linear-algebra

[winter harbor]

Oh

[zenith junco]

oh is it saying that x = b

how do i prove something is one to one?

[nocturne jewel]

injectivity is that f(a)=f(b) => a=b

or ker(f)={0}

[zenith junco]

so do i say like, suppose there exists A and B. then i do p(A) and p(B) and then something happens and we find out A=B?

[winter harbor]

Ok, so we have $P(\mathbb{F})$ the space of all polynomials over $\mathbb{F}$, what this function does is that for each polynomial $p(x) \in P(\mathbb{F})$, $\gamma$ assigns to it the evaluation function.
\
\
Meaning that for each polynomail $p \in P(\mathbb{F})$, $\gamma(p(x))$ is the function that evaluates an element $\beta \in \mathbb{F}$ of the field to $\gamma(p(x))(\beta) = \sum\limits_{k=0}^{n} a_{k} \beta^{k}$, where $p(x) = \sum\limits_{k=0}^{n} a_{k} x^{k}$

[stoic python]

MisterSystem

[winter harbor]

This can be a little bit confusing

for instance

suppose F is the field of real numbers

and we have p(x) = x^2

what gamma does

is assign to p(x) the function f(x) = x^2

this is really just a formalism

because formally we don't think of polynomials as functions at first

[zenith junco]

okay i think i understand now, thanks for explaining this

so how do i prove f(x)= f(y) then x=y

and the ker = 0

[winter harbor]

Suppose that there is a polynomial $p(x)$ such that $\gamma(p(x)) = 0$

[stoic python]

MisterSystem

[zenith junco]

ok

[winter harbor]

what does this means?

[zenith junco]

it means

[winter harbor]

\gamma(p(x)) is a function

and 0 is the constant 0 function

[zenith junco]

a_0 + a_1B + a_2B^2 + ...+ a_kB^k = 0

?

[winter harbor]

calm down

[zenith junco]

ok ;;-;

[winter harbor]

\gamma(p(x)) is a function

this means that for any \beta in the field

we have this $\gamma(p(x)) = 0 \iff \forall \beta \in \mathbb{F}, \gamma(p(x))(\beta) = 0(\beta) \iff \gamma(p(x))(\beta) = 0$.

[stoic python]

MisterSystem

[winter harbor]

Notice that we made an abuse of notation in the beginning

we are using 0 to mean the consant function 0

and 0 as the element of the field F

ok, so this means that

$\sum\limits_{k=0}^{n} a_{k} \beta^{k} = 0, \forall \beta \in \mathbb{F}$

[stoic python]

MisterSystem

[zenith junco]

ooo okay i understand that notation

[winter harbor]

Nice

This means in fact

[zenith junco]

how do i prove this though

[winter harbor]

I am just using the definition

Now we want to prove that a_{k} = 0, \forall k \in {0, ..., n}

or equivalently

that p(x) is the polynomial that is constant to 0

[zenith junco]

ok so so so we have to do that cuz B is not always 0 so we gotta prove that ak = 0 so

the polynomial that gives us that is

p(x) = 0?

and that is indeed a polynomial?

👁️ 👄 👁️

[winter harbor]

the information that we have is that p(x) is a polynomial that when evaluated to every element \beta in the field gives us 0

and we want to prove p(x) = 0 is the constant polynomial equal to 0

try to prove the converse

suppose p(x) is not constant equal to 0

then there exists b an element in the field

that when we evaluate p(b) this gives us something different than 0

that's one way to go

[zenith junco]

okay i get that now cool

now how do i prove the other thing about injectivity

[winter harbor]

if \gamma is always one to one?

[zenith junco]

yeah

f(a) = f(b) then a=b

how do i dis

https://www.youtube.com/watch?v=bjATxNZp4GI&ab_channel=TheMathSorcerer

ok imma watch this vid first

i have a midterm in 2 weeks n its not looking that good hahaahahhah

😔

[winter harbor]

We are working over vector spaces

so we can just prove ker \gamma = 0

[zenith junco]

wut how

nani

i thought we are working over a field

[winter harbor]

Let $f : V \rightarrow W$ be a linear transformation between vector spaces over a field $\mathbb{F}$, then $f$ is injective iff $\text{ker} , \f = {0}$

Try to do this as an exercise

[stoic python]

MisterSystem

[zenith junco]

ok

[winter harbor]

This is in fact even more general and works for other algebraic strucutres

[zenith junco]

ok gimmie a sec

idk lol

if its a linear transformation then its closed under scalar mult and addition

so T((a)+ f(b )) = T(a) + T(b)

OH proof by contradiction

suppose you have f != g
and T(f) = T(g)
then f = g

because lin trans

am i on the right tracklol

[winter harbor]

yeah so

if f : V -> W is linear and injective

to prove that ker f = 0 we go as follows

if x \in ker f

then f(x) = 0, but it is always the case that f(0) = 0

so f(x) = f(0) and since f is injective this means x=0

and so ker f = 0

the converse goes as follows

if ker f = 0

then if f(x) = f(y) for x,y \in V

this implies f(x) - f(y) = 0

but f is linear

so f(x-y) = 0

this means x-y \in ker f

but ker f = 0

so x-y = 0 which implies x=y

and f is injective

so to prove a linear transfomation between vector spaces is injective

we just need to check if the kernel is trivial, i.e consists solely of the 0 element

[zenith junco]

ok so a few questions i have is

why dont we check the other values?

what if f(a) = f(b) but a!=b ?

oh

also how do we know f(0) = 0

[winter harbor]

f is linear by hypothesis

so f(0) = 0

[zenith junco]

i see

thank u for explaining this

[winter harbor]

np

yeah so

the original question boils down to proving that for F = Q,R,C

If a polynomial evaluates to 0 for every element in F

Then this polynomial is constant equal to 0

We can do this using the fact that every polynomial of degree n in a field has at most n roots

And since all these fields are infinite

The result follows

For finite fields it is a bit more tricky

[limpid vine]

if a set of vectors span R^m they must also have to span R^n for 0 < n < m right

[hard drum]

No, because vectors in R^m are not in R^n

[limpid vine]

oh

oh that makes sense

[hard drum]

Eh, the most natural to me would be a projection onto the first n coordinates

Vector spaces of the same dimension over the same field are always isomorphic, so this sort of construction can always be done

[reef sleet]

How do I actually find a basis for a space?

Specifically, I have that $W$ is the orthogonal complement to the subspace generated by $(2,1,1,-1)$

[stoic python]

feather

[reef sleet]

(W is 2 dimensional)

Here's the full question

[nocturne jewel]

RREF the corresponding augmented matrix

Then you get a subspace plane of solutions

So W is just that plane

(W is called the nullspace of the coefficients matrix)

[reef sleet]

But how can I actually find a basis for that plane?

Here is the solution I got to the system

(Where x_3 = s and x_4 = t)

Could I just split this into two vectors (t-s, s-t, 0, 0) and (0, 0, s, t)?

[winter harbor]

If you take any two linearly independent vectors that satisfy (t-s, s-t, s,t) this will give you a basis for the plane W.

So in particular

You could take the point in W corresponding to t=1 and s=1

[reef sleet]

You mean independent right?

[winter harbor]

And the point in W corresponding to t=1 and s=0

Yup

Anyways, yeah, just take two linearly independent vectors that satisfy (t-s, s-t, s,t)

You can find these by substituting t and s for some values.

[reef sleet]

What do you mean by "satisfy"? What does it mean to satisfy a vector?

[winter harbor]

$W = {(t-s,s-t,s,t) \in \mathbb{R}^{4}}$
\
\
So what I mean by a vector $x \in \mathbb{R}^{4}$ satisfying $(s-t,t-s, s,t)$ is $x \in W$.

[stoic python]

MisterSystem

[winter harbor]

Satisfy in the sense of satisfying the property of being an element of W

[reef sleet]

How do I know they (two vectors x, y in W) span W? I never took a proper linear algebra course, I'm sorry :(

[winter harbor]

Ok so

[reef sleet]

Just a shitty watered down version for engineers where all we did was compute solutions to systems

[winter harbor]

Oh

it means that every element in W is a linear combination of x,y

and that x and y are linearly independent

[reef sleet]

Yes

[winter harbor]

a basis for a vector basis is a linearly independent set of generating vectors

ok so let's do this by the definition

I will take two vectors in W

and show they span W

[reef sleet]

Sure

[winter harbor]

take s=0 and t=1

we have the vector (-1,1,0,1) in W

and if we take s=0 and t=1

we have the vector (0,0,1,1)

we will show they are linearly independent

[nocturne jewel]

$[t-s,s-t,s,t] \ =t[1,-1,0,1]+s[-1,1,1,0] \ =\text{span}{[1,-1,0,1],[-1,1,1,0]}$

[stoic python]

Mosh

[winter harbor]

Yes we can do this, but i wanted to show via the definition since he doesn't know how to prove two vectors span a vector space

[reef sleet]

Don't you have to set up some generic vector then show that it can be written as a linear combination of the two vectors? And we check this by computing the determinant of the augmented matrix

[nocturne jewel]

if you have n n-dimensional vectors, yes

[reef sleet]

oh yeah

fuck

LMAO

[nocturne jewel]

since determinants only apply for square matrices

[reef sleet]

That's right

[winter harbor]

Anyways, I will take (1,-1,0,1) and (-1,1,1,0), suppose that there exists $\alpha, \beta \in \mathbb{R}$ such that $\alpha (1,-1,0,1) + \beta (-1,1,1,0) = (0,0,0,0)$.
\
\
We want to show $\alpha, \beta$ are both $0$.
\
\
This means that $(\alpha - \beta, - \alpha + \beta, \beta, \alpha) = (0,0,0,0)$ and by definition we must have $\alpha = \beta = 0$
\
\
So $(1,-1,0,1)$ and $(-1,1,1,0)$ are linearly independent.

[stoic python]

MisterSystem

[winter harbor]

Now we want to prove that they are a generating set for W

This is also clear taking the basis Mosh has shown before

but it would be a bit of more work to do using the basis I had used before

[winter harbor]

Basically, given $x \in W$ we want to show that there exists $\alpha, \beta \in \mathbb{W}$ with $x = \alpha (1,-1,0,1) + \beta (-1,1,1,0)$ but $x \in W$, so $\exists s,t \in W$ for which:
\
\
$x = (s-t,t-s,s,t)$
\
\
So $x = (s-t,t-s,s,t) = t(1,-1,0,1) + s (-1,1,1,0) = \alpha (1,-1,0,1) + \beta (-1,1,1,0)$
\
\
So if we take $\alpha = t$ and $\beta = s$ we see that $W = \text{span}{(1,-1,0,1),(-1,1,1,0)}$.

[stoic python]

MisterSystem

[nocturne jewel]

[winter harbor]

Anyways, with the set of basis we have used

Oh lol

didn't know I pinged you

[nocturne jewel]

ye reply pings unless you turn it off

[winter harbor]

No, I didn't mean to reply in the first place lol

[nocturne jewel]

[winter harbor]

Anyways, I was just trying to highlight how a proof that a set of vectors span a vector space goes

with this set we have chosen it is quite trivial

but the way to go is this

[reef sleet]

This is nice

[winter harbor]

Yeah, we first proved that they are linearly independent

and here we have shown they generate W

and so they are a basis

as an exercise

you can try taking a different set of vectors

and checking if they form a basis

[nocturne jewel]

(also make sure you understand why the span of a set of vectors is in fact a subspace of the vector space)

ie show that for $v_1,...,v_n\in V$ means that $\text{span}{v_1,...,v_n}$ forms a subspace of V

[stoic python]

Mosh

[reef sleet]

Thank you both!! Maybe I'll try that exercise when I read through a real LA textbook 🥴

[nocturne jewel]

I mean if you know the definition of subspace and span, the proof is relatively quickly, referring to my "challenge problem"

[zenith junco]

is this

an iff statement

that goes both ways?

[limber sierra]

"if"s in definitions are usually "iff"s

its... confusing

but yes, this statement goes both ways

[nocturne jewel]

is there a reason or just convention?

Cause I really dont see a reason to just not use iff

[limber sierra]

convention

"We call _ a _ if _"

[zenith junco]

thanks also

[limber sierra]

theres a sort of implicit "...and we dont call it that otherwise"

but we dont bother to state it

[limber sierra]

i agree that iff would make more sense

[zenith junco]

what about

V2 and V3

does that circle plus to R^2

[nocturne jewel]

o plus

$\oplus$

[zenith junco]

o

thanks

[stoic python]

Mosh

[zenith junco]

lol

is it?

[limber sierra]

what do you think?

[winter harbor]

take any vector (x,y) in R^2, you can write (x,y) = (x,x) + (0,y-x)

Anyways

You didn't check the case V_1 + V_2 by yourself, did you?

You would have known

I'd recommend checking these examples the book gives you

[zenith junco]

im juust reading the lecture notes

idk for that as well

😔

o k imma check

[winter harbor]

If you can't right off the bat tell this

You could try the following

[winter harbor]

Notice that $V_{2} = \text{span} {(1,1)}$ and $V_{3} = \text{span}{(0,1)}$. Since $\text{dim} , \mathbb{R}^{2}$ and since $(1,1)$ and $(0,1)$ are linearly independent and $V_{2} \ocirc V_{3} = \text{span} {(1,1), (0,1)}$, then $\mathbb{R}^{2} = V_{2} \oplus V_{3}$.

[zenith junco]

is this how

omf

omfg

im so dumb

i couldve just thought about linear indepencen

facepalm

thanks

[stoic python]

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

[winter harbor]

I mean

You could do this via definition

basically

all you want to check is that V_2 and V_3 have intersection 0

which is trivial

and if you take any vector (x,y) in R^2

[zenith junco]

oh

[winter harbor]

you can write it as a sum of an element in V_2 and an element in V_3

[zenith junco]

okay i get it now thanks

also

i have another q

this statement is so confusing

i understand it but im confused on one part

[winter harbor]

About what exactly?

[zenith junco]

1. the upper + lower matrix = a new matrix right
2. But since the union of LM and UM is nontrivial it does not fit the definition of o plus

so

wtf

[winter harbor]

?

[zenith junco]

wtf is the dif between + and o plus

[winter harbor]

So like

[zenith junco]

im so

confused

like both statements makes sense but wtf

[winter harbor]

There's a difference between sum of subspaces

and direct sum

[zenith junco]

ok

direct sum is o plus def

but like

[winter harbor]

$\forall U,W \subset V$ subspaces of a vector space $V$, we always have that $U+W$ is another subspace of $V$. We can always construct this.

[stoic python]

MisterSystem

[winter harbor]

Now, the direct sum has a slight difference

If $U,W$ are subspaces of $V$ such that their sum is the whole space $V$, i.e $V = U + W$ and at the same time $U \cap W = {0}$ we say that $V$ is the direct sum of $U$ and $W$.
\
\
We denote this by $V = U \oplus W$. We also say that $U$ and $W$ are a decomposition of $V$.

[stoic python]

MisterSystem

[winter harbor]

So you can think of the direct sum as a particular case of sum od subspaces

where the direct sum is a decomposition of your whole space

[zenith junco]

wait just to clarify W1 + W2 (both subspaces) is also a subspace because it is inside of V which is closed under scalar mult and addition ?

and

[zenith junco]

thanks this makes sense

[winter harbor]

and 0 \in W1+W2

[zenith junco]

oh yeah

btw

[winter harbor]

try proving that the direct sum is in fact a subspace

this will give you some intuition

[zenith junco]

ok give me min rn

wait it is

1. the zero vector is in it bc intersectio is 0 vector

2. it is closed under scalar mult and addition because it is inside of V

which is a vector space so we know for sure cuz W1 and W2 is inside

[winter harbor]

That's not an argument

sorry

like

your second argument is basically

''it is closed under scalar multiplication and addition because it is inside of V, aka being a subset of V''

But being a subset of a vector space is not enough for it to be a vector subspace

[winter harbor]

Also, as I have said before, the intersection of two subspaces is not necessarily the 0 vector

and sum of two subspaces is always defined even if their intersection is non trivial, i.e equal to 0

[winter harbor]

we want this to be the case for direct sum tho

[zenith junco]

btw is this correct

[winter harbor]

yeah so, I sort of understand your intuition

the idea is that you take a certain matrix in M_n \times n

and you decompose it

as the sum of an upper triangular matrix

and a lower triangular matrix

[zenith junco]

isn't that basically it tho

i get what u mean

[winter harbor]

yeah, I think that's what you meant you wrote it this way

Yeah, I think it is alright

[zenith junco]

how do i draw these two

[zealous torrent]

Is anyone from here able to help me in questions-7, I should probably have asked it here from the beginning

[winter harbor]

Let $W_{1}, W_{2} \subset V$ be subspaces of $V$ a vector space over $\mathbb{F}$ , we have that $0 \in W_{1} + W_{2}$ since $0 = 0 + 0$, where $0 \in W_{1}$ and $0 \in W_{2}$.
\
\
Now, $\forall u,v \in W_{1} + W_{2}$ and $\lambda \in \mathbb{F}$, we have that $\exists w_{1},w_{1}' \in W_{1}$ and $w_{2}, w'{2} \in W{2}$ where $u = w_{1} + w_{2}$ and $v = w'{1} + w'{2}$
\
\
So $u+ \lambda v = w_{1} + w_{2} + \lambda (w'{1} + w'{2}) = (w_{1} + \lambda w'{1}) + (w{2} + \lambda w'{2})$. Since $W{1} \subset V$ and $W_{2} \subset V$ are subspaces of $V$, we have that $w_{1} + \lambda w'{1} \in W{1}$ and $w_{2} + w'{2} \in W{2}$, so $u + \lambda v \in W_{1} + W_{2}$
\
\
With this, we finish the proof that $W_{1} + W_{2}$ is a subspace of $V$ ; $\square$

@zenith junco

Idk how much used yoi are to writing proofs

but here you go

You can read the solution after trying by yourself

[stoic python]

MisterSystem

[zenith junco]

i dont understand the 7th line

how is w1 + lambda*w_1prime inside of W1 ?

we only established that w1, w2 is inside of W1

[winter harbor]

because W1 is a subspace of V and w_1, w'_1 \in W_1

what is the proeperty subspaces have with respect to the operations on a vector space?

[zenith junco]

but W1 is a subspace closed under it's own addition and scalar mult

[winter harbor]

oh

[zenith junco]

how can w1prime exist inside of it

[winter harbor]

typo

[zenith junco]

im familiar with proofs and reading them, but writing my own is hard rip

[winter harbor]

Let $W_{1}, W_{2} \subset V$ be subspaces of $V$ a vector space over $\mathbb{F}$ , we have that $0 \in W_{1} + W_{2}$ since $0 = 0 + 0$, where $0 \in W_{1}$ and $0 \in W_{2}$.
\
\
Now, $\forall u,v \in W_{1} + W_{2}$ and $\lambda \in \mathbb{F}$, we have that $\exists w_{1},w_{1}' \in W_{1}$ and $w_{2}, w'{2} \in W{2}$ where $u = w_{1} + w_{2}$ and $v = w'{1} + w'{2}$
\
\
So $u+ \lambda v = w_{1} + w_{2} + \lambda (w'{1} + w'{2}) = (w_{1} + \lambda w'{1}) + (w{2} + \lambda w'{2})$. Since $W{1} \subset V$ and $W_{2} \subset V$ are subspaces of $V$, we have that $w_{1} + \lambda w'{1} \in W{1}$ and $w_{2} + w'{2} \in W{2}$, so $u + \lambda v \in W_{1} + W_{2}$
\
\
With this, we finish the proof that $W_{1} + W_{2}$ is a subspace of $V$ ; $\square$

[stoic python]

MisterSystem

[winter harbor]

corrected it

it should be w_1, w'_1 \in W_1 ofc

[zenith junco]

btw

for checking the subspace, we have to do under scalar mult and addition but you just combined the both right?

[winter harbor]

yeah

it is equivalent

and way faster

[zenith junco]

ok ima going to study this proof lol thanks

itll help me write my own proofs

[zenith junco]

how tf do i

do this

[winter harbor]

Just use the definitions

the definition of equality of sets

then the definition of span

and the definition of sum of subspaces

To get a bit warmed up, let us prove that
$$
\text{span}(S_{1} \cup S_{2}) \subset \text{span} , S_{1} + \text{span} , S_{2}
$$
Let $u \in \text{span}(S_{1} \cup S_{2})$, then there exists $n \in \mathbb{N}$, $v_{1}, \cdots, v_{n} \in S_{1} \cup S_{2}$ and $\lambda_{1}, \cdots, \lambda_{n} \in \mathbb{F}$ such that
$$
v = \sum\limits_{i=1}^{n} \lambda_{i} v_{i}
$$
Then, we have that $\forall i \in {1, \cdots, n}$ we have that $v_{i} \in S_{1} \cup S_{2} \iff v_{i} \in S_{1}$ or $v_{i} \in S_{2}$
\
\
Now, suppose $v_{1}, \cdots, v_{n}$ are ordered such that for some $k \leq n$ we have that $v_{1}, \cdots, v_{k} \in S_{1}$ and $v_{k+1}, \cdots, v_{n} \in S_{2} \setminus S_{1}$, then:
\
\
$v = \sum\limits_{i=1}^{k} \lambda_{i} v_{i} + \sum\limits_{j=k+1}^{n} \lambda_{j} v_{j}$
\
\
Now, we have that $s_{1} =
\sum\limits_ {i=1}^{k} \lambda_{i} v_{i} \in \text{span} , S_{1}$
\
\
Moreover, we have then that $s_{2} = \sum\limits_{j=k+1}^{n} \lambda_{j} v_{j} \in \text{span} , S_{2}$
\
\
So, $v = s_{1} + s_{2} \in \text{span} , S_{1} + \text{span} , S_{2}$ and
$$
\text{span}(S_{1} \cup S_{2}) \subset \text{span} , S_{1} + \text{span} , S_{2}
$$

@zenith junco

[stoic python]

MisterSystem

[winter harbor]

Try to prove the converse

This should give you an idea of how to do it

you just need to unwind the definitions

there's nothing so deep

[zenith junco]

thanks ill have to think about it for a bit

[zenith junco]

i dont understand how that sum is a linear combination of the union between s and sprime

[winter harbor]

I mean, it's pretty much by definition

Each u_i \in S, i particular each u_i \in S union S'

And each w_j \in S', so in particular each w_j \in S union S'

And that is a linear combination between the u_i's and w_j's

Each of which is an element of S union S'

[reef sleet]

I need to solve a system of 4 equations with only 3 variables

Idfk now man 😔

[reef sleet]

The determinant is nonzero so I think it has a solution

Idk man I am so lost

Im skipping this question fuck this

[winter harbor]

Gaussian elimination

[reef sleet]

I've been trying

Nothing works

I get solutions that satisfy two equations

But not the other two

[winter harbor]

How do you compute the determinant of a 4×3 matrix?

[reef sleet]

and I've tried like 8000 different combinations of row reduction

[reef sleet]

Well I went on MSE and it said to subtract the constants and then find the determinant of the corresponding matrix (which would now be 4x4) and then if that's nonzero then the system has a solution

[winter harbor]

What system of equations are you trying to solve tho?

Could you post it here?

[reef sleet]

I can't anymore, I'm skipping this :/ I've spent way too much time on it. Here's the system

[winter harbor]

Ok, I need to sleep now too.

Tomorrow I will take a look at it.

[reef sleet]

Sure, thank you for all your help and explanations man

Including those in #math-discussion

You've been so helpful

[winter harbor]

Np

[reef sleet]

Appreciate you too @nocturne jewel

It's so wild how the professor's homework problems are so much easier than the textbook's

[zealous torrent]

I have

[stoic python]

V4

[zealous torrent]

Where A, X_{n}, B and Y_{n} are all NxN matrices. Given a large data set of X's and Y's, how can I determine the optimal A and B that minimises the following

[stoic python]

V4

[lavish jewel]

luckily for you, you can vectorize the expression

replace the sum of frobenius norms for a sum of 2-norms after vectorizing Y_n and A X_n B

and nicely enough:

[zealous torrent]

This is amazing! Thanks so much

[lavish jewel]

separating B and A later is another issue, but yeah

cuz here you get the 2 together

[zealous torrent]

I will have to learn what vectorizing a matrix does though, but it looks like I'll be able to solve it

Do you have any thoughts on separating B and A later on?

[lavish jewel]

only up to a scaling factor

if you need the specific entries, this won't work and you'll be better of doing some alternating optimization

[zealous torrent]

Like an iterative approach?

[lavish jewel]

mhm

[zealous torrent]

Oh man. Guess I might have to do that

Was really hoping for an analytical solution

[zealous torrent]

Does it simplify things if A is in SO(3), and B in O(3)?

[lavish jewel]

so(3) means its a rotation?

this eliminates the scaling ambiguity, yes

since then you know any scaling factor is associated to B

oh, and B is also orthonormal

[zealous torrent]

yup

[lavish jewel]

yeah you can just rescale as needed

you can solve it with least squares using the vectorization stuff

then just do something like average out the blocks in the matrix resulting from the kronecker product

this'll give you a scaled version of A

rescale it so that it's in SO(3)

and then use A to get B^T from B^T kron A

the averaging isn't needed if X and Y aren't noisy

[zealous torrent]

So I'm not super familiar with vectorization and the kronecker product. It looks like vectorizing just stacks the matrix columns on top of eachother? And the kronecker combines two matrices of nxm and pxq to np x mq?

multiplying each component of matrix A by the entire matrix B, and that becomes an "entry" in the big matrix

Not 100% sure how it ties in yet. And then I'd have to find the gradient of that massive thing?

Oh actually, I guess standard least squares result would apply if I can get everything in the form of Ax ~= B? No need to compute the gradients

[lavish jewel]

yep

sounds about right. i think the result should be the average of all the least squares solutions (average of pseudo inverses)

you can double check

[stoic python]

V4

[zealous torrent]

wait no

that doesn't look right at all

[lavish jewel]

ah i guess it's not quite the usual least squares, since the variable is on the left of a matrix product instead of on the right

you can transpose the problem tho

[zealous torrent]

Like transposing the matrices before taking the frobenius norm?

[lavish jewel]

mhm

[zealous torrent]

$L(A, B) = \sum_{n} |Y_{n}^{T} - B^{T}X_{n}^{T}A^{T}|_{F}$

[stoic python]

V4

[lavish jewel]

i guess it makes more sense to transpose after vectorizing

(y - (B^T kron A) x )^T = y^T - x^T (B^T kron A)^T

now this one is least-square-ey

you get outer products of the vectors

[zealous torrent]

The x there should be vec(X), right?

[lavish jewel]

i wrote them lowercase cuz they're vectors now, yes

[zealous torrent]

oh I see, cool

[lavish jewel]

both y and x

[zealous torrent]

that makes sense

x is the variable right, doesn't the first one match (standard) least squares better? y - Mx where M = (B^T kron A)

[lavish jewel]

you said you want to find B and A, so

[zealous torrent]

oops, sorry

[lavish jewel]

B^T kron A is the variable

[zealous torrent]

getting tired haha

Yeah, definitely going for A and B

[zealous torrent]

Okay, I think I'm getting somewhere. I don't know how to undo the kronecker product. I have a matrix B^T kron A

[short sorrel]

lol

[zealous torrent]

don't laugh, I'm a newby haha

[zealous torrent]

Where can I read up on this?

You're not going to believe it Edd, it almost looks like I have to do a second optimisation problem to undo the kronecker product https://math.stackexchange.com/a/321424

[lavish jewel]

normally you would, as i had said before

but you know the structure of your matrices

that should make it easier to do something without needing optimization

[tranquil steeple]

A general question, is there a name for sum(kron(x1,x2)), where x1 and x2 are vectors?

[lavish jewel]

i can't think of any off the top of my head

looks like a scaled version of the "grand sum"

if the matrices themselves have some physical meaning, then you could come up with something from there

like a total energy or something

otherwise, kron x1 x2 is isomorphic to some linear transformation that can be represented as a rank 1 matrix, and i don't think the sum of the elements of such a matrix has any special meaning

[bronze falcon]

hey i need help, i have a set of data points temperature compared to pressure, but how do i make a formula that is non linear i have no idea how i change by points into ax2+bx+c=0
https://www.agas.com/media/2421/r507-pt-chart.pdf?fbclid=IwAR3wD-L-LneYod4jaxyLc9uEHa4jEf-ODrHJwO7BK4lOKmbCH9aA3305GR4

these are the data point i need to find a formula from barg to celcius, any1 can help ?

[limber sierra]

asking #linear-algebra for nonlinear formulas is a bold move

this is known as "curve fitting"

[bronze falcon]

haha i couldnt find non linear 😛

[limber sierra]

its a technique in statistics, excel can do it

you might hear synonyms like "regressions"

[bronze falcon]

i hate my life i cant even make a graph in excel for some reason it just shows up completely wrong all the time

[lavish jewel]

you CAN write it in matrix vector form tho

if you know the input and output values and you have good reason to believe the model is quatratic

then you can associate the inputs and outputs by making a matrix whose rows are M_n = [x_n^2 x_n 1] for each input x_n

and then each output y_n satisfies y_n = M_n [a, b, c]^T

you can stack several samples into a vector y, and build a matrix M that corresponds to it

then you can find the parameters a, b, c with your favorite way of solving $\text{arg}\min_v \Vert y - Mv \Vert_2^2, ,, v = [a,b,c]^{\text{T}}$

[stoic python]

Edd

[vivid field]

Can someone please explain how the hell my teacher just turned that into that ?

[nocturne jewel]

2x2 inverse

[vivid field]

All I know is something about 1/detA but I didn’t really understand that

[nocturne jewel]

$A=\begin{bmatrix}a&b\c&d\end{bmatrix}\implies A^{-1}=\frac{1}{\abs{A}}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$

[stoic python]

Mosh

[lavish jewel]

what didn't you understand about it?

[pliant tendon]

Yeah

[vivid field]

She didn’t explain what determinant is. She only used it in this example for 2x2 matrix and didn’t show what she did. Determinants is the next chapter. But I see it now.

[lavish jewel]

for now, you can think of it somewhat like the... "size" of a matrix. when you divide by a number, you can also represent it as multiplying by that number's reciprocal, or "multiplicative inverse"

when you take a matrix inverse, you are finding the matrix's multiplicative inverse as well, and so something like the "reciprocal of its size" shows up

[ebon lily]

if in a linear transformation we are getting the coordinates where the basis vectors for our n dimensional vector space land on the coordinates of the vector space which are described by the matrix. then what does it mean if one of the coordinate's of any basis vector is complex? how is 1 dimension enough to plot the complex number? what would the vector look like ? and how can i understand the transformation geometrically?

TLdr: what is the geometric interpretation of doing linear transformations using complex numbers?

[lavish jewel]

i dunno how far geometric intuition is gonna take you on that one

[ebon lily]

at least for a 3 dimensional space their should be some geometric idea about it

[lavish jewel]

at this point i'd rather think of vectors and matrices more abstractly, based on their definition

because just taking 1 complex number as an input and another as an output requires 4 dimensions to "visualize"

[ebon lily]

even taking eigen vectors whose span don't change and it makes it easier to describe transformations as based on the axis (eigenvalue) and the amount of rotation

[ebon lily]

how do you define the span of 2 vectors with complex numbers as it's coordinate ?

[lavish jewel]

same as always

all linear combinations of the vectors

same way you do it when the vectors are functions, or anything else

regarding your original question, the best you could do (imo) is take the real and imaginary parts of the vectors and stack them into a vector twice as long

and make the matrices 4x larger

[ebon lily]

so basically linear algebra defined in a vector space of more than 3 dimensions is studied only mathematically and not geometrically?

[ebon lily]

will try that thanks

[lavish jewel]

in this sense, C^1 is the only thing you could visualize, cuz vectors of 2 complex numbers already need 4D

[lavish jewel]

sure, though i guess you meant visually

[ebon lily]

yep

[lavish jewel]

cuz the stuff still has a geometric interpretation with angles and the like, just in a way that is difficult to visualize

[ebon lily]

i saw 3blue1brown's videos that's why i was interested if such an awesome model was there for even complex numbers

thanks @lavish jewel

[vivid field]

Okay where did I go wrong ?

[vivid field]

I was supposed to get this right ?

[nocturne jewel]

the determinant isnt 4x

you do the inverse first then multiply the 2 through.

$\abs{A}=(2x)(3)-(x)(5)=x$

[stoic python]

Mosh

[vivid field]

Got it. Thanks.

[flint hinge]

Imo matrices are simply a bookkeeping device for keeping the track of the linear transformations

[limber sierra]

"just" is a bit dismissive but yeah

thats fundamentally what they are

this is very useful though

[pliant tendon]

Matrices are for alpha males

[short magnet]

Quick question; So I'm currently trying to show (to prove another statement) that $B = \frac{1}{q} \sum_{k = 0}^{q-1}{A^k}$ is a projector. Knowing that $A^q - I_n = 0$ with A being a matrix in Mn(R).

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

The thing is, I don't know for sure if q is the order of A

so I can't justify that the application that for all A^j in <A> gives A^k A^j is an isomorphism

Is it still possible to show that B is a projector in this way ? Can B, under these conditions, not be the matrix of a projector ?

$$ dim(Ker(A − I_n)) = \sum_{k = 1}^{q} {Tr(A^k)} $$ is the statement that I'm trying to show is true

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

So the idea I had was : 1 - Using the lemma of decomposition of kernels in R^n, I show that Rank(B) = dim(Ker(A - I_n))

And if B is a projector, then it is similar to a J_r = diag(1,...1,0,..0) , thus dim(Ker(A-In)) = Rank(B) = Tr(B) = the right hand side

... if I set p = o(A), then p divides q, then the second sum becomes q/p times the sum from 1 to p ...

and I can then easily apply the isomorphism to simplify the sum..

(I think)

Yeah.. that ended up being it : )

[modern palm]

If I have two 2x2 matrices, say (2,2,2,2) and (2,3,3,3), and I want to show their linearly independent. I should show the only solution to a(2,2,2,2) + b(2,3,3,3) = (0,0,0,0) is when a = b = 0 right?

So i get 4 equations, put them into a matrix, and show it gets to an upper triangular matrix equalling 0

But what happens if the two matrices are not linearly independent, and I try the same thing?

[short magnet]

Yes

then you'll get non-null solutions

just like with vectors

[modern palm]

non-null solutions?

[short magnet]

specifically, for the two matrices

say : A and B

you'll find lambda such that A = lambda B

You don't really need to put them back into matrices to find a and b, you can just solve it by hand !

[modern palm]

right lol

[short magnet]

for the two matrices that you showed us, we find that : 2a + 2b = 0 and 2a + 3b = 0

if we substract the first equation from the second one, we find that 2a + 2b -(2a + 3b) = 0

i.e. : -b = 0, thus b = 0

and since 2a + 2b = 0, it follows that a = 0 also

[modern palm]

So if we get 2 equations like 2a+2b = 0 and 4a + 4b = 0. We'd get (4a+4b) - 2(2a+2b) = 0, then this just becomes 0 = 0. Hence, infinitely many solutions, which suggests theres infinitely many u and v, such that u(2a+2b) + v(4a+4b) = 0. Hence linearly dependent.

[short magnet]

the two equations :

2a + 2b = 0 and 4a + 4b = 0 are one and the same !

you get the second one by multiplying the first by 2

so in fact, there is only a single equation

And, as a general rule, if you have less equations than unknowns (equations that "are not similar"), then there is a non-zero solution

for the equation we end up with : a + b = 0, if we fix a value for b

then by choosing a = -b, we get a non-zero solution

(a = 1,b = -1) ...etc

Perhaps I'm being unclear and confusing :"

[winter harbor]

Is that the full statement you want to prove or are there some other assumptions? What is q here?

[short magnet]

just an integer

[short magnet]

I've already solved it ^^ thanks for your time though !

[modern palm]

Nono, it's making sense.

[winter harbor]

Oh, ok then.

Btw

If it is not that much of a deal

Could you post the whole question?

Maybe I will try to do it by myself.

[short magnet]

np dude ^^

give me a sec

Let A be an endomorphism of $$\mathbb{R}^n$$ such that : $$ A^q - I_n = 0 $$ q being a positive integer. Show that : $$ dim(Ker(A − In)) = \frac{1}{q} \sum_{k = 1}^{q} { Tr(A^k) } $$

[stoic python]

Der Gegenstand ist einfach.

[short magnet]

(It was originally posed in french..)

[modern palm]

What makes the equations "not similar"?

[short magnet]

them being lineary dependant., just like with matrices.

actually exactly like matrices !

Are you familiar with the fact that matrices represent systems of equations ?

[modern palm]

As in taking coefficients of the system of equations to be the enteries in the matrix?

[short magnet]

yep

I just suggested that solving such a system by hand is easier than putting it back into a matrix

especially with systems of two or three unknowns. It'd be quicker to solve it by hand, manually, than putting it back into a matrix and solving it using them

[wintry steppe]

yo the topic is simple long time no see

[short magnet]

ikr

haven't had the time recently

cuz trying to catch up on maff and physics.. but today is sunday so : ) kinda chill

[wintry steppe]

what did u try?

[modern palm]

okok understood.

[short magnet]

.

[wintry steppe]

maybe some induction might work?

[modern palm]

thanks

[short magnet]

?

[wintry steppe]

actually I didnt see the discussion

sry

[short magnet]

lol

[short magnet]

np

what u doing here anyways @wintry steppe

u supposed to be hanging out in abstract algebra

[wintry steppe]

idk this seems really boring

I shouldnt be here

I cant helpa nyways

[short magnet]

:" )

[wintry steppe]

;'3

wait so u solved ur problem or what?

[short magnet]

yeh

[wintry steppe]

oh ok thank you now I dont have to read all these paragraphs

[short magnet]

xD

np

we actually just finished abstract algeb

and are now in "reduction"

[wintry steppe]

wot

[short magnet]

so I'm probs gonna hangout more often here or smth

[wintry steppe]

wtf

wait do u have some list of topics ur aa classs covered?

and lin alg too?

[short magnet]

on abstract algeb ?

we spent a week and a half on it

[wintry steppe]

wait what

youve been doign aa for a while tho no?

ok just explain plssssssssss

[short magnet]

on the side yeh

but in class it's technically a week and a half

that we spent with our prof on it

[wintry steppe]

what did u cover in just a week and a half

[short magnet]

we just went full speed over multiple topics

erm :

give me a sec

[wintry steppe]

,w how long is a seconnd

[stoic python]

Results provided by WolframAlpha

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

Requested by Ledog#7947

[short magnet]

you got ur answer : )

[wintry steppe]

xD

omg

[short magnet]

• Basic group definitions and theorems : def of group, sub groups, normal subgroups, cyclic groups, generation of groups, Lagrange theorem, Poincarré theorem, index formula, simple groups, classification of finite abelian groups, group actions, cyclotomic polynomials, cyclicity of Z/nZ* , Carmichael numbers, Asymptotics behavior of prime numbers, Arithmetic functions
• Rings, Polynomials of multiple indeterminants

[wintry steppe]

can be in french

[short magnet]

here, to not keep u waiting :"..

[wintry steppe]

bruh

how do you do that in 1.5w I dont believe you

[short magnet]

idk myself my dude.. wait, I have here the exact number of online zoom classes we had :

[wintry steppe]

so whole week and a half was just this class then?

[short magnet]

It was this and physics

but starting from next week we might have engineering and CS too

[wintry steppe]

Hmm okay maybe I can see how prof goes through all of this in like 8 lectures 1.5h long

[short magnet]

2h but yeh

He kinda insane

[wintry steppe]

but I dont see how (at least me) one can understand it in that much time

[short magnet]

well, we've been working towards this since first year..

[wintry steppe]

idk feels like you have to study 6 horus a day besides attending

[short magnet]

(we usually study more though ...)

[wintry steppe]

what

I dont remember the last time I actually spent like 6 hours on learninng pure math

my brain is kinda fried after 3

[short magnet]

ik.. but we kinda don't have a choice

[wintry steppe]

[short magnet]

yeh.. that's why I usually just alternate between subjects

I'd do 3 math, 3 phys, 3 math for ex

[wintry steppe]

homie theres not that many hours inn a day

classes

sleep

eating

[short magnet]

it actually works out well

[wintry steppe]

9 hours not possible

[short magnet]

Dude.. we have this guy in class who hasn't shaved in a year

[wintry steppe]

I wish i had that much motivation

you inspire me araragi kun

[short magnet]

to save up time

We don't have motivation rly :"

[wintry steppe]

yes you do\

[short magnet]

we just want this to be over

[wintry steppe]

id go through 73 mental breakdownns

[short magnet]

rly depressing at times

and when burnout hits

it hits real..

[wintry steppe]

yeah...

I feel you

im now kidna hyped for myy semester starting

but the hype ends after 2 weeks usually

[short magnet]

^^

: (

well

[wintry steppe]

how much of this you have left?

before uni

[short magnet]

I think we have until month 4 of next year

we'll have written exams for like a month ..

[shrewd lily]

So I was asked to move this question here

This is the Leibnitz formula

[hard drum]

Start by plugging in A^T

[shrewd lily]

where

in the formula?

[hard drum]

Yes

Then you'll have to use a particular property of permutations

[zenith junco]

i dont understand this solution or what the question is askin

obviously asinx + bcosx is linearly dependent

[shrewd lily]

You have to show that sine and cosine are independend vectors

[zenith junco]

like

by itself?

span {sinx} is lin indep

[shrewd lily]

you can show that they are orthagonal for example

[zenith junco]

how can span{sinx, cosx} be lin indep when they have a non trivial solution

[hard drum]

they don't

[shrewd lily]

which

[zenith junco]

but

[lavish jewel]

the solution approach literally shows you can only get 0 if both coefficients are 0

[hard drum]

Note it's for all x

[zenith junco]

but pi/2 * sin0 + 0*cos0 = 0 + 0

b=pi/2

thats a nontrivial solution

[lavish jewel]

you showed it for x = 0

but it has to work for all x

[hard drum]

Functions f and g called linearly independent if setting af(x) + bg(x) = 0 for all x means a = b = 0

[zenith junco]

oh

[lavish jewel]

the same a and b have to work for every single x

[shrewd lily]

you have to show that

[zenith junco]

but

[hard drum]

A nice way to think about it is to forget about x in the definitionand just say af + bg = 0 (as functions!) implies a = b = 0,

so it's like anything else

[zenith junco]

1 * sinpi + pi/2*cospi = -pi/2

[shrewd lily]

you can plug in x=0 and get what b must be

you plug in pi/2 and get a

[lavish jewel]

i think you're getting mixed up in the 0 part

the 0 given there is a 0 function

you get 0 for every single x

[shrewd lily]

or this solution which is a bit more difficult to understand

[lavish jewel]

what they then do is realize that sines and cosines both cross the x axis periodically, and at different values of x

this means for some values of x, either the sine or the cosine vanishes

and then if you still want the sum to be 0, the coefficient of the sine or cosine must be 0

[zenith junco]

i dont understand why taking the limit shows linearly independence

[shrewd lily]

it'S not the limit

[zenith junco]

wait hmm can someone clarify what it means for two functions to be linearly independent again

integral*

[lavish jewel]

potato gave you the definition just now

[lavish jewel]

here

[lavish jewel]

here, as a function of x

[zenith junco]

oh

[lavish jewel]

what mxffin proposed is to show sin(x) is orthogonal to cos(x), which is also valid

[shrewd lily]

it's Inner product space

[zenith junco]

if two functions are orthogonal it implies they do not lie on the same vector right

so it would be indepdent in R^2 space?

[zenith junco]

i guess im confused because for this definition
pi/2 * sin0 + 0* cos0 = 0

pi/2 can actually be anything

this is a nontrivial solution 😦

[lavish jewel]

this is linear independence

but yes, orthogonality implies it

lemon, it isn't

because it doesn't work for ALL x

[zenith junco]

oh

[lavish jewel]

change x to 0.1, for example

[zenith junco]

really?

oh i see

so

[lavish jewel]

,w pi/2 * sin (0.1)

[zenith junco]

the a and b are the same for all x

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

no longer 0

yes, that's what i said a while ago

[zenith junco]

oops

[lavish jewel]

same a and b for ALL x

[zenith junco]

oooo

i get it now thank u

um sinc ur here can i ask another question

😔

i asked on piazzza and my professor was like its in the lecture notes i alrdy explained

but idont get it

shes so mean ;-;

anyways could someone please explain these two to me

[zenith junco]

i dont understand how to draw the venn diagram for it

why is w1+w2 the smallest

[lavish jewel]

well, the largest is all of V, so that's not it

[shrewd lily]

well

[zenith junco]

how is w1 intersect w2 the smallest

i dun get it

skjdfjdslkjflkekjsfd

[shrewd lily]

(A⊊B)∧(B⊊A) means they are disjointed sets A∩B=∅

[half ice]

W1 + W2 is a bit easier to see, I think. If any vector space contains both W1 and W2, then it contains the sum of any vectors in W1 and W2

It turns out that this rule alone is enough, and produces the smallest space that contains W1 and W2

[zenith junco]

😔

[half ice]

Any part of my explanation that isn't quite clicking? Maybe I can express that more clearly

[zenith junco]

the first sentence

second*

[shrewd lily]

what?

[zenith junco]

idk

hm

[shrewd lily]

So the first sentence makes sence

it says

[zenith junco]

" If any vector space contains both W1 and W2, then it contains the sum of any vectors in W1 and W2 " ???

[half ice]

Sorry I think Mx decided to talk about ∩ and I decided to talk about +

[zenith junco]

oh lmfoa

[half ice]

I got you confused haha

[shrewd lily]

oof

[shrewd lily]

yeah

[zenith junco]

i dont get it

[shrewd lily]

that's what vektor spaces are defined as

[zenith junco]

oh

i gwt it lol

how is that the smallest subspace tho

[half ice]

Know how Span() works? Basically, we're taking the span of everything in W1 and everything in W2

[zenith junco]

wait are utaking about + or intersection

[half ice]

Span() is following the least amount of rules that it has to, in order to make a vector space. So we get the smallest possible

[zenith junco]

hoq

how

i understand how span works

how do we get the smallest possible 😔

[shrewd lily]

if you look at the definition of ∩ you can see that it is the largest space contained by A and by B

[zenith junco]

😦

[shrewd lily]

since it is all the xs in A and in B

[zenith junco]

are we talking about the intersections now?

[shrewd lily]

well we solved A+B didn't we?

[zenith junco]

no 😦

i dont understand anything

😔

[shrewd lily]

if you have two vektors, they span a vectorspace

[zenith junco]

this makes sense but

[shrewd lily]

meaning that you can create any vector in that vectorspace by using those two vectors

for example the coordinate system is a vectorspace

[zenith junco]

ok ok

hold on

hold on a second

[shrewd lily]

yo

[zenith junco]

i think iget it now but let me summarize my understanding

W1 intersect W2 is the largest vector subspace of V that is contained in W1 and W2

This is because every value that is contained in both W1 and W2 HAS to be in the intersection

[shrewd lily]

yes

[zenith junco]

W1 + W2 is the smallest vector subspace of V that contains W1 and W2 because

the definition of a span is closed under vector addition, scalar mult.

the addition is only part of the span

[shrewd lily]

yes

[zenith junco]

my teacher is so mean

she couldnt just tell me thi

ss

😔

makes me feel so stupid rip

thanks for yalls help

[half ice]

You're fine, these aren't easy concepts to generalize and abstract

Np, feel free to ask if you have anything else

[zenith junco]

i shall, im planning on doing math all day. thanks so much yall helping me through this class 😩

[shrewd lily]

I'm still not getting this

[shrewd lily]

the ¬σ should've been σ^(-1)

[hard drum]

yeah

Ok, so did you try my advice of putting in A^T? if we suppose A = (a_ij), then what did you get for det(A^T)?

[shrewd lily]

haven't been using this one much since it is inefficient to me

[hard drum]

$\sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^{n} a_{\sigma(i),i)}$ is what you should get

[shrewd lily]

imma look what i have

[stoic python]

potato

[hard drum]

yup, exactly

Now there's one snag here obviously - we want to 'swap' where we've written σ(i) and i

However, there's an important property of permutations

(linked to the hint)

[shrewd lily]

we invert the sigma function?

[hard drum]

Pretty much. In the sum, we can let j = σ(i) and then i = σ^-1(j), so it becomes

$\sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{j=1}^{n} a_{j,\sigma^{-1}(j)}$

[stoic python]

potato

[hard drum]

Now let τ = σ^-1 and try to write everything in terms of τ

[shrewd lily]

so i is identity

itself

[hard drum]

Wdym?

[shrewd lily]

just written down differently

[hard drum]

Oh i mean yes, i've just written it in a different way, yes

you can think of this as essentially er

in the product, the terms will be $a_{\sigma(1),1)}, a_{\sigma(2),2)}$ etc, but we could instead put them in order by σ(i)

[stoic python]

potato

[hard drum]

so it'd become $a_{1,\sigma^{-1(1)}$ etc

[stoic python]

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

[hard drum]

Does that make sense?

[shrewd lily]

so we substitute for i this and get a_(σ(σ^(-1) (j)),σ^(-1) (j))=a_j,σ^(-1) (j)

oof wait

[hard drum]

Yeah sorta

[hard drum]

just relabelling yes so it becomes this

[hard drum]

Yeah

:)

[hard drum]

So are you happy with this being the same as what we originally had for det(A^T)?

[shrewd lily]

what is tau

[hard drum]

I've defined τ = σ^-1

[shrewd lily]

is it just the hint thingy

[hard drum]

Yeah, essentially

[shrewd lily]

a muchas gracias

[hard drum]

So try to write everything in terms of tau in that sum/product etc

[hard drum]

from this i mean

And lmk if you need any more help etc

[shrewd lily]

hold on a sec

forgot one sigma

[hard drum]

genau

[shrewd lily]

is it like this

[hard drum]

Not quite, I'd put it as τ^-1 in S_n

But... the important thing is

We're summing over ALL permutations σ

which means we're also summing over all possible τs

so you can just write it as the sum of τ in S_n

:)

[shrewd lily]

ah right 😄

[hard drum]

And now all that's left is the hint :)

[shrewd lily]

since it's a group they are identical

[hard drum]

Yeah, exactly

[shrewd lily]

Thank you very much

[hard drum]

np :)

[shrewd lily]

spent 5 hours solving that

[hard drum]

This is one of the fairly rare cases actually using the Leibniz formula is useful i think, lol

[shrewd lily]

i think there is an easier solution

[hard drum]

This is the standard proof I've always seen, but I reckon there's a way in terms of EROs

[shrewd lily]

yeah so it's a common question?

[hard drum]

I wouldn't say more of a question, it's more a standard proof you'd see in a linear algebra text

But I guess it was just given to you as a problem instead

[shrewd lily]

studying for my 2nd exam that is in 3 days

this was a question from a homework i had to do this semester and the funny thing is that it was only worth 1 point more than getting the sign of these permutations

[hard drum]

I reckon something along the lines of saying
-Det is multiplicative for elementary matrices (if E is elementary, det(EA) = det(E)det(A)= det(AE)
-if A is invertible, then we can write A as a product of elementary matrices and then taking the transpose of both sides won't affect the determinant, so A = A^T
-det(A) = 0 <=> A not invertible <=> A^T not invertible <=> det(A^T) = 0

[hard drum]

Lol

[zenith junco]

why is this true

[limpid vine]

$\begin{bmatrix} 9 & 0 & 0 \ 9 & 9 & 0 \ 9 & 9 & 9 \end{bmatrix} ~ \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 9 & 9 \end{bmatrix}$

[stoic python]

maximo

[limpid vine]

is this the case?

because if so the implication is that $\begin{bmatrix} 9 & 0 & 0 & 1 & 0 & 0 \ 9 & 9 & 0 & 0 & 1 & 0 \ 9 & 9 & 9 & 0 & 0 & 1 \end{bmatrix} ~ \begin{bmatrix} 9 & 0 & 0 & 1 & 0 & 0 \ 0 & 9 & 0 & -1 & 1 & 0 \ 0 & 9 & 9 & -1 & 0 & 1 \end{bmatrix}$

[stoic python]

maximo

[limpid vine]

but also: $\begin{bmatrix} 9 & 0 & 0 & 1 & 0 & 0 \ 9 & 9 & 0 & 0 & 1 & 0 \ 9 & 9 & 9 & 0 & 0 & 1 \end{bmatrix} ~ \begin{bmatrix} 9 & 0 & 0 & 1 & 0 & 0 \ 9 & 9 & 0 & 0 & 1 & 0 \ 0 & 0 & 9 & 0 & -1 & 1 \end{bmatrix}$

[stoic python]

maximo

[limpid vine]

i just realized the ~ are missing, but i'm doing row operations

[stoic python]

ThreeLives

[limpid vine]

i just realized my mistake.

[granite wren]

Can someone explain to me what exactly the determinant of a matrix is? Like I know how to calculate it but what is it?

Like yeah I can find the determinant if someone asks me but id like a bit more understanding about it

[fringe haven]

It represents how much it scales something

So if you took a 2x2 matrix and applied it to the unit square then it would increase the area of the unit square by a factor of the determinant

Or if you took a 3x3 matrix and applied it on another matrix then it would represent the change in volume

Can someone give me a start on this problem?

Im not sure how to begin

[hard drum]

I think the easiest direction to start with is to suppose B is invertible.

Pick an arbitrary, non-zero vector x and then choose y (in terms of x) such that x^T By is non-zero; this shows β is non-degenerate (since there's no x such that x^T By = 0 for all y)

The hard part is finding such a y

[past violet]

intuitively: B annoys you, but you know how to remove it: it's sufficient to yield something like B.B^{-1}

that could give a clue maybe.

to prove that B has an inverse, maybe prove its kernel is null

[half ice]

@granite wren
The problem is that the determinant isn't only one thing. One representation is scaled volume. If you make a parallelogram with n vectors, then apply T to those n vectors, you'll get a new parallelogram. det(T) is the ratio of those two parallelograms' volumes.

However, a geometric approach fails for a lot of vector spaces. I think the most useful way to see the determinant is a "multiplication shortcut". Multiplying two matricies is hard, but getting the determinant of their multiplication is much easier. That way, you can know things about a matrix product without actually carrying out a matrix product

One big goal of a linear algebra class is to find basis-independent properties of linear transformations. These are very valuable. The determinant is one of them.

[modern palm]

This is what the solution says, dont understand where the red came from?

why it's not $(a_0+b_0)k + (a_1+b_1)kx +...$

[stoic python]

42_Toxic_Potatos

[modern palm]

nvm

[limpid vine]

if i have $A$ is a 2x4 matrix and $B$ is a 4x2, and the equality $AB = I_2$, does that imply that $A$ and $B$ are invertible since $(I_2)^{-1} = (AB)^{-1} = B^{-1}A^{-1}$

[stoic python]

maximo

[hard drum]

You just assumed B^-1 and A^-1 exist

inverses only exist for square matrices

(because B is the inverse of A if and only if AB = BA = I and for those equalities to make sense A and B must be square)

[limpid vine]

yeah that's what i got from the book, so the identity (AB)^-1 = B^-1 A^-1 is only true if A and B are both invertible?

[winter harbor]

yup

it would not make sense to write B^-1 if B wasn't invertible, for example

And that formula would not make sense altogether

[limpid vine]

alright thank you both

[royal bluff]

can anyone assist me with this

[odd kite]

defn. of linear independence

[royal bluff]

I am still a bit confused

[limpid vine]

linear independence asks whether there is a solution to Ax = 0, other than the trivial solution where x = 0

there will be non-trivial solutions only if there are free variables

which of those statements would suggest there are free variables to the system?

i just realized the top says A has linearly independent columns

[royal bluff]

Does that change what you said above

[limpid vine]

it changes the answer

since we know, by definition, that Ax = 0 has only the trivial solution

let me think about it for a sec

well my understanding of the question is that A has linearly independent columns, which is to say, Ax = 0 only has the trivial solution

so simply by being linearly independent we can conclude that (E) the equation Ax = 0 never has nontrivial solutions

for a transformation to be one-to-one and onto it must have a square standard matrix right?

[royal bluff]

not sure

I'm getting help in vc if you are interested in the solution

[limpid vine]

let me know what you end up on

for your problem

[royal bluff]

we got E

[limpid vine]

good to know