linear-algebra

[vague belfry]

B is not a good one

[patent cloak]

The problem is for the eigenvalues, right?

[lone cove]

the eigenvalues only square if theyre on the same eigenvector

[patent cloak]

Ooooh

[slow scroll]

ah i have an idea

[lone cove]

yeah, i missed it at first too

[vague belfry]

I found the original problem

http://web.mit.edu/18.06/www/Spring14/quiz_3_draft.pdf

4B

http://web.mit.edu/18.06/www/Spring14/quiz_3_solution.pdf

in the solution

Maybe I missed some info that could've helped

[lone cove]

whats a singular value

[vague belfry]

nvm

I'm an idiot

[slow scroll]

what?

[vague belfry]

well the implication wasn't that if AtransposeA and BtransposeB have the same eigenvalues then A and B have the same eigenvalues

it was that if they have the same eigenvalues

then A and B have the same singular values

since you just take the square root

Sorry 😐

[lone cove]

its cool

[chilly dragon]

is this set linearly dependent ?

[native lodge]

well v1 + v2...

is v3

[quaint heart]

Lmao

[chilly dragon]

yea i know

[native lodge]

so it's....

[chilly dragon]

the thing is i can't solve this using rref

i get the 0 vector

[quaint heart]

Any n elements of $\mathbb{R}^{n-1}$ are linearly dependent

[stoic python]

Liquid:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly dragon]

ok thanks for that

but how would i prove it using the definition in linear dependence

[quaint heart]

Um you can span the zero vector with v_1 +v_2 - v_3

Like amphy said

[native lodge]

(c1 * v1) + (c2 * v2) + (c3 * v3) = 0
if c1=c2=c3=0 is the only combo that gives the zero vector then is it independent
if there is a nonzero combo of c's then they are dependent

I see that (1, 1, -1) gives zero

[chilly dragon]

yes

[native lodge]

so then they are dependent

[chilly dragon]

but how would i find (1, 1, -1)

i know that dude

[native lodge]

by inspecction

[chilly dragon]

i said i can't find the vector using row echellon form

[native lodge]

like this one was fairly trivial

[chilly dragon]

i get (0 0 0)

[quaint heart]

Lol

[chilly dragon]

😂

[quaint heart]

Do you know what row echolon form is?

[chilly dragon]

i'm confused af

yes

ok maybe not for sure, but i know gauss-jordan ie comfortable with it

[native lodge]

$\begin{bmatrix} 1&0&1\ 0&1&1 \end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[native lodge]

this is close to rref already lol

[quaint heart]

N vectors are linearly independent iff the matrix with them in it can be put into row echelon form with no zero rows

Or I guess columns actually

[chilly dragon]

$\begin{bmatrix} 1&0&1\ 0&1&1 \end{bmatrix} = \begin{bmatrix} &0 \ &0 end{bmatrix}$

[quaint heart]

If you use amphys representation

[chilly dragon]

$\begin{bmatrix} 1&0&1\ 0&1&1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$

you do that

[gray glen]

[quaint heart]

Vectors being linearly independent corresponds to their matrix being fullrank

[native lodge]

$\begin{bmatrix} 1&0&1\ 0&1&1 \end{bmatrix} \begin{bmatrix} c_{1}\c_{2}\c_{3} \end{bmatrix}=\begin{bmatrix}0\0 \end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[quaint heart]

Which I guess is a bad explanation lmao

[chilly dragon]

so $ c_1 + c_3 = 0 \ c_2 + c_3 = 0 $

[stoic python]

Wosco:

[quaint heart]

Cause you probably don't know what the rank is

[chilly dragon]

^

[native lodge]

this matrix is already in rref, and column three is a free variable so let's choose c_3 = 1

$\begin{bmatrix} 1&0&1\ 0&1&1 \end{bmatrix} \begin{bmatrix} c_{1}\c_{2}\1 \end{bmatrix}=\begin{bmatrix}0\0 \end{bmatrix}$

[chilly dragon]

why is it a free variable

[stoic python]

⚡Amphy⚡:

[chilly dragon]

i thought a free variable is when 0 = 0

which means an infinite solutions

[pallid rampart]

https://math.stackexchange.com/questions/720971/what-does-it-mean-for-the-free-variable-and-leading-variable

[chilly dragon]

^ this made things clear for me, thank y'all for your help

[pallid rampart]

You're welcome ask again if you have a question

[indigo cradle]

how may i prove this?

where P is a permutation matrix

[dusky epoch]

really it doesn't matter that P is a permutation matrix bc this holds no matter what P is

$u \cdot v = u^Tv$ for any col vectors $u, v$

[stoic python]

Ann:

[dusky epoch]

$(Px) \cdot y = (Px)^T y = x^TP^Ty = x^T(P^Ty) = x \cdot (P^Ty)$

[stoic python]

Ann:

[indigo cradle]

omg thanks

that looked familiar

and i realised i actually wrote the proof earlier in my notes

oops

[indigo cradle]

could i have help making sense of this question

why can't D be negative

the question is asking for me to come up with S

[wintry steppe]

Any matrix S?

S=[-1]

[indigo cradle]

this was the answer

i thought it was any S too

[ripe hemlock]

what is the prequisite knowledge i need before studying this?

want to knock it out of the way before i head into college in september

[slow scroll]

anybody taking linear algebra has at least had calculus, but its not a prereq per se

[grave plank]

im not really sure why calc is required for linear algebra

[hasty ether]

it's not everywhere

there's usually the one example about a group made of things and their derivatives or whatever but usually LA helps with calc not the other way around

it depends on the level of LA

[granite light]

doesnt LA help derive the half angles etc

[lone cove]

you can use it for that, using the rotation matrices

[spice void]

imo you should take proofs and calc before you hit Linear

There are a lot of things that are insanely interesting when it comes to Cramer's Rule and unless you understand vectors it's kind of over your head

Same with how cross products work imo

Linear really isn't hard, but having been presented with foundational material from other math classes really helps make it easier and more interesting

[indigo cradle]

can i have help making sense of this answer

what is "all pairs that P has in the wrong order"

[sonic osprey]

E.g.

Row 3 is lower than row 5

[bold tiger]

$X^T X$

[stoic python]

Fate:

[bold tiger]

if i understand correctly this is the square of a matrix X

so does that mean the derivative is 2X?

(im trying to understand the normal equation for linear regression)

[sonic osprey]

It's not the square, the square would be X^2

assuming your matrix is square at least

[bold tiger]

what if the matrix is not square?

im just confused over this step in the derivation

i was trying to comprehend where the 2's came from

[crystal glacier]

Is a matrix with no solution (all but one are 0 on the bottom row) automatically not in Row-echelon form?

[wintry steppe]

The 2 comes from the fact that the expression you're differentiating is a "square" in some sense.

[bold tiger]

that's wat i thought but then i was told i was wrong

[wintry steppe]

Where can I find a derivation of the rotation matrix about an axis in euclidean space?

[lone cove]

wikipedia

[wintry steppe]

It's ok I found it ty

Question came up on an old exam and it's not even in the textbook of the course

[hard violet]

yo I could use some help in #help-1 if anyone is free

[last plume]

@crimson violet

[slender yarrow]

[pallid rampart]

[cedar solar]

So keep in mind that $$X \theta$$ is a vector. Replace it with u

[stoic python]

BamBam:

[cedar solar]

$$(u-y)^T(u-y) = (u^T-y^T)(u-y) = u^Tu - u^Ty - y^Tu +y^2$$

[stoic python]

BamBam:

[cedar solar]

But since u and y are vectors, $u^Ty = y^Tu$

[stoic python]

BamBam:

[cedar solar]

This simplifies to $X^TX \theta^2 - 2 X^Ty \theta + y^2$

Take the derivative wrt theta and you get that result

[stoic python]

BamBam:

[long trench]

Do you peeps know any good YouTube channels that cover most of linear algebra?

[slow scroll]

mit ocw

comes to mind..

[half ice]

3B1B does an awesome series meant to teach intuition to somebody who has already been introduced to the concepts

[cedar solar]

I second 3B1B

Going through a theoretical textbook, nothing makes sense. The videos help a ton

[brittle fog]

Dr. Peyam has lots of linear algebra content. I believe he has several organized playlists

[hard violet]

hey does anyone have a really sytematic way to get something into RREF?

[slow scroll]

gaussian elimination

[hard violet]

no gaus jordan elimination

[sonic osprey]

same thing

[hard violet]

.

[dusky epoch]

[hard violet]

one uses row echelon form the other uses reduced row echelon form...

those are 2 different things

[sonic osprey]

ah I'm wrong here. I didn't know they had different names

Anyways, just do gauss jordan elimination what's not systematic about that

[hard violet]

I know what it is, I'm asking if there a more systematic way of performing it than just "look for what cancels"

[slow scroll]

yea. its more systematic than that

[sonic osprey]

That's not how gauss jordan elimination works

[slow scroll]

you use the pivots to delete all of the entries below it. A computer programmed to do this is not "looking" for anything

[native lodge]

Strang, in his intro linear algebra book, talks a bit about how matlab or software at the time performed elimination

Was interesting

[hard violet]

okay is there a way of determining the pivots prior to elimination maybe thats what I'm missing?

[native lodge]

Uppermost left corner contains the first pivot

That’s about the most I can determine at a glance

[sonic osprey]

That's not always true, if that entry is 0

[hard violet]

right but you can also swap rows

[native lodge]

Oh yea

[sonic osprey]

But if you do have a non-zero entry in the first column, you can swap the rows up so that the top left is non-zero

[hard violet]

so what go column by column to get the appropriate 0 values then?

using row ops

[native lodge]

Assuming you have non zero pivot, you use that to zero out all entries below it, then move to the next pivot which lies along the main diagonal, and assuming that pivot is also non zero, use that to zero out all entries below that pivot and repeat

[hard violet]

right okay yeah thats what I was looking for

thanks

[unborn wedge]

Des français ici? Hehe

[winter reef]

How do I prove Cayley Hamilton using Jordan form?

[late rampart]

Well the minimal polynomial is the least common multiple of the minimal polynomials of each block.

(And the lcm divides the product of their minimal polynomials)

[dusky epoch]

show that CH holds for a matrix that is a single jordan block

[late rampart]

Or that. And because of invariance it will follow.

[dusky epoch]

though i guess proving it in the general case requires passing to your field's algebraic closure

to ensure existence of the jordan form

[winter reef]

Whats LCM?

[spring wolf]

context?

lowest common multiple is my first guess

[dusky epoch]

least common multiple, most likely

[late rampart]

Yes.

[winter reef]

Let V be a linear space of finite dimension over $\mathbb{C}$ Assume $\phi \in \text{End}\left(V\right)$ isn't diagonalizable. Can $\phi ^2 \in \text{End}\left(V\right)$ be diagonalizable?

[stoic python]

dog:

[lone cove]

|| probably not ||

[winter reef]

yeah, unfortunately I didn't get any points for such answer

[lone cove]

assume that phi squared is diagonalizable

[winter reef]

I was thinking it has to do something with rotation since phi^2 is just phi o phi

if you dont know how to do it dont throw these at me, I tried to do that with assuming phi is diagonalizable but couldnt do it

[lone cove]

youre working in C, so roots are always acceptable to take over a diagonal matrix

since its just taking the roots of the elements of the diagonal

[winter reef]

so wat

[lone cove]

so you can try to construct a diagonalizable matrix that squares to phi^2 if phi^2 is diagonalizable

[winter reef]

but that doesnt prove anything

it doesnt really prove that phi^2 is diagonalizable iff phi is diagonalizable

[winter reef]

Let P and $Q \neq P $ be affine subspaces of $\mathbb{R}^6$ of dimension 5. Is it true that $P \cup Q$ is a hypersurface of degree 2?

[stoic python]

dog:

[sonic osprey]

Do you mean intersect

[prime escarp]

How does one find a dual basis to a given basis of R^3 space?

[winter reef]

@sonic osprey nope

also you should try MTG arena if you like hearthstone

[coarse juniper]

Is it possible to combine 2 recurrence relation if one relation is valid for even N and other is available for odd N

[sinful drum]

2+2

[sonic osprey]

You can, but it wouldn't be nice @coarse juniper

[coarse juniper]

@sonic osprey Then should I solve for N=even and N=odd separately??

[cedar solar]

@winter reef yeah a rotation matrix is a good counterexample

Try a 180 degree rotation

If you square it you get the identity matrix which is clearly diagonalizable

[hushed smelt]

What do distances in the 2D PCA space mean?

[winter reef]

@cedar solar but rotation Matrix is also diagonalizable

Right?

[dusky epoch]

not in the real case

[wintry steppe]

In R^2 there are 2 diagonalisable rotations

Nothing and 2pi

[dusky epoch]

you mean 1pi

2pi is the same as identity

[wintry steppe]

Oh yeah that too

You either have eigenvalues-1 or 1

Since no other subspace can be be an eigenspace otherwise

[cedar solar]

@winter reef oh i just googled and rotation is diagonalizable in complex. Sorry

And if 180 rotation is diagonalizable in R^2, set phi to 90 rotation

But kinda moot since i realized rotation is the wrong answer for C

[winter reef]

Fuck linear algebra, muting this channel

[cedar solar]

Sorry, i understand your frustration, in asking for help but getting the wrong answers. But people do make mistakes

Try $$\begin{pmatrix} 0&1\ 0&0 \end{pmatrix}$$

[stoic python]

BamBam:

[cedar solar]

This one i checked is not diagonalizable even in C, but its square, the zero matrix, is obviously diagonalizable since it is a diagonal matrix

[winter reef]

No i mean I had final exam and I pretty much done everything right and got shut mark

No worries

[cedar solar]

Oh, yeah linalg is really tough imo

[winter reef]

no its not

Its ez

[cedar solar]

I do like it, in the sense i appreciate how useful and universal it is. But i struggled in that class so much

I guess different strokes for different folks, shrug

[winter reef]

Its just exam being so unfair that frustrates me

[halcyon pecan]

It’s also really tough

Atleast compared other mathcourses taken around that time imo

[winter reef]

Not for me lol, other course is Real analysis

[wintry steppe]

Yeah linear algebra is easier than real analysis

But in the us they do calculus with it so I see the point

[chilly dragon]

the dimension of a subspace is the length of the set of its basis ?

for lR^2 this set {e_0, e_1} is a basis for R^2, and it's length is 2, so 2 is the dimension of lR^2 ?

[quaint heart]

Well the dimension of a space is the number of elements in it's basis yes

This is more general than just for subspaces lol

[chilly dragon]

oh the definition said for a subspace, i was just trying to explain it to myself in a non-formal way

¯_(ツ)_/¯

thanks btw

[gilded junco]

Hello, question:

So it's well known fact that for a function, f, to have a left inverse, say g, it has to be injective (and surjective is optional). I get that. Cool.

Also that for the function, f, to have a right inverse, g, it has to be surjective (and injective is optional). I don't get that, if f is not injective, then the function for a right inverse won't be a function (as two elements in co domain map to same element in domain)

[gilded junco]

omg @slow scroll, i've been necking myself

thank god

this is where i've ended up, in that a right inverse "may get you back to where you started"

Here is a post I made on my class forums - which puts this issue clearly

Or there are two possible right inverses: one such that g(C) = 3, and another one where g(C) = 4. That's where the "you may not get back to where you started" comes from

[slow scroll]

intuitively speaking, it doesn't really matter what happens in between. Only that every input gets mapped back onto itself

[gilded junco]

fuck, if I wanted someone to regurgitate the definitions

[slow scroll]

xd

[gilded junco]

do you intuitively emphasize with the situation i'm in?

Here: my problem simplifies to this

Can you write the right inverse function for that

[slow scroll]

oh i know what the problem is. you have it mixed up lol. The right inverse is injective and the left inverse is surjective

[gilded junco]

false

[slow scroll]

hmm im confused as well i guess

[gilded junco]

ok good

[slow scroll]

nonon

i knwo what im talking about. ur thing just words it differently

if gf is the identity map, such that g is a left inverse to f and f is a right inverse to g.
g is surjective and f is injective. Does that make sense?

[gilded junco]

Do you think in notation?

[slow scroll]

no. i speak in notation

agh im butchering my examples lmao

[gilded junco]

lol

thanks so much for your help

I admit i'm being tough

[slow scroll]

{1, 2, 3, 4, 5}
{a,b,c,d}
g is surjective but not injective (left inverse)
g(1) = a
g(2) = a
g(3) = b
g(4) = c
g(5) = d

f is injective but not surjective (right inverse)
f(a) = 1
f(b) = 3
f(c) = 4
f(d) = 5

verify that gf is maps every letter back onto itself

There

[gilded junco]

sorry to disregard what you just said, but I understand it now

[slow scroll]

now, intuitively speaking, this makes sense because the rightmost function is applied first, and non injective functions "lose" information about their inputs

yea. alright kool

[gilded junco]

Yeah cool

love you

[slow scroll]

[gilded junco]

@slow scroll , can I get you on this?

[amber bay]

1.5 ain't an integer mate

[dusky epoch]

yeah

1.5 is not an integer, much to the surprise of everyone involved

[frosty vapor]

lol

[gilded junco]

I realized that as I was washing the dishes haha, thanks guys

[feral mountain]

why ask questions when you can wash dishes

[gilded junco]

@slow scroll , can I get you on something real quick?

Counter example: m = 5, n=10

I.e. any example where m < n

[slow scroll]

hm one sec

[gilded junco]

nvm

I suppose q can be a negative integer

[dusky epoch]

5 = 10*0 + 5

q = 0, r = 5

@gilded junco

also this doesn't really belong in #linear-algebra

[frank gate]

Can someone explain to me why they decided to perform yet another step on the second entry in the first row on the right matrix when there is already a pivot point in the left matrix??

[terse blade]

are we finding an inverse matrix

[frank gate]

no this is find a vector x for t(x) = ax

[terse blade]

t(x) is....?

[frank gate]

This is the exact one. Question 5

why is it that they make x1 = 3-3x3

when the left one says x1 = -2 + 5x2 + 7x3

[terse blade]

so we're finding x such that Ax = b

[frank gate]

ye

[terse blade]

(A|b) ---ELO---> (A'|b')
A'x = b' after doing ELOs(elementary row operations) still holds

so we're making A' the most easiest form we can use

[frank gate]

so if the answer is x1 = 3-3x3

is x1 =-2 + 5x2 +7x3 still equivalent to that?

[terse blade]

do you mean:\
$x_1 = 3-3x_3$\
$x_1 =-2 + 5x_2 +7x_3$

@frank gate

if so yes

[frank gate]

ehm, isn't it = -2 + 5x2 + 7x3 ?

[stoic python]

chamkkae-ramyeon:

[terse blade]

this

sorry

[frank gate]

ye exactly, okay thanks

= )

[terse blade]

u confused me with all those x

yes

that holds

[frank gate]

oh sorry! Alright tyty

[terse blade]

since $x_2+2x_3 = 1$

[stoic python]

chamkkae-ramyeon:

[terse blade]

np

[wintry steppe]

How would one obtain the intersection of these 2 lines?

[cedar solar]

@frank gate it’s in RREF because we want as few free variables as possible. We want x2 to not be a free variable if possible, so x1 is a function of x3 and x2 is a function of x3

[plush galleon]

10*-2 + 15

chat updated rip

[plush galleon]

@wintry steppe can you ping me when you get an answer to your question

[jagged marsh]

Hi iam struggling here to solve this without any help

the line basically says to show what T is (you dont need to actaully do the math)

as far as I can see the only the third line changes

but with different multipliers

[dawn apex]

Since Z and A are very similar why not try T=1+X

[dusky epoch]

you mean I+X

[dawn apex]

<same thing>

And just assume most of X’s components are 0

[jagged marsh]

wait what do you mean by T = 1 + X ?

iam fairly new

[dawn apex]

maybe a important thing is notice how the second row of Z relates to the third row of A

[dusky epoch]

by 1 they mean the identity matrix

[jagged marsh]

oh yeah

[dusky epoch]

basically try finding a matrix X such that $(I+X)Z = A$

[stoic python]

Ann:

[dawn apex]

(1+X)Z=A
XZ=A-Z

yea

[jagged marsh]

i somehow gotta find a connection between the lines

[dusky epoch]

please don't call the identity matrix 1 @dawn apex

[dawn apex]

ehh it’s kinda common abuse at this point XD

[jagged marsh]

i already have the first second and fourth line 😄

its basically I

now the third need a change that applys to all lines

I cant seem to find it

[dawn apex]

Look at the second row of Z, are they all divisible by some number?
similarly the third row of X-Z
edit: A-Z

[dusky epoch]

$T = \begin{bmatrix} 1 \ & 1 \ a & b & c & d \ & & & 1 \end{bmatrix}$

oops

[stoic python]

Ann:

[dusky epoch]

there we go

you could try to do this

instead

[jagged marsh]

Iam exactly at that point right now

but I gotta find a b c d 😄

I already figured that I need 21 0.5 to reach 10.5

33 to reach 16.5

[dawn apex]

ehh

[jagged marsh]

9 to reach 4.5

17 to reach 8.5

[dawn apex]

Try considering A-Z, it’s easier that way

[jagged marsh]

$T = \begin{bmatrix} 0&0&0&0\ 0&0&0&0 \ 10 & 16 & 4 & 8 \ 0&0&0&0\end{bmatrix}$

[stoic python]

typhon:
Compile Error! Click the reaction for details. (You may edit your message)

[jagged marsh]

I tried

lol

well

A-Z

is basically all 0s and in third row 10 16 4 8

[native lodge]

$T = \begin{bmatrix} 0&0&0&0\ 0&0&0&0 \ 10 & 16 & 4 & 8 \0&0&0&0\end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[jagged marsh]

hmm

[dawn apex]

and the second row of Z is
35,56,14,28

notice how they are pretty related

[jagged marsh]

1/3,5 ?

i guess

but took me way too long honestly xDDD

[dawn apex]

10 | 35
16 | 56
04 | 14
08 | 28

^put side by side, see if you can notice anything

[jagged marsh]

oh

yeah

they actually each fit 3,5 times in it

I guess if i get such a task I should definitly write it down like this

thanks :S

😄

[dawn apex]

Yw:)

[halcyon pecan]

How do inequalities with complex numbers work? Is it just distance from origin?

[gray glen]

inequalities aren't really a thing without involving absolute values or real/imaginary parts

or arguments I guess but that's a bit weird

[halcyon pecan]

Hmmm fair enough

[subtle walrus]

yeah, you can't order the complex numbers

[gray glen]

I mean you can but it won't be pretty

[subtle walrus]

there is no total order on the complex numbers

so you can't turn them into an ordered field

[sonic osprey]

There is a total order on the complex numbers

It just doesn't respect the field operations

[gray glen]

just biject them with R for a total order

[subtle walrus]

oh yeah, fair enough

[golden ermine]

Would a well defined binary operation mean that for any two objects in some group G, a and b, with operation , ab is also an element in G?

Like is that all you need?

[subtle walrus]

yeah

to be called a binary operation, G doesn't even have to be a group

[sonic osprey]

Yes, a binary operation is just a function from GxG to G

[patent orbit]

I have a matrix whose only eigenvalue is 2 with repeated multiplicity 3.

Is the characteristic polynomial logically necessarily x^3 - 6 x^2 + 12 x - 8?

[feral mountain]

multiply the expression by a constant?

[dusky epoch]

@patent orbit is it a 3x3 real matrix

[patent orbit]

yes

[dusky epoch]

then yes, (x-2)^3

[sonic osprey]

[slow scroll]

this aint linear algebra xd

[digital hazel]

This is more suitable in #prealg-and-algebra

[quartz vector]

Hi all

Is it the right place to ask about numerical issues in some specific matrix diagonalization method

[native lodge]

Like executing diagonalization using software and the problems that arise with certain algorithms?

[quartz vector]

ah sorry just saw your answer

@native lodge yes kind of

I posted a question on SO

actual post is there: https://math.stackexchange.com/questions/3272668/numerical-stability-cannot-unitarily-diagonalize-normal-matrices

I am interested in understanding why I don't find (up to a reasonnable precision) a unitary matrix when trying to diagonalize a unitary matrix

although condition matrix of the unitary matrix is very good (=1 up to a good precision)

Because in theory that works well, but in practice, I cannot unitarily diagonalize a unitary

I haven't tested the example given in the first answer though

based on Schur triangularization

[cedar solar]

You need to apply gram schmidt to the matrix v

I looked up the documentation for np.linalg.eig and the v is eigenvectors normalized to length 1, but they are not guaranteed to be orthogonal

You mentioned in your reply that eigenvectors are orthogonal when all the eigenvalues are distinct

But when eigenvalues are not distinct, the eigenvectors dont need to be orthogonal, and you need to use gram schmidt

Think of it this way

How can you choose two vectors to span the xy plane?

They dont have to be orthogonal, right? When you have two copies of the same eigenvalue, the eigenspace is a subspace with dim 2

And to span this subspace, the eigenvectors need not be orthogonal

[pliant berry]

Okay, so I'm going through Linear Algebra Done Right, and I decided to actually verify what they told me to. And for some reason, I can't get it right?
I figured that I'd just have to apply the linear map to the different basis vectors and I'd get the new matrix, easy peasy. But I don't, I get $\begin{pmatrix} 69 & 322 \ 276 & 230 \end{pmatrix}$

I feel like I'm either missing something trivial or I've got my head backward and have no clue what I'm actually doing.

[stoic python]

TheKikko:
Compile Error! Click the reaction for details. (You may edit your message)

[pliant berry]

(Ignore the compile error, it's a minor thing with the preamble that I'm waiting for the bot to update)

[sonic osprey]

What do you mean by applying the linear map to the different basis vectors?

It's not that straightforward

Because you also have to interpret the results of that new linear map in terms of your new basis

[pliant berry]

Oh?

Okay, I've definitely missed something hahah

[sonic osprey]

Yeah, you're not far off though

[pliant berry]

Luckily hahah

but how would you interpret the results?

or would you mind either explaining what I'd need to do or pointing me toward what I should read up on?

[sonic osprey]

Well okay, what you've essentially done is figured out where the new basis vectors go

Under the linear transformation

[pliant berry]

right

[sonic osprey]

But what you get out is still written in terms of the old basis

[pliant berry]

oh

That makes sense

[sonic osprey]

You want to write it in terms of the new basis

[pliant berry]

Damn, I really need to reread the chapter on bases

Thanks for the help!

[lean maple]

The bigger the angle between vectors the smaller the dot product right?

[dusky epoch]

not really no

[lean maple]

Is there a general relationship between the size of angle and size of dot product

[dusky epoch]

"the size of the relationship"?

what the fuck is that supposed to mean

i mean you have $u \cdot v = | u | \cdot |v | \cdot \cos(\theta)$

[stoic python]

Ann:

[cedar solar]

He’s not wrong, since the angle can only be between 0 and 180

Holding the lengths constant, the bigger the angle = the smaller the dot product

[halcyon pecan]

Cos 7pi/8

Vs cos Pi/2

[slow scroll]

So what is the matrix for P_E?? All I see is a sum. Also, vv* should just be ||v||^2, right?

[dusky epoch]

no, $vv^*$ is an $n \times n$ matrix.

[stoic python]

Ann:

[dusky epoch]

it's an n × 1 multiplied by a 1 × n

it's $v^*v$ that is equal to $|v|^2$.

[stoic python]

Ann:

[slow scroll]

Oh yeah...

[slow scroll]

It’s exactly what you have there

You negated the entries on the diagonal matrix

Np

[slow scroll]

https://cdn.discordapp.com/attachments/550859457308000276/595854169823510548/unknown.png
I asked about this yesterday, but on second look, its still not really obvious to me how 3.2 follows from 3.1....

Is there some sort of derivation for this?

[winter reef]

whats v*? and is (v,v_k) supposed to be dot product?

[slow scroll]

@winter reef v* is the hermitian adjoint

and yeah, (v, v_k) is an inner product

[wintry steppe]

Wait isn't this just a straightforward manipulation of the equation? If you have Av = [some matrix]v, for any v, then A=[some matrix]

(v, v_k) = v_k^* v

So it's not just any inner product; they must've told you the definition of that particular inner product

[oak briar]

Hello, Eigen serves two modules Eigen::FullPivLU and Eigen::PartialPivLU. I had to implement something using a LU decomposition. The specifications of my assignment said nothing about permutations or pivit. However I can only reproduce the required outcome using Eigen::PartialPivLU how can I use Eigen::FullPivLU and get the same outcome? Or why can't I seem to get it done?

[pine hearth]

was reading about the Moore-Penrose pseudoinverse and wasnt sure what the R(b) and R(A) notation meant. anyone here know by any chance?

is it just the set of numbers included inside the matrix/vector?

[gritty ore]

hey guys quick question

if I have something like $R_{u, \theta} = PAP^T$ where everything here is a matrice and we are trying to solve for A how would I do this?

[stoic python]

pzezson:

[cedar solar]

If P^T = P inverse, P has to be orthonormal

Everything else is the same as usual diagonalization

[gritty ore]

ok i have everything except A so I want to solve for A?

how do I get A on one side again/

[cedar solar]

You cannot possibly get P without solving for A first

[brittle fog]

What does transpose mean? does it have an equivalent operation in the real numbers? Like a negative matricy -A makes sense, powers make sense Aᵖ, but when thinking about AAᵀ i dont know what it means geometrically or what to imagine

[fringe cave]

Well, if you look at a matrix yourself, you notice it sorta just reflects over the diagonal line

[dusky epoch]

it's a bit tough to give a geometric meaning to the transpose

however, one place the transpose does show up is in the dot product

since given two col vectors u and v, their dot product can be expressed in terms of matrix multiplication as u^T*v

[fringe cave]

yeah I can't think of a geometric interpretation except maybe some symmetry of a square on the plane

which isn't quite helpful or a nice real number operation

[brittle fog]

Yeah

Ok. Is there any other like, helpful understanding of it you've gained?

[sonic osprey]

https://math.stackexchange.com/questions/37398/what-is-the-geometric-interpretation-of-the-transpose

[brittle fog]

Oh! Thanks

[brittle fog]

So it seems that if you think of matrix action as applying isometries to a vector space, the transpose scales by the same amount, but rotates and reflects inversely.

So AAᵀ preserves the orientation of, say, the unit cube, but scales by the square of the determinant. Good?

[dusky epoch]

uh

no

not at all

first off not all linear transformations are isometries

[sturdy scaffold]

he talking about invertible ones i think so he ok

[dusky epoch]

no

ffs even diag(2,1,1) isn't an isometry

and that's definitely invertible

so not even fucking CLOSE to the truth

[brittle fog]

Oh. Ok. How do I correct my thinking here

By the link zopherus provided, I think I understood it. Matricies rotate, reflect, and scale space. And the transpose matrix rotates and reflects backwards but scales the same. So if you rotate, scale, flip, flip back, scale, rotate, you're back to your initial orientation plus two scalings. Where am I wrong?

[dusky epoch]

Matricies rotate, reflect, and scale space.

this is not a complete description of what a matrix can do

[brittle fog]

Rotate, scale, shift?

Im not trying to be dense I'm just new

What's your geometric understanding of matrix transposes

[dusky epoch]

ok so like

i guess the closest to that i can give is $(Au) \cdot v = u \cdot (A^Tv)$

[stoic python]

Ann:

[dusky epoch]

and no, shifts aren't a linear transformation

but shearing is

[brittle fog]

Right

Where do the similarities between isometries and linear transformation fail?

Cause I thought it was like

[dusky epoch]

well ok so like

[brittle fog]

You can create any linear transformation out of a isometries

[dusky epoch]

nope

[brittle fog]

A set of them applied in a row

[dusky epoch]

only ones that themselves happen to be isometries.

and the matrices they describe have the special property $A^{-1} = A^T$

[stoic python]

Ann:

[dusky epoch]

such matrices are called orthogonal matrices

[brittle fog]

Like rotate 45°, stretch out by 2, flip across the diagonal

Ohh no wait fuck your right shit no hold on

[dusky epoch]

scaling by 2 isn't an isometry 😛

[brittle fog]

Isometries are just det(A) = 1 matricies

[dusky epoch]

nope

[brittle fog]

Fml

±1?

[dusky epoch]

you can have a matrix with det ±1 yet not an isometry

simple example

[ 1 1 ; 0 1 ]

a shear

[brittle fog]

So only matricies whose transpose equal their inverse

[dusky epoch]

yes

[brittle fog]

Ok

I thought.....um...ok what am I thinking of here: if you stretch out space by a factor of 2, then the distance between points is also that far apart, so when you measure it with the stretched out metric it's the same

Why am I thinking that's not breaking a rule

[dusky epoch]

stretched out metric

lol

if you can change your metric any invertible function can be an isometry, it doesn't even have to be linear

[brittle fog]

Ok

So am I still wrong that AAᵀ doesnt restore orientation

[dusky epoch]

hrgh. yes.

in all honesty i don't think AA^T can really be given a good geometric interpretation

[brittle fog]

If (Au)•v = u•(Aᵀv), then does Au = uAᵀ?

cause maybe thats something.

Actually nvm

[chilly dragon]

$ F = < (1,2,0), (2,1,1) > $

[stoic python]

Wosco:

[chilly dragon]

what does this notation mean ?

because i'm asked to prove that F is a vector subspace

[dusky epoch]

this means the set of all linear combinations of (1,2,0) and (2,1,1)

[chilly dragon]

o i see

thank you

[dusky epoch]

also, \left< and \right>

[chilly dragon]

which is basically span({(1,20), (2,1,1)})

[dusky epoch]

yes exactly

it's a notation for span

that's exactly what it is

[wintry steppe]

i am trying to find puzzles/math challenges in linear algebra that have physical/mechanical analogues that a person could hold and attempt to solve through physical, spatial manipulations - kind of like a rubik's cube, but a actions on a rubik's cube is very much group-like more than linear algebraic persay.

:3 any thoughts?

[stone drum]

Write a 3d rendering engine.

[slow scroll]

You mean arguing that $P_E \mathbf v = \sum_{k=1}^r \frac{\mathbf v_k^* \mathbf v \mathbf v_k}{\lVert \mathbf v_k \rVert^2}$ and somehow the $\mathbf v$s cancel out?

[stoic python]

kxrider:

[slow scroll]

@wintry steppe

wait no that wouldn't even make sense..

[wintry steppe]

Nothing cancels

[slow scroll]

enlighten me

[wintry steppe]

Pay attention to how Pv is written

Use the fact that (v, v_k) = v_k^*v

Rewrite Pv using the hermitian

[slow scroll]

https://cdn.discordapp.com/attachments/540211747613704221/596811724154470468/195237017049759744.png
thats what i did here. That is what you mean, right?

[wintry steppe]

Note that vk^*v is a scalar, do you can move it to the right of vk

Now use what I said before about the some matrix stuff

[slow scroll]

hmm its not even obvious to me that $(v_k^v)v_k = (v_k v_k^)v$

[stoic python]

kxrider:

[slow scroll]

the commutative step is, but not the associative

[gritty ore]

Question is as follows. Let u = (1, 0, 0), v = (0, 1, 0), w = (0, 0, 1). Find $p \in S^2 $ and $\theta \in R$ such that $R_{u, \frac{\pi}{2}} R_{v, \frac{\pi}{2}} R_{w, \frac{\pi}{2}} = R_{p, \theta}$

I know how to do this question the "brute force" method
i.e. finding the matrix for each of the corresponding rotations and finding its composition than finding the matrix representing it algebraically
but I was wondering if there's a shortcut to do this question geometrically
Since the vectors in question are the standard basis in vectors in R^3

[stoic python]

pzezson:

[slow scroll]

what is S?

[gritty ore]

just the unit sphere

$R_{u, \frac{\pi}{2}}$ means rotating around vector u with angle theta

[stoic python]

pzezson:

[gritty ore]

where u, v, w are all vectors whose end lies on the unit sphere and tail sits at origin

[slow scroll]

oh ok i see. it would be the identity

[gritty ore]

what would be the identity?

[slow scroll]

that particular composition is just the identity. Draw a picture with the u,v,w axes and apply each transformation yourself.

so yea, no brute force necessary

[gritty ore]

so every vector ends up in the smae place after applying those 3 rotation matrices?

[slow scroll]

yes.

[gritty ore]

how are you getting this? are you just looking at a test vector?

[slow scroll]

its just a geometric thing. Look at a globe if you have one nearby. Start somewhere at the equator, move pi/2 somewhere along the equator. Then move pi/2 up toward the north pole, then move pi/2 back down perpendicular to the first two transformations. You'll end up back where you started.

[gritty ore]

ok yea so first rotation matrix you move along the equaator

shit yea

you form a spherical triangle right?

if you trace the lines

[slow scroll]

yea yea. idk if thats what its actually called, but thats what i'm referring to here

[gritty ore]

ok so in this case

p is the zero vector?

and theta is 0?

[slow scroll]

theta is definitely 0. I don't think the choice of p actually matters, but p=0 might break things I'm not 100% sure

[gritty ore]

yea

[slow scroll]

p=(1,0,0) and theta=0 is trivially the identity for example

[gritty ore]

the definition of $R_{u, \theta} = PAP^T where P = (u, v, w) and A$ is the rotation matrix in 3D, and u, v, w form a orthonormal basis for $R^3$

[stoic python]

pzezson:

[gritty ore]

so yea zero vector would screw tehings up i think

so yea p in this case by anything because regardless of what you are rotating around if you rotate 0 degrees then nothing changes

[slow scroll]

yep

[gritty ore]

alright thanks man you saved me from a lot of matrix multiplication lol

[slow scroll]

npnp

[wintry steppe]

@slow scroll why is it not? Matrix multiplication is associative

[gritty ore]

@wintry steppe what do you mean? answer is correct right?

[slow scroll]

no this is a different question

ur stuff is fine

[gritty ore]

oh ok mb ty

[slow scroll]

hmm on second thought, I take back what I said. It does make sense if you use the fact that v_k and v_k* can be thought of as matrices. I guess its just feels weird because it requires thinking of v_k as a vector and a matrix simultaneously...

[wintry steppe]

Vectors are just nx1 matrices

[soft flicker]

Hello

When I try to get the determinant of this matrix on Matlab

B=[1 2 0;3 -1 2;-2 3 -2]

I get this solution

7.7716e-16

But the solution is 0

[broken hawk]

in numerical math, 10^-16 is 0

that’s just a rounding error

anything below 10^-14 or so you just have to ignore. computers aren’t perfect

[soft flicker]

Thank you! I have been really confused because of that

[broken hawk]

it probably does a division at some point and then stores the faction as a number and cuts off the decimal binary expansion at some point cause it can’t store an infinitely long number

and boom, tiny error

[dusky epoch]

[potent horizon]

hello

could anyone provide me a good skript or free book for LA?

[native lodge]

depends on what you want really

applied or pure?

[grave plank]

what is a good pure one?

[native lodge]

lots of people start with linear algebra done right I think?

I hear about that book all the time as like an intro pure book or whatnot

have not read it myself though

[potent horizon]

@native lodge pure

[native lodge]

then perhaps linear algebra done right is a good place to start

you can get that online if you look in the right place

[potent horizon]

okay ty

is it only introduction? I need smth on first semester level

[native lodge]

I don't know really, since I've never read it

you can download it from somewhere and see for yourself then

[potent horizon]

okay ty mate

if anyone else knows a good book just go ahead

[balmy fjord]

@potent horizon http://joshua.smcvt.edu/linearalgebra/

[wintry steppe]

It's pretty good for an introduction

[cedar solar]

I personally like Lay’s textbook as an intro

[wintry steppe]

Does anyone know where I can see an introduction to linear algebra

[golden cloak]

https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

[sonic quartz]

^

gud

although consider picking up a textbook as a supplement to get some more rigor

[potent horizon]

@balmy fjord ty

[balmy fjord]

Hi, how should I approach problem 2? I assume there is a better way than attempting to apply the transformation that many times

[sonic osprey]

Diagonalization

[balmy fjord]

https://www.math.uh.edu/~jiwenhe/math2331/lectures/sec5_3.pdf This stuff?

[sonic osprey]

That does say diagonalization yes

Think about why diagonalization solves your problem too

[balmy fjord]

@sonic osprey Okay I have to study this first Don't even know what that is yet

[fervent palm]

when im multiplying by an orthonormal vectors, (these are vectors that has a magnitude of 1 and when you take a dot product you get 0) with an orthogonal matrix, then the result is also orthonormal, but that means the matrix HAS to be orthogonal (matrix * matrix_transpose = I)

but why?

usually when multiplying a vector with a matrix, i know it changes the magnitude, and direction of the vector

why is it that when we multiply with an orthogonal matrix the resulting Ax is still orthonormal>

[sonic osprey]

I don't see what you're saying?

Vectors are not orthonormal by themselves

[fervent palm]

like if say I have a set of vectors that are orthonormal and then a matrix thats orthogonal

[sonic osprey]

You can have a set of orthonormal vectors

[fervent palm]

multiplying the two gives me Ax thats also orthonormal

[sonic osprey]

Right

[fervent palm]

but only if the matrix i multiplied with orthogonal

[sonic osprey]

You can show that orthogonal matrices have determinant 1

Or -1

Which means that it doesn't change the length of vectors

[fervent palm]

im trying to see how that works I know that if a matrix is orthogonal, then M*M_Transpose = Identity and determinant M = determinant M_transpose

how does that mean that the determinant is 1 though or -1?

[native lodge]

try it out on a permutation matrix then

[sonic osprey]

Take the determinant of both sides of that first equation

[fervent palm]

ohhhh

that makes a lot of sense thank you!!

https://www.khanacademy.org/math/linear-algebra/alternate-bases/orthonormal-basis/v/lin-alg-orthogonal-matrices-preserve-angles-and-lengths another helpful tutorial! :D

[balmy fjord]

what is the correct Identity matrix for a 2x3 or 3x2 matrix? Is there such a thing as an Identity matrix for a non square matrix at all?

[stone drum]

I don't think non-square identity matrices can exist.

Since after matrix multiplication, a non-square matrix will not have the same dimensions.

[balmy fjord]

so a non square matrix is not invertable?

[sonic osprey]

It can't have a single inverse

A non square matrix can be left invertible or right invertible, but not both

[balmy fjord]

tyvm! (both of you)

so you can't use the double wide method of finding an inverse on a (A | I ) --> ( I | B ) non n*n matrix

[sonic osprey]

depending on which way you extend your matrix, you may get a left/right inverse if one exists

[dusky epoch]

@balmy fjord a non-square matrix never has a two-sided inverse

[stoic python]

twitledumb:

$ \begin{pmatrix}
a & b\\
c &d\\
e &f
\end{pmatrix}
```Compile error! Output:

! Missing $ inserted.
<inserted text>
$
l.16 \end{document}

I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[balmy fjord]

[
\left[ \begin{array}{cc|cc}
a & b & ?& ? \
c & d & ? & ?\
e & f &
\end{array} \right]
]

[stoic python]

twitledumb:

[balmy fjord]

finally. If this is possible, what is the identity matrix of the appropriate size?

[dusky epoch]

uh

what is this meant to be

what'll go in the blanks?

you can't just have a jagged array of numbers like this. it's meaningless

[balmy fjord]

the blanks are supposed to be an Identity matrix for the left-hand side

the double wide matrix method of finding an inverse on a (A | I ) --> ( I | B ) non n*n matrix

[dusky epoch]

no

not the question marks

the blanks underneath them

what are those meant to be

[balmy fjord]

more question marks I guess. Clearly I'm missing something. I left the blanks because I thought I had to keep this rule of A A^-1 = I and a 3x2 times a 3x2 is undefined so i made the A^-1 a 2x2

[dusky epoch]

well like here's the thing

you can't multiply a 3x2 by another 3x2

so the notion of a 3x2 identity matrix doesn't even make much sense to begin with

[balmy fjord]

a 3x2 having an identity matrix or a 3x2 being an identity matrix to another matrix

[dusky epoch]

"an identity matrix to another matrix" is nonsensical word salad

[balmy fjord]

😦

[dusky epoch]

you can't make the 3x2 matrices into a ring is what i'm saying

since you can't even define multiplication

and without that everything falls apart. bc you can't have a 3x2 matrix that leaves other 3x2 matrices unchanged when you multiply them with it. because you can't multiply.

[balmy fjord]

so as SamanthaCS was saying (I think) based on what you just said for a 3x2 (or any non-square )Matrix A, there exists no identity matrix I where where AI = A Or IA = A ???

[dusky epoch]

ok so like

look

letting $I_n$ denote the $n \times n$ identity matrix

[stoic python]

Ann-iversaire:

[dusky epoch]

for any $m \times n$ matrix $A$, you have $I_mA = A$ and $AI_n = A$

[stoic python]

Ann-iversaire:

[balmy fjord]

whoa

[dusky epoch]

what

[balmy fjord]

nothing . thanks finally think I got it....
if you r left multiplying use the $I_3$ and if you're right multiplying I use $I_2$ for a 3x2 matrix

[stoic python]

twitledumb:

[dusky epoch]

yes

[balmy fjord]

cant thank you enough

[balmy fjord]

@sonic osprey Can eigendecomposition also help me find 2b and 2c as an alternative to diagnalization

[sonic osprey]

Eigendecomposition and diagonalization are the same thing

Well, at least when the matrix is diagonalizable

[balmy fjord]

hmm ok tty

[cunning finch]

Can someone help explain cramers rule to me? I get how to use it, but im trying to picture how it actually works

[sonic osprey]

Search up 3blue1brown Cramer's rule

[cunning finch]

I watched it and am rewatching it again and again and im still lost

[sonic osprey]

That's as good of an explanation you're going to get

What in particular do you not understand

[cunning finch]

https://youtu.be/jBsC34PxzoM?t=514 right at that time

I dont get how det(A) is supposed to represent only the x value of the area of the transformed paralellogram

So I'm imagining what he's doing is hes getting the y value of the output and hes got the area of the new paralellogram, and he has some det(A) which is supposed to be the x value of this area and he's basically just doing the opposite to get the value of the input parallelogram

Like am i allowed to picture the areas as squares and the transformation as scaling up the squares?

[lime sentinel]

Can anybody help me with this question?

I'm stuck. This is the last question I have assigned. I don't know what to do.

[wintry steppe]

I'm guessing it goes along the lines of rewriting each as a dot product and substituting in the knowns

Use distributive property somewhere

[dusky epoch]

@lime sentinel are you familiar with the dot product?

[lime sentinel]

@dusky epoch Yeah, what about it?

[dusky epoch]

well

i suggest rewriting |a|^2 = 9, |b|^2 = 25 and |a+b|^2 = 49 in terms of the dot product.

and also |a-b|^2, so that you can find it afterwards

[lime sentinel]

@dusky epoch I got radical 19. Is that correct?

[dusky epoch]

i'd rather see your work than just the answer

but that matches what i got, so i guess it is correct

[lime sentinel]

I'll send you a picture in a bit.

I left home to go grab some food.

Give me 20'ish minutes max.

[oak briar]

Hello, simple question, I have to use a formula that says that a matrixD can be written as LU. with L and U upper and lower matrices. (an LU decomposition). However everything online has these matrices P and Q as well in them. How am I suppose to know which permutation I'm suppose to take?

In short if unspecified how should I interpret an LU decomposition?

The formula is Da + b = I , LUa+b = I so a = U^-1 and b = I - L when using the formula in my code the result is only correct for somethings, I can't find out why.

[half ice]

@oak briar
It can be impossible to take an LU decomposition, but in many cases simply rearranging your rows can fix that, and you still get the benefits of an LU decomp. In those cases, it's called an LUP decomp.

I'm not sure what P and Q are though

[oak briar]

The issue is that b and a have to be upper and lower matrices as wel

It's important that D equations work with respect to D not PDQ

Cause I'm doing a basis transformation with D somewhere else

My current "fix" is to write an non-pivot algorithm then, even though it might be unstable

@half ice Else I need to find a solution for P^-1 LU Q^-1 a + b = I with the constraint that a is upper triangular and b strictly lower.

P and Q are the pivots or permutations to ensure stable decomposition

[half ice]

What are we solving for? What's known?

Where are you getting this info from? May help if I read it, as I'm not sure of this algorithm

[oak briar]

It's from a book

I'll cut some pages sec

It's formula 11.9.22 derived from the things above it.

[slow scroll]

No clue on this one. Any hints?

[broken hawk]

well, the straightforward approach is to calculate the projection into that subspace and then calculate the length of the distance you get

you already have an orthogonal basis for the subspace

[slow scroll]

It sounds like the question is implying that I don’t have to do that

[oak briar]

Choose an arbitrary third orthogonal vector for the subspace, then find the 4th one, transform the vector to that space and compute a 2D projection

[broken hawk]

I mean that’s just computing the projection onto the orthogonal complement

which is arguably more work

[slow scroll]

Hmm at work now so I’ll try that later

[uneven tendon]

how do I change the point by which I rotate a vector around when using quaternions?

right now it only rotate around the origin

[crystal oracle]

Hello everyone. I am trying to figure out right inverses today. I understand the definition of right inverse of a linear function. Also, wikipedia provides a way to construct a right inverse of a full row rank matrix (pic attached).

1. Is this formula only for real matrices or for complex matrices as well? I am somewhat confused because there we have transpose symbol instead of conjugate transpose symbol.
2. How to formulate a similar construction for linear functions between finite dimensional vector spaces? Judging by matrix transposition, we can somehow use dual spaces and dual maps...

Hmm. Actually it seems it's very easy to construct a right inverse. However I don't have such a nice looking formula.

[sonic osprey]

Every linear function between finite dimensional vector spaces can be represented as a matrix?

After you choose some bases

[crystal oracle]

What's puzzling me is.

We have no inner product, just two finite dimensional vector spaces.

Then what meaning does A A^T have?

And why is it necessarily invertible?

[subtle walrus]

the product of invertible matrices is invertible

the transpose of an invertible matrix is invertible

[crystal oracle]

Here A is not a square matrix

[subtle walrus]

oh, my bad

[crystal oracle]

Also, I dont understand what is the point of dual spaces, dual maps, annihilators, when sooner or later we throw all that away when we start talking about inner product spaces.

[potent horizon]

the kernel of a matrix is always a linear subspace right?

[sonic osprey]

well what properties does a subspace have to satisfy

[potent horizon]

closed under addition

closed under scalar multiplication

contains the zero vector

it contains obviously the zero vektor

but I am not sure about closed under addition

[sonic osprey]

Well try it

[potent horizon]

yes I have to find an exercise

[balmy fjord]

https://www.youtube.com/watch?v=0tBdU2b8w4Q&list=PLKXdxQAT3tCtmnqaejCMsI-NnB7lGEj5u&index=30 Try this out @potent horizon

and clarification on vector space https://www.youtube.com/watch?v=-PQ-Z5K-PKg&list=PLKXdxQAT3tCtmnqaejCMsI-NnB7lGEj5u&index=25

[potent horizon]

well my problem is that lets say we have a matrix and a basis of the kernel.

The Kernel is a subspace, that also means no matter how if I make random linear combinations with the basis of the kernel I will still get an element from the kernel. Cause it is closed under addition.

But I am currently stuck to imagine this, why can I not find a linear combination which will bring me out of the kernel

[noble swallow]

Because f is linear, it distributes over the linear combination which will then be a linear combination in the 0 vector

[crystal oracle]

Ok, my question above was ill posed - the supposed theorem is false

[potent horizon]

Already find my mistake

but ty guys

[slow scroll]

For a projection matrix, I'm pretty sure its eigenspace would be the subspace the matrix projects on, which would have eigenvalue 1. But what about the complex eigenvectors and eigenvalues? On part b for example, (1,1,..,1)^T is an eigenvector with eigenvalue 1, but I don't know whether that is the only vector with eigenvalue 1, so I cannot determine geometric multiplicity. What am i missing?

[slow scroll]

well... id say its been around 15 minutes <@&286206848099549185>

[elfin schooner]

it's been around 15 minutes BWHAHAHAHA

@slow scroll What's a projection matrix?

[slow scroll]

well a vector projection is transformation that sends a vector v to a vector w in a particular subspace E such that v-w is orthogonal to every vector in E. The projection matrix is just the matrix that sends every v to its respective w

[elfin schooner]

that sends every v to its respective w

What's every v in our case?

[slow scroll]

every v in the domain of the projection. So you would have some vector space V and E is a subspace of V, i.e. the subspace we our projecting onto

[elfin schooner]

yeah and in this case its R^n right

[slow scroll]

oh err i meant to say the domain would be V, but yeah, R^n is fine for visual purposes i guess

[elfin schooner]

...and the co domain is....?

[slow scroll]

$P_E: V \to E$.

[stoic python]

kxrider:

[elfin schooner]

I'm not asking for the general definition mate

What's E in this particular question?

[slow scroll]

span{(1,1,...,1)}

[elfin schooner]

exactly

I'm gonna talk about the particular case where n=2, and then we shall generalize things alright?

P_E : R^2 -> {k(1,1)^T | k in R}

Is that alright?

[slow scroll]

yep, its eigenspace is span{(1,1)}

[elfin schooner]

ye

[slow scroll]

err well, the eigenspace corresponding to the eigenvalue 1. but is 1 the only eigenvalue? What about complex eigenvalues?

[elfin schooner]

why talk about eigenvalues now?

let's figure the matrix out

[slow scroll]

alright.
[[1,1],[1,1]]

err oops. that divided by 2

[elfin schooner]

state your reason

[slow scroll]

direct computation. (1,1)* (1,1)/||(1,1)||^2 where * denotes the hermitian adjoint

err wait uh

yea thats right it just looks weird cuz i wrote it in rows

[elfin schooner]

erm

im kinda confused as well

lemme get my pen and paper

[slow scroll]

here this is where my book talks about it

[elfin schooner]

which book is this?

[slow scroll]

linear algebra done wrong

[elfin schooner]

erm

I don't think I have enough knowledge in this

besides i woke up rn

feel free to ping helpers again

[slow scroll]

alright thanks for trying tho 🙏

[sonic osprey]

What did you get for part a)?

@slow scroll

[slow scroll]

a matrix with all entries 1/n

[sonic osprey]

So what would the geometric multiplicity of 1 be

[slow scroll]

im guessing it would be 1. But an nxn matrix has n eigenvalues (potentially indistinct) so either there must be other complex eigenvectors in the eigenspace of 1 or other complex eigenvalues, right?

[wintry steppe]

Vector projection does not have to be orthogonal

[slow scroll]

idk what you mean lach

[wintry steppe]

There are oblique projection matrices

Anything that is orthogonal to the relevant subspace will be projected to 0 because this is an orthogonal projection

So what are the eigenvalues?

[slow scroll]

huh thats true so 0 and 1. Is that all though?

[wintry steppe]

Well think about the eigenspaces

One eigenspace is the subspace and it has eigenvalue 1

Another is the orthogonal complement and it has eigenvalue 0

There are no more vectors left

[slow scroll]

oh.... true true. For some reason, I was thinking: ok there are vectors orthogonal to E and there are vectors in E, but there are also vectors that are neither orthogonal to E nor in E, so there must be other eigenvalues, but ofc thats not the case. thanks!

[wintry steppe]

There are, but they can't form an eigenspace, since the eigenspaces sum up to at most the whole space

[sonic idol]

hey uh

I have an exercise where it's just
x^2 -4xy +8y^2 +4x -4y +6 = 0 and it asks what type of conic it is

[slow scroll]

look at the coefficients of x^2 and y^2. Since they are both positive and the coefficients are not equal to each other, its an ellipse. Also, this is #precalculus

[sonic idol]

oh ok

srry

and thanks

[slow scroll]

npnp

[sonic idol]

I'll just ask for the not so direct way there to understand what you said lol

still not ded

apparently there's something that needs to be done to get rip of -4xy but I don't know

[slow scroll]

well i found a little more context: https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections

[amber bay]

you could substitute x and y by rotation matrixing them

and find the tan theta that makes the xy bit disappear

[sonic idol]

I think I found what you mean

there's a deduction I'm not gonna be reading rn and there's tg(2theta)=B/A-C

[slow scroll]

nope you don't have to do that. I don't know any of this stuff, but you can represent the equation with this matrix:

and since we have a degenerate equation:

this is under the "classification" subheading

[sonic idol]

managed to do it

it's an empty set

[slow scroll]

yeah. how did you end up doing it? By computing det A_33?

[sonic idol]

well first I did this delta which was given by B^2-4AC, which has nothing to do with the a^2-4ac btw, and that gives you it's either an ellipse or degeneration
then you do det of Aq, and since it's a positive number you go on another table and magic tells you it's an empty set

I am completely unsatisfied but oh well

[serene prairie]

How to solve this linearly?

[tight plover]

Can someone give me a pointer for this problem:

No. 47: Exponential skew-symmetric matrices
Let n ∈ R^3 be normed, let a ∈ R^3 and let σ = (σ1, σ2, σ3) be a vector of Pauli matrices.
Define a · σ := a1σ1 + a2σ2 + a3σ3 ∈ M(2, C), define the vector space su(2) := {A ∈ M(2, C)| A = a · σ, a ∈ R
3}.
Proof that exp(iα(n · σ)) = cos(α)E2 + i(n · σ) sin(α).
Now conclude that exp(i su(2)) ⊂ SU(2).

I've been on this for hours, we were told to look at the exp series and the properties of the vector space, the inner product being 1/2 trace(AB). I have very little time left to get this done...

[slow scroll]

So... I have an attempt on part a. It feels cheat-y so I want another opinion on it / tips on anything i missed.$\newline \newline$ Any solution, $x_0$, to $Ax=b$ can be written as $x_0 = \alpha + \beta$ s.t. $\alpha \in \Ker A$ and $\beta$, a non-homogeneous solution to $Ax=b$. Then it follows that $x_0$ minimizes the norm for a choice of $\alpha$ that minimizes $\lVert \alpha + \beta \rVert$, i.e. the distance from $\beta$ to $\Ker A$. By definition, this is given by $-P_{\Ker A} \beta$. This solution is unique because the projection is unique.

[stoic python]

kxrider:

[slow scroll]

dang answers here are slow lately. I'm thinking my answer is somehow wrong as I can't figure out how part b follows from this.

[sonic osprey]

what do you mean by the projection is unique

[slow scroll]

there is only one vector w = P_E v that minimizes the distance from v to a subspace E, i.e. ||v-w||<||v-x|| for all x in E x != w

[sonic osprey]

I mean if you do have that statement proven

Then yes this proof is fine

[fringe cave]

or if it's given by the book or something

[slow scroll]

yea its proven in the book, but I'm still not quite sure how part b follows from this. Attacking it in a similar fashion, $P_{(\Ker A)^\perp} x = P_{(\Ker A)^\perp} (\alpha + \beta) = P_{(\Ker A)^\perp} \beta$. And then what lol?

[stoic python]

kxrider:

[indigo cradle]

hi guys

im wondering why the elimination stops here to get the echelon form

shouldnt the 4,4 element be the next pivot?

and take row 4 - row 3 to get 0 below it?

why does it stop here?

[slow scroll]

hmm okay but i don't see why you couldn't replace r4 with r4-r3 and then divide r4 by 2.

[feral mountain]

i don't see y either

[crimson mirage]

hmm yeh.

[slow scroll]

its looks special how it is with all of the 1's and stuff. Maybe its on purpose xd

[crimson mirage]

is this from a textbook?

[feral mountain]

if we finished the algorithm it'd still have 1s everywhere

[slow scroll]

oh yea true

[crimson mirage]

if i'm not mistaken, you could continue and eliminate a few more 1s

[indigo cradle]

yes its from a textbook

or rather the question

proceeded to replace r4 with r4-3

but it meant that i'll have 4 pivots so only 2 free variables

but the question wants 3 columns in its nullspace matrix

thats what the textbook calls "special solutions"

Introduction to Linear Algebra by Gilbert Strang

please enlighten me

[slow scroll]

https://cdn.discordapp.com/attachments/488104216678760469/599525416604270603/wolf.png

@indigo cradle look closely, columns 3, 5, and 6 are free. Compute the null space. It'll be dimension 3.

[indigo cradle]

but how...

wait a sec

Could I have a third eye in finding where i went wrong?

[crimson mirage]

the very start

first row

[slow scroll]

oof

[crimson mirage]

missing neg sign

[indigo cradle]

O MY GOD

[crimson mirage]

the pain

[indigo cradle]

trust me guys i legit looked through this too many times

[crimson mirage]

too many straight lines 😄

[indigo cradle]

k thanks guys

[slow scroll]

np xd

[frank gate]

I need help! When i am trying to find inverse matrix of a 3x3 matrix, i use this method: https://gyazo.com/d31d9e16f0cf8c3d4349b1e18787786e

But when i am doing exercises, i find that sometimes, the sign of a value in the matrix is the reverse. So when the solution shows that it is -3 for example, i have it to 3 instead. Is this still wrong or no? i have double checked everything