linear-algebra

[tulip quiver]

because the lines are parallel

they have the same slope

the slop of the 2nd is 4

and the 1st is -b/2

so the slope of the 1st is 4

-b/2 = 4

[jaunty fern]

thank you

you two

[placid oracle]

Can someone pls help with this problem, I have no idea where to styart: Find an orthon ormal basis of P1 with respect to th einner product <a+bt,c+dt> = 2ac+2bd+ad+bc

[winter reef]

Whats P1? Whats t? Did you put correctly?

[quaint heart]

The vector space of degree 1 polynomials

[placid oracle]

^

[winter reef]

Ok now I understand

[placid oracle]

and yes

[winter reef]

I suppose you should use Gram Schmidt ortogonalization?

[placid oracle]

I think so but im not sure how to here

not sure what to do with : <a+bt,c+dt> = 2ac+2bd+ad+bc

[sonic osprey]

It serves the role of the dot product for normal vectors

[placid oracle]

but how come the ts cancel out

i have no idea where to start w/this problem

[sonic osprey]

The t's don't cancel out

The dot product is just describing a new function that takes in two elements of P1 and returns a real number

[placid oracle]

but the dot product of a+bt and c+dt is not = 2ac+2bd+ad+bc

[sonic osprey]

What?

You literally just told us it was

[placid oracle]

i dont undertsand what to do with this at all

is this correct: {1/root2, root(2/3)t-1/root6}

[sonic osprey]

Yeah that's right

[uneven crater]

I am supposed to say whether these three statements are either always true, sometimes true or always false. For the second one, I think it is always true since the smallest value being positive means that all the values are positive. but idk about the other two....

[winter reef]

first one is always false

2nd one sometimes true

I think

[quaint heart]

Yeah

Singular values are in descending order

So there could be a negative singular value

The last one is sometimes true

Orthogonally diagonalizable iff symettric

[uneven crater]

could you explain the reason for why the first one is always false

and if singular values are in descending order, doesnt it mean there could not be a negative value?

[winter reef]

if a vector would be in $V \ and \ V^\perp $ then $<a,a> = 0$ (because it would be orthogonal to itself) and thats impossible because of definiton of dot product ($<a,a> > 0$ for all non zero a)

[stoic python]

dog:

[uneven crater]

mk, wb the 2nd one?

wait but why are we implying that its <a,a>, what if the vector is <a,b> where only one of them is zero, doesn tit make the statement sometimes true?

@winter reef

[slow scroll]

well yea, 0 would be in both spaces

[uneven crater]

r u saying that im right or wrong

[slow scroll]

"doesn't it make the statement sometimes true"

In general, its not true. It is true only when one of the vectors is the 0 vector. This would make sense since the only vector V and V^\perp share is the 0 vector

[uneven crater]

got it got it

and kx, for the second one. they said that if the smallest singular value is positive its possible that one of them is negative... how does that make sense?

[slow scroll]

i don't know anything about singular values. Haven't studied that far in LA i guess

[uneven crater]

@quaint heart could u explain how that makes sense?

[quaint heart]

Oh, I'm stupid

I read it as the first singular value

Smh

Yeah, it has to be positive definite

@uneven crater

[uneven crater]

ok i was getting confused there haha

thanks!

[jagged saffron]

if we have N linearly dependent rows do we have eigenvalue 0 with multiplicity N-1?

or can we say anything about the multiplicity of the eigenvalue 0

if we have N linearly dependent rows

[slow scroll]

you can. if you have N linearly dependent rows, then the nullity of the matrix would be N, and the geometric multiplicity of the 0 eigenvalue would be N

[jagged saffron]

n-1

[sonic osprey]

Thats not true. At least, depending on what you mean by N linearly dependent rows

[jagged saffron]

right

ok i have N identitical rows

[slow scroll]

n-1, thats right

[jagged saffron]

same values

[slow scroll]

wait its not true?

[sonic osprey]

Consider N-1 independent rows, but then the last row the same as the first

Then, all N rows are linearly dependent, but the multiplicity of the 0 eigenvalue is only 1

[jagged saffron]

but in this case

where everything is the same

[slow scroll]

how does the last row being the same as the first make all of the other rows linearly dependent?

[jagged saffron]

zero vector?

[sonic osprey]

The set of all N rows is linearly dependent

[jagged saffron]

oh wait duh

if i have identical rows

then the rank is m-n-1

[sonic osprey]

In your case, you're right

But the phrase, N dependent rows is pretty ambigious

[jagged saffron]

yea my bad

[modest jasper]

Hello.

Does anyone know how to answer a question like this?

[dusky epoch]

do you know how to find the equation of a straight line like that given two points on it?

[modest jasper]

@dusky epoch I think so, but I just don't understand what t is.

[dusky epoch]

t runs over the real numbers

the line is the set of all points you can get from that formula by different choices of t

[modest jasper]

So t can be anything?

[dusky epoch]

well like

each value of t gets you a different point on the line

so in that sense yes

so you can pick any two real numbers you want - no, really, any two you want, as long as they're distinct - then plug them into this formula to obtain two points on your line

thereby reducing the problem to one you said you know how to solve

[modest jasper]

That makes sense I guess 😄

I'll try it!!

@dusky epoch Wait.

When you said two numbers, two numbers for what?

Where?

[dusky epoch]

two values of t

[modest jasper]

@dusky epoch Thank you.

Do you know why this might be wrong?

[dusky epoch]

probably an arithmetic fuckup so unless you show your work i can't say much

[modest jasper]

v=[-10, 1]
v=[-14, 7]

1=20/3+b
-17/3=b

That was my work basically.

[sonic osprey]

Slope is rise over run

Or y/x not the other way around

[modest jasper]

Oh yeah!

It still doesn't seem to be right though...

Wait..

I might have done it wrong.

One sec.

Yeah I still haven't gotten it.

v=[-10, 1]
v=[-14, 7]

1=-15+b
b=16

This is my working now.

[dusky epoch]

so where did you get 1 = -15 + b from

[modest jasper]

WAIT I GOT IT RIGHT

I mixed up the signs.

Thank you!!!!!

[steel coral]

can anybody link me a proof of this? im not understanding the one in the book. thanks

[slow scroll]

yea i asked this same question the other day. https://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_maps_and_Hermitian_matrices

[steel coral]

thanks

[cedar solar]

A Hermitian operator must be normal, and a normal operator has a basis of its orthonormal eigenvectors. By schur’s theorem, a normal operator can be represented as an upper triangular matrix in this basis

But since T = T* and both of them have to be upper triangular in this basis, they must both be diagonal

[steel coral]

hmmm

[slow scroll]

huh idk anything about normal operators but that sounds better

[steel coral]

i will stick to the wikipedia proof thanks

ayyy i get it now

tjhere was one specific thing i didnt understand that was bugging me but this proof addressed it

[slow scroll]

what was it?

[steel coral]

i didnt understand why the 2nd eigenvector HAD to belong to the subspace of vectors orthogonal to the first

ive seen the proof that all eigenvectors of a hermitian operator are orthogonal but it still was confusing with the original proof i was looking at

but this proof explained that the subspace is invarient meaning the characteristic equation has to output a vector belonging to that subspace

[cedar solar]

That first statement is incorrect

Eigenvectors corresponding to the the same eigenvalue belong to the same subspace

You can make them orthogonal through gram schmidt but the existence of orthogonal eigenvectors doesnt meant they have to be

[green aurora]

I have a question: If f: R^2 -> R^2 and g:R^2 -> R^2 are linear transformation, ker(f) = ker(g), Im(f) = Im(g). Then is f=g?

[dusky epoch]

well what do you think?

[slow scroll]

i think no

[dusky epoch]

i was not asking you, but in this case, can you prove that that statement is false?

[slow scroll]

consider the set of invertible $2\cross2$ matrices. They are invertible so their column space is $\bbR^2$ and their null space is ${0}$. there are clearly many different invertible linear transformations. $\begin{pmatrix} 3 & 0 \ 0 & 3 \end{pmatrix} \neq \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$ as a simple counterexample.

[dusky epoch]

👏

[stoic python]

kxrider:

[slow scroll]

yay

[astral junco]

Sry if this is dumb, but wouldn't any set of two nonzero vectors from R^n where n > 1 that aren't parallel be linearly independent?

[dusky epoch]

yes.

you need to prove, however, that two vectors that are perpendicular and nonzero are not parallel.

[astral junco]

hmm

i don't really get it. if they're not parallel then they're not scalar multiples of eachother and therefore not dependent by definition?

is that a proof?

[dusky epoch]

no

you are given v, w ∈ R^n, v != 0, w != 0 and <v,w> = 0

you need to prove that there does not exist a scalar c such that v = cw

how do you prove that there does not exist a [thing] such that [condition]?

[astral junco]

dunno really. proof by contradiction?

[dusky epoch]

yes. you assume such a thing exists, and from that derive a contradiction.

so why don't you do that here

suppose there DOES exist a scalar c such that v = cw, and yet <v, w> = 0, v != 0 and w != 0.

[astral junco]

yeah i'll try that. it's just the question looked odd to me like it was too easy. like "prove a square isn't a circle". i don't know i get confused a lot.

[dense stone]

Hello, during my exam I had to demonstrate that (u,v,f(u),f(v)) was linearly independent (question 3)

But I don't see how

Could you explain to me please ?

[brittle juniper]

Tu prends 4 réels $a_1,a_2,a_3,a_4$ tels que $a_1u+a_2v+a_3f(u)+a_4f(v)=0_{\bbR^4}$\
Il s'agit de montrer qu'ils sont tous nuls\
$f(a_1u+a_2v+a_3f(u)+a_4f(v)) =a_1f(u)+a_2f(v)+a_3f(f(u))+a_4f(f(v))\
=a_1f(u)+a_2f(v)\
=0_{\bbR^4}$\
Or $(f(u), f(v))$ est libre donc $a_1=0$ et $a_2=0$\
Donc $a_3f(u)+a_4f(v)=0_{\bbR^4}$, et encore une fois, $(f(u), f(v))$ libre donne $a_3=0$ et $a_4=0$

[stoic python]

Tuong:

[dense stone]

ah j'avais vu pour a1 et a2 et j'ai pas vu plus loin :facepalm:

merci beaucoup !

[brittle juniper]

De rien 👌

[faint lintel]

Hey what is the best way to learn Linear Algebra

Like textbooks and online resources and such

[dense stone]

In France some preparatory classes put their lessons online, maybe the same kind of ressources is disponible in english too ?

I'll search

[slender yarrow]

have you checked #books-old ?

@faint lintel

[faint lintel]

Yea but are those the best books?

I'm ok with buying a physical textbook

I'd actually prefer that and if there is a better text that's only available in print I'll get that

[slow scroll]

I'm not sure about the best book, but if you do "Linear algebra done wrong", I can help if you ever need help with something up to about the first half.

[slender yarrow]

well mit ocw is a good source but i'm also french so i don't really know physical textbooks in english either

and yeah lin alg done wrong indeed

[faint lintel]

Mit ocw + that book is good?

[half ice]

Are you looking to learn as an applied student, or as a pure math student?

[faint lintel]

I've seen MIT OCW I've just never actually used it

[dense stone]

oh j'ai oublié une question, M la matrice de f dans une base B, est-ce que M = M² prouve que f est un projecteur (désolé pour la coupure dans votre conversation) ?

[faint lintel]

Uhhh

[slow scroll]

mit ocw is quite different than LADW.

[faint lintel]

I'm going to go into applied probably because it's easier to get a job with it but I do love pure math

[slow scroll]

ocw follows gilbert strang's book, which is ehh imo

[faint lintel]

I'm in HS lol

[brittle juniper]

Oui, M²=M prouve que f est un projecteur

[slow scroll]

I just graduated HS, u should totally do LADW

[faint lintel]

Aight

[half ice]

I wouldn't fall for that myth! Jobs aren't particularly easy for anyone. Anyway, I have an engineering textbook that does applied lin alg in a chapter and makes it pretty easy

[dense stone]

encore merci Tuong x)

[winter reef]

🥖

[brittle juniper]

🥖

[slender yarrow]

🥖

[onyx tundra]

🥖

[half ice]

🥖

[slow scroll]

🥖

@faint lintel If you haven't already, I would definitely watch 3b1b's essence of linear algebra series, like 10 times over. ladw was kind of a difficult read for me at first, but it gets better as you stick to it.

[faint lintel]

Alright

I liked his calc series so yea that'll be good

[low hornet]

linear algebra done wrong is a terrible book

[slow scroll]

why do you say that?

[steel coral]

how do block diagonal matrixes work

how can a matrix have matrixes in its entries??

i have no idea whats going on

nvm i got it lol

[dire hound]

please can someone help me understand why the answer to this is so

why are x1 and x3 included here

can someone walk me through a step by step here of how to write the general solution for that ^? I can't see any FREE variables here

just one basic variable

[slow scroll]

you have x_2 = 0, and columns corresponding to x_1 and x_3 are free (don't have pivots)

[dire hound]

yes

but then why is there a 1

in the top entry for the x1 vector?

since in all the entries in the matrix there are all zeroes in the spot for x1

[slow scroll]

because its free. x_1 can be any element of the field, so that entry can take on any value that x_1 takes on

[dire hound]

why cant there be a 1 in the other two spots?

[slow scroll]

because the value that the first coordinate takes on is linearly independent from the others. The algorithm you use to find the null space gives you a basis for the null space.

[honest marlin]

I know what a linear transformation is. When would I want to use it? Could someone give me a couple of simple examples where 2D linear transformations come in useful?

[jaunty jolt]

Say you are integrating a diamond area

That's pretty difficult

A linear transformation could make it so that you are integrating a rectangle

That's pretty easy

[jolly nexus]

AB-BA = 0 implies simultaneous triangularisability

but does that mean they share the same Q in QR factorisation

provide counterexample if necessary

every diffeomorphic transformation is linear if you zoom in enough

[jolly nexus]

dw it's only true for hermitian matrices i'm rart

[dire hound]

what's an easy way to tell if a non-square matrix is invertible?

[cedar solar]

Not square -> not invertible

Try to prove this using the dimension theorem and how non-square cannot be a bijection

@jolly nexus be careful. Some non-hermitian matrices also have AB - BA = 0

[jolly nexus]

yes i know

[cedar solar]

Sorry for misunderstanding then. Which part is only true for hermitian?

[jolly nexus]

i meant like

necessary

it's necessarily true for hermitians

but not only

[astral junco]

@faint lintel I found http://joshua.smcvt.edu/linearalgebra/book.pdf as a recommendation for a free linear algebra textbook on r/math. I'm def no expert, but I can say that in my opinion it's really good. Tons of practice problems with answers.

It's free and open source, so you can simply print it out if you prefer a physical copy.

There's also links throughout so whenever they reference a theorem you can click on it and go to straight to the page where the theorem is explained and proved.

and each exercise number is clickable link to an answer. very convenient.

[faint lintel]

@low hornet why do you not like the book? because alot of people are recommending it and I think that's what i'm going to go with but I'd like to hear your opinion on the book

[low hornet]

i didnt use a book my lecture notes sufficed but when i did look at it, it didnt match anything being taught really, the notation was different, found most simple concepts quite mysterious within the text, didnt have much worked out examples if i recall

overall found it not pleasant to read at all

[faint lintel]

interesting

I don't have access to a lecture or teacher so I can't really do with just lecture notes (tho that's what i've done for calc and it work really well)

[low hornet]

that linked book right there seems alright

http://joshua.smcvt.edu/linearalgebra/book.pdf

[faint lintel]

yea

@astral junco thanks!

Imma also ask on the subreddit because I'd love to have a physical textbook but this does seem like a great resource

[low hornet]

get a book called finite dimesional vector spaces then

but ye i never used a book

[astral junco]

yeah no prob. it's worth checking out.

[low hornet]

paul Halmos

[faint lintel]

Hold up everyone here recommended Linear Algebra Done Wrong

but what about Linear Algebra Done Right?

people seem to be recommending that online

[low hornet]

any book will suffice

[faint lintel]

I mean i'd like a book that I can use that and only that if possible

[low hornet]

any will do

[faint lintel]

gotcha

[modest jasper]

Could someone help me understand why number 1 is false?

[slow scroll]

because A0 = 0

b \neq 0

[modest jasper]

Ohhhh, right.

Thanks 😄 😄

[slow scroll]

npnp

[modest jasper]

Why isn't everything true here?

[dusky epoch]

let S be that set

[slow scroll]

closed under vector addition?

[dusky epoch]

$(1,0,1) \in S$ and $(0,1,0) \in S$, but $(1,0,1) + (0,1,0) \notin S$

[stoic python]

Ann:

[modest jasper]

Ohhh, I didn't think of that.

Thank you 😃 😃

[slate hound]

https://media.discordapp.net/attachments/320078200124473344/583806313604317184/Screenshot_from_2019-05-30_16-14-04.png

A correct?

[slow scroll]

nope

[slate hound]

oof how is it D

[slow scroll]

D(3v1, 3v2, 3v3) = 3D(v1, 3v2, 3v3) = 9D(v1, v2, 3v3) = 27D(v1, v2, v3)
That's how the determinant works

[slate hound]

true

[slow scroll]

in general if you have a nxn matrix ur finding the determinant of
det(cA) = c^n det(A)

[cedar solar]

n-linearity

[modest jasper]

Why is there only a dash in question 2?

I mean only a dash where a value should be in some of the vectors.

[dusky epoch]

beats me. i think the document got miscopied or something

could be any of 4, 7 or 9 missing

[fringe cave]

I'd think it's a mistype/miscopy

[modest jasper]

Oh...okay.

[inner isle]

I need help

[slow scroll]

ok

[inner isle]

Can we go to #help-1

[modest jasper]

Just so I can be reminded, the determinant of any diagonalised matrix is the product of the diagonals, right?

[slow scroll]

yea. same for triangular matrices as well

[modest jasper]

Is it just for square matrices or any matrix?

[slow scroll]

the determinant is only defined for square matrices

[modest jasper]

Thanks 😃

[slow scroll]

npnp

[broken hawk]

(you can define it for rectangular ones to just always be 0)

(but many just leave it undefined)

[indigo cradle]

hello

i'hv got this question here

but i have no other idea what i'm suppose to learn from it

was i suppose to get the transformation matrix by trial and error? or is there something im missing

[cedar solar]

You need to figure out what the imaginary part means, and work backwards to get the matrix

[indigo cradle]

this is from the book Intro to Linear Alg by Prog Gilbert Strang

i see

so is there any special meaning to the matrix on the left

and what does the comment next to the solution mean

[cedar solar]

Have you worked with complex numbers much?

[indigo cradle]

yes

[cedar solar]

It is saying that you need to do a bunch of operations with real numbers to get the product of two complex numbers

And the matrix is a representation of complex multiplication

[indigo cradle]

but the matrix only stores coefficients?

[cedar solar]

Yes, so any complex numbers you can plug in the coefficients for it

[indigo cradle]

i see i see

let me think over it more

was i suppose to take away a method of calculating the product of 2 complex numbers using matrices

or does that matrix on the left have no meaning

[cedar solar]

The matrix on the left and the vector to the right of it, together, is a representation of multiplying two complex numbers

[quaint heart]

@broken hawk for a nxm matrix, you can define a continuous determinant like function which will be nonzero only for fullrank matrices

Using the determinant of submatrices

This function is pretty useful

[indigo cradle]

@cedar solar o rly but how does the first matrix translate to the first complext numbers

where does -B even come from

when the complex number is A plus Bi

[dusky epoch]

$i^2 = -1$

[stoic python]

Ann:

[indigo cradle]

ok so the matrix equation is only a representation

but not a method to multiply two complex numbers

amirite

[quaint heart]

Multiplying the matrices corresponds to multiplying complex numbers

@indigo cradle

[indigo cradle]

whats the point

of getting the matrix at the front

[plain fjord]

Let P be on element of K (nxn space, R or C) and P is a projection. Show that eigenvalues of P are 0, 1 or 0 and 1.

I started it with writing the claim in another form, (in this is y means eigenvalue), that "y = 0 or y = 1" can be written as y^2 -y = 0, or y^2 = y. This is what I need to prove. And then from the definition of eigenvalues in general form, we have Ax = yx, and in this case A = P, so we get Px = yx then if I multiply both sides of equation with y, we get
=> yPx = (y^2)x, but now I'm not sure how to proceed

I heard a tip that I gotta use P^2 = P somehow

[sonic osprey]

That tip you got is literally the definition of protection

[plain fjord]

Projection. Yes, but how does that help, y(P ^2)x = (y^2)x doesn't really help

[sonic osprey]

Don't multiply both sides by y

[plain fjord]

Oh, Px=yx only... hmm

But, uh, how does that mean that only eigenvalues are 1 and 0...

Px - yx = 0 doesn't really work either, I can't take the vector as common multiplier there

or... (P-yI)X where I is identity matrix.... hmm

[broken hawk]

this is overcomplicated

[plain fjord]

Oh?

[broken hawk]

look at the kernel, and at the image

and look how each behaves under P

[plain fjord]

Kernel of P....

I'm bit of a loss what you mean, how do I figure those out from what I have...

[broken hawk]

you don’t have to literally figure them out

just look at how an element in either the kernel, or the image, acts under P

[plain fjord]

Yes, kernel means those values which are mapped to zero

[broken hawk]

and use the fact that Ker ⊕ Im = V

[plain fjord]

Uh...

[swift plaza]

Or try writing out P^2x (where x is an eigenvector) in two different ways, one using that P^2 = P and one using that P^2x = P(Px)

[plain fjord]

Oh... interesting

Hitting a brick wall still :I

P(Px) = yx doesn't seem to help too much...

[broken hawk]

let A be a matrix and λ be an eigenvalue of A

can you name an eigenvalue of A²?

[plain fjord]

The same

as A'

[broken hawk]

sure about that?

let’s sya x is the eigenvector

then A²x = AAx = Aλx = …

[plain fjord]

No, wait, lambda^2

That would make the example above, P(Px) to Pyx, if we use the letters i wrote in the original claim

Then it would mean Pyx = yx, but how do i get rid of the P.... unless I don't need to?

Since, hmm, if y = 1 or 0, the equation holds

[swift plaza]

P(yx) = yPx = ...

[plain fjord]

yPx = yx, and

y^2 = yx

wait... y^2x = yx I mean

Which has solutions, 1 and 0

Hmm, wonder if that's it...

[broken hawk]

the argument works and is way nicer than the one I was thinking of. just gotta write it out cleanly

[plain fjord]

What was the solution you were thinking of? I'm always interested in alternating solutions

[broken hawk]

well, basically, the kernel of P is the eigenspace with eigenvalue 0
the image of P is the eigenspace with eigenvalue 1, as for x in the image, x=Py for some y and thus Px = P²y = Py = x

and then you need to argue that V is Im(P) ⊕ Ker(P) and thus that’s all there is

[plain fjord]

Wow

[placid oracle]

What is an example of this? in an inner product space, not necessarily R^n, there can be a nonzero vector of length 0

[quaint heart]

There cannot

That's one of the properties of norms

@placid oracle

[placid oracle]

that is what i thought but im being asked to give an example of that^

[broken hawk]

then say “no”

[placid oracle]

mistake maybe?

[quaint heart]

Can you state the exact problem?

[broken hawk]

that, or you can’t read

it’s meant to be a jab at best

not an insult

[quaint heart]

Everyone misreads a question every once in a while

[placid oracle]

it sliterally copy pasted

[broken hawk]

I call myself an idiot so often, I get to do it to others too sometimes :^)

[placid oracle]

😛 haha

[broken hawk]

(in all honestly, if you think the above is rude then I’m gonna play the “cultural differences” card cause that’d be a perfectly normal stamenet to make here)

(as a light-hearted jab)

[quaint heart]

Huh, maybe you're right

I would say that to a friend I guess

[covert dragon]

Hello

[placid oracle]

How do I find the projection of b onto col A here?

[waxen pivot]

uh

First check if the columns are linearly independent

which they clearly are

so A has col space of 2

the projection of b onto col A is just

i forgot the name of this theorem but

sheesh i need to learn how to use the bot

[quaint heart]

Is orthogonal projection when you express $\mathbb R^3$ as $A\oplus A^{orthogonal}$, write b using that basis, and then forget the 3rd coordinate?

[stoic python]

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

[quaint heart]

Or nah?

[waxen pivot]

yes

you need to have the basis be

[quaint heart]

Oh cool

[waxen pivot]

uh

right angles from each other

crap

terms are not coming up in my head sorry

[quaint heart]

Orthogonal

Lmao

[waxen pivot]

lol 🤦

oh yeahhh ok the col of A needs to be an orthogonal basis

which you get from some formula

[quaint heart]

You can use the cross product right?

To get the orthogonal vector

[waxen pivot]

I didn't learn the cross product method

but

from my lin alg class, we just used dot product to make an orthogonal basis

i guess you can use cross product to get an orthogonal vector, and dotting it with b to find the projection onto that?

yeah that should work too

but it only works in R^3

I'm not entirely certain how cross product reacts in R^n

[broken hawk]

you can’t even define it

[waxen pivot]

yeah ok that's what i suspected

[broken hawk]

there’s a generalization, but it won’t give you a vector

the wedge product between two vectors gives you an oriented area element; the cross product takes that and produces a normal vector to that area

the wedge product works in all dimensions, but it introduces a new type of objects

which can be undesirable

[waxen pivot]

Huh

[broken hawk]

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

[waxen pivot]

Can't we just say the null space of the two vectors?

[quaint heart]

?

[waxen pivot]

if what i'm interpreting is correct

[broken hawk]

what is the null space of two vectors

[waxen pivot]

uh

basically the space that a set of vectors in R^n cannot reach

[broken hawk]

to get more orthogonal vectors, take any vector outside their span, and apply gram-schmidt

[waxen pivot]

Yes that's the name of that XD

but you need to get a vector that isn't in the span in the first place though

[broken hawk]

that’s usually not too hard

[waxen pivot]

But is there a systematic way to get it

[quaint heart]

Yes

If you have an inner product

It's pretty easy

[waxen pivot]

Inner product?

[quaint heart]

I mean

Dot product

Whatever

Read about the orthogonal complement

[waxen pivot]

oh yeah i forgot about that

[quaint heart]

You just have to take the vector (x, y, z)

And dot it with all of the vectors you have already

And set that equal to 0

Obviously I don't mean (x, y, z)

I mean a vector with n variables

[waxen pivot]

Yes

A vector in R^n

[quaint heart]

Then you get a bunch of equations

Which you solve

[waxen pivot]

and it's just linear set of equations

[quaint heart]

Yes

[waxen pivot]

with a particular solution

ahh yes ok

[broken hawk]

well, many solutions

just pick one

[waxen pivot]

it is defined to always have many solutions

if i'm not mistaken

[placid oracle]

i am confusion

[waxen pivot]

Sorry sorry we deviated from your question quite a bit

[brittle juniper]

Hi confusion, I'm Tuong

[waxen pivot]

Gram shmidt or somethign like that

use that to find orthogonal basis of A and use that to find orthog proj of b on A using something similar

I might be taking a couple extra steps, but i'm not certain it works without orthogonal basis.

oh wait

I can't really remember, but i somewhat vaguely remember the least squares solution of things

Might be what you're looking for?

Yeah i'm not too familiar with that method

[placid oracle]

I think thats it

im gonna trey it and tell u what i get

[placid oracle]

@waxen pivot is it 21/4 // 15/4 // 21/4 ?

[waxen pivot]

You can check that

if you that from the original vector b, you should get the third orthogonal basis

base*

and if dotted with both of the col of A, you should get 0. if not, it isn't the orthog proj of b

[sinful hornet]

Time to learn about Non-negative Matrix Factorization

😃

I will probs have lots of questions so be prepared lmao

[waxen pivot]

lol oh gosh

[sinful hornet]

its for my CS research class lol

I have to read a bunch of papers on using it for Music Transcription and then apply it to something

it already doesnt make sense oh fuck

[sour garden]

Oh hi @sinful hornet

Were you always in this server?

I'm PTYamin if you remember me years ago when we played smash through anthers ladder

Unless I'm mistaking you

[sinful hornet]

I dont really play Anthers that much but its possible

and I only joined this server a month or two ago

[gray glen]

I maintain that anther's is a glorified chatroom

[sinful hornet]

it is

[dire hound]

how does the second step turn into the third step?

i don't see where they plugged stuff in

[quaint heart]

Do you know what the dot product is @dire hound

[dire hound]

yes

[hasty ether]

that is how

[quaint heart]

So <a, b> is simply a•b

The dot product is the standard inner product we use

[dire hound]

right but how is there a third 5 in front?

[quaint heart]

Probably you defined a different inner product

Actually

Is there context?

@dire hound post the whole page maybe

[quaint heart]

Yes

Weighted Euclidean inner product

Read the instructions and it will make sense

Lmao

@dire hound

Does that make sense now?

[dire hound]

oh yeah it does

<u v> is caluclated with that formula that defines <u v>

and it has 5 and 3 and 1 as coefficints

ok thx

[hasty ether]

you will be a wonderful engineer

[mighty cargo]

wait what's the point of defining the inner/scalar product in different ways rather than
[u_1, u_2, . . . , u_n] \cdot [v_1, v_2, . . . , v_n] = u_1v_1 + u_2v_2 + . . . + u_nv_n ?

[quaint heart]

For finite dimensional probably there's some physical interpretation

[mighty cargo]

huh

[quaint heart]

But any inner product induces the same topology for finite dimensions

For infinite dimensions it's more interesting

[mighty cargo]

wdym by "the same topology", sorry am brainlet

[quaint heart]

So you can make a norm from an inner product

Once you have a norm you can define open sets

Exactly the same way as you define it for R^n

The collection of open sets is called a topology

Every norm induces the same topology on finite dimensional real vector spaces

(which is a fun exercise)

[mighty cargo]

😮

that's actually p cool

[quaint heart]

For infinite dimensional that's not true

The fact you use to prove that the topologies are the same is that the unit ball is compact

In finite dimensions

That's not true in infinite dimensions

[mighty cargo]

what happens to the unit ball in infinite dimensions?

[quaint heart]

There are sequences which are cauchy but don't converge

Because you have an infinite number of coordinates

[mighty cargo]

so it's no longer a unit ball in a sense huh

[quaint heart]

It is

It's all the points of distance <= 1 from the origin

It's just not as nice

What sense do you mean?

[mighty cargo]

hmm I think I'm understanding wrong

ooh

a unit ball in infinite dimensions

would have an infinite number of coordinates square-summed to be equal to one

right?

[quaint heart]

So no

[mighty cargo]

oops

[quaint heart]

It only has a finite number of nonzero components

So you know the Bolzano weierstrass theorem?

[mighty cargo]

nope

[quaint heart]

So it says that for a compact space, any sequence has a convergent subsequence

Consider the sequence (0, ..., 1, 0,...) Where the 1 is in the ith slot

[mighty cargo]

okie

[quaint heart]

So that sequence doesn't have a convergent subsequence

The sequence looks like (1, 0, ...), (0, 1, 0, ...) (0, 0, 1, 0,...),... Btw

The Bolzano weierstrass theorem isn't that hard to prove

The proof for [0, 1] is pretty simple also

https://youtu.be/dfO18klwKHg

[mighty cargo]

lmao

[quaint heart]

This is cringy but also informative

[mighty cargo]

I'll give it a hot listen

wow this hurts to listen to

u really weren't lying

[quaint heart]

Yeah

But the proof is nice

The backup singers tho

[mighty cargo]

we're down with that

wait so

how does monotone convergence theorem work

I get that the bolzano-weisterstrass theorem works in that you keep on slicing the interval in half until you get a really small interval with a small peak, and that interval can converge to a bigger peak due to the monotone convergence theorem

wait nevermind

I'm dumb

I think I get it

[cyan minnow]

Hey guys, I was just doing math homework and I had a question. We are on the Vectors portion of our Calculus and Vectors course. So far every single concept has had real life applications questions. For example, in vectors we can show where an airplane should aim itself given the max speed of the airplane and the direction and speed of the wind. However, now we have been learning about planes (vector planes) for a while now. I have been looking through my workbooks and I have not found a single real world application problem for planes. IT seems purely mathematical. This may sound like a stupid question but, are there real life applications of a plane? What are planes used for in different fields?

[placid oracle]

is this always false or is there an exception? If T is a linear transformation which takes R^3 onto P_2 (polynomials), then T must be one-to-one.

[quaint heart]

It's very false

But not always

Isn't P_2 3 dimensional?

Just map the basis vectors to 1, x, x^2

@placid oracle

[placid oracle]

yes

[quaint heart]

You go to #prealg-and-algebra

[placid oracle]

it is

[wraith tulip]

Okay

[dusky epoch]

is this always false or is there an exception? If T is a linear transformation which takes R^3 onto P_2 (polynomials), then T must be one-to-one.

take the zero map

so very not injective

[quaint heart]

He was asking if it's always false

Not if it's always true

[placid oracle]

its neither tho

sometimes true?

[quaint heart]

It's sometimes true

I guess the way you phrased it would be false

In fact the set of maps which aren't injective have measure 0

So it's usually true

[placid oracle]

If a set of vectors in a p dimensional vector space V is a spanning set for V, it is a basis.

this isn't true, right?

[quaint heart]

No

Consider V

@placid oracle

[placid oracle]

what do you mean

[quaint heart]

The space itself

Is a spanning set

If you have p vectors

Then it's true

[placid oracle]

but what if u dont have p vectors though?

then its false

[quaint heart]

Yes

[placid oracle]

it doesnt tell us how many vectors we have

thats so annoying, why is this a true or false question :/\

[quaint heart]

So the statement is false

[placid oracle]

yep, thats what i said

ty

[rare barn]

how do i find the standard matrix for a lionear transformation for T: R^2 -> R^2 where T(b1)=c1 and T(b2)=c2

[quaint heart]

You know the matrix with respect to the basis b1, b2 and c1, c2

Is the identity matrix

Just transform your basis appropriately

[dusky epoch]

@rare barn express (1,0) and (0,1) as linear combinations of b1 and b2

use this to calculate T(1,0) and T(0,1)

[quaint heart]

This works too

[rare barn]

is it 1 // 2

0 // 1

@dusky epoch

[dusky epoch]

uh

i don't understand that notation

[rare barn]

if it is, let B - {b1,b2} and C = {c1,c2} , how do i find the changhe of basis matrix P_ C to B

um

[1 1]

[2 0]

[dusky epoch]

why would you need that matrix

[rare barn]

wait wdu mean

[dusky epoch]

if anything, if you want to do this via transition matrices, then at least you should be tying them to the standard basis!

since that's the basis in which you want to find the matrix of your transformation!

[rare barn]

thats a seperate question

i found the matrix whcih is ^

[wintry steppe]

Could someone please explain how the solution was derived?

[sonic osprey]

What do you not understand?

@wintry steppe

[wintry steppe]

Mostly the part where it said: |k| ≠ 1

But what puzzles me, why did it try 1, -1 and not equal 1 specifically? Is this the de factor numbers that you try?

[dusky epoch]

$|k| \neq 1$ is the same as $k \neq 1, -1$

[stoic python]

Ann:

[dusky epoch]

$1$ and $-1$ are the roots of $k^2 - 1$

[stoic python]

Ann:

[dusky epoch]

bad stuff happens when what you thought was a pivot turns out to be 0

[gray glen]

[rare barn]

Let B be the basus for {b1,b2,b3} = {2,1+t,1-t^2}. How do i find {t^2-3t+2]_B ?

[wintry steppe]

@dusky epoch Thank you.

Drawing the graph is straight-forward, but I wonder how it deduced no solution? Does all the lines have to intersect at a single point for a solution to exist?

[dusky epoch]

yes, there must be a point that all three lines have in common

[wintry steppe]

Thank you @dusky epoch 🙏

Why isn't point (1, 1, 1) included in the span?

[dusky epoch]

what two vectors

[1; 2; 3] and [2; -3; 1]?

if (1,1,1) were in the span then that system of yours would have a solution

[wintry steppe]

This is the past paper, first question.

And I am confused about which vectors they're implying.

[cedar solar]

You can think of columns in matrix A as basis vectors if they’re linearly independent

Since Ax, the vector x gives you how to combine those columns

So Ax for all possible values of x is the span

[gritty ore]

anybody can help me solve this system of equations

[native lodge]

how do you want to solve it then? plenty of ways to do it

[gritty ore]

x^2+y^2+z^2=1, x+2y+z=sqrt(6)/2, x-y+z=0

[native lodge]

$x^{2}+y^{2}+z^{2}=1$, $x+2y+z=\frac{\sqrt{6}}{2}$, $x-y+z=0$

[stoic python]

⚡Amphy⚡:

[native lodge]

unfortunately not linear, but still solvable, we have three equations and three unknowns

I suggest start with final equation and solve that for a variable, perhaps x if you want

x-y+z=0, solve that for x

[gray glen]

you can immediately get y as a number

[native lodge]

solve a system is more busy work than anything else lol

[gritty ore]

oh yea u can immediately get y from last 2 eq than sub it in to find x and z right

[gray glen]

yes

[gritty ore]

ok ty

[native lodge]

then you go from there to get the other values

[gray glen]

np

[vague belfry]

How to find the eigenvalues of a matrix A=[0, 0, 0, 1; 1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0] without finding out the charasteristic polynomial

My reasoning was that it is a permutation matrix

And then if there is an eigenvalue, 1 should be one eigenvalue since permutation matrix will not actually scale the vectors

And then since the trace is 0, I tried -1 as the other eigenvalue

both with geometric multiplicities 1

[dusky epoch]

why are you not allowed to find the charpoly

[vague belfry]

uni test

[dusky epoch]

are you specifically instructed against finding the charpoly or what

[vague belfry]

yes

[dusky epoch]

wow yikes

[vague belfry]

And well I found out that i and -i are the other eigenvalues, is there some reasoning to that or just guessing a complex number that does not scale

[dusky epoch]

permutation matrices are unitary

so all their eigenvalues will lie on the unit circle

[vague belfry]

Do eigenvalues of all unitary matrices lie on the unit circle?

[dusky epoch]

yes

[vague belfry]

I guess orthogonal matrices as well then

[broken hawk]

well, orthogonal matrices may not have a full set of eigenvalues (since orthgonal matrices are real)

but if you see orthogonal as “unitary with only real components” I suppose the answer is yes

[vague belfry]

makes sense

😄

[broken hawk]

a rotation matrix for example has complex eigenvalues

[vague belfry]

Hm, is there a way to reasonate as what does it mean for a matrix to have a complex eigenvalue, I mean in space, as in R2

I'm sorry, I don't know the names in english xD

I mean if an eigenvalue is real, it will get scaled after the transformation is made

to the same line

Is there an equivalent intuition for what it means that an eigenvalue is complex

[broken hawk]

well, complex eigenvalues by definition only make sense in vector spaces over C

the issue with that is of course that C^2 is already hard to reason about intuitively

[vague belfry]

Yup

Thanks for the help 😄

[astral junco]

Why does the order of vectors in a basis matter?

[vague belfry]

if you have a basis (1, 0) and (0, 1) that span R2, let's denote e1=(1,0) and e2=(0,1) and if you have say (1,2) = 1*(1,0) + 2*(0,1), it is the representation of vector(1, 2) with basis (e1, e2)

if you change the order of the basis with the same representation 1, 2 then you will have 1*(0,1) + 2*(1,0) which is (2,1)

so with the same representation, but the order of vectors in a basis changed you got two different points in space

[astral junco]

ah yeah that makes perfect sense. thank you

[vague belfry]

If I have a plane whose basis is b1 = [1, 1, -1, 0], b2 = [1, 0, 1, 1] and I want to create a matrix of orthogonal projection on this plane, is there a quicker way to do this than the one I will describe

I will solve a system a+b-c=0
a+c+d=0

Take one vector from the solutions of that system

then solve another system to get the fourth vector

Now I have a basis b1, b2, b3, b4. The matrix of orthogonal projection with respect to that basis is A =
[1 0 0 0]
[0 1 0 0]
[0 0 0 0]
[0 0 0 0]

Now I want to convert this matrix to another matrix A' with respect to the standard base

So I would have to find an inverse matrix of S = [b1 b2 b3 b4]

which is pretty slow

and then get the matrix A' by transforming it to the standard base

soo, is there a quicker way? xD

[broken hawk]

it is… but it’s still not linear algebra

#prealg-and-algebra is more suitable

also, we can’t see the question so…

idk how you expect help

[dapper kindle]

i was just asking what problem it was that's all

sorry i'll delete the photo now

[cedar solar]

@vague belfry the projection of x onto {y1, y2} is simply <x,y1> y1 + <x,y2> y2 if y1 and y2 are orthonormal bases

You can turn your bases vectors into orthonormal ones using gram schmidt

[vague belfry]

I'm sorry, whta does the notation <x, y1> mean

I just found out the problem is from here

https://web.mit.edu/18.06/www/Fall17/1806/exams/exam2-sol.pdf

1.c

The one that I was doing when I asked the question

[cedar solar]

<.,.> is inner product

The standard one is dot product

[vague belfry]

Welp, I have to learn that

I mean the Euclidean spaces

[plain fjord]

What about them?

[vague belfry]

I have to learn the axioms

And general stuff about it

[quaint heart]

Like the geometric Euclidean spaces

With the euclidean group acting on it?

Or do you mean real vector spaces?

[wintry steppe]

How i do algebra 1

[dense holly]

u in the wrong hood dog

[vague belfry]

@quaint heart The axioms regarding the Euclidean Vector spaces so I can extend them to abstract vector spaces I guess

And the axioms regarding norms and stuff

[vague belfry]

I have another problem. I have to prove or disprove the following statement.
The set of all 3x3 matrices of real numbers A whose kernel is the plane x1+2*x2-x3=0 is a vector subspace of 3x3 matrices of real numbers and it's dimension is 3.

Now I have proven additivity and homogenity

I constructed one matrix A
[1 2 -1]
[0 0 0]
[0 0 0]

It's kernel is the plane above

Now I'm thinking the basis are matrices like the one above, then matrix
[0 0 0]
[1 2 -1]
[0 0 0]

and matrix
[0 0 0]
[0 0 0]
[1 2 -1]

The dimension is three

so it's a vector subspace

[dusky epoch]

Now I have proven additivity and homogenity
this ⬆ should have been followed immediately by this ⬇
so it's a vector subspace

[vague belfry]

I still had to prove it's dimension 😄

[dusky epoch]

find its dimension

[vague belfry]

not prove, yeah

I mean the problem was to say true or false, so I guess once I found out it's dimension it's true

I found out it's dimension is 3 it's true, if the dimension is 3

[jagged rock]

Is the concept of "column space" only useful when the transformation matrix has a 0 det?

[dusky epoch]

define "useful"

[quaint heart]

Lmao

It's very useful always

It's the image

[jagged rock]

As in, it tells us something new which would've been inaccessible by our previous knowledge

[slow scroll]

if you have an nxn square matrix where the determinant is not zero, then the transformation is surjective and spans some n dimensional space

[quaint heart]

If you deal with exact sequences, the image is very useful

No matter if your linear map is an isomorphism or not

[rigid cypress]

col space is p useful

[jagged rock]

From what I've learned so far, it seems like "column space" is only useful to find the solutions of equations of type Ax = b, when A has a 0 det a.k.a no inverse matrix. Column space enables us to find the set of b vectors for which a solution does exist.

For others, where the transformation retains the full dimensionality of our space, there are other methods that can be used to compute solutions

[plain fjord]

Wait you can only take determinant from square matrix

[jagged rock]

correct me if I'm wrong please

[plain fjord]

But in Ax = b, the A can be other than square matrix

[jagged rock]

Oh true

Wait, yeah that's my point. For matrices where you can either a) not take the determinant b) whose det is nonzero, there are other methods to compute the solution to our equation

[slow scroll]

the determinant doesn't help you find solutions, but if the determinant of a matrix is non zero, then yeah, the image of its transformation is exactly its co-domain

[plain fjord]

if A is square and invertible, and Ax = b, then x = (A^-1)*b

[jagged rock]

No no, I'm not saying the determinants helps us find solutions. But it does narrow down our options of finding the solution, if the det of the transformation matrix happens to be 0 i.e. when it squishes all of space into a lower dimension.

[cedar solar]

@vague belfry your matrices are wrong. The kernel is the plane, means that A dot (1 2 -1)T = 0

[jagged rock]

And that is when the column space concept is really useful

[slow scroll]

right, i.e. the transformation is not surjective. However, depending on how you define the determinant, it is also true that matrices with zero determinant can still be surjective.

[jagged rock]

That's why I asked..

[quaint heart]

So linear algebra is used as a tool in many other areas of math

[slow scroll]

i.e. you could have a map R2 to R that is surjective but has a zero determinant (if you extend det to non square matrices).

[jagged rock]

oh, interesting

[quaint heart]

Maybe you're right from the point of view of systems of linear equations

[jagged rock]

Let me rephrase my question then: Is the column space concept still as useful when A, our transformation matrix, is full rank?

[quaint heart]

That's impossible to measure

[slow scroll]

"still useful" is probably not really what you mean to ask :p

[quaint heart]

I would say yes though

[plain fjord]

hmm if you have A= [[1,1],[1,1]] * [[x1],[x2]] = [[1],[1]] (b) - that has infinite solutions, determinant of A is zero, but A= [[1,1],[1,1]]*[[x1],[x2]] = [[1],[0]] has no solutions, and determinant is zero too

if you understand the notation, i mean, A is matrix with 4 elements, all 1, and is a 2x2 matrix

and b is a 2x1 matrix

damn i dont know why discord changed the text i wrote like that

[jagged rock]

Dw, I understand it

[plain fjord]

Yeah that's the notation wolfram alpha uses,

[jagged rock]

Wait, so what are you trying to say with that example?

[plain fjord]

That determinant can be zero, but the whole equation can still have no solutions or infinite amount of solutions

[jagged rock]

Yeah, that's why the column space concept is useful, right? We can determine all the possible b vectors that lie in the column space of A for which the equation has a solution

and this is despite the fact that the det of A is 0

which I can see how it could be useful in applications

For full rank A, we can simply take the inverse and muliply by b to find the solution set

[plain fjord]

Ax = b has solutions if and only if b is an element of Im(A)

[jagged rock]

I'm not familiar with image

agh I think I'm talking out of my ass here. I should probably study some more

[plain fjord]

Don't worry man

I'm not that far in my studies either

[slow scroll]

Finding the determinant of a matrix is a good way to gauge the invertibility of the transformation, and if a transformation is invertible, then it is surjective (the column space spans all of the co domain). That's pretty much all there is to it.

[plain fjord]

If and only if Ker A = {0}, the solution to Ax = b is unique.

[slow scroll]

unique is the word :p

[plain fjord]

Hahah yeah sorry 😃

Image is always a bit hard to understand to students, and thus, for me too

[vague belfry]

@cedar solar Hm, yeah but if we have x1 + 2*x2 - x3 = 0, then x3 = x1 + 2*x2, so we have a vector [x1 x2 x3]T = [x1 x2 x1+2*x2]T = x1*[1 0 1] + x2*[0 1 2] where x1 and x2 are real numbers so the plane is actually span([1 0 1]T, [0 1 2]T) right?

So when we multiply A*x where x is from that span we always get 0

[plain fjord]

uh, you mean [[1,2, -1]] for the first row?

wait, uh

[vague belfry]

Uhm, sorry for bad writing, but if you scroll up you have the problem 😄

And since A*x = 0 for every x from that span, that span is the kernel

[plain fjord]

I don't really understand at all. But if you have 3 linearly independent vectors, they form a basis in 3x3.

But above, you have the same row in three matrices?

[vague belfry]

Well yes, the span of those matrices

since I have proven that the subset is a subspace

now I needed to find the dimension of the subspace

And the subspace is the set of all matrices whose kernel is that plane

the dimension needed to be 3, and if you have a span over those three matrices, every time the rank will be 1 and the dimension of the kernel is 2

and the plane of the kernel is the one that is asked for every time, I mean should be I'm not sure xD

[plain fjord]

Whatta hell 😃

If you want to find a basis, the vectors must be linearly independent. Meaning they can't all be the same vector...

[vague belfry]

They're not the same vector

since they are a matrix

three matrices

and those three matrices are linearly independent

[plain fjord]

Hmm, can't say anything yet to that...

[cedar solar]

@vague belfry oh yeah youre right, my bad

[indigo cradle]

hi could i get some help understanding this section of my textbook

i don't understand the part where "The elimination steps are taken by an invertible M" and why this means that MA has a zero row.

[cedar solar]

So elimination is done by a sequence of elementary operations, which can be expressed as elementary matrices

M is the product of all the elementary matrices needed to perform elimination on A

So if A doesn’t have n pivots it will have a zero row after elimination. M is the transformation for elimination

But M must be invertible since it is a product of invertible matrices

[indigo cradle]

i c

then why does it conclude that C must be the inverse of A

[dusky epoch]

because CA = I

it satisfies the definition of an inverse

[indigo cradle]

okie thx i think im thinking too much

[tacit sigil]

How does one get this power of a matrix?

[sonic osprey]

By the induction hypothesis

It says assume that the statement is true for n = k

And that's what they used there to get that power of the matrix

[dusky epoch]

@tacit sigil are you familiar with induction?

[tacit sigil]

Lamda^k is understandable but
1^k = klamda^(k-1)??

[dusky epoch]

...

no

do you not know how matrix multiplication works

hint: it's not just multiplying entry by entry

nor does raising a matrix to a power equate to raising each entry to that power

@tacit sigil

...

[indigo cradle]

help i dont understand the answer given for 7a. why does the answer give for the case for Ax= (1,0,0) when the question asked to explain for Ax = (0,0,1)

[dusky epoch]

typo probably

yeah

typo

[indigo cradle]

so why isnt there a solution

i was looking to the answer for help 😅

[dusky epoch]

replace (1,0,0) with (0,0,1) and 0 = 1 with 0 = -1

[indigo cradle]

thank you very much

so i was suppose to look it as a system of equations

[tacit sigil]

@dusky epoch Thanks

[dusky epoch]

i did nothing that would merit gratitude but you're welcome ig

[unborn wedge]

are there french people in here PLEAXE

[slender yarrow]

non

on est pas français ici

[unborn wedge]

Lol je pense que je te crois psk un français aurais plutot dit "nous somme pas français"

are you french or nah?

don't be thrown off by my profile picture i'm a smart cookie

[dusky epoch]

pourquoi as-tu besoin d'une personne française ?

[slender yarrow]

je le suis

[unborn wedge]

ah cool, c'est juste pour demander un tuteur du coup psk je vais passer un examen d'algebre lineaire dans quelques semaine et j'en ai vraiment besoin de quelqu'un pour me suivre haha

[slender yarrow]

ah

bah je suis un peu occupé ces temps-ci

(genre passer le code quoi)

[unborn wedge]

ah je vois haha y as pas de soucis ❤ bonne chance quand meme

@dusky epoch et toi!

[dusky epoch]

ben moi je suis pas française mais je le parle

mais quelques semaines ça fait combien ?

[slender yarrow]

genre 3 mois ou 10 jours lel

[dusky epoch]

mdr

mais j'ai mes propres examens

[unborn wedge]

like two weeks haha, i can speak english fluently so i can probably translate

LOOOL

ah bah non alors concentre toi sur tes examens haha ❤

[dusky epoch]

je pensais pouvoir t'aider après qu'ils finiront

[unborn wedge]

tant pis je trouverai qelqu'un d'autre merci quand meme haha ❤

c'est dans cb de temps tes exos?

[slender yarrow]

bha je peux t'aider de manière ponctuelle éventuellement, si t'as des questions pose-les là (ou en MP ça me gène pas)

[dusky epoch]

euh voyons... ils sont le 15 et le 28 juin je pense

[slender yarrow]

mais attends pas une réponse forcément rapide quoi

[unborn wedge]

les miennes commence le 17 dans tant pis haha ❤

@slender yarrow oui oui je te poserai des question de temps en temps haha

du coup peut etre je pourrai te dm j'ai une demande un peut particuliere

[slender yarrow]

ya Tuong qui a fait prépa aussi

[unborn wedge]

si je te derange pas biensur

[slender yarrow]

parce que je me démerde en alg linéaire mais je suis pas un génie non plus

[unborn wedge]

lol pareil

[slender yarrow]

ça serait genre des trucs de 1ère année ou plus tard?

[unborn wedge]

premiere année c'est tres simple mais enfaite j'ai raté presque tout le deuxieme semestre donc je suis un peu dans la merde quoi lol @slender yarrow j'avais des problèmes médicales et familiales mais bref

[slender yarrow]

ah oof

[dusky epoch]

[slender yarrow]

ouais ça devrait aller alors

[unborn wedge]

cool!

[slender yarrow]

et le reste ça va?

parce que bon je suppose que t'as pas que de l'alg lin ce semestre

[unborn wedge]

pas du tout LOL je vais en rattrapage psk j'ai meme rater mes examens a cause d'un accident de voiture ....

mais genre tout quoi lol

[slender yarrow]

tfw life goes shit

[unborn wedge]

literally life just went to shit all of the sudden

mais bon je pense pouvoir avoir des notes passable au rattrapage et compenser avec le premier semestre psk j'ai bien travailler

[faint lintel]

is LADW supposed to be a first or a second book in Linear Algebra

cause some people are saying LADR is a good second book in Linear Algebra

[broken hawk]

they‘d both be fine first books if you’re okay with getting thrown a bit into the deep end, rigor-wise

[faint lintel]

Hmmm

when people say "rigor wise"

what does that mean

[broken hawk]

well, basically

there’s two main ways to start a linear algebra course

[faint lintel]

cause I've never studied math past AP calculus (which I just finished)

idk if that counts as "rigorous" (but I assume it doesn't)

[broken hawk]

1. This is a matrix. This is matrix multiplication. This is matrix-vector multiplication. Oh look we can think of this as functions. (and then go from there)
2. This is a vector space. This is a basis. This is a linear function. Oh hey, we can write down how linear functions act on the basis in a rectangle, neat. (and then go from there)

the latter is more “honest” in a sense, but perhaps a bit harder to graps cause it starts with a bunch of very general abstract definitions

and tends to not use matrices a whole lot cause you’ve got the less annoying and more powerful linear functions

[faint lintel]

gotcha

[broken hawk]

the former is often more computation-focussed

and less abstract

[faint lintel]

which one is regarded as the better text? LADW or LADR?

I want to study math (idk what kind but I want to)

so if that makes a difference

[broken hawk]

the former will probably also stick to the “easy” vector spaces ℝⁿ and ℂⁿ, the latter will use the full generality of “let V be a vector space over a field 𝔽”

[faint lintel]

So probably go for LADW?

[broken hawk]

I honestly don’t know what the two do exactly

[faint lintel]

I've literally done zero Linear Algebra

oh ok, gotcha

I'm just coming from a perspective of "I've heard it's useful and interesting so let's look at it"

[broken hawk]

we used Friedberg/Insel/Spence in class, which looks good from what I’ve seen and goes very much the “fuck matrices, more generality” way

I was completly fine with it, but I‘d worked with vectors a bunch in physics classes before

if you have zero experience with vectors maybe not

[faint lintel]

I mean i've done some basic basic basic basic vector stuff in calculus

like vector valued functions and stuff

but it was more of like "can you take the derivative? can you take the integral? ok cool"

[broken hawk]

my “this may not work for you but if it does it’ll be awesome” is:

1. Watch 3blue1brown’s series on LinAlg. This’ll give you plenty of intuition
2. Take a book that goes the rigorous route (first chapter should either define vector spaces or be on abstract algebra) and work through it carefully

[faint lintel]

ok

[broken hawk]

if 2) doesn’t work for you get another book, or maybe get a book on mathematical rigor / proofs first

and then try again

[faint lintel]

I'll probably do LADW because the description of LADR says it's a good 2nd course in Linear Algebra so yea

thanks!

[upbeat plank]

Can anyone tell me how to approach this problem?

[brittle juniper]

use the definition

it's the very first question, it's just here to check if you know what "linear" means

[dusky epoch]

^

in all honesty, i'd rather do the general case and then instantiate c with [1,2,1]^T for (1)