linear-algebra

[sick sandal]

[glad acorn]

yes you can set up a homogeneous system to proof it

[marble lance]

Do you perhaps already know that A is invertible?

[glad acorn]

no it's not given

Seems these would be enough to imply

A+aI_{n} is trivally invertible for all A

[lavish jewel]

any A whose eigenvalues are all -1 will not work

[glad acorn]

AHH

[lavish jewel]

or with any eigenvalue of -1, really

so that's not true

well, -a, rather, since it's A + aI

[glad acorn]

so far I can only determine two cases that A=O and A=-a

[golden reef]

is the cross product of 2 vectors, a new vector?

[winter harbor]

Yes

[golden reef]

and its perpindicular to both vectors correct?

[winter harbor]

Yup

[golden reef]

thanks!

[dusky epoch]

by hyperplane do you specifically mean a subspace of codimension 1

just to make sure

hm

let's see

so our hyperplane is of the form $$\curly{X \in F^{n \times n} \mid \sum_{i,j} a_{ij} x_{ij} = 0 }$$ where $a_{ij}$ are fixed coefficients

[stoic python]

Ann

[dusky epoch]

does that help at all

i was thinking of like

saying something relying on a nonzero a_ij and letting its corresponding entry vary or something to that effect

but that might fall apart

[jolly tapir]

Can someone help me with this

[median ocean]

number 2 part a please

[wraith badger]

doesn't the rules channel say "this is not a homework help server" or

[winter harbor]

Notice that hyperplanes over $\mathcal{M}{n}(\mathbb{K})$ are the kernel of rank 1 linear maps $\psi : \mathcal{M}{n}(\mathbb{K}) \rightarrow \mathbb{K}$. Moreover, notice that $\text{tr} : \mathbb{M}{n}(\mathbb{K}) \rightarrow \mathbb{K}$ has rank $1$ and we can easily construct invertible matrices with trace $0$. What kind of relationship do we have between general rank $1$ linear functionals $\psi \in \text{Hom}{\mathbb{K}}(\mathcal{M}_{n}(\mathbb{K}), \mathbb{K})$ and the trace ?

[stoic python]

MisterSystem

[winter harbor]

I think this is the way to go.

[gray dust]

this is a timed graded quiz?

[fallen slate]

How do you interpret this? does this mean both vector lines intersect and are perpendicular to each other?

[hollow finch]

i think this might be what theyre talking about

mmmmm beautiful....

i think in the diagram P is the topmost point on the triangle, L1 is the horizontal line at the bottom, and F is whats being pointed to.

not sure though someone can correct me

[lavish jewel]

looks right

[hollow finch]

https://puu.sh/IiFiL/e12c3081be.png

or that yeah sure

[lavish jewel]

so yes, they're saying the 2 lines are perp

and presumably the question is to find what exactly point F is

[fallen slate]

ohhh

alright

thanks

[fallen slate]

follow up - ive tried using the r=a+td and r=a+sd thing to find coords of F but I just cant seem to find the answer. Am I missing something?

for a

one step has something to do with finding t = -1

[lavish jewel]

what i would do is

make a direction vector P - F, let's call it d

then the second line is of the form P + sd

and as you said, this point is also on the other line, so there is some value of t for which F = (3,4,9) + t(2,-1,1)

and lastly, (2,-1,1) dot (P - F) = 0

[fallen slate]

I did as
Let vector P->F be (a, b, c)
Then dot product d and P->F = 0
getting 2a - b + c = 0

but thats all I could make out from the fact of the two vectors being perpendicular

[lavish jewel]

ah, how about this

let's make a plane whose normal is the direction vector of r

and it contains the point P

and then we find the intersection of the line r with that point

that should work

make a unit vector out of (2,-1,1)

and all it n

then t = n^T(P - (3,4,9)), and F = (3,4,9) + t(2-1,1)

hmmm

i might've messed up the plane's normal

no, i think this is right

might as well test it out in octave

[fallen slate]

oh wait

I think I mightve found an easier way

nvm

what I did was

[lavish jewel]

i had an issue with the lazy normalizations i made

lemme test again really quick

yeah

as you said, t = -1

the procedure i gave you works by first defining the plane of all vectors perpendicular to the line r

and then finding the intersection of the line with the plane

one finds t and uses it to find F

and $t = \frac{ -\frac{d^T}{\Vert d \Vert_2} (r_0 - P) }{ \frac{d^T}{ \Vert d \Vert_2 } d}$

[fallen slate]

since they both intersect, r1 = r2
r = (3 +2t, 4-t, 9+t) and L2 is (3+as, 2+bs, 1+cs)
since they both intersect, their respective x,y and z would be equal to each other
3+2t = 3 + as labelled as 1
4-t = 2+bs labelled as 2
9+t = 1+cs labelled as 3

I multiplied label 1 with 2 and rearranged all 3 e.q. to let t as the subject
4t = 2as
t = 2 - bs
t = -8 + cs

add all e.q. to get
6t = 2as -bs +cs +2 -8
6t = s(2a - b +c) -6

since 2a + b + c =0

its just 6t = -6

[stoic python]

Edd

[fallen slate]

I did find the answer using your method

just realised that there was an easier way to solve it after I solved it

[lavish jewel]

should be the same thing, just with geometric intuition vs solving a system of equations

[fallen slate]

yeah

thanks tho

[lavish jewel]

no prob

[wintry steppe]

Where my linear algebra pals at ?

[dusky epoch]

...do you have a linalg problem you'd like to get help with?

[proper owl]

Hi, in my bilinear algebra class we were studying the notion of the dual space. Could someone explain me what the point of a dual space is ? Why is it useful?

[lavish jewel]

the linear maps V -> F also form a vector space

so if you have a vector space V over F, it immediately has an associated vector space called the dual space

and if treating for the vectors in V and these linear maps as vectors (the linear maps would be in some vector space V*), there is a bilinear map V x V * -> F that relates them

[proper owl]

thank you very much!

[lavish jewel]

the most straightforward example is vectors in R^n

for which the dual space is transposed vectors in R^{1 x n}

or rather, the transposed vectors are linear maps

and they are associated with other vectors also in R^n via the transpose, and this second set of vectors is the dual space

[forest quiver]

Yo is a 1x1 matrix just a scalar?

I haven’t had vector spaces unit yet or anything I don’t know the formal definition

Online says it’s not

But Dr. Peyan vid says it is

Peyam

[stable kindle]

no, as you can't multiply a 1x1 matrix and a 2x2 matrix

[forest quiver]

Bruhh

[stable kindle]

but you can multiply a scalar and a 2x2 matrix

it's like

they're different operations

[forest quiver]

Y’a i see what u mean

Anyway

[stable kindle]

matrix multiplication vs like scalar multiplication, or scalar-matrix multiplication

[forest quiver]

det ( 1x1 matrix ) = the entry

Is det(scalar) undefined ?

I presume so it’s fnctn mapping matrix to R

So why is det( 1x1 matrix ) = the entry…. So to evaluate this I take the entry and multiply it by its cofactor

Which is just nothing

So I guess the nothing cofactor is defined to be 1 XD idk

[gleaming knot]

scalar and 1x1 matrix are the same thing ~~even in abstract linear algebra because the set of endomorphisms of any 1-dimensional vector space is canonically identified with scalars, although you could argue that doesn't make them the same thing, only that the canonical map from the field of scalars to End(V) is an isomorphism ~~

A scalar is not a function mapping matrix to R

[forest quiver]

?

I don’t know what morphisms are really

[gleaming knot]

You don't have to read the crossed out part, it's for a different audience

As for why det of a 1x1 matrix is the entry, depends on what your textbook takes as the definition of determinant

[forest quiver]

I bet this guy thinks he’s cool using that jargon

Confusing noobs

[gleaming knot]

[forest quiver]

Now I have to look up hypercube

And orientâted length

[gleaming knot]

I have a good video about that

https://www.youtube.com/watch?v=eGguwYPC32I

18 million views !!

[forest quiver]

I watched this do they talk about hyperspace ?

[gleaming knot]

they talk about 4 dimensions and I believe he uses a hypercube as an example

[forest quiver]

Oh wait

A hypercube in 3d is a cube

[gleaming knot]

ya

[forest quiver]

A hypercube in 2D is a sqr

[stable kindle]

sort of

[forest quiver]

Same for 1d is a line

0d is a vertice

Correct?

[gleaming knot]

yes

[stable kindle]

eh, i'd just be a little more precise

[forest quiver]

?

[stable kindle]

a zero-dimensional hypercube is a point, a one-dimensional hypercube is a line, a two-dimensional hypercube is a square, a three-dimensional hypercube is a cube

[forest quiver]

That’s what I said no?

I think I get it anyway

So a 1x1 matrix say in R2 is a just a point right ?

[stable kindle]

no

i think what you said might be mistakable for saying what the intersection of an arbitrary hypercube with three/two dimensions is, which can vary massively

[stable kindle]

what does it mean to have a 1x1 matrix in R^2

[forest quiver]

A point on x axis I presume

[stable kindle]

no

[forest quiver]

Wait it wouldn’t even be in R2

Because there is no y component

[stable kindle]

yeah

[forest quiver]

Like it would be off the R2 plane into a different place

R1 then

[stable kindle]

i mean the thing is matrices aren't in R^n anyway

[forest quiver]

Yeah yeah

[stable kindle]

matrices are maps from R^n to R^m

or m to n i don't remember

[forest quiver]

It doesn’t matter

Yeah I get it

[stable kindle]

it kinda does but ok

[forest quiver]

So what does this map accomplish

It’s weird bc it’s a vector

[stable kindle]

a 1x1 matrix is a map from R^1 to R^1, ie. just R to R

it's linear, so it's just a multiplication

x -> ax

[forest quiver]

Yeah but it’s a vector at the same time

[gleaming knot]

Remembering definitions in long term memory is the key to not getting confused in math

[stable kindle]

i don't think it's a vector

[forest quiver]

So you can have A^2 Is a mapping

Its a vector

[stable kindle]

why

[forest quiver]

You can draw it on R1

[stable kindle]

it's not a member of R

[forest quiver]

Like a length

[stable kindle]

it's a member of linear R -> R,

it can be represented by a single number

[forest quiver]

It can also be member of R

[stable kindle]

but it's like

[forest quiver]

No?

It can be both

[stable kindle]

ehhhhhh

[gleaming knot]

$A\in (R\to R)$

[forest quiver]

Also R

[stable kindle]

i mean it has to be linear so i was being imprecise

but yeah it's still not

[stoic python]

Icy001

[stable kindle]

no, it has to be linear, so it's not really like that but

i mean the thing is matrices and vectors and scalars should always be distinct

[forest quiver]

$A\in R$

[stoic python]

Tim O'Brien

[forest quiver]

Also

[stable kindle]

things'll start breaking if you have an exception for literally only when the dimension is 1

[forest quiver]

Wait that’s so weird to think

[stable kindle]

it'll just be a hassle

[forest quiver]

So it’s a constant

[stable kindle]

keep them separate

[forest quiver]

I suppose

Do you know the Laplace extension of a determinant

[stable kindle]

... laplace expansion?

[forest quiver]

Yeah cofactor

How is that defined for this

[stable kindle]

i mean the determinant is simply the single entry of the matrix

clearly

[forest quiver]

But use Laplace extension

This thing

[stable kindle]

...

no

[gleaming knot]

This is actually on point

kind of like how the empty product is 1

[forest quiver]

And 0! = 1

[gleaming knot]

ah-ha

[forest quiver]

Reminds me of that we learned it in clsss the other day

[gleaming knot]

0! is the number of ways to arrange 0 objects, which is 1

[forest quiver]

Yeah I guess that’s kinda weird to think about

[gleaming knot]

and the nothing matrix is the unique endomorphism from the point to the point, which is the identity

[stable kindle]

sure

[gleaming knot]

hence has determinant 1

[forest quiver]

What is endomorphisme anyway

[gleaming knot]

linear map from a vector space to itself

[forest quiver]

I will just drown in jargon if I search these things up

[stable kindle]

an endomorphism is just a homomorphism from an object to itself

[forest quiver]

Yeah well played bro

Thanks for explaining it to me

[stable kindle]

endo- meaning internal

endoskeleton vs exoskeleton

[forest quiver]

Ok thanks

[stable kindle]

it goes to itself

[forest quiver]

So T: R2 ~> R2 is a endomorphisme

[stable kindle]

yes

[drifting seal]

hello, sorry if this is not the right place for this! i'm struggling a little with vector spaces. i can't imagine or understand really what summing subspaces are like? why do you sum them and what does it achieve?

[stable kindle]

summing is basically like

the sum of any random two lines is the plane that contains them both

the sum of any random two planes, or a line and a plane, is the three-dimensional block that contains them both

[drifting seal]

and the sum of subspaces?

[gleaming knot]

Algebraically it is pretty straightforward

The sum of $W_1$ and $W_2$ is the span of ${w_1+w_2\colon w_1\in W_1,w_2\in W_2}$

[stoic python]

Icy001

[stable kindle]

it's the least space that contains both the subspaces

[gleaming knot]

Actually do we need to say span

That set is already a subspace too

[stable kindle]

yes

wait

no

[gleaming knot]

$W_1+W_2\coloneqq{w_1+w_2\colon w_1\in W_1,w_2\in W_2}$

[stoic python]

Icy001

[drifting seal]

thank you so much, is it ok if i ask a few more questions later? as im reviewing old content i didn't rlly understand before

[stable kindle]

if channel's not in use, if it's reasonable to ask, sure

[lofty topaz]

Find out if the above statement is true - if so, prove it; if not, find a counterexample.
\ Let $ (G, *) $ be a group and $ x, y \ in G $. If $ x * x = y * y $ holds, then $ x = y $.

[gleaming knot]

Try to find a counterexample first before proving it

whenever you're stuck like this

[lofty topaz]

I have tried but unsuccessfully...

[gleaming knot]

Share your attempts!

[lofty topaz]

I stuck on that that I don't know which elements set G contains

[gleaming knot]

The statement must have been for all groups G, right?

[lofty topaz]

Yes

[gleaming knot]

Then in a counterexample you get to pick whatever group G you like

[lofty topaz]

Hmm

Okay, I'm going to try something else...

[little scarab]

Exercise 2.6.23

If x /= 0 and y are vectors in R^n, show there is a linear transformation T such that T(x) = y

Isn't T just a matrix [0 0, 0 1]?

seems too simple so I think I'm just not parsing the question right?

[gleaming knot]

the letter x and y are names, not designating the standard basis vectors

Although I don't even think your matrix works for (1,0) and (0,1) anyway

[lofty topaz]

I didn't find any counterexample, I think it's true

[gleaming knot]

How thorough were you in your counterexample search?

Don't want to be gullible and led into believing something is true that is false, after all

If you truly believe it's true you can start trying to prove it

[lofty topaz]

I believe it's true but I have no idea how to prove it

[gleaming knot]

Ok so start with an arbitrary group G and arbitrary elements x and y

You have the one hypothesis x*x = y*y that you can use

and from there you must derive x = y from the group axioms

for example you could multiply both sides by x^-1

or by y^-1

And try other stuff

If nothing you try seems to work, maybe it's actually false

then you search for counterexamples with renewed vigor

back and forth between counterexamples and proof

Until you solve it

[stable kindle]

lol why is this in #linear-algebra

it's not linear

but yeah good advice from icy

[gleaming knot]

Maybe the examples of G he knows are GL(n)

[stable kindle]

even so

[gleaming knot]

me not serious

[stable kindle]

yeah

[worldly bear]

If i have the set V = M_22 and W is the set of all 2x2 matrices with integer entries, then W is not a subspace of V because it fails closure under addition and scalar multiplication right

since a real number + or *an integer does not always result in an integer

or am i only concerned with rather an integer is closed under addition and scalar multiplication

[lofty topaz]

I tried this, but x * x * x^-1 gives x * e, but what about y * y * x^-1

[sick sandal]

you mean subspace of V right

[worldly bear]

yes subspace sorry

I guess i’m just not sure if i check it with W having integer entries and an arbitrary matrix K having integer entries

or if K has real entries

[sick sandal]

the set would not be closed under scalar multiplication since the scalar can be real as the field worked with is R

correct?

even if it passes the vector conditions

[worldly bear]

how do i know i’m in R

[sick sandal]

because m22 is a vector space over R no?

[worldly bear]

like i don’t understand how to know if i let the scalar in this case be a real number or an integer

[sick sandal]

it comes down to the field at hand i presume

[worldly bear]

,rotate

[sick sandal]

i assume m22 is indeed the set of all 2x2 matrices with real entries correct?

[worldly bear]

i mean that’s kinda what i was figuring

but it’s not define that way so idk

if that’s the case then W is not a subspace of V

[sick sandal]

if thats the case then any subspace of V will be a subspace over the same field i presume

[worldly bear]

same field as in real numbers?

[sick sandal]

they did say W is a subset of V

its not a subspace in that case correct

[worldly bear]

no, since it fails closure under scalar multiplication

[sick sandal]

here is the definition of a subspace
let V be a vector space over field F, A subspace of V is a subset of V which is itself a vector space over "F" with vector and scalar multiplication on V

so yes the same field of real numbers

[sick sandal]

correct

[worldly bear]

so i’m assuming V is over the field of real numbers unless otherwise stated

[sick sandal]

yep i think thats the case

[worldly bear]

okay, thank you for the clarification

[sick sandal]

[golden reef]

what does it mean to determine cos theta as a function of t

would it be f(t) = cos(t)

[nocturne jewel]

Not linear algebra, but I suppose so.

[golden reef]

its related to vectors sry didnt know the appropriate channel

are angles between vectors unique? or can they be mutliples of each other such as pi/3 is the same as 2kpi + pi/3 ?

[vivid field]

I determined that a vector is in the nullspace of a matrix. What does it mean to write it as a linear combination of basis for null(matrix) ?

Like just a [vector 1] + b [vector 2] ..... ?

Like that ?

[sick sandal]

we can look at the canonical basis (1,0,0)(0,1,0)(0,0,1) and how it spans R^3 by noticing that any vector in R^3 can be written as a linear combination of this basis
any vector B in R^3 is a linear combination of this basis since there exists scalars c1,c2,c3 in R^3 such that B=c1(1,0,0)+c2(0,1,0)+c3(0,0,1)

the 3 vectors being (1,0,0) ,(0,1,0) ,(0,0,1)

and those vectors can be written as columns of a matrix ofcourse

dk if that answers your question

[marble lance]

Your second line, the zero column has four entries when it should have three.

[vivid field]

I understood everything that you just said and by all means it was a great explanation.....but.....hear me out....

Either you did answer my question indirectly and I am not smart enough to see that. Or---you didnt answer my question. Ill let you be the judge of that cause I cant make that choice.

[marble lance]

This is what it means

[little scarab]

how come you can just take factors out of rows like that? like he took a factor 3 out of row 2 but not the other two rows?

and a factor -1 out of row 3 but not row 1 and row 2

[marble lance]

You know how you can evaluate the determinant by expanding over any row?

[vivid field]

Uhmmm....are you sure ? I calculated it by hand and calculator and got 4 0s. Its being multiplied by a 1x4 vector

[little scarab]

yeah

[marble lance]

It's a 3x4 matrix times a 4x1 matrix, she be 3x1. @vivid field

[marble lance]

Okay, so imagine doing that and then you can just factor out 2 from every term

And then what's left is the determinant of the matrix where that row is divided by 2

[little scarab]

2?

[marble lance]

Or 3

[little scarab]

ah, I see

[vivid field]

Mhmmmm it seems like you are indeed correct. Apologies for the incompetence 🙂

[little scarab]

but how can you then combine them?

[marble lance]

1 sec, I will write it out

[vivid field]

Okay So what I did was how you are supposed to do it ?

[marble lance]

@little scarab

Here is a proof for the 3x3 case

[little scarab]

I see, I guess that's considering the det of the first row, but then they factored out -1 from row 3 too (I guess considering the det of row 3). I now see how you'd get either of the transformed det formulas independently but not how you'd combine them together?

[marble lance]

You could expand the determinant across any row

And do the same thing I did there

So expand the determinant across the third row instead of the first and take out whatever factor

So the idea is you can take a factor out of any row

[little scarab]

ahh. I wrote it out, think I see what you mean now

ty

[marble lance]

👍

And for combining it

Just first take out a factor from the one row

And then do the other

[versed yew]

hi

can someone help me solving this question

I dont quite get it

[sick sandal]

https://math.stackexchange.com/questions/88301/finding-the-basis-of-a-null-space

might be a better explanation as i dont feel too confident on the idea

[vivid field]

Wait a minute I think I got it... What I think its asking me to do is

Since I already found the basis for Null space of my matrix, and any linear combination of the basis for null space will span "something" (not sure about the terminolgy)

And since we just figured out that the new vector is in the null space

That means I think its asking me to find the vector that will produce that vector when gon through a linear combination of the basis

I know what I just said made no damn sense but ill shall test this theory and return 😆

[vivid field]

I just had those types of questions last unit

Its a 2x2 matrix so i think you would just

2a + b = 20
a - b = 7
c - d = 3
c + d = 7

and just solve for the variables

[versed yew]

thanks!

[vivid field]

Yeah just throw it into a matrix, row-reduce, and then back-substitute

[versed yew]

thank you I got it now

[visual raft]

(N-kM)v = Nv - kMv and work with that

Just my first thought

[median ocean]

Can anyone help me with number 2 part b

[half ice]

So for example, M(3,2) = 3b1 + 2b2

What's M(1,0)?

Let M =
a b
c d
What happens if you actually carry M(1,0) out as a matrix multiplication?

[hazy forum]

Hi I have a question I was hoping someone could answer. It's about how one shows that a linear transformation T: R^n -> R^n is surjective. For the problem statement, I'm asked to show if T(u)T(v)=uv for all u,v in R^n, then T is an isometry. I've already shown that T is injective, but it's the surjective part I'm still not sure about.

please ping me if responding

[winter harbor]

Well, since T is a linear transfomation between finite dimensional vector spaces of same dimension, injectivity is in fact equivalent to surjectivity!

That's a classic result

And can be proved immediately from rank-nullity / first isomorphism theorem

If you have already seen these

I encourage you to give a full proof.

@hazy forum

[hazy forum]

thanks for the reply. hmmm I'll have to think about this some more. My professor had told me that injectivity and surjectivity aren't equivalent in this case since while the cardinality of the domain is the same as the cardinality of the codomain, that cardinality is not finite.

[winter harbor]

Uh?

Maybe you misunderstood him?

[hazy forum]

nope, that's word for word what he said

[winter harbor]

We have an operator T : R^n -> R^n that we know is injective

this readily implies it is surjective (via rank-nullity)

and we have to use that this is a linear operator (on a finite dimensional vector spaces)

this is not true for general functions

[hazy forum]

Hmm, I see. Okay, I don't think he knew what theorem I was talking about then. He must've been thinking about arbitrary functions g: X->X where X is an infinite set.

Thanks again

[rich gate]

how would i graph y>x-3y>-x-1?

[winter harbor]

This doesn't fit this chat...

[rich gate]

Oh my apologies.

[winter harbor]

[rich gate]

which chat would it fit in? I seem to be a little lost haha

[hazy forum]

any of the question channels

[rich gate]

Appreciate it, thank you.

[rich gate]

Odd response but okay, stay... awkward I guess..

[storm canyon]

nvm im dumb

[worn hawk]

Hello. I have a friend who has a linear algebra exam tomorrow until now she only knows row reduction and some basis/span. How can i help her in the remaining time?

[rich gate]

just tell her you like her already

jkjk

[wintry steppe]

tell her to get as much sleep as possible and pray the exam consists mostly of that

[gleaming knot]

Advise her that if she encounters a problem she has no idea what to do for, it's possible all she has to do is read the question and answer the question

👌

[astral sedge]

Hi, how to formally define a linear map that takes a kxk-dimensional matrix and includes it in an nxn-dimensional vector space (n>k) by padding with zeros?

Is it just like:
$T:F^k \times F^k \rightarrow F^n \times F^n$ such that $A \mapsto ???$

[wintry steppe]

are you asking how to write it down?

[astral sedge]

yes

[stoic python]

RiesZ

[wintry steppe]

if you don't want to use words you could just write it explicitly

[astral sedge]

I know how to define it but not sure how to write down how it operates

sorry, I'm not sure if I;m being clear enough..

[lavish jewel]

kinda looks like you either explain it in words or show it explicitly

A |-> A' such that A' blah blah

[astral sedge]

Explicitly as writing a full matrix and the padding with zeros

?

[lavish jewel]

or define A' as a block matrix

[astral sedge]

oh nice idea

[lavish jewel]

then you specify the 0 blocks and its dimensions

or smth like that

[stoic python]

TTerra

[lavish jewel]

that's pretty compact

[astral sedge]

nice, thanks

If I define another map which does the same to a vector (i.e. F^k --> F^n, (x1,...,xk) \mapsto (x1,...,xk,0,...,0)), is there a way to link between the two or should I just define them separately?

[lavish jewel]

wdym does the same

[astral sedge]

I edited

[lavish jewel]

zero padding?

[astral sedge]

yeah

[lavish jewel]

i guess you mean f^k to f^n

[astral sedge]

oh yes sorry

[lavish jewel]

you can be lazy and say f^k is isomorphic to f^k x f^1 and use the same def

or just say like v' = [v^T 0_{k-n}^T]^T

[astral sedge]

lol okay got it

Thanks!

[wintry steppe]

or just say what you already wrote

[lavish jewel]

yeah lol

[astral sedge]

I'm considering it 🙂 btw is it valid to call these maps inclusion maps?

[wintry steppe]

they're not technically inclusion maps but no one will hurt you for using that terminology

[astral sedge]

Because in the image the matrix/vector dont keep their original form?

[lavish jewel]

hmm why would it technically not be?

[wintry steppe]

when someone says inclusion map i think of actual subsets

[lavish jewel]

i guess as a set it is, but not with the structure of the operations

[astral sedge]

I tend to agree

Great

[lavish jewel]

tend to agree as x-> infty

[wintry steppe]

[sick sandal]

so i understand this theorem but i dont think i see nor understand why this is "one of the most important results in linear algebra" according to H&K

[half ice]

Yeah this really is up there. Rank and Nullity are two pretty important concepts, each for their own reasons. It turns out that knowing one, gives the other

(As long as you also know dimV, but you often do)

[sick sandal]

sweet

guess it'll be more clear the more i progress why each are imp

[lavish jewel]

it's related to being able to decompose vector spaces into orthogonal subspaces

this sort of stuff is important, for example, when solving differential equations

idk if you've seen that for a diff eq, the solution is the particular + the homogeneous sol

[sick sandal]

nah im taking it this sem

it'll be easier to see then

[stoic python]

james_ash_

[sick sandal]

found this lovely explanation

didnt even realize this till now

does nullity has its own unique application or is it just used for sake of rank as well

[lavish jewel]

right. the names they're given when dealing with matrices are the row space, null space, and column space

wdym?

i think i can give u an example, gimme a sec

[sick sandal]

like is nullity that important on its own or is it just a way to find the rank via theorem

[lavish jewel]

it's super important

say we have a linear transformation T: V -> V that has a nontrivial kernel, so the nullity includes stuff other than the 0 vector

importantly, the null space is a subspace

so we can pick a basis for it

let's call the nullity N and pick some vectors as its basis

and then keep adding vectors to extend that into a basis for all of V

if you then use gram schmidt, you have an orthonormal basis for V, and some of these vectors in the basis of V are a basis for N

[fallen slate]

https://cdn.discordapp.com/attachments/890473264126767115/899221620600750100/unknown.png

[lavish jewel]

and then you get that the orthogonal complement of N, let's call it C for complement, is the whole of V

now the reason this is important is that when T is rank deficient, it's not just that associated problems can have no solutions

when they do have solutions, they can have infinitely many

and the way you characterize these solutions rely on C and N

[fallen slate]

So I know that the normal of the plane is parallel to the plane Im finding. I can make a line equation from B and n.
but how do you find the normal of the plane Im supposed to find?

[lavish jewel]

let's say we take now a matrix A that carries out the action of T, and vectors n in N and c in C

and we want to solve a system like Ax = b

if we find one solution Ac = b

it turns out that any A(c+tn) = b is also a solution

(for a scalar t in the field)

and so in order to describe the solution set for the problem, you actually NEED to know what the null space looks like

as a trivial example

[sick sandal]

so the null space kinda fills the space or rather dimension left unfilled

or am i misreading

[lavish jewel]

say $A = \begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix}$, and say we want to solve $\begin{bmatrix} 5 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 && 0 \\ 0 && 0 \end{bmatrix} x$

[stoic python]

Edd

[lavish jewel]

you can immediately see that x = [5;0] is a solution

but we also know that n = [0,1] makes it so that An = 0

and so the full solution is of the form x = [5;0] + t[0,1] for any value of t

what you usually see as free parameters when doing row reduction

those are vectors in the null space

[sick sandal]

OHHH

[lavish jewel]

and you need to specify all of them

[sick sandal]

that makes things much clear now
ile do more examples to get it fully

thank you so much

[dusky epoch]

why double ampersands

[lavish jewel]

hmm

what does it look like without?

more square

more purrty

[fallen slate]

still need help😕

[lavish jewel]

as you say, there's more than one plane that can do this

you can pick a vector on the given plane, and use that as a normal

unless your book specifies some other way of defining orthogonal planes

[fallen slate]

for any plane equation, the coefficients for x, y and z cannot be zero right?

[narrow pilot]

Note that the equation $x+2y-3z-4=0$ has normal vector $i+2j-3k$

[stoic python]

Captain_Mat01

[narrow pilot]

Knowing this information, you can easily find the perpendicular plane using the cross product

[sick sandal]

is non singular just an old school word for injective or is it used for another context
T the linear transformation is non-singular if Tv=0 --> v=0 according to H&K

cause T non singular if and only if T is 1:1

[dusky epoch]

depends on what you mean

if you mean that we can't have all three of them be zero at the same time, then yes
if you mean that none of them can ever individually be zero, then no, that's too stringent and excludes planes such as z=1

[earnest sigil]

can someone say how to do this sum

nd my ans is incrct coz my approach for the sum was incrct
nd now i dont know how to approach it

[nocturne jewel]

expand the RHS

[earnest sigil]

ok i did tat then

[nocturne jewel]

$A^2-6A+8I=0$ how might you introduce $A^{-1}$ into this?

[stoic python]

Mosh

[earnest sigil]

s tats exactly were im stuck

[nocturne jewel]

$IA^{-1}=A^{-1}$

[stoic python]

Mosh

[earnest sigil]

ok

[nocturne jewel]

so $(A^2-6A+8I)A^{-1}=0$

[stoic python]

Mosh

[nocturne jewel]

rest should be straightforward

[earnest sigil]

ok lemme try it

@nocturne jewel thanks i got the ans

the ans is i

[nocturne jewel]

I, yeah

[earnest sigil]

im not able to understand this qn

wont A-1 exist evrywhere

or is there any value for wich matrix wont exist

[dusky epoch]

i think what they meant is $A^{-1}$ but then somebody screwed up the typesetting.

[stoic python]

Ann

[earnest sigil]

ohh ok now it makes sense

[heavy vine]

How do i prove that this is a vector subspace i've just started with this part of algebra and dont quite get it

idk if it is called vector subspace in english tbh i've just literally translated it, lmk if it's wrong

[sudden narwhal]

a0, a1 and a2 are real numbers ?

@heavy vine

[heavy vine]

it is a subspace of R3

ups sorryu, it says of R2[x]

[sudden narwhal]

let call it A
firstly you show that A ≠ ∅

maybe by showing that the neutral element of R2[X] is in A

after all this, you show that A is stable by linear combination

[nocturne jewel]

dont be an annoyance

[prisma sail]

i tried b c d but was wrong

[wide yew]

why does full rank imply surjective?

[fallen slate]

?

[nocturne jewel]

what?

[fallen slate]

the message you were replying too was my message when I clicked on it

it seeemed

[wide yew]

means tou can write linear combination of y,w as linear combination of z for first entry

every subsequent entry means something similar to that

[nocturne jewel]

it was not

[fallen slate]

oh ok

Ill carry on then

[wide yew]

w/2 + 5/4y = z

so that is true

[prisma sail]

oh

[wide yew]

for b)
-3x+4y=z ,w=6x+3y-6x+8y

and x=x

so b is true

c is true for similar reasons

wait nvm

d is true too i thinks

since w corresponds to 11y

so all of them are true

wait

not c my bad

so a b and d

not c because w=11y

and doesnt depend on x

so span x y isnt equal to span w

[prisma sail]

ah

thx

[golden reef]

If (x+yi) = re^theta(i)
Does -(x+yi) = -re^theta(i)

[wide yew]

no

your first equivalence isnt clear since you need to define theta and r

r is defined as sqrt(x^2+y^2)

theta defined as atan(y/x)

[worthy hinge]

divide both sides by x

[wide yew]

r stays the same

but theta changes

[sudden narwhal]

it's true

[golden reef]

If it’s -(x+yi)^1/2
What should I do with the negative outside the bracket

Would it be tan theta = -y/-x ?

[sudden narwhal]

well yes

if im not wrong...

[golden reef]

Well I’m getting 2 different answers lol from both of you

[wide yew]

bruh moment you arent wrong

i am partially wrong

[golden reef]

So it’s wrong to say that the exponential form of the negative would just have a negative infront

[wide yew]

no its right im wrong

[golden reef]

So if the exponential form of let’s say (x+yi) is e^ (pi/4)i
Is -(x+yi) = -(e^(pi/4)i)

[wide yew]

yea

x+yi=r(cos theta + i sin theta) = re^i theta

[golden reef]

If let’s say it was ax+ ayi

Can I do this
a(x + yi) = a( r(cos theta + isin theta) = a( r(e^i theta))

[sudden narwhal]

consider a linear form and find the kernel

[wide yew]

yea

thats why you are right

[bold sun]

hey - for this question do i have to multipy the matrix a and b then after c and then bc and multiply that by a??

like do i expand it all ??

cuz its taking agesssss - is there an easier way or something im missing here?

[nocturne jewel]

(AB)C is AB left multiplied to C

so you need to compute AB then left multiply it to C

[bold sun]

so basically expand it all fully?

cuz compute ab is same as multiplying it right?

and wdym left multiplied by c?

like this is what i did so far but got really lost cuz it wentt suppppper long winded and confusing?

[wintry steppe]

my condolences that you have to do this question

[bold sun]

cuz also i dont think for the second one it is equal either like when done?

[bold sun]

lol yh is there an easier way??? i keep getting numbers mixed up when writing it and it wrong way round cuz i cant read properly.....but im teliing myself rn that its good practice with matrix multiplication??

[winter harbor]

You will have to work it out...

:(

Well

With a bit of more generality and abstractness

We can see why this should be the case

[bold sun]

okk well i finished it but it didnt give me theyre equal ?

[bold sun]

?? becuase multiplication is commuative ?

if thats the word

[winter harbor]

This is from Serge Lang

The proof that matrix multiplication is associative is done in one line

[bold sun]

whats it called

[winter harbor]

Wdym?

[bold sun]

lol i was a little stuck on that notation like what to put at the top and bottom ?

[bold sun]

book title its a book right?\

[winter harbor]

Yeah, it uses sigma notation idk if you are familiar with that.

That's why I said it would be a bit more abstract

[winter harbor]

It is called Algebra, by Serge Lang.

It's a Graduate Abstract Algebra book

But it has a chapter on linear algebra too.

[bold sun]

yh familiar with it but its bit confusing ngl like in this case? but the notation is used in lect notes too i think but it has k= and a top no so i didnt know where they got that from

[bold sun]

aah not graduated yet or no where near.......lol

but his name sounded familiar

[winter harbor]

He has some undergraduate books too

Even HS math books

He was really prolific.

Wrote down a bunch of math textbooks.

[bold sun]

like here he used the notation but was 5 at the top i didnt know if that was cuz 5 rows or 5 colums and where thry got 1 frm

[bold sun]

ooh okk

[bold sun]

uhuh

[winter harbor]

But it should be kind of straightforward, just use the definition of matrix multiplication given here. Then, verify if you didn't make any calculation mistakes, and you are done.

[bold sun]

but i dont really get it like what they did

or how it works....

[nocturne jewel]

check your notes on how to do matrix multiplication

the (i,j)th entry of the product is the dot product of the 1st matrix's ith row and the 2nd matrix's jth column

[bold sun]

these r the lect notes i usully print and annotate it.....but i dont have the sigma notation explained...

[nocturne jewel]

the 1st sigma is just how you write a dot product, the 2nd one is just the generalization for matrices

A_{1k} is the (1,k)th entry of A.

[bold sun]

oh......

i think im gnna have to start all over agian with this mltiplication and expanding and adding 2 see if its equal its just not working out for me

i think its gnnna be the easiest/ only way right??

[nocturne jewel]

wdym

what's "it" in that?

[bold sun]

the (ab)c=a(bc) thats the q

[nocturne jewel]

Yeah, you're verifying associativity

[bold sun]

here

[nocturne jewel]

ik

[bold sun]

ooh thats wat it was called i thought it was commutivity lol

[nocturne jewel]

commutativity is AB=BA, which isnt true for matrices

[bold sun]

ooooohhhh okkk

anyways thank youuuu for the helppp......im gnna have another shot at it all over again (my 5th attempt) and see if it workkss

[bold sun]

okk guess what i fianllly did it and it worked- thank youuu!!!

[winter harbor]

Was this done by you?

If so, what program/software did you use?

I mean, if you can intuitively deduce / recall the rank-nullity theorem by this diagram, then it is ok man.

As long as you can picture it via this diagram.

[golden reef]

-sqrt(1+i)
How do i incorporate the negative to find the exponential or polar form of this value

[winter harbor]

That diagram is really lovely

I should learn how to use inkscape at some point.

[wise meadow]

How can I prove that a 4D matrix with 2D hermitian basis’ has real diagonals?

[whole cedar]

Guyysss

[glad acorn]

Does anyone has ideas on how to decide whether A+aI is invertible or not?

[glad acorn]

guess: since we cannot determine the result in the form of (A+aI)B=-bA for completely unknown matrix A,B, we need to add something to both sides in AB+bA+aB=O

that can result in kI in one side, where k is the non zero real scalar and (A+aI) appearing as a factor on the other side

(A+aI)*something=kI

I don't know how can I relate this guess to the matrix B

Is this enough to deduce?

@marble lance @lavish jewel I have asked this question last week and everyone say there is not enough information to deduce

[marble lance]

@glad acorn that looks good yeah

[glad acorn]

it's merely only one step after (A+aI)B=-bA to implicate the result... we were so close last week

[wintry steppe]

why is the dual basis defined like this?

[glad acorn]

orthogonality

[wintry steppe]

ah

[winter harbor]

Uhhh idk if that's a good way to describe it.

Notice the following

With we have a basis $ \mathcal{B} = {v_{1}, \cdots, v_{n}} \subset V$ of a certain vector space $V$ over a field $\mathbb{K}$. Then for any vector $u \in V$ we can write it as:
$$
u = \sum\limits_{i=1}^{n} \alpha_{i} v_{i}
$$
For some constants $\alpha_{1}, \cdots, \alpha_{n}$. Then a very natural question in this setting is the following, given that we have a vector $u$ and this ordered basis, how can I calculate the i-th coordinate of $u$ with respect to $\mathcal{B}$?
\
\
This motivates us to introduce coordinate functions on $V$, meaning functions
$$
\pi_{i} : V \rightarrow \mathbb{K}
$$
Which takes a vector $u$ and sends it to $\alpha_{i}$, its i-th coordinate with respect to $\mathcal{B}$. Turns out that these coordinate functions are preciely the dual basis of $\mathcal{B}$ !

[stoic python]

MisterSystem

[winter harbor]

To see this, notice that if we have $u \in V$ given in this way, then we want $\pi_{i}(u) = \alpha_{i}, \forall u \in V$.
\
\
Now, let us see what happens when we apply the dual basis $v^{i}$ to $u$.
\
\
We have
$$
v^{i}(u) = v^{i}\left(\sum\limits_{j=1}^{n} \alpha_{i} v_{i}\right)
$$
But $v^{i}$ is linear, so we have:
$$
v^{i}\left(\sum\limits_{j=1}^{n} \alpha_{j} v_{j}\right) = \sum\limits_{j=1}^{n} \alpha_{j} v^{i}(v_{j})
$$
But notice that $v^{i}(v_{j}) \neq 0$ only for $i \neq j$. So we actually have:
$$
\alpha_{i} v^{i}(v_{i})
$$
And $v^{i}(v_{i}) = 1$ by definition. So we indeed have that $v^{i} = \alpha_{i}$ and the i-th dual basis gives us the i-th coordinate of a vector with respect to the basis $\mathcal{B}$.

[stoic python]

MisterSystem

[winter harbor]

And that's pretty much why we define the dual basis like this

They are the coordinate functions with respect to the basis B.

Which is a very natural thing to ask for, what is the coordinate of a vector with respect to this basis.

@wintry steppe hope this makes things a bit clearer

[wintry steppe]

oh that make sense

thanks

[vocal isle]

Hey folks, I'm trying to find a 3d parametric equation of a disk of radius R. All I know is the normal vector n, and the point to the vector of the center of the circle P0. I need to parameterize the disk using two variables, r & theta. Any help? 🙂

^this picture shows the work I've done so far, et1 and et2 are tangent unit vectors on the plane of the disk. But I'm unsure how to find them

[dusky epoch]

i suppose you can take e_t1 to be n × whatever, then normalize it and take e_t2 = e_t1 × n

make sure the 'whatever' you pick at the start doesn't happen to be parallel to n

[vocal isle]

Hmm, i need this to work regardless of any n

it should be possible to solve using 6 equations:

1. n . et1 = 0
2. n . et2 = 0
3. et1 . et2 = 0
4. sqrt(et1x^2 + et1y^2 + et1z^2) = 1
5. sqrt(et2x^2 + et2y^2 + et2z^2) = 1
6. sqrt(nx^2 + ny^2 + nz^2) = 1

By contrast this is trivial to do in 2D.

[x,y] = [Pox; Poy] + [r cos(theta); r sin(theta)]

[dusky epoch]

theres no avoiding some casework i think

[vocal isle]

hmmm

Here's another approach i just thought about

[dusky epoch]

ok wait so like

[vocal isle]

Q . n = 0 and Q = X - Po

[dusky epoch]

are you trying to do this programmatically

[vocal isle]

therefore X . n = Po . n and so now we have the equation of our plane

[dusky epoch]

or is this for some manual computations

[vocal isle]

yeah pretty much. I'm trying to plot a shaded in disk in 3d using a program called Manim. To do that I can shade it in using 2 variables; r and theta

[vocal isle]

where X = [x;y;z]

[vocal isle]

I figured it out:
et1 = [n3; 0; -n1] / sqrt(1-n2^2)
et2 = [0; n3; -n2]/ sqrt(1-n1^2)

these are tangent vectors of the disk described in terms of normal vector components.

Now X = P0 + r cos(theta) et1 + r sin(theta) et2

the only problem is this breaks when n1 or n2 = 1 (implying the disk is perfectly in the x or y direction). Is there a way to ensure this doesn't break?

[wintry steppe]

Use Latex you dang Republican, how is anybody supposed to read that???

[dusky epoch]

there was no need to make such a rash assumption of their political affiliation, was there...

[sonic plank]

can anyone help me please?

[winter harbor]

With what exactly tho?

This is just a list of common norms of matrices.

Which uuuuh

Don't seem to be matched correctly

So are you asking for the correct match?

The frobenius norm is given by:
$$
|A|{F} = \left(\sum\limits{i=1}^{m} \sum\limits_{i=1}^{n} |a_{i,j}|^{2}\right)^{\frac{1}{2}}
$$

[stoic python]

MisterSystem

[winter harbor]

The maximum norm is given by the maximum of the absolute value of the entries of your matrix

Meaning

$|A|{\text{max}} = \text{max}{i,j}(|a_{i,j}|)$

[stoic python]

MisterSystem

[torn hornet]

these are all equivalent norms I think

[winter harbor]

Yeah, they are norms on a finite dimensional vector space (complex, real vector space) so via some basic Bolzano-Weirstrass argument we prove that all such norms are equivalent

But I suppose the question she sent is asking us to match which norm is which

[lavish jewel]

you could argue with equivalence of norms that they're all equivalent, sure

[winter harbor]

In any case, the spectral norm is given by the square root of the spectral radius of AA^t

And the norm I have never fucking heard before

Must be that one which is the sum of the singular values of your matrix.

That one must be the Ky-Fan K norm

[lavish jewel]

sounds about right

[winter harbor]

Edd, this is offtopic, but are you doing a Phd in which area?

[lavish jewel]

signal processing

[sonic plank]

yeah we need to match them

[winter harbor]

Alright, then just take a look at what I just described above

[sonic plank]

I got it thank you very much!

[wise meadow]

This is a question I would very much like the answer to, if anyone can point me in the right direction.
"How can I prove that a 4D matrix with 2D Hermitian basis’ has real diagonals?"
For example, take a matrix
$ \begin{pmatrix}
s_0 & 0 & 0 & 0 \
0 & s_1 & 0 & 0 \
0 & 0 & s_2 & 0 \
0 & 0 & 0 & s_3
\end{pmatrix} $

[stoic python]

Bryan

[wise meadow]

With basis vector
$ \begin{pmatrix}
I \
\sigma_1 \
\sigma_2 \
\sigma_3
\end{pmatrix} $
Where $\sigma$ is a pauli matrix

[stoic python]

Bryan

[wispy pewter]

Bryan do you mind if I ask a simpler question?

[wise meadow]

Go for it

[winter harbor]

Wait, wdym by a hermitian basis in this case?

and a 2d basis?

[wise meadow]

Honestly, not too sure. Let me run you through my logic real quick and see if it makes any sense.

So, I am going to define a Hamiltonian operator as a linear combination of Pauli matrices and the unit matrix.
$H = s_0 I + s_1\sigma_1 + s_2\sigma_2+s_3\sigma_3$

[stoic python]

Bryan

[wise meadow]

Now, I thought since this is a superposition of linear independent matrices that can form every element of our operator, that I could form a matrix.

[stoic python]

Bryan

[wise meadow]

Now, I was hoping that there would be a nice set of relationships between this matrix and the pauli matrices that I could use to prove that s_0, s_1, s_2, and s_3 have to be real.

Maybe since this is a linear combination of hermitian matrices that therefore this matrix must also be hermitian?

But, I am unsure.

x,y,z are interchangable with 1,2,3: my apologies.

[winter harbor]

Oh, ok this seems to be more well defined now.

Are s_{0}, s_{x}, s_{y}, s_{z} taken to be real?

No, we want to prove they are real

[wise meadow]

^

[winter harbor]

Yeah, this totally fails

these can be complex numbers, ofc.

[wise meadow]

Well

If the overall matrix is hermitian then they cannot be complex

Since H=H-conjugate

[winter harbor]

But a complex linear combination of hermitian matrices is not necesarily hermitian

[wise meadow]

Mm

[winter harbor]

this is basically a consequence of the fact that the complex conjugate satisfies $(\lambda A)^{\ast} = \overline{\lambda} A^{\ast}$

[stoic python]

MisterSystem

[winter harbor]

So notice that if \lambda is real

then it is equal to its conjugate

and we can thus prove that the set of hermitian matrices form a real vector space

but they are not a complex vector space

thus is basically the only reason why they fail to be a complex vector space

since the complex conjugate is an anti-homomorphism, we have (A+B)* = A* + B*

and the sum works fine

but the fact that the conjugate transpose of a scalar multiple of a matrix brings out the conjugate

they do not form a complex vector space under these operations

[wise meadow]

How is this true?

The rest of what you said, I believe I understand and makes sense. That would be the way to prove it.

[winter harbor]

Ok, I will write down the standard proof that the set $\text{Her}{n}(\mathbb{C}) = {A \in \matchal{M}{n}(\mathbb{C}) , \vert , A = A^{\ast} }$ is a REAL subspace of $\mathcal{M}{n}(\mathbb{C})$.
\
\
We have that $\forall \lambda \in \mathbb{R}$ and $A,B \in \text{Her}{n}(\mathbb{C})$:
$$
(A+\lambda B)^{\ast} = A^{\ast} + (\lambda B)^{\ast} = A^{\ast} + \overline{\lambda} B^{\ast}
$$
Since $\lambda$ is real, we have that $\overline{\lambda} = \lambda$ and so $(A+\lambda B)^{\ast} = A+ \lambda B$ which thus implies $A+\lambda B \in \text{Her}{n}(\mathbb{C})$. Meaning it is a real subspace of $\mathcal{M}{n}(\mathbb{C})$ we could naively try to mimick this proof and conclud that it is a complex subspace of the space of $n \times n$ complex matrices too, but this can't be the case, because for $\lambda \in \mathbb{C} \setminus \mathbb{R}$ we never have $\lambda = \overline{\lambda}$.

[stoic python]

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

[winter harbor]

So for example, a very trivial example is thinking about the identity matrix

it is hermitian

but we don't have that i * I_{n} is hermitian

since the conjugate transpose of such a matrix is - i * I_{n}

[wise meadow]

Ahh, fair enough.

[winter harbor]

(4,-6,0) = 2*(2,-3,0)

then use that T is a linear transformation

(0,5) = 2*(-1,1) + 1*(2,3)

then use T is a linear transformation

[boreal spindle]

hey can someone help with part b

I found the rref

[maiden swan]

Observe that **any point **in the plane containing A , B , a nd C ???

points containing points?

[boreal spindle]

OH

LOL

[boreal spindle]

thank you lmao

I'm fried it's 3am omg

[maiden swan]

right

i'm learning linear algebra and just found this in the intro

[boreal spindle]

God I'm literally brainless

[maiden swan]

ok i'll stick to being a taxi driver

[boreal spindle]

wait nvm I still don't get it actually

How the hell do I express "one of the columns"

as a LC of these

<@&286206848099549185>

[naive valve]

Consider a linear combination of the columns. You should know (and prove) that these columns are linearly dependent.

Knowing that they are linearly dependent, how can you apply this knowledge to a linear combination?

[boreal spindle]

Do I assign coefficients?

[naive valve]

You know what a linear combination is, right?

[boreal spindle]

yes

But I just learned it so I’m a bit confused

vectors are linearly dependent if they can get the zero vector only when the coefficient is zero

[naive valve]

Vectors are linearly dependent if and only if there is a nontrivial linear combination of these vectors equal to the zero vector.

[boreal spindle]

yeah

[naive valve]

Then, what does it mean that the columns are linearly dependent?

[boreal spindle]

Ohh lol, so I multiply them with any real nr and that counts as the answer

Okay now it makes sense in my brain

or I could show the zero vector too

[naive valve]

You are told that the columns are linearly dependent.

It means one can be expressed as a linear combination of the others.

You only need to find the coefficients of the combination.

[boreal spindle]

So, I found -1/3 x1= x2 = x3

Can I just write -1/3u + 0v = w

this is what I’ve done so far

[naive valve]

Why 0v?

[boreal spindle]

Ohhhh, I’m literally so dumb

okay okay got it.

thank you for your patience 😅

[naive valve]

yw

[boreal spindle]

:))

[stoic ore]

hey could anyone here help me with some problems involving vectors?

[wintry steppe]

ask

[stoic ore]

how would i go about solving this

[torn stag]

@charred root You can solve $Ly = b$ for $y$ easily working from $y_1$ to $y_n$ sequentially.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Same for $U$

[stoic python]

IlIIllIIIlllIIIIllll

[sweet cliff]

does anyone know a quick way to tell if 3 2d vectors are in the same hemisphere?

[dusky epoch]

the same hemisphere?

@sweet cliff what do you mean by that

[sweet cliff]

nevermind i got it

semicircle*

[wispy panther]

Can someone ELI5 the relationship between PCA and Fourier transform? https://math.stackexchange.com/questions/2383585/whats-the-difference-connection-between-pca-and-inverse-fourier-transform

[lavish jewel]

PCA and the fourier transform both try to see if some hidden structure of the data becomes easier to interpret in a different basis

for special classes of matrices, the PCA is the same as the fourier transform

i.e. you diagonalize them with the fourier basis

[still lodge]

does row reducing a matrix before finding it’s eigen vals change the values themselves

conceptually i dont think it would but idk

[lavish jewel]

it changes the eigenvalues

otherwise RREF would be the same as eigenvalue decomp 😛

[teal grotto]

if only it was that easy lol

[still lodge]

gah

pain

ok well i have this 3x3

-6  -7  -3
6    6    2```

there has to be a better way to find the eigenvalues of this than the tedium of finding the determinant of the thing thing

[lavish jewel]

idk if better by hand, but there are many other ways

determinants for 3x3 mats are pretty easy tho

sarrus moment?

[little scarab]

just wanna clarify what this is asking... are x and y specific vectors and it wants a general form of A that maps x -> y, or is it basically just asking for T(x) to be another vector in R^n ('y')?

[stable kindle]

fix x, fix y; find a matrix that maps x to y

[little scarab]

that's what I assumed it meant, but I don't think we've covered the content for that

[stable kindle]

have you done matrices

[little scarab]

yeah

well, we did addition, subtraction, multiplication, inverses, and some on linear transformations

[stable kindle]

do you know how they're defined wrt. bases

[little scarab]

no

[stable kindle]

oh

[little scarab]

for context, definition 2.5 is this

[stable kindle]

ok so you haven't done matrices from a proper linalg perspective?

[little scarab]

guess not

[stable kindle]

urgh

[little scarab]

is this definitely the only interpretation for what the Q is asking to find?

[stable kindle]

95% sure

[marble lance]

A linear map T from R^n to R^n can always be represented as T(x) = Ax where A is an n by n matrix and x and T(x) are written as column vectors

Also if you define a map T(x) = Ax then it is always linear

So there is a correspondence between the linear maps from R^n to R^n and the n by n matrices

So basically for a given x and y you have to find a matrix A such that Ax = y and then define your map as T(x) = Ax

[little scarab]

so A is just
[ a1, a2, ..., an ] ?

and T(x) is just x1a1 + x2a2 + ... + xnan ?

[marble lance]

Yes

All linear transformations can be written like that

[little scarab]

hmm, but then what are we showing here? isn't it just repeating the definition they gave?

[marble lance]

You have to choose a specific a1 to an so that for a specific vector x, T(x) will be equal to y

For example, if x = (1, 4, 5) and y = (2, 5, 10), you need to choose the a_i's so that T(1,4,5) = (2,5,10)

[little scarab]

im assuming there exists a general form for the a_i's then?

[foggy coyote]

i have a weird situation which i don't understand

so im using quaternions for rotation in a game engine

and this happens (the camera is not meant to have any roll)

what seems to be happening is when i calculate the quaternion from euler rotation the x axis rotation is applied before the y axis rotation

but if i swapped it around then the rotation of other objects won't work

so it only applies to the rotation of the camera

the rotation of the camera is calculated using a matrix of the pos, look and up vectors

the up vector is the problem - its tilting to the left or right because of the above problem

however when i convert the quaternion back to euler, the z component (or the roll component) is 0

[foggy coyote]

if i swap x and y like this then the z component is some random number i don't know what it means

[wintry steppe]

To what extent is matrix multiplication actually a form of multiplication?

[winter harbor]

Well, there's no "universal" definition of what multiplication means in a really general setting.

But it is multiplication in the sense that it satisfies most properties we expect from a multiplication to have

Namely, if we restrict ourselves to n×n matrices they form a unital ring with respect to matrix multiplication.

So it is associative, has a unity as distributes over sum.

[wintry steppe]

ic

It is curious how one does not expect "multiplication" to also commute

[wintry steppe]

true, all the forms of multiplication should be renamed: multiplication with any scalers should be called "repitition" in that case

whatever multiplication is lol

[winter harbor]

Uh, multiplication of real numbers is not even iterated sum.

This intuition we have for multiplication of natural numbers doesn't extend to most stuff we call multiplication

[wintry steppe]

exactly

matrices for example

but for real numbers it fits

[winter harbor]

Not really

How do we define sqrt(2)* \pi as an iterated sum?

[wintry steppe]

I would expand their decimal expantion to start with

Then I would realize that the decimals are fractions

and I will simply add fractions of the other number

[winter harbor]

That's not how we define it.

There's tons of ways to define multiplication o real numbers

The two more standard ones are via a certain operation on dedekind cuts

and the other one via a certain operation on equivalence classes of cauchy sequences of rational numbers

Ofc this is indirectly related to sum of natural numbers

[winter harbor]

Because, for instance, in order to define multiplication in this way we need to define multiplication of rational numbers first

and multiplication of rational numbers is defined in terms of multiplication of integers, which is itself defined in terms of multiplication of natural numbers.

but we can't directly define multiplication of rational numbers nor real numbers as iterated sum.

At least we usually don't do this.

[wild epoch]

Can anyone help me understand point 2? Whats the difference between a matrix transformation and a transformation induced by a matrix? Im kinda lost

[north anvil]

matrix transformation and a transformation induced by a matrix

I believe these are the same

I could be wrong, however.

[wild epoch]

thats what i thought, am i just misunderstanding 2? Is it saying any linear transformation can be written like that?

[marble lance]

1 tells you that linear transformations and matrix transformations are the same thing. 2 is telling you that such transformations are given by a unique matrix and how to find that matrix for a linear transformation.

[north anvil]

If a transformation is linear, then we can say for sure that this transformation occurred via the multiplication of a matrix.

If a transformation occurred via the multiplication of a matrix, then we can say for sure that the transformation is linear.

[wild epoch]

Ok thats fine thank you, that makes it clear

[north anvil]

This is the if and only if part

[wispy panther]

So we have the matrix $X \in \R^{m \times n}$ as our data where $m$ is the number of datapoints and $n$ is the number of features.
The SVD decomposition is $X=Y \Sigma V^T$, where $V$ is the PC matrix.
The DFT is $S=WX$ where $W$ is the DFT matrix \url{https://en.wikipedia.org/wiki/DFT_matrix}

for special classes of matrices, the PCA is the same as the fourier transform

If $X$ has translational invariance, then $S = V$ ? I don't think that's right. For once, they have different shape. $S \in \R^{m \ times n}$ while $V \in \R^{n \ times n}$ . But I'm sure that's what you meant. But I just don't know what you meant instead.

i.e. you diagonalize them with the fourier basis

I don't know what this means...

[stoic python]

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

[wintry steppe]

it may not be how we define it, but it is a way we can evaluate it

[wintry steppe]

I'm sure there are multiple ways to define multiplication, this is simply a way we can evaluate it. btw, why do you even care about a definition if you can look at it from different perspectives? I think it is better to give an intuition and then get into the different perspectives (aka definitions)

[wintry steppe]

I would not use something like this to explain multiplication to someone who has never heard of it before. Actually, this is why I am asking. One way to look at multiplication of numbers, is as repeated addition, this is logically an area. Ofc it is weird to use this as a template for fractions, but at least the area perspective helps. However I find it unclear how matrices relate to any of these interpretations (or maybe I am missing one). Ik you say that multiplication is a flexable term to use, but when a simple dude thinks of multiplication, this is what they think, so one should at least be able to say matrix multiplication has some relation to this.

[wintry steppe]

I think we missunderstood eachother. I meant this:

sqrt(2)*pi=1.414....x3.14159....=(1+4/10+1/100+4/1000.....)x(3+1/10+4/100+1/1000.....)

and then multiply those

with the sums I was not referring to multiplication, I was referring to how the decimal expansions of sqrt(2) and pi can/are interpreted as infinite sums

one needs this to multiply

[torn stag]

@charred root It's basic algebra. Write out the matrix equation as a system of linear equations. You will see that $y_1$ is immediatley found. Then $y_2$ is found from $y_1$, then $y_3$ is found from $y_1, y_2$, etc.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

This is assuming you are solving $Ly = b$ for $y$ where $L$ is lower triangular.

[stoic python]

IlIIllIIIlllIIIIllll

[midnight elbow]

Hello! Would anyone calculate this by hand? I started and quickly realised that it would take me a year and way too many errors

Also if there is a way to reduce the problem since I think we are expected to do it by hand!?

[nocturne jewel]

I mean, laplace along bottom row is easiest first step if there's no nicer way to do it

[limber kiln]

Halp

where do i start with this

[nocturne jewel]

what have you tried..?

[main badger]

Hint: T is a linear transformation