linear-algebra

[lavish jewel]

what the heck

[wintry steppe]

Oh my gosh. I have these dumb moments every day.

it happens

[wintry steppe]

Okay, the last part is what I'm having issues with.

Am I doing this wrong?

[half ice]

,w inverse of {{1,2},{1,1}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Kaynex#7520

[half ice]

You don't have the correct inverse matrix. It looks like you put A inverse in there insead

@wintry steppe

[wintry steppe]

Okay, I solved it@

Thank you so much!!

[half ice]

Np. Feel free to ask if you have any others!

[misty storm]

How would I go about finding the intersection for 2 subspaces?

I've found a couple of resolutions online]

for small subspaces

I have a subspace of 3 vectors and one of 2

they're R4, if that's relevant

[wintry steppe]

what form are the subspaces given in?

[wintry steppe]

what's p(x), can it be smth like 1/x ?

[misty storm]

oh, sorry

p(x)=x^2-x-1

so the last vector would be x^3-x^2-x

[wintry steppe]

convert it to column vectors first (to make it easier), e.g., where coordinate 0 is coefficient of x^0, ... , coord. 3 is coef. of x^3
then you have two linear combinations, one of vectors from A and one from B
set them equal to each other and solve to coefficients in front of vectors of one of the sets

[misty storm]

how come they're equal to each other?

[wintry steppe]

because you're looking for intersection

i.e. all points that can be expressed as both linear combinations

[misty storm]

A * v1 + B * v2 + C *v3 + D * v4 - E * v5 - F * v6 - G * v7 - H * v8 = 0

is that it?

[wintry steppe]

you set them equal to each other, then when you move those columns from B to the left, (negating them) and you get 4x8 matrix to solve

[misty storm]

A * v1 + B * v2 + C *v3 + D * v4 = E * v5 + F * v6 + G * v7 + H * v8

right

now pass them to the left like I did

A * v1 + B * v2 + C *v3 + D * v4 - E * v5 - F * v6 - G * v7 - H * v8 = 0

[wintry steppe]

yes, now solve for either ABCD or EFGH

[misty storm]

I have to solve for both at the same time, no?

I only have that equation right now

[wintry steppe]

yes, I meant, you only need one of two, sorry

[misty storm]

just to make sure I'm getting the idea here, doing for the first row I would have that either a=-1 or h=-1, yes?

the bottom letters being the coefficients multiplying the entire vector

the equation for the first row being: a * -1 = 1 * h

rather

-a=h

[wintry steppe]

wait, i thought botton letters just labels

[misty storm]

I meant them as coefficients multiplying the entire vectors

[misty storm]

that's why I wrote this equation out

[wintry steppe]

ah yes

I ust never bothered myself to write them out, always had to remember where's what

unless they span the same space (or one is a subset of another), there should appear a restriction on coefficients

[misty storm]

what does "a restriction" mean in this context?

it shouldn't be solvable?

[wintry steppe]

it should

every equation is a restriction

it seems both A and B(the sets) span the entire space of cubics?

[misty storm]

what does that mean?

I'm not american so terminology often doesn't match

[wintry steppe]

I mean, they seem to be the same subspace already, just expressed differently

[misty storm]

it is possible that that is the case, but I can't tell

[wintry steppe]

I mean, when you find intersection, you will probably end up back where you started, not because something's wrong, but just because intersection of a space with itself, is that space again

[rigid plover]

so I just started reading about this group theory stuff

so XOR is considered an "abelian" or commutative group in the boolean space, but this link also brings up the concept of a "boolean ring" in contrast to a "boolean algebra" structure

The answer is that it is equivalent to the structure of a Boolean algebra (using AND, OR, and NOT) or equivalently a Boolean ring (using XOR, NOT, and AND). Specializations of this structure include
https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
but are all of these rings considered abelian, and why are they called "rings"?

[wintry steppe]

I think you should ask in #groups-rings-fields

[limber sierra]

a ring is a different object than a group

both are studied in abstract algebra

rings come with 2 operations, not 1

[rigid plover]

hmm

[limber sierra]

one typically doesnt say "Abelian" for rings

but instead "commutative"

[rigid plover]

ok...

[limber sierra]

but yes, a boolean ring is a commutative ring

[rigid plover]

which two operations are in XOR then

[limber sierra]

XOR is a single operation

the other operation in a boolean ring is AND

(to use ring theory jargon, XOR is the addition operator and AND the multiplication operator)

[rigid plover]

so the ring consists of {0,1} and the operations XOR and AND?

basically those operations are our defined allowed operations?

[limber sierra]

technically a boolean ring is a more general object

but yes

[rigid plover]

ok...

[limber sierra]

the boolean ring consists of {0, 1} with the operations XOR and AND

this ring goes by other names

[rigid plover]

so there is only one boolean ring?

[limber sierra]

like "The integers modulo 2" or "the field of 2 elements"

no, its just the most common example of a boolean ring

[rigid plover]

oh ok

so here

Boolean algebra (using AND, OR, and NOT) or equivalently a Boolean ring (using XOR, NOT, and AND).
what is the difference

it uses XOR instead of OR but...

[limber sierra]

your "default operations" are OR and AND

instead of XOR and AND

now, as it turns out

you can write XOR purely in terms of OR and AND

[rigid plover]

oh

[limber sierra]

and similarly, you can write OR in terms of XOR and AND

so these end up being equivalent structures

[rigid plover]

I see...

[limber sierra]

but at their "core" theyre defined using different operations

and you need to do a bit of work to show you can go from one to another

[rigid plover]

interesting

what do you mean by core?

[limber sierra]

heres how you write XOR using AND and OR

(and negation)

[limber sierra]

i mean the operations that make it a boolean algebra/a ring, respectively

this would perhaps be more clear if youre more familiar with other examples of rings

[rigid plover]

what's another example

[limber sierra]

well, i said that you can think of a boolean algebra as arithmetic modulo 2

[rigid plover]

oh right

[limber sierra]

arithmetic modulo n forms a ring

for any integer n ≥ 2

it wont be a boolean ring unless n = 2, though.

another example of a ring is all the integers under the operations + and *

[rigid plover]

*?

[limber sierra]

multiplication

times

[rigid plover]

oh

ok

[limber sierra]

or you can replace "integers" with "rationals" or "real numbers" or "complex numbers"

[wintry steppe]

there's a video by michael penn about types of rings where he shows some simple examples

[limber sierra]

these are fairly familiar examples

[rigid plover]

integers when multiplied or added will always yield other integers so this is what is meant by a ring?

[limber sierra]

but there are some weirder rings out there

[rigid plover]

but you can't include division

[wintry steppe]

you define A/B as A x (B^-1) where B^-1 is multiplicative inverse, not all rings have multiplicative inverse so not all rings have division

[limber sierra]

basically a ring is a set two operations, commonly denoted + and *.

• (AKA "addition", though it doesnt always act like standard addition) makes the set an abelian group, so its associative, commutative, has identity, and has inverses.

• (AKA "multiplication", though it doesnt always act like standard multiplication) is associative and has an identity, but may not be commutative, and may not have inverses.

• and * are related by the property of distributivity: a*(b+c) = a*b + a*c, and (a+b)*c = a*c + b*c

if * is commutative, we call it a "commutative ring". if, furthermore, * has inverses as well, we call it a "field"

these are still rings

just specific types of rings

[rigid plover]

so essentially these properties

associative, commutative, has identity, and has inverses.
are dependent upon whether or not the result is also within the defined set

[limber sierra]

as an example, ℚ (the rationals under standard + and *) is a field, but ℤ (the integers under standard + and *) is not - its just a commutative ring

because we can "divide by rationals" and get another rational

but we cant "divide by integers" and get another integer

[rigid plover]

right

sorry but what is a field

[rigid plover]

what are inverses

[limber sierra]

ah, i thought you were familiar with groups

[rigid plover]

nope, I literally just started

[limber sierra]

an inverse is an element that you can add/multiply by to get your identity

[rigid plover]

oh

[limber sierra]

in this case, something you multiply by to get 1

[rigid plover]

to get your identity element

[limber sierra]

so the inverse of the rational number 5 is 1/5

under standard multiplication

and the inverse of -11/3 is -3/11

[rigid plover]

so in addition the identity element is 0 if we use our integer example

[limber sierra]

to use a slightly more esoteric field

[limber sierra]

right

and the multiplicative identity is 1

now lets take a slightly weirder example

to demonstrate this

the integers {0, 1, 2, 3, 4} modulo 5

[rigid plover]

hmm

[limber sierra]

this actually DOES have inverses

obviously 1 * 1 = 1

[rigid plover]

sure

[limber sierra]

but 2 * 3 = 6 = 1

3 * 2 = 6 = 1

4 * 4 = 16 = 1

(modulo 5)

[rigid plover]

huh

[limber sierra]

the only element without an inverse is 0

in general, 0 (the additive identity) will never have a multiplicative inverse

[rigid plover]

so in that ring(?) 2 and 3 are inverses

[limber sierra]

2 is an inverse of 3

3 is an inverse of 2

4 is an inverse of 4

1 is always the inverse of 1

[rigid plover]

nice

[limber sierra]

but to take another example

lets say instead we were working mod 4

so {0, 1, 2, 3} modulo 4

as it turns out, 2 lacks a multiplicative inverse here

we can check:

[rigid plover]

so back to the field definition

[limber sierra]

2 * 0 = 0
2 * 1 = 2
2 * 2 = 4 = 0
2 * 3 = 6 = 2

[rigid plover]

what makes it a field

[limber sierra]

so the integers modulo 5 ARE a field

but the integers modulo 4 are NOT

[rigid plover]

because they have inverses such as 2 and 3?

[limber sierra]

EVERY element in the integers modulo 5 (except 0) has an inverse

which makes it a field

[rigid plover]

even though there is an EXCEPTION?

[limber sierra]

0 will always be an exception

we exclude 0

[rigid plover]

by convention?

[limber sierra]

0 * anything = 0 in any ring

[rigid plover]

ok

[limber sierra]

in the case of the integers modulo 4, 1 and 3 actually have inverses

1 * 1 = 1 and 3 * 3 = 9 = 1

but 2 doesnt have an inverse

[rigid plover]

so it's not a field

[limber sierra]

so the integers modulo 4 arent a field

[rigid plover]

wow

[limber sierra]

(in fact, one can prove that "the integers modulo n" is a field precisely when n is prime)

(5 is prime, but 4 isnt)

[rigid plover]

ok, thanks for your help, I gtg, we can talk here later

[limber sierra]

see ya.

[wintry steppe]

howcome when they were finding the determinant they just got rid of the first column and the first row

i dont understand

Same thing here

[limber sierra]

expanding along the first column

if we expand this along the first column

it becomes: (will take a min to type)

[stoic python]

Namington

[limber sierra]

but 0 times anything is 0

[stoic python]

Namington

Namington

[wintry steppe]

Ohhhh

[limber sierra]

a similar thing is done in your second image

[wintry steppe]

Howcome you chose the first column?

or they did

[limber sierra]

using a row/column with lots of 0s makes the laplace expansion easy to compute

so they looked for a row/column they could fill with a lot of 0s

the first column was convenient for this

in theory they couldve chosen another instead

doesnt make much difference.

[wintry steppe]

expanding on rows or columns doesn't make a different in the end right?

like if you chose the first row to compute

[limber sierra]

it doesnt

[wintry steppe]

okay, thank you!

[limber sierra]

wiki example

(this follows from the determinant of the transpose being the same)

[wintry steppe]

whats naive gauss elimination?

[limber sierra]

i believe thats the term for gaussian elimination implemented with a fixed order, usually something like:

• divide the first row by its first entry to make that entry equal to 1
• for each other row, if denotes is the first entry of that row, subtract x * (the first row) from that row
• repeat the above steps, replacing first row/entry with the next row/column in your matrix, until you reach the end of your matrix

im not addressing potential division by 0 because theres no standard way to handle it

its "naive" because its typically not the most efficient order to do gaussian elimination in

since you just apply the algorithm in the same order every time, without using any shortcuts

(the standard version doesnt even swap rows!)

but it ALWAYS works

itll always get you a row reduced matrix

so you can implement it in a computer without worrying

(just add some handling for division by 0 - say moving a pivot row to the bottom if its pivot entry is 0)

@wintry steppe

[wintry steppe]

Does this proof work?

[stoic python]

n/c

[wintry steppe]

yes, this is fine

Thanks

[wintry steppe]

Suppose V is finite dim and T in L(V,W). Prove that there exists a subspace U of V such that U cap null T = {0} and range T = {Tu : u in U}

Can someone give me a hint for this problem?

[dusky epoch]

fix a basis of null(T), then extend to a basis of V?

[wintry steppe]

Wouldn't that mean that the basis of null T would intersect everywhere with the subspace?

[dusky epoch]

??

let me clarify

for ease of notation, let n = dim(V) and k = dim(null(T))

take a basis v_1, v_2, ..., v_k of null(T)

extend it to a basis v_1, v_2, ..., v_n of V

and take U = span(v_{k+1}, ..., v_n)

this is what i meant.

[wintry steppe]

But then we don't have that range T = {Tv : v in V}

For example Tv_k is undefind

But Tv_k in range T

[dusky epoch]

what

since when is Tv_k undefined

[wintry steppe]

Never mind sorr

[dusky epoch]

v_k is the last vector in null(T), so Tv_k = 0

[wintry steppe]

I got confused but it's clear now

Thanks

I was making it too complicated

[wispy thicket]

just wanna know if there's a flaw with my proof

[dire thunder]

what is E(z)

[wintry steppe]

ez pz

[wispy thicket]

E(z) is this

[wintry steppe]

guys i dont understand this

https://gyazo.com/ea6ec2c55f25ee73abd72c345a098677

can someone explain pls?

[lavish jewel]

looks like the vector [a11, a12, a13] is an eigenvector of the matrix with eigenvalue 0.521

meaning 0.521 is a root of the characteristic polynomial, as confirmed in the line below

that's about as much as i can say from a random image with no explanation

[nocturne jewel]

and that the eigenvector is in the kernel of the bottom matrix

[wintry steppe]

yeah, but what i mean is

-0.171a11?

shouldnt be 0.35?

[lavish jewel]

that's why i said it's a root of the characteristic poly

what they did is take the matrix from above, call it M, and take M - lambda I

where I is a 3x3 identity matrix

[wintry steppe]

yes, i see now

mmm okey

ty

[lavish jewel]

or at least that's what i'm guessing, i didn't actually subtract the numbers. let's see

,w 0.35 - 0.521

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Edd#5182

[lavish jewel]

yeah

[nocturne jewel]

oh they just made the A-tI matrix

[wintry steppe]

do u know how to get the principal components?

I thought thats the way but i may be wrong

[lavish jewel]

for data arranged in a matrix M, you can get the sample covariance as 1/N M M^T

[wintry steppe]

the matrix given is already the logs of the covariances

[lavish jewel]

if the data was zero-centered, all you need after that is an SVD or EVD

[wintry steppe]

idk if it is zer-centered

[lavish jewel]

so you already start with a symmetric mat of log covariances?

[wintry steppe]

Thats my matrix

and those are the eigenvalues and eigenvectores, respectively

shity maple has to write 0i xd

[wintry steppe]

so i thought the principal components were calculated like here

where 0.521 is an eigenvalue

[lavish jewel]

i'm not sure S is a covariance mat

[wintry steppe]

well, assume it is. Maybe it is wrong but thats what teacher gave us

it is not symetric tho, as u pointed

maybe she confused, but

[lavish jewel]

anyway

my understanding is that, given a covariance matrix, the PCA is the same as the EVD, which for symmetric matrices is the same as the SVD

but since this isn't symmetric, idk

[wintry steppe]

isnt this the algorithm to calculate the principal components?

[lavish jewel]

that's an EVD of a symmetric matrix, as i said

the other 2x2 matrix you gave isn't so nice

[wintry steppe]

yeah xd

the first component is 0 😄

[lavish jewel]

do you have the original task so that i can read it?

[wintry steppe]

it is a maple worksheet

can u open it?

or i can screen shot

[lavish jewel]

screenie

[wintry steppe]

i havent calculated the covariance matrix, so it may be wrong too

[lavish jewel]

me da la impresion que ese es el problema

dejame probar en octave

logaritmo en cual base?

D:

[wintry steppe]

😄

XD

[lavish jewel]

ah, base 10

wth

[wintry steppe]

I dont know honestly

that was given

[lavish jewel]

mira

symmetric, as we'd expect

cuz what you have atm is pretty cursed

[wintry steppe]

so the matrix is bad

sec , ill fix it

also...

wait

why u have 0.35

and teacher 0.06

she used other base?

[lavish jewel]

i have no idea wtf they did to get the third matrix

i found the cov mat of the log data

which SEEMS to be want they want to do

but they don't say

what they have is definitely wrong tho

[wintry steppe]

sec, lets find the base

[lavish jewel]

maybe they subtracted the mean?

[wintry steppe]

by basic maths XD

[lavish jewel]

the base is 10

[wintry steppe]

i mean the base she used

[lavish jewel]

you can see my "ldT" matrix matches the log(X) matrix

that's what i mean

[wintry steppe]

log_a b = c means a^c = b?

xD

[lavish jewel]

yea

lemme try subtracting the means

to see what's up

[wintry steppe]

can u, on octave, get the eigenvalues and eigenvectors easily?

[lavish jewel]

yep

[wintry steppe]

could u?

[lavish jewel]

of which matrix?

[wintry steppe]

the covariances

the S

' on octave means transpose?

[lavish jewel]

yes, ' is transpose. complex conjugate transpose, to be fair, but these are real numbers, so it's just transpose

here, this e-18 number is actually 0. it's just a numerical precision issue

[wintry steppe]

so

why we get different results

ah noo no

[lavish jewel]

the vectors might be backwards

since you said you got 0 first

swap the columns in Q and the diagonal elements in D

[wintry steppe]

values are 0.078 and 0.0498

with S symmetric

[lavish jewel]

ooooh wait

i made a small mistake

or did i? one sec

i forgot to divide by 6 earlier

this is my cov mat

[wintry steppe]

why divide by 6 lel

[lavish jewel]

in the EVD i sent earlier, i used the S matrix as your teacher gave it

[wintry steppe]

ye i know, i noticed

[lavish jewel]

give me one sec to do something else

yeah no, i still don't know what your teacher did

anyway

this is the covariance

i divided by 6 cuz of the 1/n

[wintry steppe]

ah okey

i see

[lavish jewel]

we have n = 6 samples

[wintry steppe]

ye ye

okey so idk either. I just made her matrix to be symmetric

[wintry steppe]

and got this

now i am supposed to get the principal components. But i need an algorithm, so i am doing (S - eigenvalue1 * Identity) = 0

is this correct?

[lavish jewel]

i would just follow the procedure as done in wikipedia

[wintry steppe]

could u take a look at some example?

[lavish jewel]

what do you get out of doing that though?

[wintry steppe]

mmm, i am just following the example she gave on a pdf, since what i saw on the wiki is more complicated

[lavish jewel]

i would either just do an EVD, or do it as on wiki

i'm not sure where you're going by taking S - eigenvalue1 * identity = 0

do you wanna solve for the eigenvalue?

[wintry steppe]

thats what the pdf sais to get the principal component

[wintry steppe]

^

[lavish jewel]

ah, so find the roots of the char poly and then find the eigenvectors

this is just an EVD

[wintry steppe]

well XD

something is going wrong hahaah

[lavish jewel]

[wintry steppe]

S - eigenvalue1 * identity = 0 this resulting matrix has det = 0

[lavish jewel]

yes

[wintry steppe]

well, thanks anyway, i wrote her

[woven fossil]

I need help in matrices and vector subspace question but for some reason I am not able to send it here. Someone who can help me? Please dm

[wintry steppe]

<@&681260374879633482>

[hearty citrus]

anybody help me with this question

[wintry steppe]

<@&286206848099549185>

[jagged pendant]

that's not abstract algebra or linear algebra

1. you're in the wrong channel
2. you're using the pings wrong
3. read #❓how-to-get-help

why didn't it link

#❓how-to-get-help

there

[native rampart]

#848786850994585651

[wintry steppe]

none cuz real numbers don't exist

[wintry steppe]

@lavish jewel I tag you, because you're the only one who knows German, hope that's not a problem.

May I do such a thing?

[lavish jewel]

i'm not sure what exactly R[x]<=1 is

[wintry steppe]

All functions whose degree <= 1

[lavish jewel]

ah

[wintry steppe]

at maximum 1.

[lavish jewel]

not sure why exactly you did it like that tho

lemme see

what i would've done is take f( 3(x+2) - 6(1) )

then you get 3f(x+2) - 6f(1) = 3(-2x+1) - 6(x+3)

[wintry steppe]

Okay. So we just used what we have.

[lavish jewel]

that's what i would say. maybe someone else has another idea tho

[wintry steppe]

Why am I not allowed to this?

It's strange.

But just came to my mind.

[lavish jewel]

hmm

the thing is the addition is an affine transformation

maybe i'm overthinking it

[dusky epoch]

no, saying f( (3x-2)+2 ) = -2(3x-2)+1 is just blatantly wrong

f doesnt take real numbers as input, f takes polynomials as input

[wintry steppe]

isn't 3x+2 a polynomial?

you know.

I'm just trying to figure it out.

[dusky epoch]

yes, but you cannot just substitute polynomials into other polynomials and expect that to factor through f like you just did

you're overthinking it and getting distracted by extra structure present on your vector space

f(x+2) = f(x) + f(2) = f(x) + 2f(1)

-2x+1 = f(x) + 2(x+3)

f(x) = -2x + 1 - 2(x+3) = -4x - 5

f(3x) = 3f(x) = -12x - 15

that's it

it's no different than if you were given a linear map g from R^2 to R^2 about which you are told that g(1,2) = (-2,1) and g(0,1) = (1,3) and asked to find the value of g(3,0)

[lavish jewel]

hmm yeah, what you had done was kinda like... trying to put polynomials in a straight line

which doesn't make sense

[dusky epoch]

yeah as i said, they were getting distracted by things other than the vector space structure of R[x]_\leq1

[lavish jewel]

thanks for the help

[wintry steppe]

Really thanks for the help.

[lavish jewel]

what we did here is pretty much the trick of figuring out what a transformation is doing by feeding it vectors in a basis

one normally does this, e.g. using the canonical basis

here, they gave you examples with an order 1 poly and a constant, but it works the same way

you then try to express all the polys as a linear comb. of those "vectors" in the basis, and use all the nice properties of linearity

[stoic python]

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

[latent ledge]

Let $A \in SO(n)$ and denote the corresponding linear transformation on $R^n$ by
$L_A$. Show that if $L_A$ preserves a subspace $V \subset R^n$, then it also preserves its
orthogonal complement $V^{\bot}$ .

[stoic python]

亜城木 夢叶

[torn hornet]

what did you try @latent ledge

[latent ledge]

i don't have anything so far

[torn hornet]

hmm, well we are talking about orthogonal right

so i say consider the inner product, and how an orthogonal matrix interects with it

[latent ledge]

<u,v>=<Au,Av>

[torn hornet]

mhm try using this to prove the statement

[latent ledge]

i don't see how that connects to the orthogonal complement

[torn hornet]

well whats somethings orthogonal complement

[latent ledge]

$V^{\perp}={v\in R^n,|,<u,v>=0, u\in V}$

[stoic python]

亜城木 夢叶

[torn hornet]

yeah

[latent ledge]

0=<u,v>=<Au,Av>

[torn hornet]

yeah and now try reasoning out the statement from this

[latent ledge]

thank you for the help

[torn hornet]

np

what did you try(assuming you are asking for help)?

[misty storm]

can someone help me out with this one?

I need to find the intersection

I've been knocking my head against the wall for hours now

[wintry steppe]

does the notation [blah] mean subvector space gerated by blah?

[misty storm]

they're both subspaces

probably should've mentioned that

[wintry steppe]

what did you try

[misty storm]

I've transformed it to echelon form

and tried figuring out the equations from there

not only have I got nowhere, I'm not even sure of what I'm looking for

I suppose another, smaller (or at most equal), subspace

[wintry steppe]

equations might help

if you have equations for U and other for W then joining them together is for intersection

[misty storm]

well, I have the equations for both of them

I suppose that, since I have both of them

intersecting

it would be

W=U

v1 * c1 + v2 * c2 + v3 * c3 = v4* c4 + v5 * c5

right?

v1 * c1 + v2 * c2 + v3 * c3 - v4* c4 - v5 * c5 = 0

alternatively

[wintry steppe]

uh

i dont think its only 1 equation for either

[misty storm]

well, I'm just saying that to illustrate that, the way I see it, there shouldn't be equations for either, but rather for both

those at the bottom are c1/c2/c3/c4/c5 which multiply each of the vertically alligned vectors up at the top

my camerawork needs a wee more practice, I'm afraid

c4 gets cut out cause it equals 0

I just don't know where to go from now

[wintry steppe]

sry i cant see well but if you get a system of equations for U and another for W

then join the equations

you should be done

cuz if u have x belongs to U iff p(x) and x belongs to W iff q(x) then x belongs to intersection iff p(x) and q(x)

[misty storm]

I've got a system of equations for both

since they're intersected

I suppose that the system on the left equates the system on the right?

like, I have that:

c1 + c2 + 2c3 = 0

and c1= -c2 -2c3

I can't really get anywhere from here

[wintry steppe]

um

systems dont equal things (at least in the sense youre thinking)

each system is a bunch of equations

[misty storm]

| 1 3/2 3/2 | | 1/4 -1 |
| 0 1 -1 | und | 1/4 -2 |
| 0 0 0 | | 1 0 |

this is the echelon form*

well

of both combined

feck

forget about it, I give up

I'll just hope this doesn't show up on the test

[wintry steppe]

uh sry im in phone on bed hard to answer in detail

im sure some1 else can help

[misty storm]

don't worry about it

[glad finch]

Can anyone help me with this? I'm so lost 😭

[lavish jewel]

try doing substitutions with what you're given

[glad finch]

What I'm given is b (rref A), but it asks about A column

[twilit anvil]

no, youre given A and its columns as well

it is the first matrix that appears

the rref is the second one

i hope im not misreading something

[zealous junco]

2questions: 1. is dual problem equivalent to finding KKT pts or is there some difference and 2. can the lambda and v here be complex

[lavish jewel]

the dual problem is related to maximizing wrt the lambdas and vs in what you have written

iirc they can indeed be complex, but depending on whether x is also complex, one has to be careful with differentiation

the lagrangian may not actually be differentiable, but one can use wirtinger calculus

what?

(left null space)

lol

[olive halo]

does R refer to image here?

[lavish jewel]

if you did the previous one, you should have that x is in the null space of T

should be due to the definition of the (hermitian) transpose of T

[olive halo]

I thought it just represented the possible values of T*

[lavish jewel]

the upside down "T" there means orthogonal complement

the * should be the conjugate transpose

[zealous junco]

but are they both finding possible minimizers?

of the original constrained opt problem

[lavish jewel]

yeah

iirc the dual problem with the lagrangian is to maximize wrt the extra variables that incorporate the constraints, followed by minimization wrt the original variable

finding a saddle point of the unconstrained lagrangian

[zealous junco]

thanks

and yea i think the intuition may be the maximum of the infimum being the exact minizer

[wintry steppe]

Let A be complex n x n matrix and ⟨ , ⟩ be a standard scalar product of complex vectors. Let $F(A) = {⟨Ax, x⟩; x \in \mathbb{C}^n, ||x|| = 1 }$. What does F(A) look like if A is hermitian? Also, show that all eigenvalues of A are included in F(A) (even if A isn't hermitian).

does anyone have an idea how to approach this

[stoic python]

Matejp1

[dire thunder]

well for eigenvalues you have

<Ax,x>=<ax,x>=a<x,x>=a where a is eigenvalue

[wintry steppe]

wait which x do you take

do I have to take x to be a normed eigenvector?

@dire thunder

[dire thunder]

well yes

[wintry steppe]

ok thanks how about the other part where i need to describe F(A) for hermitian matrices?

[dire thunder]

well hmm

<Ax,x>=<x, Ax>

[wintry steppe]

yes that's all I can figure out

then i'm stuck

[dire thunder]

prolly a way to go is just first constructing a couple of examples

like trivially for identity matrix F(A)={1}

ok, i have an intuition that F(A) has to be connected with eigenvalues of A

but not sure how to prove it

[wintry steppe]

hmm okay thanks for help

[lavish jewel]

the eigenvalues part should be straightforward

[wintry steppe]

yes i figured it out now

but not the other part

[lavish jewel]

not sure what they meam by what it looks like

[wintry steppe]

it says I need to describe it

I suppose there is an elegant way to write what it looks like idk

I have absolutely no idea how to approach those problems from the last chapter

Can you help me with this one: Let A be a real symmetrix n x n matrix. Show that complex matrices $I - iA$ and $I + iA$ are invertible. Also show that matrix $(I-iA)(I+iA)^{-1}$ is unitary.

[stoic python]

Matejp1

[lavish jewel]

i think they might want something more elegant, but

[dusky epoch]

spec(A) ⊂ R lol

[lavish jewel]

let's say we have this symmetric A. its eigenvalues are real, and so when we multiply by i, we get imaginary eig vals

then it's simple to show that eigenvectors of A are also eigenvectors of Ai + I

where the eigenvalues are 1 + i*(whatever eigenvalue A had, which are real, and so they only modify the imaginary part of the eigenvalue that always has a nonzero real part)

or -, if the iA matrix is subtracted

(this is the same as ann pointed out)

[wintry steppe]

why do we get imaginary eig vals if we multiply A by i

[lavish jewel]

because of this

symmetric matrices are also hermitian, and then it's straightforward to show that they have real eigenvalues

[wintry steppe]

I don't see why multiplying A by i gives us eigenvalues multiplied by i

[lavish jewel]

and a real number times i is imaginary, with real part necessarily 0

well, just use the properties of multiplication

namely, associativity

we have iAx

[stable kindle]

imaginary numbers are complex

[lavish jewel]

just do i(Ax) = i(lambda x) = i lambda x

[wintry steppe]

ok I see it now

ok I get the first part now

but how does that imply invertibility

[lavish jewel]

all the eigenvalues are nonzero

[wintry steppe]

oh i didn't know that implies invertibility

[lavish jewel]

the determinant is the product of all the eigenvalues

[wintry steppe]

oh and if determinant is nonzero it is invertible +

[lavish jewel]

i can't see the second part off the top of my head

[velvet basin]

is it the determinant of A or B that i am trying to find thats equal to 0

A?

can u find determinants of nxm matrices or does it need to be n x n

[nocturne jewel]

det is a strictly square matrix property

if Ax=b doesnt have a unique x, then det(A)=0, and A is 3x3

[velvet basin]

alright thx :)

[wintry steppe]

How do i know which row operations is the best to do to achieve row echelon form using gaussian elimination?

Is it just trial and error?

[half ice]

You can always do it by

• choosing the leftmost column
• Dividing everything to get 1s and 0s on the column
• Subtracting rows from eachother to have exactly one 1 left
• Rearrange the rows to put it in the correct spot
• Repeat for next column

[limber sierra]

your goal is to make every entry below your pivot variable 0

so for each column

identify your pivot

find the easiest way to do that

move to the next column

repeat

[wintry steppe]

is that the easiest way of doing it? It seems as tho it is

[half ice]

No haha

It's a sure way to do it

[wintry steppe]

but u can only do operations for the whole row

[half ice]

You can likely do it faster if you're willing to take advantage of 0s and stuff

That comes with practice

[wintry steppe]

afaik i only need to make the 4 in row 3 a zero as well for it to be in normal row echelon form right?

[half ice]

I suggest swapping row 2 and row 3, as you don't want a 0 in the middle there

Then divide that by 4, you can put a 1 in the middle

Then finally divide the last row by -3

[wintry steppe]

ahh yeah because if theres a zero in the middle it would have to be in the bottom right?

because that row has the most zeros

or idk

[half ice]

1 1 1 | 1
0 1 2 | 1/4
0 0 1 | 2/3

[wintry steppe]

the row with most zeros have to be at the bottom right

or is it only if the row is all zeros

[half ice]

If there's a 0 where you think you can put a 1, swap it out

You'll be bringing empty rows down

[wintry steppe]

why cant there a be zero in the middle?

[half ice]

Because you can't get rid of that 1, and you don't want that 0 above it

1 1 1 | 1
0 0 1 | 2/3
0 1 2 | 1/4
Like, you could get this if you don't rearrange, not row echelon

[wintry steppe]

what do you mean?

oh im just trying to do regular row echelon form, it says i dont have to have the 1's

is there a requirement for the second pivot in the second row to be the next element of the column or not?

if u understand what i mean

like the picture above, the second pivot is in the third column not in the second

does that matter?

[half ice]

A pivot is always to the right of the pivot above it

[wintry steppe]

cause as far as i can see there is no pivot in the second column

[half ice]

1 1 1 | 1
0 0 1 | 2/3
0 1 2 | 1/4
Breaks that rule, and is not row echelon form

A row swap fixes it:
1 1 1 | 1
0 1 2 | 1/4
0 0 1 | 2/3

If you don't need 1s for pivots, then you don't have to divide like I did. But you still do need the row swap

[wintry steppe]

okay but does the second column always need a pivot thats nonzero?

cause i understand that the pivot is always to the right of the pivot above it

but is it one element to the right or just to the right in general?

[half ice]

This is also echelon form
1 1 1
0 0 1
0 0 0

[wintry steppe]

ah so it doesnt matter where to the right the next pivot is

as long as its to the right of the one above it

[half ice]

Yaya

[wintry steppe]

i see, i thought they had to be like
pivot 1 1
0 pivot 1
0 0 0

[half ice]

They often will be, but not always

[wintry steppe]

okay thanks man

[zinc wasp]

Anyone able to help figure out what equation I need for this. I am stuck

[dreamy iron]

Is this rank nullity?

This stinks of rank nullity.

[native rampart]

In a sense,yes

[topaz jetty]

can anyone suggest me a good playlist on linear algebra ?
i really need that plz

[lavish jewel]

gilbert strang's content

[dreamy iron]

1. From the Axe man himself:

http://www.youtube.com/playlist?list=PLGAnmvB9m7zOBVCZBUUmSinFV0wEir2Vw

2. Robert Won, University of Washington:

https://youtube.com/playlist?list=PLoxJTbDttvt7ny0WEJHWw6-0Sjx7EImIQ

3. Erin Pearse, Cal Poly SLO:

https://youtube.com/playlist?list=PLBUiHiRFQhsI--yc2PoCcK17fUR_mNJNH

3.a. Part 2, continuation of the above:

https://youtube.com/playlist?list=PLBUiHiRFQhsJg3fWe17OBx-PZYOKasoSA

4. Aviv Censor, Technion University, Israel:

https://youtube.com/playlist?list=PLW3u28VuDAHJNrf3JCgT0GG_rjFVz0-j9

5. I don’t know the dude, nor the institution. My best guess is Berkeley. From LADR 2nd Ed.

https://youtube.com/playlist?list=PLflMyS1QOtxwiN5oOuyY4W_8fZlTTnRcF

6. For something different: Math the Beautiful. It’s more “slice of life.”

https://youtube.com/playlist?list=PLlXfTHzgMRUKXD88IdzS14F4NxAZudSmv

7. of course has a legendary playlist. I myself have found it useful for building intuition.

Enjoy!!! Most of these are lecture series based off of Axler.

2, 3, 4, 5, are really enjoyable and in my opinion very useful.

1 is added for completeness (but not recommend).

6 and 7 are fun.

@ mods, I think that set of playlists is sufficiently useful to merit being pinned. I hope y’all agree.

[topaz jetty]

@dreamy iron
thank thank you so much
it's mean a lot to me
again thank you

[topaz jetty]

I'm agree

[dreamy iron]

Are you going through LA as a first pass, or as a second pass?

[topaz jetty]

second pass

[astral fern]

hola
I've got a rotation matrix that needs to be offset (it's object movement, the offset is for transferring it into a software with different coordinates), yet applying this offset through Matrix.Rotation(radians(-90.0), 4, 'Y') @ Matrix.Rotation(radians(-90.0), 4, 'Z') results in a value jump (as seen on the second pic). Is there a name for it? I'd like to avoid its occurence

[lavish jewel]

wdym by it needs to be offset

and what do the axes in the plots represent

[crude falcon]

it is correct to say, that the reason a vector space can have multiple basis, is because you can find multiple sets of linearly independent vectors that spans the whole space?

[dusky epoch]

it's tautological

so calling something a 'reason' for itself is a bit... eccentric at best

[crude falcon]

thats all it came to mind haha, maybe theres a proof or something but idk

[dusky epoch]

aside from some edge cases, it's very easy to construct two different bases for the same vector space

take one basis, and then construct another by taking the same basis and doubling one of the vectors

[wicked palm]

or just multiplying it by any scalar, so many options!

[dreamy iron]

https://cdn.discordapp.com/attachments/540211747613704221/843749418996858940/162017091996745728.png

[wicked palm]

I'd try and reconstruct v1, v2 and v3 from linear combinations of v1+v2, v2+v3 and v3

[wintry steppe]

b-but what about the vector space {0}!!!!!

[wicked palm]

?!?!???????!!!!!

[dusky epoch]

i said

aside from some edge cases

i didn't feel like covering every single one, but {0} is one of those edge cases

[dire thunder]

even zero scalar?

[wintry steppe]

What does it mean exactly when a systems of equations have solutions?

[lavish jewel]

it means it was possible to solve it exactly

[wintry steppe]

like using gauss jordan for example

i mean how would i find out how many solutions a systems of equatiosn have when its in matrix form?

[dire thunder]

find rref of matrix

[wintry steppe]

rref is unique right so that means one solution exactly?

[dire thunder]

rref is unique

and if rref is identity matrix then exactly one solution

if you arrive at zero row in rref of matrix and nonzero corresponding entry in b vector then no solution

otherwise infinitely many solutions

[lavish jewel]

you could also find it has no solutions

if a row has all 0s but ends in something nonzero

[dire thunder]

tha is what i said

[wintry steppe]

but to know how many solutions then i would have to find rref ? It is not substantial to just find ref?

[lavish jewel]

i read too fast, oops

[dire thunder]

well there are cases when you can find set of solutions' size w/o rref

but in general approach is find rref and it says you the number of solutions

or you can use determinant

[wintry steppe]

So i have an example here of a systems of equations, there is a question that says "show that the systems of equations doesnt have a solution for a = 1/4"

So i would find the rref given that a = 1/4 and so i would apply the rules that you wrote before right

?

[dire thunder]

yes, but it seems for me that in that particular case it may be possible to avoid rref

i am not sure

but like your second and third equations should be equal then

[wintry steppe]

is solutions also what we find when just doing row echelon form and then doing back substitution?

regular gauss elimination

[lavish jewel]

they should be equivalent

you'll run into any inconsistencies eventually when back substituting

[wintry steppe]

so the best way is to just do rref right

[lavish jewel]

whatever you find easiest. and vimes said, in some cases it's pretty to just "see it" without even doing anything

in others, you need something that reveals the "rank"

[wintry steppe]

okay thanks

whats a free variable exactly?

Is that like the variable "a" in the picture above?

[lavish jewel]

it's a variable that always ends up multiplied by 0

so its value does not change whether the solution is reached

[wintry steppe]

when is that the case?

[lavish jewel]

when you get a row of all zeros

[wintry steppe]

ah

is there any reason to know which variable is free and which is basic? As in why would i need to know that when performing gauss elimination

[frosty vapor]

a row?

i thought it was a column

[wintry steppe]

or whats the importance of it

[frosty vapor]

or maybe same thing

cus transpose or wtv

[lavish jewel]

not reasons, but rather, doing those operations will tell you which variables are free

so it's backwards

and yeah maybe it's columns, i'm eating and probably making many mistakes

[wintry steppe]

yea i misworded myself, but whats the importance of knowing it after doing rref?

[frosty vapor]

when constructing the solution to some system free variables are given as arbitrary constants

[lavish jewel]

if variables are free, you have infinitely many solutions (if the system is solvable)

knowing which variables are free lets you express the general form of ALL the solutions

[reef sage]

anyone know how to take a subset of a matrix in Armadillo for c++?

pop is my vector

[lavish jewel]

try #computing-software

[reef sage]

ah

didnt see

[dire thunder]

ye, it is columns

row would mean just that codomain has dimension larger than domain

[lavish jewel]

but say you have a 4x4 mat with 2 rows 0, and 2 lin indep rows

you have infinitely many parameterized sols

[dire thunder]

well prolly me also picked wrong words

but point was that zero row does not guarantee by itself free variable

3x2 matrix is an example

[lavish jewel]

i did put that it was in the case you already know it has sols, then 0 rows blah blah

some details missing 😛

[dire thunder]

btw why you are not mod

you were mod iirc

[lavish jewel]

i was never a mod

[wintry steppe]

does leading 1's in every row mean one solution?

[lavish jewel]

for square matrices, yes

[wintry steppe]

and a complete zero row with the b also being 0 is infinite solutions

[dire thunder]

not necesarily

2x+3y=1
x+y=0
has a solution (-1,1)

and if we extend this system to be
2x+3y=1
x+y=0
x-y=-2

then it would still have exactly one solution tho there will be zero row

[wintry steppe]

how can i idenfity an infinite solution matrice then?

[dire thunder]

2x+3y=4
6x+9y=12

find rref of this

or better example x+y+z=1
x-y+z=2

[north anvil]

Are variables that represent unknown matrices usually capital?

[wicked palm]

yur...

[north anvil]

Thanks - is this by convention?

[wicked palm]

not sure

I've never seen a lowercase matrix though

[nocturne jewel]

yes, usually

similar to side lengths being lowercase, angles being greek letters

$A=[a_{ij}]$

[stoic python]

Mosh

[north anvil]

👍

makes sense to me

sometimes with the law of sines, i remember angles being represented as uppercase letters

and the sides being lowercase letters

[nocturne jewel]

yeah, the vertices are capital letters

[north anvil]

RIGHT

the points are

given that in geometry, a triangle is labeled ∆ABC

using capital letters

[nocturne jewel]

you also sometimes get capital greek letters for matrices in some factorizations

[north anvil]

Don't think I'm there yet, but I'll cross that bridge when I get there.

[nocturne jewel]

$M=P\Sigma Q^*$

[north anvil]

Appreciate the help - respect 👊

[stoic python]

Mosh

[twilit anvil]

sanity check: i have a problem that says the 3x3 matrix

0 1 1
1 0 1
1 1 0

has 1 as an eigenvalue.

this is incorrect, and probably shouldve been -1, right???

[quartz compass]

yeah

[silent sandal]

If a $n$ by $n$ real matrix $A$ is diagonalizable and has eigenvalues $a_1 \leq a_2 \leq ... \leq a_n$. It's not hard to show that $||A||_2 \leq \rho \sqrt n $ where $\rho = a_n$ is the spectral radius of $A$ (assuming I haven't made any mistakes here). Is this true in the case $A$ is like the above, but possibly not diagonalizable?

[stoic python]

phao

[silent sandal]

$||A||_2$ in there is the induced operator norm from the usual Euclidean vector 2-norm in $R^n$.

[stoic python]

phao

[silent sandal]

To show that inequality, I used $||Ax||_1 \leq \rho$ for $x \in R^n$ with $||x||_1 = 1$. Then I used that $||x||_1 \leq \sqrt n ||x||_2$.

[stoic python]

phao

[silent sandal]

So that $||Ax||_2 \leq ||Ax||_1 \leq \rho ||x||_1 \leq \rho \sqrt n ||x||_2$.

[stoic python]

phao

[silent sandal]

The issue is that the way to show $||Ax||_1 \leq \rho||x||_1$ relies on $A$ being diagonalizable.

[stoic python]

phao

[silent sandal]

And... $||A||_1 \leq \rho$ isn't true in general...

[stoic python]

phao

[silent sandal]

Btw... $a_1$, above, is positive. Forgot to say that.

[stoic python]

phao

[silent sandal]

(not sure if this is the right place to ask though)

[wispy pewter]

Is there a professor Leonard YouTube series equivalent for linear anyone recommends?

[nocturne jewel]

3B1B's essence is a good series

certainly not like Leonard's stuff

[edgy tree]

Will i get different results for changing the order by which i multiply the matrices ?

like for example if i have 2 3x3 matrices A and B

Is AB not equal BA ?

[lyric swallow]

in general yes

[edgy tree]

Alright thanks

[nocturne jewel]

Yes, matrix multiplication is not commutative

[wintry steppe]

what if it was

[limber sierra]

meme: the world if

[sci fi utopia picture]

if matrix multiplication commuted

[wintry steppe]

does free variables always mean infinite solutions?

[dire thunder]

yes, if your system results in free variables you have infinitely many solutions

unless working in finite field

[wintry steppe]

would this matrix be considered in row echelon form ?

so when theres free variables it also means that you cant put the matrix into reduced row echelon form

because the answer has one solution only

[native rampart]

Yes

You can convert it to RREF ,tho

Do R_1-R_2->R_1

[wintry steppe]

isnt that reduced row echelon form

[native rampart]

Yes

[wintry steppe]

oh u answered the picture question

but wouldnt the third column be a free variable

[native rampart]

Third column is not a free variable

You have no free variables here

[wintry steppe]

columns that dont have a leading value right they must have a free variable

or am i wrong

[feral iron]

No idea what to do here

I think that matrix im solving for is going to be D*R

okay I get it now actually, the rotation part confused me. I didn't get what it meant by rotating A into B. It just meant rotate A toward B

[wintry steppe]

but reduced row echelon form seems to be able to have free variables as well, isnt that contradicting, cause that would mean that there is infinite solutions. Does that mean i intepreted it wrong and RREF there can be infinite solutions as well?

[sage ibex]

I don't think you need choice at all. The span of the v_i's has the v_i's as a basis. Then you can get such a map on that span, and extend this map to the whole V by setting everything outside to 0

[weak needle]

It does need choice to expand vi’s to a basis

[sage ibex]

I'm not doing that

I'm taking their span

[weak needle]

You can’t set outside things to zero exactly, you have to set outside basis elts to 0

[sage ibex]

Oh you're right

[weak needle]

Things outside will still be nonzero, (v0+b) for some b in the basis

Also if raghuram still cares, I found out you do need some choice to do this but i’m not sure how much. There is a construction of vector space that has a dual equal to 0 in a different system (has no choice obviously)

[sage ibex]

Nice

Apparently you need the full strength

What I was saying gives you an equivalence between existence of such a function on (V,{v_i},v_0) to extendibility of a function from span({v_i}) to V. This thing is implied by the existence of a complementary subspace of span({v_i}), because if such a space exists then you could define an extension to be f(v+w) = f(v) where v is in the span and w in the complementary space. Also if such an extension exists then it gives a complementary subspace of span(v_0) by taking the set of zeroes. The existence of a complementary subspace of the span of a singleton for any singleton gives existence of complementary subspace of any subspace by taking intersections of [nvm this part doesn't work as I thought it would]

But we do get that the existence of such a function is equivalent to the existence of a complementary subspace for any singleton

Existence of complementary subspace for any given subspace is known to be equivalent and I feel that the singleton case could somehow be reduced to that

[dire thunder]

rref does not mean that solution is unique

rref may have free variables

{{1,0,0},{0,1,1}} is in RREF

[wintry steppe]

alright

but we can agree that this REF has a free variable in column 3?

[dire thunder]

yes

[wintry steppe]

to turn it into rref i would just have the subtract row 1 and row 2 so i get a zero in the middle top row but why is it favorable to have it in rref form?

[dire thunder]

well

for ref you would have to do backwards substitution

for rref there is no need

[lavish jewel]

as far as yesterday's discussion goes, btw, this will have no columns with all zeros, but it's equivalent to a square matrix with a row of zeros

[crude falcon]

is it true that a set of n vectors in R^m are linearly independent if n > m?

[native rampart]

No

Take {(1,0,0,...),(2,0..),..}

[crude falcon]

also if I have V and U subspaces of R^4 with dim(V) = 3 and dim(U) = 1, if U and V are disjoint then V+U = R^4?

I think so because if they are disjoint then the vectors of its basis would be all linearly independent after grouping them right?

[hushed grove]

Hi all. I can't seem to remember what taking a matrix $A^{25}$ even means.

[native rampart]

Yes

[crude falcon]

you need to find the diagonal matrix using those eigenvectors, and then A^25 will be elevating the elements of the diagonal

[hushed grove]

Thanks for the hint @crude falcon. Let me work off of that

[dusky epoch]

@hushed grove A^25 means multiplying together twenty-five copies of A

mind you, "what it means" != "how to find it", and jackieto answered the latter question instead of the former

[crude falcon]

true, also one silly question about diagonalization: why do I need to compute D = P^-1AP? I mean, when you compute the eigenvalues and its eigenvectors, you can check if the matrix will be diagonalizable or not, and if so, why can't you just find D? I mean you already know how it looks like already right?

[dusky epoch]

you don't "need" to do anything

but if you wish to find a power of A then you will need the eigenvectors (and hence P) anyway

[crude falcon]

yeah I'm talking just in general, I've been taught to find the diagonal matrix that way, but just realised I know what D is before using that formula

[dusky epoch]

the diagonal matrix holds the eigenvalues

i guess you could do P^-1 AP as a way to check yourself

[old flame]

Quick question, if we know that S is nilpotent where $S^{6}=0$, then how does it imply that $S^{3}=0$. Is it because $S^{6}=(S^{3})^{2}$.

[stoic python]

Otoro

[dusky epoch]

S^6 = 0 does not imply S^3 = 0, what are you talking about?

[old flame]

Oh so we're trying to prove that $T(z_1,z_2,z_3)=(z_2,z_3,0)$ does not have a square root. So we assume that there is $S^{2}=T$

[stoic python]

Otoro

[dusky epoch]

why didn't you say so outright...

this is an important detail you omitted

[old flame]

So I found that $T^{3}=0$, so we know that T is nilpotent

[stoic python]

Otoro

[old flame]

Sorry about that

[dusky epoch]

so what you really have is a 3 by 3 nilpotent matrix

any 3 by 3 nilpotent matrix has nilpotence index at most 3

while your sqrt(T) would have nilpotence index 6

[old flame]

Wait why is it 6 ?

[dusky epoch]

okay i mean maybe that's a bold statement

it'd be better to say that (sqrt(T))^4 = T^2 != 0

and sqrt(T)^3 must be 0 by virtue of sqrt(T) being a nilpotent operator on a space of dimension 3

[old flame]

Ahhhhhh okay thanks

Wait but how to we determine that sqrtT ^{3} must be 0 ?

[dusky epoch]

a nilpotent operator on a space of dimension 3, raised to the power of 3, must necessarily become 0

[old flame]

So you're saying since T has at most nilpotence of 3, sqrtT also has at most 3 ?

Ahhhhhhh okay, yup, N^{dim V}=0, for N being nilpotent

Gotcha thanks

[wintry steppe]

Suppose V and W are finite-dimensional and that U is a subspace of V. Prove that there exists T in L(V,W) such that null T = U if and only if dim U ≥ dim V - dim W.

If we first assume that it does exist, then just note that by the fundamental theorem of linear maps: dim ker T + dim img T = dim V, so dim U = dim V - dim img T ≥ dim V - dim W, since img T is a space of W.

Now if we assume that U is a subspace of V such that dim U ≥ dim V - dim W, then let T in L(V,W) be defined by Tu1 = 0, Tu2 = 0, ..., Tun = 0, where u1,...,un is a basis of U, and define Tvi = w1 if vi not in {u1,...,un}

Then dim U ≥ dim V - dim W implies dim null T ≥ dim V - dim W, and again by theorem of linear maps: dim null T ≥ dim null T + dim img T - dim W gives us dim W ≥ dim img T, which is clearly true since img T is a subspace of W

Is this proof valid?

[viral flint]

Wow so I didn't need to lose hope on getting an answer

[viral flint]

oof

[viral flint]

IDT the existence of the extension implies the existence of the complementary subspace. Aren't you just stuck with finding a complementary subspace of span(v_i) in ker(functional)?

[viral flint]

Yes. I realised you can first take the quotient by span(v_i) and so assume there are no v_i, only v_0.

Maybe if any singleton has a complementary subspace we can prove Choice?

[wind mountain]

Are there any sources which derive the formula for elements of the inverse of the Hilbert matrix in a elementary way?

I was asked to compute elements in the exam but I couldn't figure out a general formula

[sage ibex]

Yeah thats what I reduced to

But not sure if thats possible, maybe I will go over the usual proof and see if it can be modified for singletons only

[wintry steppe]

Can anyone comment on the proof I gave above?

[lavish jewel]

the second direction looks a bit shady to me

lemme read it again

yeah it looks like you assumed both that U is the null space and that dim U >= dim V - dim W

[wintry steppe]

@lavish jewel What do you mean?

[viral flint]

Wait what's the usual proof?

Does it not need Choice even with complementary subspaces?

[nocturne jewel]

I think Edd meant you assumed U was the nullspace/kernel, as opposed to a general subspace

[wintry steppe]

@nocturne jewel Null space of what?

[nocturne jewel]

Im just explaining what Edd said.

[lavish jewel]

of T

someone please correct me if i'm wrong, as i often am. saying dim U >= dim V - dim W means also that dim V >= dim W. then in the worst case, dim U = 0 (i.e. when dim V = dim W) and an invertible (full rank, with im(T) = W) linear transformation T can be constructed, so that only T(0) = 0. otherwise, dim V > dim W, and then we can construct U' so that U dir. sum. U' = V, and a T can be constructed so that T(u'_i) maps to w_i in W and T(u_j) maps to 0. maybe something like that?

[wintry steppe]

But I'm defining T @lavish jewel

[lavish jewel]

i guess my issue was mainly with the wording

[wintry steppe]

Which part?

[lavish jewel]

idk if implies is the right word there

more like you defined T so that U = null(T)

i think i'm just nitpicking

[wintry steppe]

@lavish jewel Implies where?

[lavish jewel]

Then dim U ≥ dim V - dim W implies dim null T ≥ dim V - dim W,

the way you defined T, dim (U = null T) >= dim V - dim W

i think you should just ignore my comments 😛 everything else looks ok to me

[wintry steppe]

@lavish jewel But that's the definition of null T

With the way I defined T

[lavish jewel]

as i said, it was a nitpick on the wording

[broken sun]

Hello. The range of $T-\lambda I$, where $\lambda$ is an eigenvalue, is $T$-invariant?

[stoic python]

MathPhysics

[wintry steppe]

Suppose f and g are linear maps from V to F that have the same null space. Show that there exists a constant c in F such that f = c*g

I need a hint for this problem please

[broken sun]

<@&286206848099549185>

[timber cargo]

?

[broken sun]

Hello. The range of $T-\lambda I$, where $\lambda$ is an eigenvalue, is $T$-invariant?

[stoic python]

MathPhysics

[lavish jewel]

V is a vector space over F? is that all you're given?

[wintry steppe]

Yeah lol

[lavish jewel]

that doesn't look true to me in general, i think i'm missing something

[wintry steppe]

It's easy to prove that there exists a constant c dependent on v

But I don't know how to show that there's a constant that works for all v

[lavish jewel]

right

[wintry steppe]

So I am stuck

[verbal pivot]

it could be useful to write a basis of V, what happens to this basis under the transformations f and g?

[wintry steppe]

I tried that