linear-algebra

[north sierra]

oh yeah now i know why you said its better to use a counter example

okay i will try to do this now

[half ice]

Try it with something that is a vector space lol

So you can see why it works to prove it

[gleaming topaz]

@half ice Would it have been correct to write it like this and show that for (x,y)=(0,0) they're not equal? https://gyazo.com/0396f197c7a19b3e8a8eb3ebb1b29216

[vast torrent]

You cant write it as a mtx if you dont know its linear

[half ice]

You could show there's no matrix that satisfies it

[vast torrent]

I guess

[half ice]

But this is pretty obviously not linear lol. No need for that much work

[vast torrent]

If that helps you somehow

Just test T[0]=0'

[north sierra]

$T\left(\left(x1,y1\right)+\left(x2,y2\right)\right)=\left(x1+x2+y,y-x1-x2\right)$

[stoic python]

The-Elite:

[north sierra]

$T\left(x1,y1\right)+T\left(x2,y2\right)=\left(x1+y1,y1-x1\right)+\left(x2+y2,y2-x2\right)$

[stoic python]

The-Elite:

[north sierra]

okay so im doing this now

was wondering if im doing this right

i know i can show it as a matrix and thats it

but i wanna practice the addition property lol

[half ice]

Nop. Take this example:
T(a,b) = (a + b, b - a)

[north sierra]

i guess im not allowed to order them like how i did in the first step?

[half ice]

In this one I mean. What's y?

[north sierra]

$T\left(\left(x1,y1\right)+\left(x2,y2\right)\right)$

T\left(\left(x1,x2\right)+\left(yz,y2\right)\right)

am i allowed to turn the first into the second?

[half ice]

What's , between the two vectors?

[north sierra]

o i meant to put a , one sec

[stoic python]

The-Elite:

[north sierra]

$T\left(\left(x1,x2\right)+\left(yz,y2\right)\right)$

[stoic python]

The-Elite:

[north sierra]

yeah am i allowed to do that?

[half ice]

Swap a first component for a second component? No

[north sierra]

lol true

ok so that was my first mistake

[half ice]

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)

[north sierra]

$T\left(x1,:y1+x2,:y2\right)$

[stoic python]

The-Elite:

[north sierra]

true

this the above right?

i guess i should provide more detail

$T\left(\left(x1,x2\right)+\left(x2,y2\right)\right)=T\left(x1,:y1+x2,:y2\right)$

[stoic python]

The-Elite:

[half ice]

No not that

I'm just adding the first components together, and the second components together

[north sierra]

omg what am i doing

lol

one sec

okay wait

lets start over

lol

so i did:

$T\left(\left(x1,y1\right)+\left(x2,y2\right)\right)$

[stoic python]

The-Elite:

[north sierra]

wait for the last question we did

you added the vectors like usual right

so for this i can do T(x1+x2,y1+y2)?

[half ice]

Yes

[north sierra]

true

[half ice]

Only one way to add in R²

[vast torrent]

Is that.true

You cant have a vector space on R2 with any other defn of v+w ?

Other than componentwise?

[half ice]

Hmmst. I don't know! Maybe you could

But we're definitely not working with that one here

[vast torrent]

[half ice]

Almost definitely something to do with
a⊕b = a + b + ab

[north sierra]

then i would get

$\left(x1+y1+x2+y2,:y1-x1+y2-x2\right)$

[stoic python]

The-Elite:

[north sierra]

?

[vast torrent]

Well this assumes the second coordinate is positive

But im sure you can adjust it

[north sierra]

are you talking about my question?

[vast torrent]

No, mine

[north sierra]

oh

[half ice]

Yeah that's right

[north sierra]

thanks for all the help @half ice

[narrow mortar]

ohhhh

helloooo

@half ice

remember meeeee

[half ice]

Ohai. What's up?

[north sierra]

if im testing T(x+y) and T(x) + T(y)

and theyre the same except the order is switched

is that still showing that its linear?

so for T(x+y), i have (x1+x2),(y1+y2)

but for T(x) + T(y) I have:

(y1+y2),(x1+x2)

[half ice]

Those aren't the same thing no

@north sierra

[north sierra]

@half ice can you check my answer

[inland gull]

can someone help me understand how to solve for t? do i need to ln both sides?

[wintry steppe]

Hey can anyone help me figure out where I've gone wrong here? I've asked in another math chat and didn't really get any further. It's been giving me some serious grief and I need to understand how to properly solve a system like this with repeating eigenvals

[native lodge]

you got two separate eigenvectors?

[wintry steppe]

it should be the same, but using the same eigenvector to form the fundamental matrix didnt work for me so I tried solving it using a generalized eigenvector, but probably did that wrong.

[native lodge]

if you didn't then you will have to use the matrix exponential

when it comes down to it, you will be using the generalized eigenvectors

[half ice]

You mean (x1+x2, y1+y2)
You don't mean (x1+x2),(y1+y2)

@north sierra

[north sierra]

oh right @kayen

@half ice

other than that is it correct?

[half ice]

No, T isn't applied in the first case

[north sierra]

what should it be?

the first box i circled should be (y,x)?

im confused

[wintry steppe]

i've missed tons of class (health) so let me just ask.. with the generalized eigenvector, do I find that by augmenting A-(lambda=1)*I with the eigenvector (-1,1)?

[half ice]

T switches the first component for the second, and the second for the first

[north sierra]

oh

true

so it should be the opposite?

[half ice]

Yus

And so you've proven T splits over addition

[north sierra]

@wintry steppe is that calculus and linear algebra?

[wintry steppe]

@native lodge how does the matrix exponential work? I don't think its required here but thought I'd ask

[opal plaza]

thx for the help this term @vast torrent

[wintry steppe]

@north sierra its an engineering course: linear algebra with differential equations

[north sierra]

😫

disgusting

and scary

[wintry steppe]

lol its actually rlly elegant. one of my homework assignments was to do a cryptography problem

but I'm lost

[north sierra]

woah Cryptography

nice sounds fun

[wintry steppe]

it technically counts as an ODE credit, so it's not too advanced

[native lodge]

matrix exponential is e to the power of a matrix

$e^{At}$

[stoic python]

⚡Amphy⚡:

[native lodge]

you have a chance at evaluating it by using the Taylor expansion of e^x

[north sierra]

@wintry steppe ^

[native lodge]

it works out nicely for diagonal matrices and nilpotent matrices

[half ice]

Could you also define the sine of a matrix?

[wintry steppe]

Oh ok, I got confused until I realized the problem itself uses e^t.

@half ice I bet you could. Maybe by diagonalizing the matrix, then taking the Taylor expansion for sin(x) of the remaining elements and sticking that in there?

[inland gull]

do you need to use ln to solve for t?

[wintry steppe]

@inland gull what kind of a problem is this? does it have a part a?

[inland gull]

i just need to isolate and solve for t

[split condor]

https://i.imgur.com/RXrZ1gS.png

im not sure how to show A must be symmetric and have only positive eigenvalues for this problem

[dusky epoch]

what does it mean for $\ang{\mathbf{x},\mathbf{y}}=\mathbf{x}^TA\mathbf{y}$ to be an inner product?

[stoic python]

Ann:

[split condor]

as in <x,y> is commutative, is linear in the first argument

and <x,x> >= 0, and is = 0 iff x is the 0 vector

[dusky epoch]

well

can you write those out

in terms of A

[north sierra]

is this true?

[solemn lotus]

yes

for an n x n matrix

[north sierra]

woah never knew that

thx

[quartz compass]

det(bC) = det(bIC) = det(bI)*det(C)

introducing an identity matrix makes it clearer

since bI has b on the diagonal, so its determinant is b^n

[north sierra]

ohh

that makes more sense

thxx

[slow scroll]

Also, it comes from the fact that scaling a row or column scales the determinant by the same amount. i.e. let v1, v2, .. , vn be the column of a matrix A, and let det(A) = D(v1, v2, ..., vn) denote the determinant of A. Then
D(v1, v2, .., cvi, ..., vn) = c*det(A)

so when you scale every row/column by some amount (i.e. when you are taking a scalar multiple of a matrix) you get the identity det(cA) = c^nA for an nxn matrix

[north sierra]

is the answer -20?

[gray dust]

show work?

[north sierra]

getting phone second

@gray dust

[gray dust]

[north sierra]

what

[gray dust]

you're aware A is the matrix AFTER the row ops were done on B?

[north sierra]

what

A is obtained from B yeah

[gray dust]

walk me through the effect of each row op on the det

we start with B where det(B)=10

[north sierra]

yeah

then first row of B is multipled by 1/2

so the determinant is divided by 1/2

[gray dust]

er

[north sierra]

then the second operation has no effect

[gray dust]

hold up

[north sierra]

third one changes the det sign

[gray dust]

1st row op doesn't do that

[north sierra]

so what does it do?

[gray dust]

tell me again what 1st row op is

[north sierra]

1/2R1

well the determinant is multiplied by 1/2 then

but if i want it back id have to reverse

it

[gray dust]

what's the det of the resulting matrix after 1st row op?

[north sierra]

5

[gray dust]

ok, continue

[north sierra]

then the second operation doesn't do anything to the detemrinant

third operation changes its sign

[gray dust]

so det(A)=?

[north sierra]

lol

-5

[gray dust]

👍🏽

[north sierra]

i was interpretting the Q so bad lol

[leaden fiber]

what does the second sentence mean/ask for? I know they are linearly independent so they span the entire 3 dimensional space for the first part :o

[nimble egret]

a[1, 1, 0] + b[1, 1, 1] + c[0, 1, 1] = [b1, b2. b3]

find possible a, b, c

in terms of b1, b2 and b3

[median mantle]

to check independence just check whether the determinant is 0 or not

If they are independent, since the number of column vectors is equal to the dimension, it must span the space

[leaden fiber]

ahh thank you!!

[half osprey]

Is this true?
translation:
If two operators are diagonalizable, then their composition will be too.

[half ice]

I'm going to say no. I'm yet to get a counter example though @half osprey

Nvm that was easier than I thought
https://math.stackexchange.com/questions/1084522/is-the-product-of-any-two-invertible-diagonalizable-matrices-diagonalizable/1084537

[half osprey]

ooooh

I was thinking of a diagonal matrix instead of a diagonalizable

thats why I couldn't find a counter examplel

[half ice]

The product of diagonal matricies is diagonal

But not diagonalizable no

[half osprey]

yeah exactly

sure now I get it, thanks!

[half ice]

Cool cool! Feel free to ask if there's anything else

[winter siren]

Is there any oblique projection formula that can be used to find the transformation matrix for a 4D hypercube to a 3D hyperplane?
If that makes any sense.

[north sierra]

if we're finding the inverse of a 3x3 matrix or higher

and Im using the

[A | I]

way

and solving till i get to the identity matrix on the left side

are we allowed to swap rows?

will that mess up the inverse ?

[slow scroll]

you can do that. It won't mess up the inverse. You can do any operation you want as long as it has a corresponding invertible matrix associated with it.

There is a matrix that permutes the rows of any matrix its composed with, and since permutations can be undone, it is definitely invertible. You don't have to actually have to think about that all of the time, but it helps to know what is going on in the back end.

[north sierra]

okay thanks so much

yeah cause i kinda had to swap rows to get an identity matrix for this question im doing right now

i dont really get what that second sentence you wrote means

cause i dont know what permutations is

[slow scroll]

I just mean a re-ordering of the rows. In this case, you are really only switching (transposing) rows. If I switch two rows, then if I switch those same two rows again, I get back to where I started. So that process is invertible.

[north sierra]

true

ok

thx

[slow scroll]

npnp

[north sierra]

,rotate

[north sierra]

so i have that matrix

and i wanted to get the eigen basis

anyways, did i do it right

[half ice]

That one matrix should read
[2x2]
[x2]
[x3]

When you add vectors, you get another vector. Never a matrix

[north sierra]

ok but is my basis right ?

and the parametric form or whatever you call it

there's two free variables so i should get 2 vectors for the nulA

[half ice]

The final answer looks right

[north sierra]

ok thanks bro

so like my professor put the basis vectors in a matrix

and he got this

idk how

it dosn't make sense

he already made 2 mistakes today that i pointed out

so this could be the 3rd but idk

[slow scroll]

well the column space of that would be the eigenspace. A weird way to write that tho

[half ice]

Also putting the basis vectors in a matrix is a what

They're very unrelated

[north sierra]

well he was trying to do the PDP^-1 thing

so P had all the eigen basis vectors

[half ice]

Oic

[north sierra]

yeah

okay my prof was wrong lol

😦

hes making me doubt myself 2 days before my linear algebra exam

this is no good

[native lodge]

your exam is on sunday?

[quartz compass]

what are you doubting?

it's pretty common to put vectors in a matrix if that's what you're doubting

$Av_i = v_i \lambda_i$

[stoic python]

Merosity:

[quartz compass]

for instance these equations one for each i can be written all together as 1 matrix equation if you put the vectors in a matrix and the scalars on a diagonal of a matrix

$AP=PD$

[stoic python]

Merosity:

[quartz compass]

the columns of P are the eigenvectors v_i

[north sierra]

no im just doubting myself cause the answer that i get (which is correct) is different from the profs answers (which is wrong

so when i compare my answers to his, im like huhhuhhuh

and i start doubting myself

"did i do this right?"

"am i supposed to do this instead"

[native lodge]

just check using symbolab then

[north sierra]

yeah i checked using one of those types of websites

and i also emailed him and he said he made a mistake

so yeah im relieved

[north sierra]

i remember asking help for a) before

so 2 eigen values would be -1,1?

for a)

[quartz compass]

which eigenvectors have eigenvalue 1?

like qualitatively, what are they

[north sierra]

that dont reflect itself?

like they dont move

[quartz compass]

like if I had a mirror, where would they be

[north sierra]

in the same spot

[quartz compass]

which is where, perpendicular to the mirror?

[north sierra]

um

[quartz compass]

what spot relative to the mirror are the vectors that don't move

[north sierra]

the vectors that are on the line

the line that run through the origin

[quartz compass]

which line?

there are infinitely many lines that go through the origin

[north sierra]

bruh

the line that we drew lets just say

lets say i drew y=x okay

thats the line

im talking about

[quartz compass]

so the points (1,1) and (-1,-1) are on that line

[north sierra]

yeah

[quartz compass]

but that has eigenvalue -1 if we reflect

and it's stuck on the line

you said it has eigenvalue 1

[north sierra]

oh it gets reflected on the line?

so it moves

but on the other side of the line?

[quartz compass]

idk you're being too vague

[north sierra]

if we have a vector that's on the y=x line

where would it reflect to

[quartz compass]

where's your mirror?

[north sierra]

what

[quartz compass]

if you're reflecting

you have a mirror

where is it

[north sierra]

lets say y=x is the mirror

and the vector is on y=x too

[quartz compass]

ok so does the point (1,1) reflect to (-1,-1)

[north sierra]

idk thats what im asking

no idea

[quartz compass]

imagine the mirror

think

draw a picture

[north sierra]

i think it would say at (1,1)

cause thats what my TA said

[quartz compass]

sad

you're gonna remember something your TA told you about mirrors?

think for yourself!

you know how a mirror works

if your mirror is along the line y=x

and you pick the point (1,1)

where does this reflect to?

draw a picture if you have to

[north sierra]

i did

[quartz compass]

so where does the point go

[north sierra]

i drew the cartesian plane with y=x

and then a vector on y=x

[quartz compass]

so (1,1) ends up where?

[north sierra]

idk

i still think it stays at (1,1)

[quartz compass]

good

you're right

[north sierra]

whatttttttt

why were acting like it was wrong then

what im so confused now

[quartz compass]

because you asked such a braindead simple question

you have no confidence

[north sierra]

yeah i know

i dont

😦

[quartz compass]

so be confident

[north sierra]

but thanks bro

[quartz compass]

that's all you need

[north sierra]

u made me more confident yeah

yeah

[quartz compass]

don't ask where mirrors reflect points

lol

[north sierra]

LOL

im laughing so hard

[quartz compass]

obviously eigenvalues are 1 and -1 haha

good good

[north sierra]

when you said "sad" after i said the "TA said this"

my heart stopped cause the TA is super smart

and like hes insane at Linear Algbra lol

cause i was like

[quartz compass]

yeah it was only sad because you were trying to argue by authority rather than your own mind, that's all

[north sierra]

omg he got something wrong

yeah

oh but the thing is

so -1 would be the vectors that are not on the actual line then im pretty confident in saying that lol

okay but i dont get what the eigen vectors are tho

or what the eigen vectors would be

oh wait

so for the eign value 1 id be all the points of the mirroe?

or idk

(1,1) is what im thinking for the eigen value = 1

[quartz compass]

keep at it

I will neither confirm nor deny

[north sierra]

LOL

arghj

[quartz compass]

imagine the mirror

be the mirror

look through the mirror

ok don't do that

[north sierra]

i guess we have to define a line first tho right?

cause there are infinite lines through the origin

[quartz compass]

it's probably better if you draw a picture, then draw the eigenvectors in the picture

[north sierra]

yeah true

ill will send a pic

soo for the eigen value of 1, i guess itd just be (1,1) in this case

and for eigen value of -1, itd go from (-1,1) to -1(1,-1)

so vector would be (1,-1)

[quartz compass]

yeah good

how about (-1,-1) is that an eigenvector or not

[north sierra]

yeah it is

[quartz compass]

yeah, good

[north sierra]

corresponding to eigen value 1

[quartz compass]

all scalar multiples of an eigenvector forms your entire eigenspace

[north sierra]

true yeah

[quartz compass]

makes sense hopefully

[north sierra]

yeah it does~

!

i dont get b) though lol

[quartz compass]

I don't remember, repost the question

[quartz compass]

got kinda buried lol

[north sierra]

lol yeah

R^3 this time 👀

[quartz compass]

well we live in R^3 roughly speaking

[north sierra]

true yeah

[quartz compass]

if you're sitting in an office chair that spins

stick your arm out and rotate

[north sierra]

LOL

[quartz compass]

where do you have to stick your arm so that it doesn't move?

[north sierra]

up or down

[quartz compass]

imagine your arm is a 1 dimensional vector lol

yeah

so there you go

[north sierra]

OMG

lol

[quartz compass]

now is it possible to have more eigenvectors?

seems like in 3D space we should have 3 eigenvectors yeah?

can you rotate in a way so that you have more than 1 eigenvector

[north sierra]

umm

so when my arm is up or down that would correspond to an egien value of 1?

i mean the line could be anywhere so maybe not

[quartz compass]

well, that corresponds to the eigenvectors with eigenvalue 1 yes

all vectors along the axis of rotation are those eigenvectors

[north sierra]

thats the z axis right?

[quartz compass]

there's no such thing as "the z axis"

you can call it the z-axis

but you can call it anything

[north sierra]

oh okay

[quartz compass]

we don't even have to rotate around a single axis, it's just easier if we rotate in a basis where one of the axes is being rotated around

rather than some kind of mix between stuff, but it depends on what you're trying to do or describe

[north sierra]

i cant think of any other eigen value

[quartz compass]

people who do animation have to deal with 3D model rigs with creatures with many different joints that can rotate in certain different angles and the math has to be able to accomodate whatever in any coordinate system

just as an example of where this is applied

[north sierra]

would the eigen value have a multiplicty of 3 or what

[quartz compass]

yeah, for an arbitrary rotation that's about it, but no

the other 2 eigenvalues are complex numbers

[north sierra]

ohh

[quartz compass]

there are special cases though

if you rotate 180 degrees for instance

then you have a 2D eigenspace with all -1 eigenvalues

since everything in that 2D plane normal to the axis of rotation get mapped to their negative

[north sierra]

that is so hard to imagine

[quartz compass]

just imagine a 2D plane

draw a circle and then draw a dot on the circle

then draw where that dot ends up after rotating 180 degrees

do the same for several more dots I suppose to convince yourself

although you really only have to do (1,0) and (0,1) to see that

well if you're having trouble thinking about this, I think I can help

[north sierra]

so what are we trying to do exactly?

find our other eigen value?

[quartz compass]

well you said it was hard to imagine

[north sierra]

yeah but i dont know what the objective of this is

[quartz compass]

so idk I was just trying to help you see how to look at the eigenvalues and eigenvectors for a 180 degree rotation as a special case

[north sierra]

the lifting hand up analogy was really good

was lost after that

lol

[quartz compass]

well what do you have for your answer to this question

is it done or is there more to say

they say at least 1, so I think you're good

[north sierra]

eigen value of 1

eigen vector is hard to think about

in 3 dimension

[quartz compass]

?

the eigenvectors are along your arm up and down

it's the axis of rotation

[north sierra]

yeah

so (0,0,1)?

[quartz compass]

it helps to think about what vectors don't move or are only scaled

could be (1,sqrt(2), 7)

you can rotate around any axis

[north sierra]

but wouldn't at least 2 of the entries be 0?

[quartz compass]

eigenvalues are not dependent on your choice of basis

[north sierra]

(1,sqrt(2), 7) how did you get that

[quartz compass]

(0,0,1) would be if you rotate around that axis

I could rotate around (0,1,0)

well not in my chair but lying in bed

[north sierra]

yeah

[quartz compass]

(1,sqrt(2), 7)

[north sierra]

lool

[quartz compass]

is just another axis of rotation

I can even say that's "up"

[north sierra]

which axis would that be

aren't there only 3

[quartz compass]

ok

is north in the +y axis or x axis direction?

[north sierra]

+y

[quartz compass]

wrong

it's a fake question

[north sierra]

lol

[quartz compass]

you could say it's diagonal along the line y=x is north

[north sierra]

i mean i just imagined a cartesian plane

[quartz compass]

yeah you are imagining a specific coordinate system on the world that doesn't really exist

you can say north is the positive z direction and east is the positive x direction

or weirder than that

you can say (1,sqrt(2), 7) is north

[north sierra]

yeah i dont get that

[quartz compass]

it's all just a matter of choice of how you draw coordinate lines

[north sierra]

i get you can say north is the positive z direction and east is the positive x direction though

[quartz compass]

there's no coordinate lines in the real world

[north sierra]

what do you mean by that

so like i shouldn't think about the cartesian plane when thinking about the real world?

[gray dust]

the coord system you work with is entirely what you choose

[north sierra]

oh

i see

yeah that makes sense

but this also is true in linear algebra?

well that was a dumb Question

i guess it is

[quartz compass]

I mean if we want to get real fake

the earth isn't even flat

[gray dust]

shocker

[quartz compass]

so it is kind of not even really real to begin with

[north sierra]

i never thought about this stuff so abstractly. this is cool

my mind is always stuck in one thing

lol

thanks for making me thing abstractly

its soo cool

[quartz compass]

I guess it's more complicated topic than I give it credit for

[north sierra]

thanks for all the help

[quartz compass]

yep

[north sierra]

is a 0 dimensional basis a thing?

or is that not even a valid thing to say

[quartz compass]

I don't know, I'd have to like see a specific definition of basis first to see if it can be weaseled in or not

[north sierra]

i see

[quartz compass]

like the zero vector itself I think forms a 0d subspace but maybe I'm wrong

[north sierra]

but like a basis needs to be linearly independent too

and a 0 vector isn't

[dusky epoch]

the zero space has a basis

the empty basis

[quartz compass]

I could see how that is like, vacuously true

idk not interesting to me personally but probably something to know if you're gonna be taking a test soon lol

[gleaming knot]

the empty set is the only logical basis for the 0 space

the 0 space is 0 dimensional. The basis must have 0 elements

[dusky epoch]

the empty set is LI

and it spans {0}

[gleaming knot]

Ll?

[north sierra]

linearly independent

[dusky epoch]

^

[gleaming knot]

I read it as $L\ell$

[stoic python]

Icy001:

[dusky epoch]

what kind of savage do you take me for

[gleaming knot]

an Ll one

[quartz compass]

lol like when people write IR

[dusky epoch]

ew

[north sierra]

could someone explain why this is True?

[dusky epoch]

for any x in R^n considered as a col vector, T(x) = Ax

🤷

[feral mountain]

I was thinking maybe due to this

[dusky epoch]

not quite

[north sierra]

how is this False?

actually nvm

[dusky epoch]

gram schmidt tho

[gray dust]

how is it that every day i see a couple new variations on the thonk emote?

[dusky epoch]

because it's discord

thonks are integral to every big server

just think

what would we all do

if we had no thonks

[quartz compass]

thonking is definitely at the pinnacle of discord culture

[north sierra]

yeah gram schimdt lol

[gray dust]

society would collapse without thonk emotes. ig i was ignorant to question that

[gleaming knot]

The original thonk

[feral mountain]

cropped

[prime needle]

for B would I apply the transformation to all the vectors in B

and then convert TB to basis B'?

and then that will give A' so it's in [T(v)]B'=A'v

[wintry steppe]

Hi guys, noobie question here: But how is the vector $c = (sqrt(2)/2, 0, sqrt(2)/2$?

[stoic python]

McNulty:

[viral mason]

is part a) just asking for standard basis?

[north sierra]

so it says not to use the pythagorean theorem so i can't use the dot product or what?

[half ice]

I don't see how you'd use the Pythagorean Theorem lol

[north sierra]

lol yeah true

[half ice]

Unless they're orthogonal

[north sierra]

isn't the pythagorean threoem when you find the length ?

[half ice]

Pythagorean theorem says, for any orthogonal u,v:
|u + v|² = |u|² + |v|²

And that holds for general inner product spaces

[north sierra]

oh

so what happens if theyre orthogonal

what can i use

[half ice]

You can still find u + v and then find its magnitude to get |u + v|

[gentle oasis]

You could use the dot product if you wanted as |u|^2 = u.u

[north sierra]

||u+v||^2 = ||u||^2 + ||v||^2 is only true when u*v are orthogonal ?

[half ice]

Yes

And u,v are orthogonal when that's true

[north sierra]

okay thanks

[dusky epoch]

@north sierra "u*v are orthogonal" makes no sense

did you mean "u and v are orthogonal"?

[north sierra]

yeah true lol

yeah i meant that

[dusky epoch]

* is not a substitute for and.

[north sierra]

LOL

sorry

[north sierra]

so you know if I have an orthogonal basis, lets say {u1,u2}

and you have a vector y that you're trying to write as a linear combination of the orthogonal basis vectors

does this formula only work if the vectors are orthogonal ?

[dire delta]

Is there a periodic equvialent of the characteristic polynomial of a tensor?

I'm representing the unit cell of a crystal as a variable size tensor and I'm trying to see what alleys exist to encode it in some fixed dimensional space

[topaz plover]

how do u multiply it if your missing numbers

[slow scroll]

@topaz plover you can find the missing numbers

[topaz plover]

how

[slow scroll]

so... do you know how the adjoint is computed?

[topaz plover]

yea

cofactor

then you transpose

[slow scroll]

yea so i guess you could just plug x, y, z into those question marks and solve the system you get

[topaz plover]

wym solve

like get the adojint

adjoint

with the variables

[slow scroll]

yea, but you don't have to compute the entire adjoint or anything. Just compute the cofactors that involve the missing numbers. Then they are equal to some entry in the adjoint. For example, the cofactor you get by deleting the top row and last column is equal to the bottom left entry in the adjoint.

You can make 3 different equations like that.

Err well, you can actually make more than 3 equations. It seems that some of them are linearly dependent

[topaz plover]

ty

[slow scroll]

npnp

omg im dumb @topaz plover Im pretty sure there is an easier way

[topaz plover]

oh what is it

how would u do this

youd have to do dot product

and cross product

but this is lines not vectors

[magic slate]

lines are defined by the direction vector

hence the name

so you can just take the dot product of the two direction vectors to determine if they are parallel

you dont need to worry about cross product since the dot product will give enough info

[topaz plover]

but its lines not vectors

[magic slate]

yeah but the direction the lines are facing is all that matters

which is entirely encoded in their direction vectors, the vector next to s and t

[topaz plover]

so the left part doesn't matter

u just need the right two

[magic slate]

you could choose any vector laying on the line and put that in the left vector

[topaz plover]

the dot of the S and T

is = to 0

so its perpendicular right

[magic slate]

yes

[topaz plover]

but what about parralel

[slow scroll]

their direction vectors would be linearly dependent if they were parallel

[topaz plover]

wait the dot product is -6

cause its -3 + -3

right

[ebon gate]

Yeah

[topaz plover]

so its not perpendicular

[ebon gate]

So they're not perpendicular

And since there's no λ such that λ(3,1) = (-1,-3), they're also not parallel

Alternatively, the cross product is non-zero, so they're not parallel

[slow scroll]

so about the question from earlier: another thing you can do (probably better) is to compute A * adj(A) with generic entries on the bottom, and solve for the missing entries so that you get a diagonal matrix in the end.

A * adj(A) should be diagonal because we know from the inverse formula that A * adj(A)/det(A) is the identity.

So for the second question, you don't have to do any work, because the determinant is the factor you have to divide by to get the identity (ones in the diagonal)

[topaz plover]

ty

[ionic dust]

Good morning everyone

I don't have a question about linear algebra, but I will be taking it next semester. With that in mind, does anyone have a personal favorite book on linear algebra with proofs for undergrads?

I would like to start reading up on this subject to prep before the semester

[dusky epoch]

linear algebra done right

or linear algebra done wrong

[ebon gate]

not linear algebra done partially right?

[dusky epoch]

that's not a real thing iirc

[ebon gate]

Nope

[wintry steppe]

good morning xD

[ionic dust]

I just got that book

it assumes that I have some pre-existing knowledge of linear algebra (which I do in terms of applied context)

I've only done proofs in discrete mathematics prior

[dusky epoch]

what book

LADR? or LADW?

LADW doesn't assume that iirc

[ionic dust]

LADR

[dusky epoch]

get LADW.

[ionic dust]

Will it be easier to grasp

I got 3 weeks before this term comes

[dusky epoch]

i guess so

[quaint pelican]

Hi Guys ! I'm struggling to understand this one

[brittle juniper]

These two questions are meant to test you in your knowledge of the definitions

[quaint pelican]

For the question (a) i understand that we can prove W is a subspace of R^3 by proving three things :
Show it is closed under addition.
Show it is closed under scalar multiplication.
Show that the vector 0 is in the subset.
Is it right ?

[brittle juniper]

That would be sufficient

[quaint pelican]

But what i'm struggling to understand here is that, if we can prove that W is a subspace of R^3 then set of points W must lie in a plane or a line right ?

[brittle juniper]

a vector of W, when it is ≠0, does indeed lie on the line generated by itself, but that's stating the obvious

[quaint pelican]

But i wonder one more thing is that, by given W above how can we know that every points in W is in a plane, because by taking any a, b in R i think there may be chance that some points in W don't lie in a same plane(subspace). If this happens, W can not be a subspace because points in R^3 are not closed under addition.

[gray dust]

you can check this by finding how many vectors form an appropriate basis for W

[quaint pelican]

You mean dimension, right ?

[brittle juniper]

there's a quite natural linear map f:R²->R³ such that Im(f)=W

and the dimension of Im(f) can be at most 2

[quaint pelican]

I still don't get it, can you give me some more information, please ?

[brittle juniper]

linear maps

have you heard of them?

[quaint pelican]

No, i haven't studied about it yet

[brittle juniper]

you'll learn about them soon then

[quaint pelican]

Ok then , guess i'll have to read about linear maps now 🙂 !

[brittle juniper]

no need to rush

[quaint pelican]

But can you please tell me is there any chance to that set of points don't lie in a plane ?

[brittle juniper]

I know for sure that there exists a plane of R³ that completely contains W

[quaint pelican]

You know this by linear maps right ?

[brittle juniper]

Yes

[quaint pelican]

Ok thanks a lot !!!

[brittle juniper]

You're welcome

[wintry steppe]

Anyone know how to do this?

[lost comet]

hi guys it might be a stupid questions but

why dont they just write a matrix with a=1 instead of writing a in the matrix and next to it "a=1"

[brittle juniper]

who knows

maybe they change the parameter a in the next questions or something

[vast torrent]

Probably want to emphasize that the determinant becomes a function of a

Maybe

[lost comet]

nah this is one single questionh

this matrix never shows up again

lol

also

i got 11

or 1/11

which is the determinant

when finding the inverse do you divide by the determinant

[gray dust]

divide what by the det

[lost comet]

ok

thanks

also this question

what do they mean by contained but not equal to

its not in my lecture notes

[brittle juniper]

what don't you understand ?

it's nothing special

completely normal words

[lost comet]

wording

hang on

they mean

[brittle juniper]

contained just means contained

not equal to just means not equal to

[lost comet]

v1 is in v2 or v2 in v3 etc...

and then v1 =/= v2 etc...

right?

ok so this statement is true then

[brittle juniper]

are you sure it's true ?

a quick look at the dimensions seems to kill it

[lost comet]

dimensions?

what does that have to do with this

[brittle juniper]

if E is a finite dimensional vector space and F is a subspace of E that's not equal to E, you can say something about dim(E) and dim(F)

[lost comet]

idk man

[brittle juniper]

dim(F)<dim(E)

[dusky epoch]

have you not yet covered the concept of dimension of a vector space

[mystic mango]

We've to prove this before getting into the completeness relation, but I'm really stuck as to where I should start.

[mystic mango]

Could someone help me out?

[slow scroll]

,rotate

[slow scroll]

@mystic mango The part after "Show that" isn't parsing. Is "c" supposed to be a scalar constant? Then how can a scalar = bra vector?

[mystic mango]

After show that: Sigma{1;n} |a> Ca

[slow scroll]

equals 1? is that a 1?

[mystic mango]

Yes

Sorry for the bad pic

[slow scroll]

a linear combination of vectors equaling 1

[mystic mango]

I think its supposed to mean the outer product of |a> and Ca, and then substitute some stuff to figure out Ca

And instead of 1 its the identity matrix?

[scarlet ermine]

i'm having some trouble understanding the logic behind this

i understand the matrix, and i accept how the eigenvalues were calculated

but i don't understand how to read them or how the eigenvector matrix works, especially with the is

[wintry steppe]

hii, someone can help-me with a question in #help-2 ?

let eigenvector as v and eigenvalue as a

a eigenvector and a eigenvalue of a matrix satisfies

$A*V = aV$

[stoic python]

r41nm4ker:

[scarlet ermine]

so to find the long term distribution

i'm not quite sure what to do

i see the math that is done in the 2nd picture but it's not really making much sense

[wintry steppe]

hmmmm

those eigenvector and eigenvalues seens numerically calculated

on complex

[scarlet ermine]

yeah they used matlab, but my class doesn't use matlab so im a bit in the dark as to whats going on

[wintry steppe]

i undestand

long term distribution is a term to linear combination ?

[scarlet ermine]

?

i am very dumb rn

[wintry steppe]

you are smart xD

we just does't talk the same "language"

differents areas

[scarlet ermine]

yeah : /

well im a math major but my class is just... apocalyptically bad

my only saving grace is that the practice exam is identical to the final exam

and that the problems are all stolen so i can google them to figure out how they were done

im just so far behind in this class that most explanations don't really click

[wintry steppe]

exams

that is a pain

i have a linear test tomorrow

my friend told me how to do the question

and i can't do alone

huehuehu

but about eigenvectors

basically is

A.ev = C.ev

where C is your eigenvalue

[scarlet ermine]

so my A matrix times the eigenvector is the same as the eigenvalues times the eigenvector?

[wintry steppe]

yes, but matrix multiplication is different

is done in other way

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

take a look here

the idea is

Line * Columns

example

[scarlet ermine]

i understand matrix multplication

i'm not that far behind, i sort of fell off the proverbial cliff at diagonalizing

so pdp-1 especially

i get eigenvalues but not eigenvectors

[wintry steppe]

about diagonalizing

if you can write $A = PDP^{-1]$ is diagonalizable

[stoic python]

r41nm4ker:
Compile Error! Click the reaction for details. (You may edit your message)

[wintry steppe]

there is tests

about geometrical multiplicity and algebric multiplicity

basically P is formed by eigenvectors

and D have all eigenvalues in diagonal

[scarlet ermine]

alright yeah i remember that

and the eigenvalues i use our eigenvalues() command

eigenvectors i don't think sage can handle, or at least i've not figured out how that works

we're allowed to use sage on our final

[wintry steppe]

try do a simple matrix

a 2x2 to test

the concepts

[scarlet ermine]

right so i have the matrix in

i have the eigenvalues

but im not sure how to handle the eigenvalues bc they're complex

[wintry steppe]

yeah, that is a problem too

[slow scroll]

complex eigenvalues are fine

in general, real matrices do not have to have real eigenvalues or real eigenvectors

you find the eigenvectors exactly as if the eigenvalues were real

-while following the rules of complex numbers of course

[scarlet ermine]

alright

so i have each eigenvalue

and i have the identity matrix

so what i do next is i just take each eigenvalue * the identity matrix and then subtract that from A?

[slow scroll]

yea. if $\lambda$ is an eigenvalue, then the basis for $\ker (\lambda I - A)$ gives the eigenvectors corresponding to $\lambda$. They will also probably be complex.

[stoic python]

kxrider:

[rose grotto]

I need help with 10c

how do I get

third side

[scarlet ermine]

@slow scroll whats ker?

[slow scroll]

Null space, or kernel. Whatever y’all call it

[scarlet ermine]

we call it the null space

[gleaming knot]

We call it kernel

[scarlet ermine]

so i rref the result

[gleaming knot]

Analysts and applied mathematicians call it null space

[scarlet ermine]

hm nvm that didn't work

[gleaming knot]

I think you could just use Law of Cosines

[scarlet ermine]

alright so i have my 3 eigenvalues, i have the identity matrix, and i have the original matrix

first i multiply each eigenvalue by the identity matrix

then i subtract that from a

then i row reduce each result? i think i screwed up somewhere

also the solutions seem to say they only use the adult vector? so im not really sure what to do with that

[slow scroll]

The adult vector? What’s that loll???

[scarlet ermine]

so the original problem im trying to solve is described here with its solution

i'm trying to figure out what exactly they did

[magic slate]

Oh

You're doing Markov chains

[scarlet ermine]

i thought markov chains were for the weather or something

[magic slate]

It's just any stochastic matrix that describes a linear dynamical system

[scarlet ermine]

i see yeah i guess

anyway i've done the first step and i got the eigenvalues

[magic slate]

Yeah, I think the goal is to determine the steady state of your system

[scarlet ermine]

now it wants me to make a eigenmatrix using only the adult vector (?) but that i'm confused by

[magic slate]

Basically, find when A_k+1 =A_k

[scarlet ermine]

yeah, so the long term distribution

i'm just trying to figure out the sequence of steps to do that and im falling v flat

[magic slate]

Yeah, unfortunately I'm not home right now so it's hard for me to work through the math just on my phone

I believe you could accomplish that by diagonalozing the matrix then raising it to a higher power

[scarlet ermine]

so in order to do that i need to calculate the eigenmatrix right

[magic slate]

Well the eigenvectors and eigenvalues

[scarlet ermine]

im v weak w eigenvectors

i'm trying to read how thats done rn and its not quite clicking

[low widget]

i need help with angles

;/

[magic slate]

One other way to solve

Is just find the eigenvectors of the matrix corresponding to eigenvalue 1

Because that fullfils the requirement that Av = v

[low widget]

where is the channel for angles

[scarlet ermine]

well rn im v concerned that i 100% forgot how eigenvectors are calculated

[scarlet ermine]

even though i did so fine just 2 weeks ago

[low widget]

for stuff like this

[magic slate]

Yeah, so you if you know your eigenvalue then you can use the equation $Av=\lambda v$ which is the same as $(A-I\lambda)v=0$

[stoic python]

l spacebaR l:

[scarlet ermine]

i don't remember settng it to 0 previously

[magic slate]

Well that's just me subtracting the right on both sides

Then factoring out the v

[scarlet ermine]

right so i want to set it = to v i think

[magic slate]

You end up just having to find a basis for the null space of $A-I\lambda$

[stoic python]

l spacebaR l:

[magic slate]

It just comes down to gaussian elimination

With all zeroes on the right side of the augmented matrix

[scarlet ermine]

right so i row reduce = to 0 but i don't remember doing that previously

i guess i'll see if i can tack on a 0 to the end? i've never learned how to do that so im v confused

[magic slate]

Well that's how I've always been taught to find eigenvectors

I'm sorry if I'm confusing you

[scarlet ermine]

its fine im not operating at 100% rn

[magic slate]

Haha fair enough

I got to go for like 30 mins but if you're still stuck I'll help when I get home

[rose grotto]

can someone help with 10c

[half ice]

Depends which two sides

v - w could be another side of the triangle

[rose grotto]

I think vw

as in

w-v

[half ice]

Also could be v + w or -(v + w) lol

Depends which two sides

[magic slate]

Well not really

If you assume that they are both vectors with tail at the origin

It will just be the magnitude of their difference verctor

Vector*

[rose grotto]

so how would I solve this

[magic slate]

$|v-w|$

[half ice]

That's fair, and a good assumption

[stoic python]

l spacebaR l:

[rose grotto]

then wat

[magic slate]

That's your answer

Just take the difference

[rose grotto]

the answer is 3squareroot of 14

so I do that then get the magnitude?

[magic slate]

Yeah find the vector produced by their difference

Then find the magnitude of that vector

Which you can find using Pythagorean theorem

[rose grotto]

kk got it

ty

: D

[magic slate]

Yeah np man

[rose grotto]

15c

idk

what to do

how do I check

[half ice]

What does it mean for a vector to be in span(x,y), for example?

[rose grotto]

idk

[half ice]

So a vector is in Span(x,y) if you can write the vector like ax + by for some scalars a, b

[magic slate]

yeah so choosing a to be 2 and b to be -3 you can see that it is true

w is in the span of u1 u2

[rose grotto]

kk ty

[north sierra]

forming a matrix using u1, u2 and on the augmented side putting (2,-3) would work?

[native lodge]

You have to consider all three basis vectors

[north sierra]

oh right

[scarlet ermine]

so for this here its the fibonacci sequence but i don't know how to prove it or what its looking for

k(1)=2, k(2)=3, k(3)=5, etc

[viral mason]

@half ice just had my final and wanted to say thanks for all ur help this term

[half ice]

Yeah, no problem at all. Feel free to ask if there's ever anything else

[obsidian rapids]

Hey All. Got a question here and not really sure where I went wrong. Not really sure where i'm going wrong.

now when I take that A^-1 mod26 and multiply it by A i'm not getting an identity matrix like my instructor is saying i should.

[half ice]

Note that's A inverse only in mod 26

[obsidian rapids]

Well that's what he said will work. but it isn't lol

[half ice]

,w matrix multiplication {{10,18,1},{17,1,3},{4,7,21}}{{0,23,19},{23,14,13},{1,12,18}}

[stoic python]

Display results online and refine query

Requested by Kaynex#7520

[obsidian rapids]

damn that's cool

ohhh

gotta mod26 it again

[half ice]

,w matrix {{415,494,442},{26,441,390},{182,442,545}} reduce mod 26

[stoic python]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by Kaynex#7520

[rustic panther]

If $x_1, ... x_k \in \mathbb{R}^n$ are vectors and none of them is a multiple of any other, then $x_1 ... x_k$ ..

1. must be linearly independent
2. may or may not be linearly independent

[stoic python]

glee:

[slow scroll]

To be linearly independent, none of the vectors are allowed to be linear combinations of each other

[rustic panther]

I'm leaning towards 2. may or may not be linearly independent

cause there might be a case where despite it not being a multiple of each other, it potentially might be linearly dependent

[slow scroll]

Yes, but why?

[obsidian rapids]

@half ice Thanks for that. You got me there. I had to reduce mod 26 again and it got me to the identity matrix. Appreciate it.

[rustic panther]

@slow scroll why I'm not sure but I can probably cough up an example of 2 vectors that is linearly dependent

[slow scroll]

Well, what if v1 = v2 + v3? None of these vectors are multiples of each other, but this is a valid linear combination demonstrating linear dependence

[rustic panther]

the method I know is uhm to write it in REF then solve for $x_1, x_2$ and so on

[stoic python]

glee:

[rustic panther]

from there I can tell if it's dependent or independent, I'm not remotely good enough to see it from the equations like you showed there

also no clue about linear combination 😅

are you free now @slow scroll I have a couple questions tbh

[slow scroll]

Well take an example like this: is the set {(1, 0), (0, 1), (2, 1)} linearly dependent?

Without writing these in a matrix and reducing and stuff, it’s not hard to see that since (2, 1) = 2(1,0) + (0, 1) that this is linearly dependent

That’s all you’re checking for

[rustic panther]

are you available for a couple more LA questions?

[slow scroll]

Yea

[rustic panther]

If $A, B \in M_3 (\mathbb{R}) ; and ; det A = -7, det B = 4$, what is $det (A^{-1} B)$ to 2 decimal places?

(don't tell me the answer just point me in the right direction)

Is $M_3$ here a 3x3 Matrix?

[vast torrent]

Can someone recommend a good video teaching QR decomposition? There are many results when i search for it and i don't know which videos are best. Rigor is important

$\mathcal M_n(F)$

[stoic python]

glee:

gfauxpas:

[vast torrent]

Is the notation for n by n matrices with entries from F

It's a set

Of all such matrices

[scarlet ermine]

i'm slightly confused by dimensions