linear-algebra

[wintry steppe]

yea it contains 0

[spare widget]

A full set of equations of independent rows leads to the trivial solution {0}

Conversely 0 equations lead to the full space

e.g. 0 dot x = 0

Stuff inbetween, i.e. A with r independent rows leads to a solution with n-r dimensions

Notably if you have Ax = 0

And Bx= 0

And you put those in a system

the solution space os the intersection of the solution spaces of Ax= 0 and Bx= 0

Conversely if you have the basis vectors being the rows in A and the second set being the basis vectors in B

then putting those together and removing linearly dependent vectors gives you a basis for the sum

So sum the sum of bases has as a counterpart the intersection of the solution spaces

[wintry steppe]

hmm

interesting

[wintry steppe]

is this true? and if yes, how do you prove it?

and whats the volume for points (x1,y1,z1) till (xn,yn,zn) in that case?

[wooden iris]

For 5a are you performing gauss jordan once you put it in matrix form?

[lime zinc]

Just make A Matrix with coffiecents of given system of linear equation x is you known vector and b is the vector with constant parts

[zinc timber]

yes it's true

[serene tide]

is there a standard way to approach a problem like this

would you just solve for k in a typical system or is elimination better

tried both and both seem really messy for part b

[dusky epoch]

just do Gaussian elimination, no?

[lavish jewel]

you could do row reduction

[dusky epoch]

you'll eventually end up at a quantity that you need to divide by

[hollow finch]

i think you could also take the determinant. if det(A) is nonzero youll have a solution regardless of b.

yeah the determinant works just fine. i checked and theres nothing you get from b you dont get from ensuring det(A) is nonzero

[lavish jewel]

that seems like double the work though

[hollow finch]

really? the diagonal method turns it into ~4 arithmetic operations

[lavish jewel]

what i mean is that you then anyway have to go ahead and analyze the 0 det solutions to distinguish infinitely many sols from no sols

[hollow finch]

take the determinant and youll see its very obvious. its not like a quadratic or something, its something you can solve by inspection

[lavish jewel]

yes but 0 det doesn't immediately mean no sol

or many the augmented matrix is really easy to simply inspect? is that what you mean?

cuz just looking at the det ignores the vector b

[hollow finch]

well if the det is zero it wont be a unique solution, it will necessarily be infinitely many. but row reduction is the way to go if you also want to do part b since that will give you both

well

actually you could just plug in the bad k value and see if its consistent. but that would require row reduction anyway so w/e

[lavish jewel]

that's what i meant

you anyway have to do (r)ref at the end to check whether it was no sol or infinitely many

so double the work

[spare widget]

sum of triangle areas

One has to be careful about the winding order though and to make sure the setis convex or something

[wintry steppe]

ordering as in (x_i , y_i) has to be connected to (x_i+1 , y_i+1) and also (x_1 , y_1) has to be connected to (x_n , y_n), or something else?

[spare widget]

I think that's already implicit in the problem

also since you take the absolute value I guess the winding order doesn't matter

the issue is if you have something non-convex

although maybe that isn't an issue either

it isn't if the non-convex parts get subtracted out due to the sign I guess

you have to draw some examples though

basically form the triangles (0,0), (x_i, y_i), (x_{i+1}, y_{i+1})

the determinant multiplied by 1/2 there gives you the area

but do check the non-convex cases

just draw some examples

[leaden tide]

Hey, I have a quick question with inner product spaces.

Let E be an inner product space with its associated norm, let $(v_i){1\le i\le n} \in E^n$ such that $\forall 1\le i \le n, |v_i| \le 1$. Let $(p_i){1\le i \le n} \in [0,1]^n$. Is it true that $|\sum_{i=1}^n p_iv_i| \le \sqrt{n}$ ?

[stoic python]

Syst3ms

[zinc timber]

no not necessarily

it's true however if you assume $\sum_{i=1}^n p_i = 1$

[leaden tide]

Crud

[zinc timber]

which is a direct consequence of triangle inequality

[leaden tide]

Okay, let me post the full problem

[zinc timber]

like for easy contradiction, take v_i = e1 and p_i=1

[leaden tide]

right

With the previously introduced notation, let $v = \sum_{i=1}^n p_i v_i$. Furthermore, for $J \subseteq [![1,n]!]$, let $V_J = \sum_{i\in J} v_i$.

[stoic python]

Syst3ms

[leaden tide]

Show that there exists $J \subseteq [![1,n]!]$ such that $|v-V_J| \le \frac{\sqrt{n}}2$

[stoic python]

Syst3ms

[zinc timber]

what's that notation? [[1, n]]? any sequence from 1 to n?

[leaden tide]

[[1,n]] is the set of all integers between 1 and n, inclusive

[zinc timber]

yes ok

[leaden tide]

Some indications and previous questions

The previous question was that, given n vectors of norm 1 there exists $(\varepsilon_i){1\le i \le n} \in {-1,1}^n$ such that $|\sum{i=1}^n \varepsilon_iv_i| \le \sqrt{n}$

[stoic python]

Syst3ms

[zinc timber]

there's no restriction on pi's?

[leaden tide]

Between 0 and 1, and the v_i have norm ≤ 1

Moreover, apparently the intent of the problem is to show that probabilities are a method, because they can crop up in places where it seems like they don't belong.

So supposedly a neat solution to this problem involves probabilities (and associated theory). For the record, we only studied discrete probabilities.

[zinc timber]

why do I have a feeling that this may not be true

like take pi = 0.5 and

and vi = e1 then

[leaden tide]

e_1 being some vector of norm 1 ?

[zinc timber]

max value of |v-Vj| = n/2 |e1| = n/2 >sqrt(2)/2 given large enough n

[zinc timber]

ye

I still think you need the restriction sum pi = 1

[leaden tide]

Well no, you can just choose J of cardinality n/2 (ish if n is odd) and that will work

The point is to show that there exists a J

[zinc timber]

did you understand this example?

[leaden tide]

Sec

[zinc timber]

actually

[leaden tide]

There's no problem here

If J is of cardinality n/2, |v-V_J| = 0

If n is odd and J is of cardinality (n+1)/2, |v-V_J|=|e1/2|= 1/2 ≤ √2/2

[zinc timber]

ye but let's change e1 to ei's

say for R3, e1 e2 and e3

[leaden tide]

Sure

And then, p_i=1/2 again?

[zinc timber]

yes

[leaden tide]

Then v is the center of a cube of side length 1

[zinc timber]

you can take R^n and ei's to be std basis and pi=1/2 again

[leaden tide]

With the various V_J being the vertices of that cube

the distance from the center to a given vertex is √3/2

So not only is the inequality true, in this case it's true for all choice of J

[zinc timber]

for n=3 yes

[leaden tide]

Well, the distance is still √n/2

Taking J=[[1,n]] for example

$|V_J-v|^2=|\sum_{i=1}^n \frac12e_i|^2 = \frac14\sum_{i=1}^n |e_i|^2 = \frac{n}4$

[stoic python]

Syst3ms

[leaden tide]

Taking the square root you end up with |V_J-V| = √n/2

[zinc timber]

hmm

[leaden tide]

This is an oral exam problem

So i sure hope it is true

[zinc timber]

now I've started to believe it as well

[leaden tide]

In other words, if you consider balls of radius √n/2 around each vertex of the parallelotope, they cover it entirey

Anyway, if probabilities are to be involved, I surmise either Markov's or Bienaymé-Chebyshev's inequality will be used at some point

[subtle gust]

gave up on doing this question

how do i do it

tried column operation but didn't work /:

[wintry steppe]

what's on the right of the screenshot ?

you can try decomposing by multilinearity, that's way simpler than computing

[fringe fjord]

Split it into a sum of two determinants along the sum in one of the columns, then use column operations on each of those to see that it must equal the determinant on the right.

[subtle gust]

there isn't anything

that's the whole question

[subtle gust]

confused af rn

how can we split it into a sum of two determinants if det(A+B) =/ det(A)+det(B)

[fringe fjord]

You have (in simplified ad-hoc notation) on the left hand side |b+c, c+a, a+b|. Split that into |b, c+a, a+b| + |c, c+a, a+b| and do column operations on each of those.

[subtle gust]

ummm what's ad hoc notation /:

we only covered 3 things

[fringe fjord]

Notation that I invented for the purpose of writing that post.

[wintry steppe]

I think you could just repeat that step with the two new determinants and end up on the result, using the fact det() is alternating (a lot of the terms are = 0)

[subtle gust]

ok so i think i'm only allowed to do 3 things

because those are the only things we covered so far

row operations

column operations

swapping two rows/cols

[fringe fjord]

Haven't you learned that a determinant is a linear function of each column separately?

[subtle gust]

with those 3 things how can i prove this.....

[subtle gust]

ummmm nope ?

[fringe fjord]

Curious. What is your definition of determinant in the first place?

[subtle gust]

a scalar that is associated with square matrices ....

[fringe fjord]

That's not a definition.

[subtle gust]

this is a midterm soo... we still haven't covered a lot of the content

maybe it will be introduced more formally later on

[fringe fjord]

You can't possibly be asked to do stuff about determinants without having a definition of it.

[subtle gust]

well we only use the properties....

and how to evaluate it

nothing more

[subtle gust]

here for ex we maybe could use the fact that det(A)=det(A^T)?....

[wintry steppe]

you can do it without linearity

[subtle gust]

idk i'm DONE

my midterm is in exactly an hr :/

[wintry steppe]

you have |b+c, c+a, a+b| and notice that performing column operation, you can get a+b - (b+c) = a-c, yielding |a-c, c+a, a+b|. Now you can do a-c + c+a giving you 2a, so far your det is |2a, c+a, a+b|. Now factor out the 2 and use the first column (a) to get rid of the other a's in the 2 last columns

[fringe fjord]

Cool.

[wintry steppe]

that does kinda use linearity by factoring out the 2 though

[fringe fjord]

Scaling a column sometimes counts as a "column operation" too.

[subtle gust]

wait what column operation did we use

-C1+C2--->C2?

[wintry steppe]

C3 - C1 -> C1

the first one

then C1 + C2 -> C1

[subtle gust]

yeah got it

[wintry steppe]

doesn't it lead to a sign mistake though.....

[subtle gust]

tysmmmm @wintry steppe and @fringe fjord appreciate the help

[subtle gust]

huh

[fringe fjord]

C3 - C1 -> C1 has a sign change in col 1.

[wintry steppe]

oh okay so the sign is fixed by swapping the c and b columns in the end 🙂

[leaden tide]

@zinc timber I managed to work it out

And indeed, probability yields a rather neat answer

The key observation is that the p_i can be seen as probabilities of a bernoulli distribution

[zinc timber]

can u share the sol

[leaden tide]

Sure

[stoic python]

Syst3ms

[leaden tide]

Where the arrow means "follows the law" (it's the notation here)

These variables are taken independant

[zinc timber]

looks like endomorphism

[stoic python]

Syst3ms

Syst3ms

Syst3ms

Syst3ms

Syst3ms

Syst3ms

Syst3ms

[leaden tide]

But notice that for all p ∈ [0,1], p(1-p) ≤ 1/4

[stoic python]

Syst3ms

Syst3ms

[leaden tide]

(Notice the values that S can take are precisely the V_J)

@zinc timber Done and cleaned up

[zinc timber]

yes reading

[leaden tide]

slight err, the last inequality is large

[grave kettle]

just learnt that matrices describe linear transformations

[leaden tide]

Yeah, that is quite helpful for many things

[grave kettle]

what's the use of studying non-square matrices?

[leaden tide]

I guess studying transformation between spaces of different dimensions?

But to be blunt we barely see non-square matrices in class

Anyhow, this probability proof is rather elegant, very neat

My prof did this a while ago, he solved a random integration problem with probability

[dusky epoch]

maps between spaces of different dimension

[wintry steppe]

square matrices are usually more interesting anyways

-you can compose them with themselves; linear dynamics
-the theory of modules over PIDs apply to them; jordan, rational canonical, diagonal forms
-eigenvalues, determinants for sweet geometric interpretation
-we like to approximate by square matrices anyways

[lavish jewel]

svds, singular values, low rank approx and rank 1 decomps with overcomplete frames

😌

[zinc timber]

you can svd, singular value for square matrices as well

+1 for square

[lavish jewel]

these generalize nicely to n-way arrays that are often used to represent multilinear transformations, too

these are more general, ryu, so i'd call them a special case

[zinc timber]

non linear dynamics also uses square matrices apparently

[spare widget]

what is that about approximations with overcomplete frames?

[lavish jewel]

one can set up rectangular matrices A of size m x n with m << n so that in special cases, there exist unique solutions x to Ax = b

(given special restrictions on x)

[karmic oracle]

Is there a standard name for B^T A B (where I'm sort of conjugating, by by a transpose instead of an inverse?)

[ember vessel]

wdym standard name?

[spare widget]

Wouldn't this be similar to change of basis for a (2,0) or (0,2) tensor depending on what B is

[lavish jewel]

off the top of my head, for the general case, no

if B and/or A have any special property, perhaps

[spare widget]

If B is a change of basis matrix then B^{-1}AB gives you the transformation for a (1,1) tensor A

Where A is square

If it's symmetric you get the B^T A B thing, it's similar if it is a (2,0) tensor (e.g. metric tensor), then: u^TGv = p^TB^TGBq, where u = Bp, v = Bq

[subtle gust]

Just got back from my linear algebra midterm 💀💀

I feel like i kinda did most questions correctly ?

But there was this question that i spent literally thirty minutes staring at

Let X be a 3x3 mateix
If XX^T is skew symmetric prove that X=O3×3

I could only prove that XX^T is 0.....

[grave kettle]

why are the requirements of linear transformation the way they are

[spare widget]

I am guessing motivated from R^n and then generalized

[lavish jewel]

so you have that XX^T is 0, moash

then you consider that the diagonal entries are equal to the 2-norm of the rows of X

so that all the rows of X are 2-norm 0

the 2 norm is positive definite, so that each row is then 0

wdym by requirements of linear transformation, chromium?

like the definition? the conditions that linear maps satisfy?

[grave kettle]

definition

T(a + b) = T(a) + T(b)

c T(a) = T(ca)

[lavish jewel]

yeah, it's a more general and abstract way of expressing the nice properties of R^n, so that it also applies to other stuff

i seem to recall systems of equations being one of the original motivation behind early linalg tools, so that seems right

[grave kettle]

why are they defined like this

[lavish jewel]

because they have nice consequences

[karmic oracle]

I would say, fundamentally, a vector supports two operations: addition and scalar multiplication. So a linear transformation is a map which respects both of those operations (property one you list is about vector addition, and property two is about scalar multiplication). Now, you might rightly argue this is just pushing the question back to another question which is more fundamental: Why are those the two operations we consider on vectors?

[grave kettle]

please do explain

[karmic oracle]

I think when you get to "Why do we define vectors with these two operations?" you get a lot of different answers which are likely all very unsatisfying. The "because it ends up being useful"/"because it ends up developing a rich theory" sorts of explanations are one flavor, or the historical flavor of "because we thought of these things from physics where both are very natural" will come up to. But ultimately, all mathematical definitions are arbitrary. You can define them however you want. Which ones survive is a function of Darwinism more than anything. And linear algebra is about the end of subjects where mathematicians actually agree how to define things. Pretty quickly after that you'll find that definitions start being more... impressionistic.

So in short, I think it is a very good question, but I also think it is a question to which no very good answer exists. "In math, we don't understand things, we just get used to them," as the saying goes.

[lavish jewel]

often times it goes something along the lines of "oh, these things share something in common and have nice properties. does this generalize?"

and this was one thing that did nicely

[karmic oracle]

the completely ahistoric explanation is that the definition of linear transformation mimics the properties of matrix multiplication. But linear transformations predate matrices.

[grave kettle]

3b1b said the intuitive way to understand it is

‘it’s defined such that all lines remain lines and the origin is fixed in place’

it’s a bit goofy and doesn’t seem related with said properties

[lavish jewel]

well

that makes sense if by "line" you mean a line passing through the origin

then in this sense, a line is a scalar times a vector, and its image under linear transformation is the same scalar times a new, transformed vector

so, a new line passing through the origin

this only makes sense if your "vector" has an associated "direction", which in general is not required

so it only makes sense for vector spaces that inherently have this additional structure

so this seems like a special case of preserving scalar mult and vector addition

[grave kettle]

so it’s simply designed around preserving these?

or are there more

[lavish jewel]

that's a direct consequence of the definition

f(cu + v) = cf(u) + f(v)

[grave kettle]

what is

[subtle gust]

Ummm i don't think we covered that

We didn't study norms

[lavish jewel]

did you see dot products

[subtle gust]

I'd say try to prove it using what we covered /:

[lavish jewel]

the preservation of scalar mult and vector addition

[subtle gust]

Nope

[lavish jewel]

really? no dot products?

[subtle gust]

We only covered the basic stuff

[lavish jewel]

you did see matrix multiplication though?

[subtle gust]

The things you'd see in 4 weeks of lin alg

[subtle gust]

Yeah

[lavish jewel]

how did you see matrix mult and not scalar products

[grave kettle]

the below definition of what

[subtle gust]

I could only prove that AA^T=0

[subtle gust]

We didn't do scalar products....

[lavish jewel]

i guess you can still show it just by writing it out as a sum

[subtle gust]

I think we'd seethis later in the course idk

Ok so here's what i did

[lavish jewel]

you can see that the diagonal entries of XX^T have a special form. if X has a row r with components r1, r2, r3,...

[subtle gust]

I didn't use sums 💀💀💀

[lavish jewel]

there is a diagonal entry in XX^T of the form r1^2 + r2^2 + ...

and since XX^T = 0

you have a sum of squares that equals 0

[subtle gust]

How did we get that

Literally spent like half an hr to reach AA^T=0

[subtle gust]

Oh yeah

[lavish jewel]

well, the definition of skew symmetric is that W^T = -W

[subtle gust]

I thought abt this but i ignored it 💀

[subtle gust]

Yeah

[lavish jewel]

and (XX^T)^T = XX^T

so XX^T = -XX^T

[subtle gust]

True

Again true

[lavish jewel]

so XX^T must be 0

since matrix equality is elementwise

[subtle gust]

Oh ..

Well

[lavish jewel]

if c = -c, either 1 = -1, or c = 0

[subtle gust]

I moved -Xx^T to the other side

[lavish jewel]

same thing

[subtle gust]

Yeah

I ended up with the same result

XX^T=0

But i couldn't continue

[lavish jewel]

at any rate, the trick is to look at the diagonal elements of XX^T

those contain only sums of squares

you can use this to argue that each row of X must be 0

[subtle gust]

Yeah all of them were a^2+b^2+c^2

Maybe i'd get half the marks... 💀

Well i'm just glad it's over

[grave kettle]

this is a definition?

[lavish jewel]

it's a consequence of the 2 properties of linear maps

[grave kettle]

i still don’t really get this

[lavish jewel]

a line is of the form cv for a scalar c and a vector v in R^n

if we have a linear function f, then f(cv) = cf(v)

and f(v) is another vector

so cf(v) is a new line

and due to linearity, this generalizes to sums of scaled vectors

[grave kettle]

what about the f(u + v) = f(u) + f(v) thing

[lavish jewel]

i already explained that above

[grave kettle]

where

[gritty swift]

here

if you have a line between vectors v,w [a is a scalar]
av + (1-a)w
after the transformation T
aT(v) + (1-a)T(w)

its still a line just between new vectors now

[grave kettle]

between?

[gritty swift]

two vectors define a line, right?

[grave kettle]

?

tip of the vectors?

[gritty swift]

yes

[grave kettle]

isn’t that c (v - u)

[gritty swift]

no because all vectors are rooted at the origin

[grave kettle]

??

what does that have to do with origin

[gritty swift]

the difference vector is rooted at the origin

if you want the line between v and w you want cv + (1-c)w

think of the vectors like points

since they're rooted at the origin in LA

[grave kettle]

why isn’t this also an ok representation

[gritty swift]

https://www.desmos.com/calculator/g2jaypmmf4

[grave kettle]

just scale the (u - v)

[gritty swift]

see here, mine is the green line, yours is purple

to make your method work you'd need to add an offset
v + c(u -v)
which would simplify to
cu + (1 - c)v

[grave kettle]

oh

[gritty swift]

does this make sense now?

[grave kettle]

i still don’t understand whether this is yes or no

or what

[gritty swift]

wdym "designed"

they're equivalent

[grave kettle]

?

[gritty swift]

"all lines remain lines" and T(av + bw) = aTv + bTw are the same thing

[grave kettle]

original question was why requirements for linear transformations are they way they are

[gritty swift]

because... that's what linear means

I mean, you could have an affine transformation that doesn't fix the origin

[grave kettle]

what does it mean

[gritty swift]

but affine transformations aren't as nice as linear ones because the whole theory of nullspaces doesn't work (at least, not by default)

there are lots of mathematical transformations, after studying all of them linear ones tended to lead to especially nice theorems. hence linear algebra

[grave kettle]

so what does it mean lol

i’m now unconvinced that linear algebra is called linear because it ‘studies linear equations’

[gritty swift]

a transformation T is linear if T(ax + by) = aT(x) + bT(y) for scalar a,b and vector x,y

[gritty swift]

linear equations and linear transformations are the same thing

linear transformations are the same as matrices which are the same as linear equations

[wintry steppe]

Given the basis for two subspaces I need to find a basis for their intersection.
I solved for linear combination of the basis of first subspace = linear combination of the basis of second subspace and got some relation between the coefficients, I also know that the dimension of intersection is 2, now should I just pick two linearly independent vectors from the solution set in order to get a basis?

[gritty swift]

for example: if A is a matrix, x is a vector and b is a vector
Ax = b
is a system of linear equations, but A is also a linear transformation

this will always be solvable if Ax = 0 implies x = 0 for example

[gritty swift]

yes

[gritty swift]

finish the 3b1b series hopefully it makes more sense, https://youtu.be/TgKwz5Ikpc8 might explain it to you

[wintry steppe]

what if the dimension of the intersection is n, how can we pick n independent vectors from the solution set?

or will they automatically be independent?

[gritty swift]

they won't be automatically independent, you've got to pick them that way

in matrix form if you have Ax = b and want n solutions for x, you can pick a particular solution x_p all the other solutions will be x_p + x_n for x_n in the nullspace. then the problem is reduced to finding a basis for the nullspace then adding x_p

though you can probably do it directly for small n

actually you can do it more directly i think i havent done computational LA for a bit

put the system in reduced row echelon form then i think you can eyeball independent solutions

[gritty swift]

actually for 2d they will be

since Ax = A(cx) = c(Ax) is only true if c = 1 or Ax = 0

I think

[wintry steppe]

I need to look up nullspaces and stuff then

haven't studied those yet

[gritty swift]

for your case you can eyeball it

and check manually that they're independent

[wintry steppe]

yea but I think a more general approach would be better

for 2d its still only independent vectors, but still

I'll read about nullspaces and stuff

[gritty swift]

https://youtu.be/VqP2tREMvt0

[wintry steppe]

thanks

[fringe fjord]

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

[viral magnet]

@grave kettle you should do this problem
https://cdn.discordapp.com/attachments/903481105888452609/949104624751747072/Screenshot_29.png

spoilers
#help-22 message

sauce
#help-22 message

[slow silo]

hey im a bit confused on the start of this question it says we are given a collection of $n^2$ matrices $E_{ij}$ whose components are given by $(E_{ij}){kl}=\delta{ik} \delta_{jl}$, what would a matrix of this type look like, im finding it hard to visualise :x

[stoic python]

Wingardium Matrix-Leviosa

[gray dust]

E_ij is the matrix with 1 at the (i,j) slot & 0 elsewhere

[slow silo]

is that just an identity matrix?

[gray dust]

no, the identity has 1 on the whole diagonal

[slow silo]

what do you mean (i,j) slot?

[gray dust]

row i, col j

[slow silo]

so for a 3x3 matrix

we would have a 1 where?

im so confused :x lol

[gray dust]

u need to specify i,j first

[slow silo]

row 1 , col 2

[gray dust]

E_23 is the matrix with 1 at the (2,3) slot & 0 elsewhere

[slow silo]

oh righttt

and $E_{12} is the matrix $\begin{pmatrix}
0 & 1 & 0\
0 & 0& 0\
0& 0& 0
\end{pmatrix}$

[stoic python]

Wingardium Matrix-Leviosa
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gray dust]

yes

[slow silo]

so whats the extra part

we have our matrix (E_ij)_kl whats the "kl"

[gray dust]

used to define the matrix entries

for each matrix A u need 2 indices to define its entries

usually theyre ij, so like A_ij

but ij is already used to specify where the 1 goes

[slow silo]

so now we are specifying where other 1's go?

[gray dust]

so they use kl, (E_ij)kl

u need to know what the deltas mean

[slow silo]

yeah kronecker delta

[gray dust]

yeah so when defining E_ij

[slow silo]

1 if i=j 0 otherwise

[gray dust]

ij are fixed

[slow silo]

yeah

[gray dust]

then u go thru each combo of kl

[slow silo]

right right

so we need to have basically diagonal entries for this to work?

to satisfy i=j and k=l?

err

[gray dust]

u lost me

$(E_{ij}){kl}=\delta{ik} \delta_{jl}$

[stoic python]

RokabeJintaro

[slow silo]

yeah mb i was thinking of it wrong

[gray dust]

when is this 1? 0?

[slow silo]

so for our case

say E_23

we get delta_2k delta_3l

yeah im lost

how does a matrix = this kronecker delta

k has to be 2 and l has to be 3

[gray dust]

for it to be 1

[slow silo]

yeah

[gray dust]

what if k!=2 or l!=3

[slow silo]

what do you mean by !=

[gray dust]

not equal

[slow silo]

oh right

yeah its 0

oh its picking out the matrix entries of the matrix E_ij

[gray dust]

E_23 is the matrix with 1 at the (2,3) slot & 0 elsewhere
do u see how it matches this

[slow silo]

is that right?

[slow silo]

oh its picking out the matrix entries of the matrix E_ij that we have chosen*

does that make sense?

[gray dust]

i mean in general this is how one compactly defines matrices

for a matrix A

u take 2 dummy indices, usually ij, in order to define each entry A_ij

eg each entry of the transpose of $A$ is $A^T_{ij}=A_{ji}$

[stoic python]

RokabeJintaro

[slow silo]

right right

[gray dust]

each entry of the sum of matrices $A,B$ is $(A+B){ij}=A{ij}+B_{ij}$

[stoic python]

RokabeJintaro

[gray dust]

same thing here

its just we already fix ij

so we use kl instead as dummy indices when defining E_ij

[slow silo]

that makes sense

[gray dust]

so when u work out the definition of E_ij in general (similar to how u do E_23) u should get

E_ij is the matrix with 1 at the (i,j) slot & 0 elsewhere

[slow silo]

ok that makes sense

So we have say

for n=3

$E_{ij}=\begin{pmatrix}
E_{11} & E_{12} & E_{13}\
E_{21} & E_{22}& E_{23}\
E_{31} & E_{32}& E_{33}
\end{pmatrix}$

[stoic python]

Wingardium Matrix-Leviosa

[gray dust]

E_11 is a matrix not a number

[slow silo]

but going through each combo of k and l we have to pick that they are the same as i and j or we get 0

oh

oh yeah

i think im too tired for matrices lol

[gray dust]

as said before

E_ij is a matrix

(E_ij)kl is the (k,l) entry of E_ij

[slow silo]

right right yeah sorry

a completely out of the blue question but how does this build a basis for gl(n,k)?

[gray dust]

write matrices in terms of the E_ij

[slow silo]

does the lie algebra gl(n,k) contain matrices that have det non zero?

[gray dust]

yes

[slow silo]

so we just have to write a matrix that has det non zero using (E_ij)_kl?

[gray dust]

again, E_ij is the matrix

[slow silo]

then how does a matrix with 1 entry have det non zero?

[gray dust]

oh this is gl

gl isnt a vector space

[slow silo]

$\mathfrak{gl}$

[stoic python]

Wingardium Matrix-Leviosa

[slow silo]

this one

wait it isnt a vector space?

[gray dust]

invertibles arent closed under addition

[slow silo]

i thought if we have GL(n,K) we can go to the tangent space of the lie group

which is a vector space

[gray dust]

ok yeah its a lie group of dim n^2

[wintry steppe]

\frak{gl} is the set of all n by n matrices

GL is the invertible ones

[gray dust]

for a sec i was thinking of all nxn matrices, which the E_ij IS a basis of

[slow silo]

ahhh ok

how did you know that its the set of n x n matrices

is there a way to work this out?

[wintry steppe]

of course

[slow silo]

great

how do we do it 😄

[wintry steppe]

use the matrix exponential

[slow silo]

oh yeah

do you mean the inverse map

of the matrix exponential?

because the exp: g --> G

[wintry steppe]

no

i just mean the matrix exponential

the matrix exponential on M(n, R) doesn't even have an inverse

[gray dust]

$e^A=\sum\frac{A^k}{k!}$

[stoic python]

RokabeJintaro

[wintry steppe]

that one

if you're doing anything with matrix lie groups you need to know this

[slow silo]

yeah yeah i know this one 😄

[slow silo]

is this because of the 0 matrix?

[slow silo]

and we just get 0/0 undefined?

[wintry steppe]

the 0 matrix isn't in GL(n, R)

[gray dust]

wdym

[slow silo]

i thought you said the matrix exp on M(n,R)?

[wintry steppe]

the matrix exponential is a map M(n, R) -> GL(n, R)

to show that $$\mathfrak{gl}(n, \bR) = M(n, \bR),$$ just show each inclusion. the only non-trivial part is showing that any $A \in M(n, \bR)$ is showing that there's a curve in $GL(n, \bR)$ through the identity whose tangent vector is $A$. the matrix exponential gives this curve to you

[stoic python]

TTerra

[wintry steppe]

that curve is \exp(tA)

[slow silo]

right right

this is ringing bells

right yesss

GL(n,R) is a contained in M(n,R) and its an open set so its tangent space at any g in GL(n,R) is just M(n,R)

i think i need to sleep, im getting confused on basic matrix stuff

thank you so much for your help TTera and Rokabe

[wintry steppe]

this also works lmao

but the matrix exponential thing is good to keep in mind, because you're going to want to use it or something similar when computing lie algebras of other matrix groups

[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.

[slow silo]

yeah i will keep it in mind, thank you so much!:)

[hardy inlet]

Hello again Kings 👑
I'm trying to think what T² = I means

[wintry steppe]

it means T \circ T is the identity

[hardy inlet]

thats the definition of T² ye. But what does this say about T?

I dont have any idea how to make the identity from a square, other than the identity

but whats this -1 EV having to do with anything

i like half proved that if it is an EV that T²=I, but idk what to do about not -1

[wintry steppe]

it being an eigenvalue does not imply that T^2 = I...

it implies that on the corresponding eigenspace, sure, but not everywhere

one way i can think of is to figure out what the eigenvalues of T can be, and to diagonalize it (why is it diagonalizable?)

it's hard to suggest a way to proceed with this kind of question without knowing what you know

because there are quite a few ways to approach it imo

the fastest proof i can think of uses the minimal polynomial

[fair siren]

quick question about matrieces what does the curly braces mean

is just multiplication?

[fallen karma]

Is cross product an example of a more general type of operation?

[wintry steppe]

in a few ways

it's a lie algebra structure on R^3

there's a more general "cross product" of n - 1 vectors on R^n

wedge product

to name a few

[fallen karma]

Oh ok

Neat

Cross product. It's not associative. I want to say it's anti commutative

Iirc

[wintry steppe]

correct

[patent tendon]

2+2= airplane

[hardy inlet]

I found an outline of the proof online, fluffed it up to show the missing steps, and got this. Dunno how we can split T^2-I tho

[wintry steppe]

i can't follow this after the "for some w"

[fallen karma]

Eigenspaces🔥

[wintry steppe]

have you tried doing the exercise yourself?

[hardy inlet]

since (T+I)[(T-I)v]=0, then either (T+I)v = 0 or (T-I)v = 0

[wintry steppe]

ah, nevermind, i see what you're getting at

you're saying that if (T - I)v is non-zero, then it's an eigenvector of T with eigenvalue -1, a contradiction

[hardy inlet]

I guess I didn't explicity say "Case (T-I)v = 0

[wintry steppe]

this proof is fine logically

presentation could be improved but it works

T^2 - I = (T + I)(T - I) is a trivial equality

[hardy inlet]

well since (T+I)w = 0 implies that lambda = -1, then we know (T-I)w MUST equal 0

and the last temp of
Tv = Iv, why are we allowed to say T=I? Is that some identity too?

[wintry steppe]

if Tv = Iv for all v then T = I

[hardy inlet]

ahhh i have to specify FOR ALL v

okay

ima type it up real quicky

[gray dust]

in general thats how we define function equality

f=g if f(x)=g(x) for all x

[wintry steppe]

||the minimal polynomial of T divides x^2 - 1 = (x - 1)(x + 1) and cannot be divisible by x + 1, so x - 1 annihilates T, meaning T = I||

[hardy inlet]

it isn't divisible by x + 1 because if λ = -1 then (-1) + 1 = 0 and DIV0 error?

[wintry steppe]

what

[hardy inlet]

u said that poly wasn't divisible by x+1

[wintry steppe]

because -1 is not an eigenvalue

(not a root of the minimal polynomial)

[hardy inlet]

x+1 → x = -1 okay yeah i see

how long have u been doing this 😳

[hardy inlet]

wait so does difference of squares apply to transformations?

[wintry steppe]

as you should show, since you're uncertain, yes

i took linear algebra three years ago

[hardy inlet]

i guess u can just distribute it right, T²+TI-TI-I² = T^2 - I

[wintry steppe]

T^2 - I.

[hardy inlet]

ah

so is -I² saying -(I²)

wait thats basic math

nvm

xDDD

[wintry steppe]

[gray dust]

theres also the implicit use of the fact I commutes with every map

[hardy inlet]

is there a name for the equality?

its not difference of squares is it

[wintry steppe]

it is.

[gray dust]

try it for (A+B)(A-B)

[hardy inlet]

yeah it ends up just being T²+TI-TI-I² with replacing A and B

[gray dust]

try again

[hardy inlet]

the fact that its on a map feels weird

[gray dust]

think of em as matrices

[hardy inlet]

(A+B)(A-B) = A² - AB + AB -B² = A² - B²

[gray dust]

no

the inner term is BA

[hardy inlet]

oh shoot associativity

not a thing

[gray dust]

COMMUTATIVE

matrix multiplication associates but doesnt commute

[hardy inlet]

sorry i always get them backwards

anyways, A² - AB + BA -B²

can that be simplified?

[fallen karma]

Who wants to talk about dual spaces

[gray dust]

in general no

if A,B commute then BA=AB

u get A^2-B^2

[hardy inlet]

ok and since identity is commutative (able to commute/move freely) then it holds on I

[gray dust]

we dont say I is commutative

[hardy inlet]

TI ?

[gray dust]

we say I commutes with every matrix

[hardy inlet]

ah that makes more english sense

[gray dust]

this is a lesson in algebra where things dont always commute

[hardy inlet]

[wintry steppe]

go on

[hardy inlet]

is this sound?

[wintry steppe]

ABv = 0 doesn't imply that one of Av or Bv is zero

i suggest not copying proofs you find online

[hardy inlet]

that is a good suggestion

Does it imply that (T-I)v = 0 or (T+I)[(T-I)v] = 0

i need the composition right

[wintry steppe]

it certainly implies (T + I)((T - I)v) = 0, and that's all you need for the proof

er, (T - I)((T + I)v) = 0, but this is alright since T - I and T + I commute

[hardy inlet]

(T + I)((T - I)v) = 0 \land (T - I)((T + I)v) = 0

im not sure how thats enough for the proof

[wintry steppe]

read over the proof you found online, then

because it's either wrong or you misunderstood it

[hardy inlet]

i dont like that one D:
was using it as a starter piece

[fallen karma]

The dual map of T

If T has matrix representation A

Then T' has A^T

Because dual map T': W'-V' is...

[wintry steppe]

[hardy inlet]

(T + I)((T - I)v) = 0 \land (T - I)((T + I)v) = 0 so either ((T - I)v) = 0 and ((T + I)v) = 0? and the first equation violates our assumption of no -1 ev, so solving the second we get T=I right?

[sick sandal]

T* *!

[wintry steppe]

NO!

ABv = 0 DOES NOT IMPLY Av = 0 or Bv = 0!

[hardy inlet]

right but i thought since we had commuted the T+I and T-I to form both that would work

[wintry steppe]

even if A and B commute, NO

[fallen karma]

Oh yes, T'(p)=p o T

So I input a functional on W, and it outputs a functional on V

[wintry steppe]

that's its definition, yes

[fallen karma]

But dual spaces

Are they all that

[hardy inlet]

im trying to understand how

it certainly implies (T + I)((T - I)v) = 0, and that's all you need for the proof

[wintry steppe]

i don't know what you mean by this

[wintry steppe]

i'm not trying to imply the proof follows immediately from this or anything

i'm saying this is all you know

and you have to use to prove the statement you want

(the point of the exercise)

[gray dust]

@past yew is this a test?

[past yew]

Yup yesterday test

[hardy inlet]

i cant tell if I repeated the same mistake

[past yew]

A & B are correct according to me

[wintry steppe]

it's very sloppy, but looks a little more correct

[hardy inlet]

could u please elaborate on what causes it to be sloppy

[wintry steppe]

"...which contradicts our assumption"
what assumption? this isn't clear

[hardy inlet]

ahh okay

"our assumption that -1 is not an eigenvalue"

[wintry steppe]

it seems like you're also assuming that w is non-zero

are you?

[hardy inlet]

yeah... thats not needed is it. since T0 = 0 for any T

[wintry steppe]

well if you're using w as an eigenvector corresponding to -1 to get a contradiction, it better be non-zero

and if that's not what you're doing

then it's not clear

[hardy inlet]

sec i think im onto somethign

i think i've gotten stuck again

i feel like I cant say w=0

[wintry steppe]

this works

[hardy inlet]

this was the bottom of what i had that i was too embarrased to show; I beat my head some more and the mistake i made at the very end was substituting W into itself

[wintry steppe]

you're getting there

[hardy inlet]

its this whole \neq0 thats screwing me up

[wintry steppe]

well T and I certainly agree when v = 0 as well.

[hardy inlet]

T0 = I0 right

but I technically only proved for not equal, so would I just throw in a sentence about T0=I0 therefore Tv=Iv forall v

[wintry steppe]

if you want

[hardy inlet]

the mona lisa v2:

[wintry steppe]

the logic works out

[hardy inlet]

warmup = over;

is P the polynomial thing P(T)?

[wintry steppe]

no

it's a linear operator on V

[hardy inlet]

so its no different that T \in L(V)

[wintry steppe]

if you like writing them as T, sure

[hardy inlet]

the \align at the end feels questionable; especially the part where I dropped the dimentions

[zinc timber]

[hardy inlet]

[zinc timber]

what's fundamental theorem of linear maps

[hardy inlet]

dimV = rankT + nullity T

T \in L(V)

[zinc timber]

I don't think we call it the "fundamental theorem"

[hardy inlet]

axlers textbook does

[zinc timber]

Ok I'll accept it

[zinc timber]

these steps feel sus

[hardy inlet]

the align* steps or the whole thing?

[zinc timber]

align one

since you know V = N(T)+R(T) you can also show their intersection is trivial and so it's a direct sum

[hardy inlet]

I was using

there has to exist some map where the intersection is non trivial right

[zinc timber]

ya

[hardy inlet]

which is why we had the P² = P so that we cant apply the transformation once, get something in the nullspace, then apply such and get 0

[zinc timber]

unfortunately I can't think of an easy example now

[zinc timber]

idk what it exactly means but yes

your intuition is correct

[hardy inlet]

I think i have an example

[zinc timber]

ig it works

[hardy inlet]

which means (0,0,1) is in the intersection

thus non trivial and cant take a direct sum

ok I think i cleared it up better:

oops now I left out the direct sum part

[slow scroll]

yea idk this doesn't make much sense to me. Its not true that "for all v in V, (P - I)v = 0" (this would imply that P = I), and that does not imply that null P = {(P - I)v : v in V}

your v = Pv + (P - I)v = Pv + 0 = Pv is almost the right idea to show that null P + range P = V, but again (P - I)v != 0 in general.

[hardy inlet]

yeah... 😬 thats a good point in the first part. idk how to derive the nullP then

[slow scroll]

I don't think you really need to. while you can't say null P = {(P - I)v : v in V}, you can say null P \supset {(P - I)v : v in V}

[hardy inlet]

wait wouldn't it be a subset

[slow scroll]

if you have (P - I)v in {(P - I)v : v in V}, then P((P-I)v) = 0, so (P - I)v is in null P

so its as a wrote it

[hardy inlet]

so at least everything in (P-I)v is in the null?

[slow scroll]

ye

[hardy inlet]

im trying to figure out how there can be more

wait nvm its not saying every vector in V; its saying every vector acted on by (P-I)

so there could be some w \not \in range(P-I)

such that Pw = 0

[slow scroll]

yes

the other inclusion might hold actually, but its not necessary for this

in fact i think it does hold

yea it does

the other inclusion is one of those things that are so trivial its easy to miss. But anyway, as far as correcting this proof, what you have here

v = Pv + (P - I)v
is almost correct, but Pv + (P - I)v = 2Pv - Iv, which need not be v. There's a simple way to fix this so you get the right thing

[hardy inlet]

yeah i saw that and flipped the sign but idk what to do with it

[slow scroll]

i.e. v = Pv + (I - P)v?

[slow scroll]

ah yea, so that fixes that

[hardy inlet]

i dont think the range is right tho

like thats just the definition of range

[slow scroll]

yeah, idk why you have that

the correct conclusion from v = Pv - (P - I)v is that V = range P + null P

[hardy inlet]

plus null P?

[slow scroll]

ye

null P is closed under scalar multiplication, so (P - I)v in null P implies -(P - I)v is in null P as well

[hardy inlet]

but theres a negative sign

oh okay

[slow scroll]

if you can show that null P = {(P - I)v : v in V}, then your proof that null P \cap range P = 0 works. But there's also a different way to do that

and then once you have V = range P + null P and null P \cap range P = 0, that's it. You don't need the "fundamental theorem of linear maps" for this.

[hardy inlet]

yeah just the definition of directsum

[slow scroll]

In fact, you shouldn't use it, because it only works in finite dimensions

[hardy inlet]

true

but like, null P's equality

😬

like idek if i can say this

[slow scroll]

ye, that's perfect

[hardy inlet]

ah right nullP is a superset

so (P-1)v is definitely in it

aha

my brain was still thinking the subset

but that still doesn't solve the issue of nullP's exact form

like I can't set an equality on it to compute the intersection can i

[slow scroll]

im not sure what you mean by this. Your proof that null P \cap range P = 0 uses that null P = {(P - I)v : v in V}, so if you want to use that proof, then you'll need to figure out the other inclusion (which i verified does hold). You also don't have to use it. There is another way to see that null P \cap range P = 0 as well.

[hardy inlet]

well how'd u show the equality of null P = (P-I)v that sounds difficult

[slow scroll]

here's a hint, but it basically gives it away: || let v be in null P. v = -(P - I)v (why?) ||

oops fixed it

[hardy inlet]

i thought scalars didn't matter

[carmine echo]

Pv = 0 => v = (P - (P - I))v = Pv - (P - I)v = -(P - I)v

:o

[slow scroll]

wdym?

[hardy inlet]

u swaped from v= (P-1)v to v = -(P-1)v

[hardy inlet]

oh... that is quite trival

i just dont have the brain

[slow scroll]

well scalars could matter as far as making an equality hold. btw, i should have written
v = (P-1)(-v). Now its really clear that v in null P implies v is in {(P-I)v : v in V}

[hardy inlet]

it took me like 3 minutes to notice its trivialness

[slow scroll]

the other inclusion is one of those things that are so trivial its easy to miss
hence what i said earlier

[carmine echo]

for me to have figured it despite really sucking at vector space and being an absolute beginner to it, yes it has to have been trivial

dim V = dim null P + dim range P though, is clearly not trivially intuitively visible to me ._.

the proof makes sense tho, so yeah

[hardy inlet]

shouldn't this have a negative coefficient now

[carmine echo]

$\mathcal{L}$

[stoic python]

Ansh

[slow scroll]

you would write v = (P-I)(-v)

[carmine echo]

what is this symbol for?

L(V) a transformation defined on a vector space V?

[slow scroll]

linear operators on V, yea

[hardy inlet]

T \in L(V) means T is in the set of linear maps from vector space V to itself

L(V,W) is V to W

(the set of all)

[carmine echo]

Ty

[slow scroll]

does this make sense matt? if v is in null P, there exists an element of V, i.e. w = -v such that v = (P - I)w

[hardy inlet]

vaguely?

[carmine echo]

I mean: v = Iv = Pv - (P - I)v = (P - I)(-v) = (P - I)w

[slow scroll]

well, what does it mean for w to be in {(P-I)v : v in V}? It means exactly that there exists v in V such that w = (P - V)v

and what we've shown is that v = (P - I)w for w = -v in V

[hardy inlet]

is it necesary to create a w?

[slow scroll]

Not necessarily. Some proofs of existence use non constructive methods like contradiction.

[hardy inlet]

this is the current

[carmine echo]

uhh, null P = {x | Px = 0, x \in V}

and you figured that (P - I)v for all v in V satisfies P(P - I)v = 0

hence, (P - I)v must've been a subset of null P

:o

[hardy inlet]

i removed this -

:/

and i had this which I think is sufficient

[zinc timber]

[knotty hazel]

ryu

[carmine echo]

share all the Qs with solutions lmfao

I believe questions and solutions suit me better than grinding my brains with the theory :|

[hardy inlet]

i aint got the solutions

[carmine echo]

after you're done then..

[hardy inlet]

yeah....

the goal is to finish; but no guarantees on that one

this one feels easy so i wrote the definitions...

[hardy inlet]

not sure how to finish this or if its a good direction

so it appears that I have the magnitude of c is 1, but I need c to be exactly 1 right

then u = cv → u = v

[slow scroll]

i agree

[hardy inlet]

i feel like theres something super basic with this but idk

[dusky epoch]

<u,v> = <cv, v> = c <v,v>

@hardy inlet

[hardy inlet]

ohhh yeah those properties + the substitution

3am brain mode be like

ty

wait does that finish it tho..

[lavish jewel]

it should

you have that u = cv from C-S, and that c = 1 from that

[hardy inlet]

what about complex tho?

[dusky epoch]

<u,v> = 1

and <v,v> = 1

[lavish jewel]

c*1 = 1

[hardy inlet]

couldn't c * 1 = c?

[dusky epoch]

[lavish jewel]

[hardy inlet]

wut

ohhh

wait

sry its really late

i see...

[lavish jewel]

you should stop doing math so late

you'll make mistakes and won't retain the knowledge

[hardy inlet]

yeah I know i need to start sooner,
but heres the pwoof

not sure on the syntax here, I feel like I should have <e_1, e_2> but the problem doesn't have that

oh it says function not inner product; thats why. Should I give it a name like phi?

[lavish jewel]

i wouldn't say this is a contradiction, more like a counterexample

you didn't use the assumption that it was an inner product at any point

[hardy inlet]

fair enough

would an obscure counterexample even suffice; or is this sort of an existance problem

if lambda is not an eigenvalue then its invertible and by the rank nullity theorem f(lambda) = dimV; and if it is an eigenvalue; then its not invertible, hence nullity \geq 1, and thus f() <= dimV-1

i guess we're assuming the dimention a nonnegative integer

[zinc timber]

why not do this

$\ip{u,v}=\ip{v,v}\implies \ip{u-v, v}=0$

[zinc timber]

and ..

[lavish jewel]

which one are you addressing

[zinc timber]

the u=v one

ig I'm too late

[zinc timber]

[lavish jewel]

how would you do the last one, ryu

[zinc timber]

which one? not continuous one?

[zinc timber]

or this?

[lavish jewel]

the not cont one, yea

[zinc timber]

in C, we must have one eigen value λ amd f(λ) <= n-1 but for all points near to λ, f(λ)=n so we cannot make |f(x)-f(λ)|< eps for any δ>0

[lavish jewel]

would you have accepted using a limit

[zinc timber]

or or, if f were continuous then f(C) would be connected in N but it isn't, so f couldn't have been continuous

[lavish jewel]

or eps delta preferred?

[zinc timber]

I prefer the topology one 🙈

[zinc timber]

limit feels like easier

[lavish jewel]

should work, right?

[zinc timber]

ya

[outer goblet]

is there a quick way to tell if a matrix is diagonalisable

[dusky epoch]

no

[outer goblet]

:(

[lavish jewel]

if the matrix has special structure, yes

otherwise, you try and see if it works

[outer goblet]

mm cuz seeing if the characteristic polynomial splits can be pretty hard without a calculator

[wintry steppe]

for example, any diagonal matrix is diagonalizable

[zinc timber]

if the minimal polynomial is separable then it's diagonalizable

it's iff condition

[outer goblet]

yeas ik lol

[wintry steppe]

have fun computing the minimal polynomial

[outer goblet]

its just hard to see lol

[zinc timber]

compute the char poly

[outer goblet]

from determinantnt

yes

but then i have to factor it lol

[wintry steppe]

clench your teeth and get computing

[outer goblet]

hehe

[zinc timber]

yes

wait for it

[outer goblet]

yea algebra is so hard

[wintry steppe]

sometimes you just have to do a little bit of work

[zinc timber]

for each factor compute p(A) and see if it's zero

[outer goblet]

[zinc timber]

just use a computer

[outer goblet]

im not allowed on exam lol

[wintry steppe]

doesn't mean you can't

[outer goblet]

lemme pull up with my gaming setup in the exam building lol

grindset

[brave cliff]

Im looking at part b for this question.

If a matrix is linearly dependent, can it be said that it doesn't span the certain dimensions

[zinc timber]

or you can compute SNF to get the minimal polynomial

[zinc timber]

not necessarily

u need to be more specific about what you mean

example there could be 3 LD vectors of lR² that spans lR²

[brave cliff]

Oh, that's not my work, that's my prof's solution sheet

[steep sinew]

does this chat include algebra 2?

[zinc timber]

#groups-rings-fields

[hardy inlet]

i had a problem in my old version since i said <u,v> = 0 instead of <v,v> = 0
This works for showing a property doesnt hold, correct?:

[zinc timber]

$\overset{?}{=}$ is not a good practice. calculate them separately and show they are not equal

[dusky epoch]

@zinc timber \overset?= is fine

[quiet wren]

Hi could anyone explain to me why this is true?

aj and w are n dimensional real vectors

Why is normalisation of w leads to the inner product being the distance to the hyper lane with normal vector w ?

[hardy inlet]

ok thanks

^ryu

So rn im stuck on knowing that (T- λI) is invertible if λ is not an eigenvalue

[gray dust]

@quiet wren heres a sketch

[quiet wren]

Thanks. Just a few things to clarify my understanding

Here you have written the projection of aj onto H right?

Is it a formula?

[lavish jewel]

that's the projection onto w, not H

[hardy inlet]

i dont know if I can say this (red arrow)

[lavish jewel]

you can use geometry and the definition of the dot product to show how/why this works

your last 2 sentences break down

f(x) = x also satisfies what you wrote there

[quiet wren]

right so the projection of aj onto w gives the distance to H right?

since w is a normal vector

[lavish jewel]

that's what you wrote in the original picture you sent, sure

it's not enough that it's a normal vector though

or what do you mean by normal here?

w is normal to H

[quiet wren]

dist(a_j, H) = norm(proj_w(aj))

would it be correct?

[lavish jewel]

that'S what you were originally given in the image

(for unit norm w, anyway)

[quiet wren]

yeah a normal vector to the plane

[lavish jewel]

you're saying exactly what is written in the image you sent

so yes

this is not the answer to "why" though

that is in rokabe's drawing

[quiet wren]

yeah he also provide a formula for the projection of aj onto w which I was missing

[lavish jewel]

all right

[hardy inlet]

does this imply the others are fine. Why it (T- λI) invertible/surjective tho?

[lavish jewel]

hmm?

why is T - lambda I not surjective?

[hardy inlet]

(T- λI) = (T - TI) = 0

[lavish jewel]

what are you asking me?

[hardy inlet]

since λ ev

im asking why it is surjective for λ not an ev

[lavish jewel]

i didn't understand what you wrote in what you asked me

ah

same reasoning as you did rn

take (T - lambda I)v

[hardy inlet]

(T-λI) != 0?

[lavish jewel]

this gives Tv - lambda v

Tv \neq cv

[hardy inlet]

ah right bc its not an eigenvalue

...

thanks lol

[lavish jewel]

right

so you get a linear combination of 2 vectors

and you know for a fact they are not scalar mults of each other

you can show that for 2 vectors, linear dependence means one is a scalar mult of hteo ther

[gray dust]

no its onto w

[lavish jewel]

this means you cannot get 0

[quiet wren]

Yeah thanks