QLQ^T,where D_ii = 0 if L_ii !=0 and D_ii !=0 if L_ii = 0

[fringe fjord]

Ah, so this is not the context of the discussion anymore?

[spare widget]

it is, in the sense that I want to understand whether this is the only case that this holds

[spare widget]

that is, if I require this

Then is the only possible B that satisfies those requirements B = QDQ^T

[fringe fjord]

Eh, I think I'll just shut up; I'm lost.

[spare widget]

Basically what is the form of B given these requirements is the question. I apologise for formulating it poorly.

My guess was that it is B = QDQ^T

And there are no other possibilities

A is symmetric with A = QLQ^T

[fringe fjord]

Wait can't you choose some other (possibly different) nonzero values in place of 1 on the diagonal of D and that should work just as well?