linear-algebra

[low halo]

Can you please provide full explanation on sheet

[wintry steppe]

i'll tell you the cayley hamilton theorem but im not gonna solve the problem

[stoic python]

TTerra

[wintry steppe]

(I_n is the identity matrix of size n)

nix told you how to check invertibility. to check that the inverse of A actually has that form, all you need to do is compute...

[silver heath]

How do I do this without a ton of computations?

I believe there was somethign relating to diagonalalizing a matrix

and it would gi veyou the matrix with respect to bases... but you cant diagonalize a 4x3 matrix or a 2x4 matrix???

wait

would it be fien to do

2x4 multiplied by the 4x3 matrix?

[lavish jewel]

that's pretty m7ch what you have to do

you need a few changes of basis, but you can actually avoid the R4 one if you pay attention

[silver heath]

yeah

i was really discouraged (procrastination) because it looked like a lot of work before i realized that the R^4 was unnecessary

[haughty berry]

Is this a valid proof that two basis of the same (finitely spanned) vector space have the same size? (Sorry if my english isn't great, I didnt learn this in english)

We will prove that given a vector space $V$, with two basis (linear
independent spanning sets):

\[
B_{w}=\{w_{1},\dots,w_{n}\}
\]

\[
B_{u}=\{u_{1},\dots,u_{m}\}
\]

We must prove that $dim(V)$ is well-defined, ie. $n=m$.

Since $span(B_{w})=span(B_{u})=V$, we see that $B_{w}\subseteq V=span(B_{u})$.
So:

\[
w_{i}=\sum_{j=1}^{n}\alpha_{ji}u_{j}=\alpha_{1i}u_{1}+\dots+\alpha_{n1}u_{n}
\]

We also know that since $B_{w}$ is linearly independent:

\[
\beta_{1}w_{1}+\dots+\beta_{m}w_{m}=0\iff\forall i:\beta_{i}=0
\]

So:

\[
\beta_{1}(\alpha_{11}u{}_{1}+\dots+\alpha_{n1}u_{n})+\dots+\beta_{m}(\alpha_{1m}u_{1}+\dots+\alpha_{nm}u_{n})=0\iff\forall i:\beta_{i}=0
\]

\[
(\beta_{1}\alpha_{11}+\dots+\beta_{m}\alpha_{1m})u_{1}+\dots+(\beta_{1}\alpha_{n1}+\dots+\beta_{m}\alpha_{nm})u_{n}=0\iff\forall i:\beta_{i}=0
\]

Because $B_{u}$ is linearly independent, this is only true if each
coefficient of $u_{i}$ is 0.

So we get the following system of equations:

\[
\left\{ 
\begin{array}{cc}
\beta_{1}\alpha_{11}+\dots+\beta_{m}\alpha_{1m} & = 0 \\
\vdots & \\
\beta_{1}\alpha_{n1}+\dots+\beta_{m}\alpha_{nm} & = 0 
\end{array} 
\right. \]

Which can be represented as the following matrix equation:

\[
\left(\begin{array}{ccc}
\alpha_{11} & \dots & \alpha_{1m}\\
\vdots & \ddots & \vdots\\
\alpha_{n1} & \dots & \alpha_{nm}
\end{array}\right)\left(\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{m}
\end{array}\right)=0
\]

We'll define:

\[
A\coloneqq\left(\begin{array}{ccc}
\alpha_{11} & \dots & \alpha_{1m}\\
\vdots & \ddots & \vdots\\
\alpha_{n1} & \dots & \alpha_{nm}
\end{array}\right)
\]

Since this only has a solution if $\forall i:\beta_{i}=0$, so this
homogeneus system only has one solution:

\[
\left(\begin{array}{c}
\beta_{1}\\
\vdots\\
\beta_{m}
\end{array}\right)=0
\]

That means that $CF(A)$ must have a leading coefficient in every
row. So $rank(A)=m$. Since $rank(A)\leq n,m$
this means that $m\leq n$. Without loss of generality we can prove
that $n\leq m$, so $n=m$.

[stoic python]

Slurp

[haughty berry]

I came across this when trying to prove that if $U\leq V$ then $dim(U)\leq dim(V)$ (apparently my proof wasn't good though, because I didnt realize I had to prove that $U$ has a basis)

[stoic python]

Slurp

[haughty berry]

If you could tag me that would be great

(The part that's cut off is $\forall i: \beta_i = 0$)

[stoic python]

Slurp

[silver heath]

Does anyone know what the highligted part of the sentance means?

"left multiplication of A"???

[haughty berry]

Multiplying on the left

As opposed to the right, because matrix multiplication isnt commutative

[silver heath]

multiplying what of A?

[silver heath]

Or is it just refering to A?

[haughty berry]

I assume it means the transformation T(v) = Av

[severe ermine]

hi {{3,1,1,1,0},{5,-1,1,-1,0}}

so im aware its a homogenous system

thats about it

[wintry steppe]

so you know it has a trivial solution

[severe ermine]

what does that mean

[lavish jewel]

that if you set every variable to 0, it satisfies the equations

[wintry steppe]

all homogenous systems have at least one solution

which is what Edd said

[severe ermine]

yeah but how do I tell if its one or infinite

[wintry steppe]

this is when you have to use row operations

but actually I think this is infinite off the bat

[severe ermine]

I dont see any row operations

[wintry steppe]

since there are more "variables" than equations

[lavish jewel]

you have 4 variables, 2 equations

[severe ermine]

2 pivots at most and 4 variables

[lavish jewel]

this means 2 variables are "free variables"

you can set them to some generic parameter

e.g. z = t, w = s

[severe ermine]

so pivots are equivalent to equations

[lavish jewel]

and then solve for x and y in terms of t and s

no

[severe ermine]

I dont get it

[lavish jewel]

you can have fewer pivots than equations

[severe ermine]

like uh

[lavish jewel]

pivots is more related to the number of linearly independent equations

for example if i tell you x + y = 0

and 2x + 2y = 0

you'll have only 1 pivot if you put that in matrix form

[severe ermine]

ye

[lavish jewel]

because the two equations are linearly dependent

[severe ermine]

so if you have 3 equations

and 4 variables

you automatically assume one is free

?

[lavish jewel]

yes

at least 1

because there is no way you can have more than 3 pivots

[severe ermine]

ok that would have been helpful for my prof to explain

[lavish jewel]

your matrix has size 3 x 4

the pivots have to be in the matrix, idk how else to put it

[severe ermine]

ye I know what you mean

I was unsure why you were using the equations as a proof

[lavish jewel]

wdym?

[severe ermine]

he didn't explain that m - n = free variables

probably like walked around it

[lavish jewel]

because that is not true

that is AT LEAST the number of free variables

but there can be more

which is why i gave you the example

say we have 3 vars, x y and z

[severe ermine]

yeah I know exactly what you mean

[lavish jewel]

and i tell you x + y + z = 0 and 2x + 2y + 2z = 0

m-n = 1, but we only have 1 pivot

and so 2 free vars

[severe ermine]

so I reduced it to {{5,-1,1,-1,0},{0,8,2,8,0}}

[lavish jewel]

mhm

2 pivots

[severe ermine]

which one of those do I like do tho

I have 2 different starting points

[lavish jewel]

wdym

[severe ermine]

so like if I did w = t for t in real numbers

then do z = s for " " "

what do I do next

-y + s - t = 0

[lavish jewel]

doing this is equivalent to adding two rows to your matrix

[severe ermine]

or

im asking if it matters what row I use

[lavish jewel]

wdym which row

[severe ermine]

{{5,-1,1,-1,0},
{0,8,2,8,0}}

[lavish jewel]

you have to use both

[severe ermine]

what is next tstep

[lavish jewel]

put in the s and t there and solve for x and y

[severe ermine]

I dont understand what you want me to do

[lavish jewel]

dude

we added 2 equations, yeah?

w = t, and z = s

this means that 8y + 2s + 8t = 0

and that 5x -y +s - t = 0

solve that for x and y

you can either do this by substitution, since you conveniently already have an equation for y

or make a larger matrix and keep doing gaussian elimination

[severe ermine]

this is new

before it was in echelon form

[lavish jewel]

what is new?

this is still in row echelon form

[severe ermine]

oh god

I am very tired

I apoligise

[lavish jewel]

and when you do row echelon form, you then have 2 options

either do back-substitution, or keep going into reduced row echelon form

[severe ermine]

I have been doing math for like 12 hours

[lavish jewel]

go sleep cuz rn you're not working efficiently

[severe ermine]

I will let you know in a minute after my attempt

I cant

I have hw I didnt know about due tommorow

[lavish jewel]

[severe ermine]

I literally did the weeks hw

you know tutorials

there are bloody pre tutorial questions

you have to do

@lavish jewel tell me this right

[lavish jewel]

the idea is right

idk if the arithmetic is

[severe ermine]

good enough

oh god

there are more questions

{{1,-1,3},{2,-2,k}}

I need to find no, one and infite solutiions

so infinite is k = 6

[lavish jewel]

mhm

[severe ermine]

I get stuck after that

I have a vague idea that one of them is k = 0

and no could be 5

[lavish jewel]

idk ifv theres a way to get 1 sol for that tbh

[severe ermine]

evil questions

[lavish jewel]

you can just rref it and go from there

u get 1,-1,3 and 0,0,k-6

k=6 is inf sols

k=/=6 is no sol

[severe ermine]

it is one of those questions

that doesnt tell you when to stop looking

so you figure it out on your own

[lavish jewel]

presumably the only kind of problem you will find in your exams

[severe ermine]

I dont understand why I am getting electical engineering problems

[lavish jewel]

for you to exercise your abstraction

[lavish jewel]

learning how to write problems in known forms is important

[severe ermine]

is this A) f1 + f4 = 1000, B) f1 + f2 = 1100, C) f2 + f3 = 700 and D) f3 + f4 = 600

[lavish jewel]

i'm not gonna read that

[severe ermine]

read what

Im just asking you to look at the image

not the text

[lavish jewel]

looks ok

[sleek sundial]

if every column of a square matrix is linearly independent from one another, does that imply invertability?

[native rampart]

Yes

[sleek sundial]

which of these are subspaces of r2: f(x) = 2x, f(x) = e^x, f(x) = absolute value(x), f(x) = x^2

I'm leaning towards just the 2x but idk

e^x doesn't include the origin, I don't think x^2 supports scalar multiplication, and idk about absolute value

am I completely off here?

[lavish jewel]

just test the definition

[dire thunder]

what is r2

[forest quiver]

2 dimensional plane of real numbers

What if I have 3 vectors that are multiples of each other though?

[cyan zenith]

What exactly is your question? :)

[forest quiver]

How is this linearly dependent?

Wtf

it's not in a straight line

I mean I kind of understand the theorem

[cyan zenith]

That is not the definition of linear dependance! :)

[forest quiver]

if we have Ax=0, then there will be a free variable

that's what I have been thinking this whole time

[marble lance]

Two vectors are linearly dependent if they are on the same straight line. Three vectors are linearly dependent if one of them lies on the plane spanned by the other two vectors. In this case any two of the vectors spans the whole R^2, so your third vector is on the plane

[digital gulch]

If you add two vector they produce other one so that why they are linearly depended

[marble lance]

Your geometric intuition is off

[forest quiver]

let me think about what you said for a moment

that's kind of weird to me

[cyan zenith]

What about it is weird?

[forest quiver]

I think I get what you mean

look at this

In R3, a plane is considered linear dependence

But if a vector leaves that plane, then it's not linear anymore

is that kind of what you mean?

Like it means different things with different dimensions/amounts of vectors

im sorry my intuition is fucking horrible

[cyan zenith]

It's ok :D we all start somewhere

Do you know the concept of span?

[forest quiver]

yeah

[marble lance]

What you are describing would be: A set {v1, ..., vn} is linearly dependent if vi = a vj for some i ≠ j. But that's not what linearly dependent means. That is a weaker version. The set is linearly dependent if for some i,
vi = a1v1 + ... + a(i-1)v(i-1) + a(i+1)v(i+1) + ... + anvn.

So vi can be a linear combination of any of the other vectors, it doesn't just have to be a multiple of one of the vectors.

A nongeometric explanation.

[forest quiver]

Ok so

that basically means

that in a set if at least 1 vector is a scalar multiple of another, there will be linear dependence for that set?

[lavish jewel]

sounds right

[cyan zenith]

In that case, a set would definitely be linear dependent

But there are more cases!

[forest quiver]

[cyan zenith]

Take a look again at the definition Luna sent

[marble lance]

Yes, that is what you were describing and what my first definition says. But my first definition is not the real definition.

[lavish jewel]

sometimes you can't easily see it, so gaussian elimination and pivots are a thing

[forest quiver]

what is anvn?

[marble lance]

A scalar a_n times a vector v_n when luna is lazy to do subscripts

[lavish jewel]

$a_n \boldmath{v}_n$

for a scalar a_n, and a vector v_n

[stoic python]

Edd ✓

[lavish jewel]

whatever

[forest quiver]

ohh

a linear combination

of the other vectors

[lavish jewel]

that's precisely what linear dependence means

[cyan zenith]

A vector that can be written as a linear combination of other vectors "doesn't add anything new", you cannot produce more vectors with it, that makes the system linear dependent

[forest quiver]

ohhhhh

[cyan zenith]

You could cut one of the vectors and still end up with the same span :D

[forest quiver]

Thats why if there is a 0 vector

it all becomes linear dependent

[cyan zenith]

Exactly! The zero-vector does not add anything

[lavish jewel]

the 0 vector is a special case of the scenario you were talking about first

[forest quiver]

This is the definition my book gives

[lavish jewel]

it's any of the other vectors, scaled up by 0

that definition is the same luna gave

they just moved v_i to the rhs

[forest quiver]

yeah it was a bit confusing

in text

ok I get it

I wish there was a better geometric intuition for this

I mean I like the stright line stuff

but that doesn't always work

[cyan zenith]

well

[lavish jewel]

the reason it is defined abstractly is so that you can apply it to things that don't have a geometric representation like that

[forest quiver]

line R4?

like*

[lavish jewel]

like functions

functions can be "vectors" and form vector spaces

[forest quiver]

😨

[lavish jewel]

other things too as long as they behave nicely enough

linalg is not about putting lists of numbers in brackets

[forest quiver]

yeah

[lavish jewel]

those things just happen to sometimes have nice geometric representations

[forest quiver]

yeah that's what I was thinking too

like at the end of the day, vectors are just points

[cyan zenith]

If you have n linearly independent vectors, then the span will be an n-dimensional subspace.
But if you have n-vectors, and the subspace spanned by them is less than n-dimensional, you have linear dependance.
That's a geometric interpretation that can work for you, albeit very general

[forest quiver]

they arent arrows

they arent movements

[lavish jewel]

nope

vectors are elements of a vector space 😛

[cyan zenith]

Vectors are elements of vector spaces. What you interpret them to be is a different question

@lavish jewel lol :D

[lavish jewel]

hah

[cyan zenith]

Pure Mathematics student?

[forest quiver]

alright well thank you all again

I will stick with this definition

along the way I guess I forgot what linear indep/dependence really meant

I was relying too much on teh straight line

ty

[lavish jewel]

get the matrix to a triangular form and take the product of the main diagonal

or the product of all the eigenvalues

thought that second one is kinda circular, since you need the char poly

you could also use minors... several times

[teal grotto]

@hexed plaza you could use the formula $\det(A)=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{j=1}^n\mu^j(Ae_{\sigma(j)})$ where $\mu^j$ is the $j$th dual basis vector coming from the standard basis

[stoic python]

coycoy

[lavish jewel]

you could also do an SVD and use the singular values for it

idk if you mean by hand, btw

cuz in general finding determinants is nasty

[north hedge]

i think this was how it was defined

for me

[teal grotto]

@north hedge there is a really nice generalization for this when you try to find the dimension of the space of alternating k forms over V^n, where V is a vector space

[cyan zenith]

You're correct :)

For the sake of this task, using the fact that row rank = column rank would also suffice, but naturally det(A) = det(A^t) works

[coarse sandal]

How do find the kernel of T?

I know the kernel is the null space of T

but im not sure exactly the process for finding it for a transformation

[lavish jewel]

for this type of problem, if you don't see it immediately, one trick is to vectorize everything and try to construct the matrix that represents the transformation

for example let's say we vectorize the matrix as [a, b, a, b]^T and represent the polynomial with [a,b]^T

the transformation that does this is
1 0
0 1
1 0
0 1

this is rank two, and since it's full column rank, the kernel contains only the 0 polynomial

[coarse sandal]

yup

[lavish jewel]

(you could've also seen that from the original transformation, since having a = 0 and b = 0 was the only way to get a 0 mat)

[coarse sandal]

so we set [[a,a],[b,b]] = [[0,0],[0,0]]?

[lavish jewel]

that's what being in the kernel means, yeah?

if a + bx is in the kernel of T, then T(a + bx) = 0 element in codomain

[coarse sandal]

cause it maps T(a + bx) = zero vector

okay that makes sense

[arctic beacon]

Why do we care about the transpose of a matrix, especially when M=M^T?

[lavish jewel]

do you know about dual spaces?

[arctic beacon]

Kinda

I know that when you take the inner product of a vector and it's dual vector you get a scalar

That's about it

[cyan zenith]

@arctic beacon Taking (v,w) (for standard inner product (.,.) on a vector space V ) and taking v^T w is computationally identical; we say v^T is an Element of the dual space V*, which allows to formalize "the function that takes the scalar product with v" by giving it a vector space to live in.

[arctic beacon]

So orthogonal matrices make the computation easier or is there more to it?

[forest quiver]

I know that every matrix transformation is a linear transformation, but is every linear transformation a matrix transformation?

That just seems unlikely to me

I want to say False to this one, but I'm not quite sure

oh found it on stack exchange nvm

[cyan zenith]

Uhh that's another topic isn't it? So what exactly is your question? :)

[last holly]

share how you interpret it now, it'll help the discussion

[forest quiver]

it'sjust how my textbook defined it

[last holly]

it's contextual though, i've seen books say it's true (but it's only if it's a specific set of mappings... and they don't mention this)

for your class this may be good enough

[forest quiver]

im not in a class

but do you know

about linear transformations?

[last holly]

i'm not an authority

but basic stuff sure

[forest quiver]

Ok well I am trying to show this

And this is what I did:

[last holly]

oh that's a classic one

[forest quiver]

don't spoil it

let me show you

what I did

[last holly]

yea i won't dw

lol

[stoic python]

Tim O'Brien
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[teal grotto]

“badness 10000” thanks compiler

[last holly]

lol

[forest quiver]

But anyways

[lavish jewel]

some typos but that works

[forest quiver]

Do you see waht I did?

Does it make sense?

I don't know how contradictions work

that well

[lavish jewel]

you mentioned some stuff you didn't use

[forest quiver]

Ugh

[lavish jewel]

e.g. with the first part you mentioned, you could've done T(x +y ) =/= T(x) + T(y)

[last holly]

also m*0+b ≠ 0b

[teal grotto]

i got the idea. shorten it tho, just get to the point.

f(0) = m(0) + b = b != 0

[lavish jewel]

i assumed that was a typo

[last holly]

oh that would make sense

[forest quiver]

I did mention this

[teal grotto]

no need for it. it distracts from the point you’re trying to make

[lavish jewel]

no, you mentioned linear functions do that, but then didn't use it

i used it to show T isn't linear

[forest quiver]

Oh righttt

Oh I see

this goes against property (ii)

not property (i)

[lavish jewel]

right. you didn't use (i)

it still works

[last holly]

informally it's just enough to know there's no 0 vector and work from there?

[forest quiver]

Yeah I don't knwo why I put that relaly

I was just tyring to make sure I got everythign down I guess

But is this how I can use contradictions?

I assume something is true

then do something with that

[teal grotto]

if you just show that 0 doesn’t map to 0, then it’s not linear. (the linear maps we are all familiar with are group homomorphisms. identities have to be mapped to identities, and the image of a group homomorphism should still be a sub-group of the codomain. the one that Tim is working with is affine)

[last holly]

right

[forest quiver]

and if some things don't hold up, then I can say that it's not true for that reason

[lavish jewel]

i would call this a counterexample

[forest quiver]

oh right

this is kind of fun tbh

Oh yeha I did do a type on the before last line

[wintry steppe]

[lavish jewel]

tteppa

[frosty vapor]

affine tteppa

[wintry steppe]

what is the application of the stuff in linear algebra 2 like the symmetric product?can you give some examples of how the advanced linear algebra stuff is related to "the real world"?

[nocturne jewel]

Parseval Identity holds for any norm, so long as the linear combination is of an orthonormal set right?

$\norm{\sum_{i=1}^nc_iv_i}^2=\sum_{i=1}^n\abs{c_i}^2$

[stoic python]

Mosh

[severe ermine]

how do I solve
{
{1,0,0,1 | 1000},
{1,1,0,0 | 1100},
{0,1,1,0 | 700},
{0,0,1,1 | 600}}

is this a infinite solution one

[severe ermine]

<@&286206848099549185>

[wintry steppe]

differential geometry makes very heavy use lot of "advanced linear algebra" stuff

multilinear algebra is the cornerstone of tensor analysis

[severe ermine]

why no one help 😦

its a simple question

[sudden narwhal]

what is it ? a matrix ?

[severe ermine]

yes

how do I format it correctly

@sudden narwhal

augmented matrix

[flint jackal]

It's not in RREF form yet

[severe ermine]

how do I convert it

I tried to but I just changed R4 to R1

[sudden narwhal]

RREF stands for ?

[severe ermine]

reduced row echelon form

[flint jackal]

Reduce Row Echelon form

[sudden narwhal]

thx

[flint jackal]

You understand the concept of RREF, correct?

[tropic moss]

Hello, How i can get the laplace transform of this:

[severe ermine]

yeah

[flint jackal]

So how would you get rid of the 1 in the position R2, C1?

[severe ermine]

what is C1

[flint jackal]

Column

And row

[severe ermine]

R2 <- R2 - R1

[sudden narwhal]

$\left[{\begin{array}{cccc|c}1&0&0&1&1000\1&1&0&0&1100\0&1&1&0&700\0&0&1&1&600\end{array}}\right]$

lmao

[severe ermine]

so I can just do rref and it will work

[stoic python]

Herels

[severe ermine]

I thought because of how it was formatted it wouldnt

ok I gtg now @flint jackal can I ping you later? if Ineed to

[flint jackal]

Nope

[severe ermine]

:9

😦

[fleet orbit]

How do I go about this problem?

[broken stream]

man why did they make the numbers so annoying lol

this reminds me of an algorithm I found on stack overflow for determining if a point is inside a known triangle

https://math.stackexchange.com/questions/51326/determining-if-an-arbitrary-point-lies-inside-a-triangle-defined-by-three-points

[broken stream]

idk though determinants are weird

I don't know what these equations/symbols mean in this paper:

https://arxiv.org/pdf/2006.00988.pdf

anyone know what they're talking about?

Is U a matrix of dimensions N x D?

and apparently the T means it's transposed

but what form is it transposed from?

[last holly]

the T is how some books write vectors, since if it's just horizontal it's a row vector

[broken stream]

I guess the NxD superscript is throwing me off

and like why is it on the R

I thought R just means real numbers

[nocturne jewel]

$\mathbb{K}^{n\times m}$ is the space of all n by m matrices with entries from $\mathbb{K}$

[broken stream]

hmm

[stoic python]

Mosh

[broken stream]

so this vector U is a part of all NxD real number matrices?

[nocturne jewel]

yes

[broken stream]

das weird

[nocturne jewel]

U is a N by D matrix with real entries

[last holly]

but look at what N and D are

[broken stream]

ey

[last holly]

the text tells you

[nocturne jewel]

not really.. it's standard notation for the matrix spaces

[last holly]

is this a machine learning text?

[broken stream]

yeah

attention based word embedding

whatever that means

kek

I just realized that kek must have come from people mis-typing lel

[last holly]

i think the problem can be trying to digest multiple disparate things

ahhahaha

kik

yea

it's worth it to just spend a couple of weeks with matrices again if it's rusty

maybe your previous courses didn't cover this stuff

and linear algebra is deep

[broken stream]

yeah

[last holly]

i know mine didn't cover enough of what i needed

[broken stream]

I got a book on it off amazon but bro the questions in it have no answer keys

[last holly]

so i had to do (and still have to do) a lot of self study

which book?

[broken stream]

sheldon axler linear algebra done right

[last holly]

oh yea great one, maybe not the best for matrices

[broken stream]

right cause I think he puts all the common tools at the end of the book

[last holly]

you might want to see if you can find something complementary, maybe check in #books-old

and find some library where you can program the stuff

like matlab or that other one

where it's quick to just try out ideas

[broken stream]

oh cool

usually I use C and manually code all of the operations

[last holly]

also because lin alg theory and practice are two different things, some ways are more efficient (to the computer) to make calculations

[broken stream]

because numpy is scary and I don't knwo what it's doing

[last holly]

i mean C is great but not for learning matrices

i get that lol, i'm learning assembly and it's great

but you know like they say, right tool for the job

hold on i think there are some free resources

let me look

[broken stream]

true the C syntax is not well suited to linear algebra

I thought Fortran sounded cool

but I looked at the syntax on that and it's like bruh

[last holly]

you'd have to write your own library, but then you'd have to know linear algebra to do it

hahaha

[broken stream]

are there more cool things in linear algebra

cause I've done matrix inversion, matrix vector multiplication, 3d perspective projection, quaternion rotation

[last holly]

yes linear algebra is deep

[broken stream]

I did the thing where you color in a triangle

[last holly]

it's a great way to enter abstract algebra

and that has other applications in cs if that floats your boat

like a common way to introduce groups is to talk about 2x2 invertible matrices and work from there

[broken stream]

I would love to rewrite some of these machine learning papers to make them more readable

oh yeah and I was wondering today

how the hell did someone come up with cross products

and also is there a derivation for higher dimensions

or the determinant

like the determinant as it's taught is just this arbitrary set of operations on a matrix

[last holly]

Lagrange, it's usually motivated by other needs

[broken stream]

but is there a way to figure out from scratch how to do a determinant on any matrix

[last holly]

yes

what do you mean?

from scratch

[cyan zenith]

You can define a cross product for any n dimensional vector space using n-1 factors iirc

[broken stream]

yeah like cause every time I've needed one of those things I just go on google images and find something like this

lel

[last holly]

that'll only work for 2x2

but also it's noteworthy to note that it's a formula for the area of a parallelogram

[broken stream]

hmm this is interesting:

[last holly]

and that's the formula for the area of a parallelepiped

[broken stream]

I did see there's like a geometric proof for the parallelogram

[last holly]

and for the n-volume in general

in n-dimensions

[last holly]

yea

[broken stream]

dis shit crasee

[last holly]

what's crazy is euclid didn't use numbers

[broken stream]

what did he use?

sticks?

[last holly]

yea basically

unit lengths

[broken stream]

my guy was out here with a ruler and stuff

[last holly]

so the euclidean algorithm was made with just unit lengths lmao

i think about that every time im about to complain

[broken stream]

you're talkin bout dis?

looks like nicomachus also figured it out

[last holly]

yea i think that, i just heard Borcherds talking about it

and i was like my man... what a g

[cyan zenith]

Correct, just adding the determinant is the oriented n-volume of the span of the row-/column-vectors of the matrix, it can be negative, too.

[last holly]

ah yes of course, that's a good thing to add

my first course it was just a number we had to calculate and that's that

sad uni

[cyan zenith]

Calculating det really is a pain truth that

Luckily, it has great properties that are almost trivial to remember once you know the geometric interpretation :D

[latent tulip]

Any good linear algebra Tutorial? (I mean videos)I will have linear algebra class in the fall. Wanna study it at home before the fall semester

[quaint sage]

I like Khan academy. IMO though, the best way to learn is to do examples from the textbook

[latent tulip]

My English reading skill is not good enough 😭

but listening is fine

[quaint sage]

Good practice, if you can talk about Math in English you can talk about anything

[latent tulip]

English is not my mother language

I like watch videos

[quaint sage]

What language do you speak?

My question:

My brain isn't working today... How do I confirm if 4 points lie on a plane?

[dusky epoch]

do you have names for your points or is it ok to call them A, B, C, D

[quaint sage]

Yeah that is fine

[dusky epoch]

check if the vectors AB, AC and AD are linearly dependent

if they are then your four points are coplanar

if they are independent then the points are not

[cyan zenith]

3blue1brown's "Essence of Linear Algebra" playlist on YouTube is awesome

I'd recommend it to anyone who ever deals with LinAlg

[latent tulip]

👀

[latent tulip]

Mandarin 👀

[quaint sage]

Yeah, that sounds like a challenge

[quaint sage]

Okay new dumb question: My google fu tells me that you can check for this if the determinant of AB is non zero... Correct?

If I am in R3 I can't do a determinant (as AB would be rectangular)

Do I need to do the determinant of (ABC)? Or am I missing something completely here?

[dusky epoch]

you're missing something completely

if you want something to take the det of, it's the matrix [AB, AC, AD]

[quaint sage]

ahh, perfect

[quaint sage]

you sure about that?

A: 1,0,0
B: 4,0,0
C: 1,1,0
D: 0,0,1

should not be coplanar

The [AB,AC,AD] ends up being:

 3     0     0
 0     1     0
-1     0     1

and det (at least according to MATLAB) is 3

[peak garden]

Can someone help me understand what contravariant, covariant vectors are and what it means to lower/raise indices?
I can't find a straightforward source....

[deft apex]

I asked for you in the physics discord

[broken stream]

well for 4 points

you take 3 of those points

and define the plane by those three

so with points a, b, c and d

the plane is defined as a + x(b-a) + y(c-a)

so then you see if there exists a solution to the equation a + x(b-a) + y(c-a) = d

where x and y are scalars

@quaint sage

[teal grotto]

does anybody have any hints on this? i think you have to use Cauchy Schwartz inequality:

Let $V,W$ be finite dimensional inner product spaces over $\mathbb{K}$, where $\mathbb{K}$ can represent $\mathbb{R}$ or $\mathbb{C}$ and $T:V\to W$ is a linear map. Show that there is a constant $c\in\mathbb{K}$ such that for all $v\in V$, we have $$||T(v)||_W\leq c||v||_V$$

[stoic python]

coycoy

[slender relic]

is T meant to be continuous here?

[teal grotto]

no, i meant to add that V and W are finite dimensional inner product spaces, hence the subscripts on the magnitude operators. i dont believe you need T to be continuous, but it is anyways. pretty sure all linear transformations over finitely generated inner product spaces are continuous

[deft apex]

What I meant was; I posted you question in the discord server for you

[slender relic]

Im a bit new to this stuff, but I think T being continuous is necessary

either way, if it is then you can prove it by using continuity about 0

[teal grotto]

hmm. why do you say that

[wintry steppe]

all linear operators on finite dimensional spaces are continuous, and this inequality is used to prove it iirc

[slender relic]

ah okay

thats something I've not met then

[wintry steppe]

if you already know continuity of T then you can use the extreme value theorem to prove it

but if you don't, then cauchy-schwarz does it

[teal grotto]

would you mind expanding? cant really see how to apply it in this case

[wintry steppe]

i was thinking about the T: R^n -> R^m case at first so my answer is going to be a bit more complicated than just cauchy schwarz

but im sure CS pops up somewhere

give me a moment to make it self-contained (not referencing any lemmas)

[teal grotto]

thanks so much

[wintry steppe]

i lied im going to reference a lemma

[teal grotto]

lol okay

[stoic python]

TTerra

[wintry steppe]

if we were working with maps between euclidean spaces, then you could apply cauchy schwarz to prove that |v^1| + ... + |v^n| <= M|v| (here, |v| is the euclidean norm) for some M pretty easily

oops

[stoic python]

TTerra

[wintry steppe]

so in euclidean space it really does follow immediately from cauchy schwarz, but you need to do just a tiny bit of work to make it go through for general finite dimensional normed spaces

[teal grotto]

id imagine just isomorphism from V to R^n

[wintry steppe]

there's probably an easy way to do this and im just overcomplicating it

lmao

but this works (i hope)

[teal grotto]

yea this looks like it works. ill try and go off of this, thanks!
im sure theres some slick way to do this tho lol

[wintry steppe]

"all norms on finite-dimensional vector spaces are equivalent" might take a tiny bit of work to prove. admittedly ive never read the proof i just took it as a fact

[teal grotto]

same lol

[wintry steppe]

now you can use this to prove that linear maps between finite dimensional normed spaces are continuous

[stoic python]

coycoy

[teal grotto]

i was hoping that there would be a generalization of this, but im not sure

[wintry steppe]

i do not know of one

in fact this problem's result is new to me

[wintry steppe]

If $0\in{v_1, v_2,\dots, v_n}$, then ${v_1,v_2,\dots,v_n}$ is linearly dependent.

This is a property that is proved in my LA book but I don't understand the proof. It says:

If we write a linear combination with $1$ as the scalar that multiplies the zero vector and 0 for the rest of vectors, we obtain the vector zero and the vectors are linearly dependent.

It proves it using this situation:

$0 = 0v_1 + \dots + 1\cdot 0 + \dots + 0v_n$

I don't understand why that proves the statement, because I don't see why that would hold also for the case where you have things like $0 = 3v_1 + 7v_2 + 1737382v_3 + \dots + 1\cdot 0 + \dots + 18823v_n$

[stoic python]

Elfire

[wintry steppe]

you only need one non-trivial linear combination (i.e., one where at least one scalar is non-zero) equal to zero to be linearly dependent

if there's a linear combination that works then you don't need to worry about other ones, the thing's linearly dependent

So you can say that the set is linearly dependent because in this situation:

$0 = 0v_1 + \dots + 1\cdot 0 + \dots + 0v_n$

Instead of putting a 1, you could put any other scalar and you would also get the vector 0, and as it is not a linear combination where each coefficient is 0, you would get that it is linearly dependent, right?

[stoic python]

Elfire

[wintry steppe]

any other non-zero scalar multiplying the 0, yes

Yeah, I forgot that

Well, I think I understand the proof now, thanks

@wintry steppe you are a very good helper :)

[teal grotto]

ty TTerra

[swift frost]

anyone know of a group that is/would like to work through a linear algebra book together?

[silver heath]

im working thru one rn xD

Does anyone know how to "solve" this?

it seems obviously wrong....

*false

but maybe i am missing something?

[wintry steppe]

have you tried computing the eigenvalues

if you think it's false you just have to find one counterexample

[teal grotto]

@wintry steppe what do you think of this argument:

let $e_1,\dots,e_n$ be an orthonormal basis for $V$. Then
\begin{align*}
\lVert v\rVert^2
&= \langle v,v\rangle\
&=\sum_{i=1}^n\langle v, e_i\rangle^2\tag{simplify}
\end{align*}
so we have
\begin{align*}
\lVert T(v)\rVert_W&=\left\lVert \sum_{i=1}^n\langle v,e_i\rangle T(e_i)\right\rVert_W\
&\leq \sum_{i=1}^n|\langle v,e_i\rangle|\cdot \lVert T(e_i)\rVert_W\
&\leq \left(\sum_{i=1}^n\langle v,e_i\rangle^2\right)^{1/2}\cdot\left(\sum_{i=1}^n\lVert T(e_i)\rVert_W^2\right)^{1/2}\tag{C.S.}\
&=\lVert v\rVert_V\cdot \left(\sum_{i=1}^n\lVert T(e_i)\rVert_W^2\right)^{1/2}
\end{align*}

[stoic python]

coycoy

[silver heath]

right.... i thought it said can have...

[sharp tusk]

bro wtf.

jesus what math even is that?

[teal grotto]

horrible inner products and LA

[sharp tusk]

half those symbols are foreign

and where are the numbers????

i thought this was math?????

[teal grotto]

numbers?? i laugh

[sharp tusk]

[wintry steppe]

it's just linear algebra with a minor touch of analysis

[wintry steppe]

i'll look at it after i shower

[sharp tusk]

brain gone

jesus so this is college math?

i was excited to take linear algebra senior year of highschool from the name

[wintry steppe]

just because it looks complicated doesn't mean it is. i'm sure that once you learn linear algebra things like this will seem simple to you

math has the funny ability to seem extremely complicated if you're not familiar with notation

linear algebra is fun

[sharp tusk]

notation is stupid

makes my brain hurt

[teal grotto]

non-symbolic proof gang wya

[sharp tusk]

precalc to a proof class was not fun

[wintry steppe]

if so, whatever you do, don't look up differential geometry

[sharp tusk]

where are the numbers man???????

[teal grotto]

there was 1/2 in there somewhere lol

[sharp tusk]

differential topology looks worse

[nocturne jewel]

you squared the norm

[sharp tusk]

i tried reading a few papers and i couldnt evne understand the background knowledge

[teal grotto]

i did wot now

[nocturne jewel]

you had the number 2 too

[teal grotto]

oh yea

[copper fox]

alguien sabe hacerla????

some know how solve it?

[last holly]

hi, this is induction, you can ask this under #proofs-and-logic or maybe even #discrete-math

[wintry steppe]

the argument is correct, but this only works when the norm comes from an inner product

[sinful acorn]

Yeesh.. is that summation with imaginary numbers 🤔 😳 😭

[nocturne jewel]

no...

the summation variable is just i

[last holly]

no, i is sometimes used as an index

[nocturne jewel]

[sinful acorn]

Its terrifying

[last holly]

one sometimes hears, the i'th term

etc

it looks scarier than it is

[nocturne jewel]

not really...

[last holly]

@copper fox no sé si me entendiste en ingles, este canal es para álgebra lineal, tu pregunta cae más hacia #proofs-and-logic o #discrete-math

[wintry steppe]

an english heads up, you should translate the questions you post for us if you want us to help with them. there may not always be a speaker of your language active

[last holly]

i can translate, but not in this channel lol

[teal grotto]

so, you can have vector spaces where the inner products are just not at all related to the norms?

[wintry steppe]

any inner product induces a norm but not conversely (e.g., sup norm on R^n)

that's the specific relationship

for a norm to come from an inner product on a (real or complex) space you need the parallelogram law to hold. if that holds, then the polarization identities (different ones for R and for C) define inner products which induce the given norm

[teal grotto]

does it even make sense for this problem if the norm doesnt come from the inner product?

hypothesis of the problem statement were just that V and W were inner product spaces, not normed spaces, so i would assume that the norm coming from the inner product would be the one thats being used

but you're argument seems to work in more generality

[wintry steppe]

oh, if they're inner product spaces, then you should assume that norm refers to the one induced by the inner product

and then it's easier (and as i remarked actually does follow immediately from cauchy schwarz, as you've proven)

but yeah i guess i proved it holds more generally for just normed spaces

[teal grotto]

cool. thanks for fleshing that out with me

[wintry steppe]

if the parallelogram law |x+y|^2 + |x-y|^2 = 2(|x|^2 + |y|^2) holds for your norm, then the first (resp. second) formula here defines an inner product if your space is complex (resp. real)

[wintry steppe]

(clarifying this)

it's a fun little fact

and of course any norm coming from an inner product is going to satisfy the parallelogram law, so you can use it as a test to check whether or not a norm comes from an inner product

[teal grotto]

noted. this seems like a neat exercise

[copper fox]

Gracias!

[nocturne jewel]

speaking of norms.. does Parseval Identity hold for all norms or are there ones where it breaks, obviously assuming we have the orthonormal set required

[forest quiver]

Super quick question

A rotation of pi/2 rads counter-clockwise is the same thing as a counterclockwise rotation of 3pi/2 right?

i.e -3pi/2

[hazy forum]

yes

[forest quiver]

In terms of the matrix for my linear transformation

Ok that's what I was thinking too

[twilit anvil]

this is one exercise i never got around to doing... it was never assigned to me but when i saw it in a book, i thought it was really cool

[hollow finch]

i want to make sure im understanding normal matrices. if someone can tell me where im right or wrong that would be helpful.
the definition is that A and its conjugate transpose (adjoint) A* commute if and only if A is normal.
theyre are all unitarily diagonalizable. so theres a matrix U for which its adjoint U* is the inverse and A=UDU* for a diagonal matrix D.

hermitian (symmetric), unitary (orthogonal), and skew-hermitian (skew symmetric) are special cases of these kind of matrices corresponding to

hermitian ($H^*=H$): eigenvalues with imaginary part zero

skew-hermitian ($Q^*=-Q$): eigenvalues with real part zero

unitary ($U^*=U^{-1}$): eigenvalues with magnitude $1$ (so there's some real $\theta$ for which $\lambda=e^{i\theta}$)

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

and the spectral theorem tells us that if $v_1,\ldots,v_n$ are the columns of $U$ (making them an orthnormal basis of the eigenspaces) then

$A=\lambda_1v_1v_1^+\ldots+\lambda_nv_nv_n^$

[stoic python]

nix (@ me for the love of euler)

[dusky epoch]

yes that's correct, everything is as expected

they're coplanar when the det is 0

[wooden oracle]

How do I find the rank of a linear transformation that looks like $A \rightarrow AXA$, where A and X are $n \times n$ matrices over a field F?

[stoic python]

NupurJ

[forest quiver]

Khan is shit

[teal grotto]

https://math.stackexchange.com/q/4212567/803927
this you?

[wooden oracle]

Oh wow, nope

[teal grotto]

lol, well, it depends on the properties of A

[wintry steppe]

is there any proper justification for saying a vector is not the span of two other vectors?

Like I'm pretty sure its not, but i'm not sure how to explain it

Other than, theres no way these can all add up properly to make this

[lavish jewel]

put them as columns of a matrix and show the matrix has column rank = number of vectors, meaning they are lin indep

[teal grotto]

let $X\in\ker(T_A)$. Fix $w\in\text{im}(A)$. then there exists a $v\in F^n$ such that $w=Av$. Since $X\in\ker(T_A)$, then $AXAv=0_{n\times n}v=0_{F^n}$ which implies $XAv=Xw\in\ker(A)$, hence if $X\in\ker(T_A)$, then $X\text{im}(A)\subseteq\ker(A)$. Show that the converse is also true so that you obtain the relation $X\in\ker(T_A)$ iff $X\text{im}(A)\subseteq\ker(A)$. this should be a good start i think

[stoic python]

c squared

[teal grotto]

intuitively, if X is in the kernel of T_A, then X has to map r = rank(A) linearly independent vectors into n - r = null(A) linearly independent vectors, and these matrices X have a specific form, which will tell you the dimension of the kernel of T_A

[pearl plaza]

Hi, I don't understand the final part.

This part to be precise:

I don't see how proving Norm(Ax) > Sigma(K+1) * Norm(x) implies that x belongs to the span of {v1...vK+1}.

[forest quiver]

I messed around with this algebraically for a bit and I got a linear transformation

But I am confused why this is linear

I guess I associate percentages with "exponential" but this is wrong I presume

I guess the entire change over multiple years isn't linear

but individual years are linear

because there are just values being added

Is that right?

[last holly]

the system can be written linearly (the percentages add up to 1). then you can exponentiate the matrix to project for years

but what does it mean to raise an entire matrix to a power?

it's basically operating on itself over and over

so still linear?

simple example

ignore the silliness of the premise

[forest quiver]

The whole thing isn't linear

Like let's say from 2000 to 2003 the populations don't grow/decrease linearly

But each single year

there is linearity

you are just adding percentages (values)

I suppose

[last holly]

exp curve is not linear, sure

[forest quiver]

right

but the increase of each single year is

[last holly]

but applying a linear transformation x times is still linear

[forest quiver]

Yeah because it's just 1 time

[last holly]

like if you rotate something 3 times

[forest quiver]

Apply -> increase/decrease linear
Apply again -> increase/decrease linear again etc...
Overall -> increase/decrease exponential

[last holly]

hm but what is the exponential you're looking at

[forest quiver]

Take city for the example

each year there are less and less people leaving

From 2000 to 2001, 18000 people leave the city

From 2001 to 2002, 16560 people leave the city

[last holly]

yes

[forest quiver]

It's not the same increase/decrease everytime, so it's not linear overlal

But each individual step is linear

[last holly]

but i'm saying, A^x is not always exponential per se

it can have other behaviors

[forest quiver]

I didn't learn A^x

yet

[last holly]

i mean raising A to a power

sometimes it just cycles, or does nothing

[forest quiver]

A matrix?

A^x

[last holly]

yea

[forest quiver]

Yeah not familiar

next chapter I willg et back to you XD

[last holly]

haha

[modern salmon]

what book are you two following

@forest quiver and @last holly

[forest quiver]

#books-old

best books should be in there

Mine is the one by David C Lay

[pearl plaza]

Sorry to chime into the conversation...

But any transformation that can be written as a matrix is linear.

Linear doesn't refer to the shape of the curve, it refers to the set of operations that consist of addition and multiplications only.

The above problem you mentioned is a markov process... The matrix defines the dynamics of the system. So it is indeed a linear system...

[pearl plaza]

Can anyone please help me with this...

[last holly]

Axler and Leon

[lucid glacier]

yea raising a matrix to a power is just multiplying it by itself that many times

and still linear

cuz composition of linear transformations is linear

[last holly]

Another way to think about it, is graphing the the results is always linear

Easy to see with 2x2 matrices

[lucid glacier]

the transformation itself is linear if you fix the exponent I mean, if you're considering A^n as a function of n is then that's not linear

[edgy tree]

How did the equation with the red arrow become the one with the green arrow ?

[lavish jewel]

the inner product distributed over summation

[frail flame]

And then that fraction was taken out of the product, since it is a scalar

[lavish jewel]

indeed

i would say it's kinda sloppy because they only say inner product space, when it seems they need it to be over the reals (they ignored some stuff) or some other field that doesn't require complex conjugation

[edgy tree]

now i got it 👍🏻

[edgy tree]

but how can i do this ?

shouldn't the fraction be common between both vectors in order to take it out of the product ?

like if the fraction is a

[lavish jewel]

let <x2,x1>/<x1,x1> be some generic constant c

[edgy tree]

okay

[frail flame]

For inner products over reals its linear in each of the arguments, like the dot product

[lavish jewel]

then you have <x1, x2 - c x1>

if no complex conjugation is needed, this is linear in the first and second arguments

so you get <x1,x2> - <x1, c x1> = <x1,x2> - c <x1, x1>

then subs c back in

<x1,x2> - <x2,x1>/<x1,x1> * <x1, x1>

and cancel the <x1,x1> in num and denom of the right term

since x_i are a basis, that shouldn't be a problem, since it would mean <x1,x1> is nonzero

[edgy tree]

I do understand the process

but what i can't understand

[edgy tree]

is this part

where we take out c

is this correct ?

taking c from one vector

[lavish jewel]

you will need to go back and study what linearity means

because we already told you twice

[edgy tree]

and multiplying it by the whole inner product ?

[edgy tree]

A great way to end a conversation

[wintry steppe]

[lavish jewel]

linear tteppa

[still lodge]

is it ok to think of columnspace as like a mini span

[wintry steppe]

what

what do you mean by mini span

[still lodge]

that's definitely not good wording

well span is all the vectors that can be obtained by linear combinations of some basis vectors right

and columnspace is all the vectors that can be formed by linear combinations of the columns of some matrix

idk im trying to combine 3b1b and the strang lectures in my head

worth noting that i havent watched the 3b1b vid on colspace

[lavish jewel]

just remove the mini

[limber sierra]

yes, column space is precisely the span of the columns

so a columnspace is one particular span

[gloomy nebula]

T is an nxn matrix. If I use 1 as lambda and set the matrix for characteristic polynomial, is that why the sum of the entries in each column in that new matrix B is 0?

[hard drum]

you initially have a matrix where each column sums to one and then you subtract 1 from each column by subtracting the identity (assuming that's what you mean by 'using 1 as lambda') ig?

[gloomy nebula]

you subtract 1 from each diagonal. What you said yeah

[nocturne jewel]

https://www.youtube.com/watch?v=O8gsWPzxVeQ Problem from Michael Penn if people want to try it

[gloomy nebula]

Does this have anything to do with rank?

[frail flame]

Rank is probably the nicest way of arguing that. Summing all the rows shows they aren't independent, so the rank is smaller than the number of rows.

That depends on properties / definitions of rank as number of non-zero rows in RRE and dimension of row space.

[forest quiver]

Thanks

[still lodge]

poggers

oop sorry for ping, just ty

[last holly]

nice i love michael

[silver heath]

Does anyone know what this notation means for a vector space?

What does C[0, 1] represent???

I understand its a funciton but what do the 0 and 1 mean

wait nvm

didnt read the whole problem

🙂

[dusky epoch]

i think youve cut something off

Let C[0,1] be the vector space of continuous functions on the interval [0,1]

is that what it actually says?

[soft abyss]

maybe easy question, but what is the formula for calculating the dimension of the intersection of k linear subspaces

for two linear subspaces its the elementary formula

but is there a nice general formula for an arbitrary list of subspaces

[verbal hedge]

This is kind of recurrence problem but how to find determinant of tridiagonal matrix?

(nxn)

[lucid glacier]

Feels like the formula would generalize in a way similar to the inclusion-exclusion principle but not 100% sure

[hollow oar]

it does not

[lucid glacier]

Well it's similar enough

It shouldn't be hard to do a few examples and derive the formula by induction

[silver heath]

What does it mean to be a subspace of a set exactly?? Can the subspace be bigger than the original space example: if B was the identity matrix. The Nul B would be the zero subspace. But if A was a zero matrix, nul AB would span R^n???

[lucid glacier]

A subspace of a vector space is just a (nonempty) subset of the vector space that is closed under operations (That is, addition and scalar multiplication). Note that in particular if two vector spaces (Meaning things you already know are vector spaces), and one is a subset of the other, it's definitely a subspace

Also means that if something is not a subset, it can't be a subspace

In this case you know the null space is always a vector space, so try to reason (Or find a counterexample) to how one is contained in the other

[lucid glacier]

Your counterexample seems to work, it's exactly disproving the fact that one is not a subset of the other so in particular not a subspace

Also slight correction but Nul(AB) Would BE R^n, not just span it

[wintry steppe]

I have a question

[stoic python]

jswatj

[wintry steppe]

so it must be that each one is invertible

but is this argument valid? correct?

it's true for finite dimensional spaces, but false for infinite

what makes a linear transformation infinite?

infinite-dimensional spaces, i mean

i think (i had an example in mind to show it's false for infinite-dimensional spaces but realized it doesnt work)

Suppose $R:\mathbb{R}^2\rightarrow\mathbb{R}^2$ same with $T$

[stoic python]

jswatj

[wintry steppe]

would that mean its false?

i'm not sure what you mean

Nvm

thats finite

wait im so confused

What makes it finite/infinite?

[lucid glacier]

Are they necessarily endomorphisms?

[wintry steppe]

are you working only with R^n -> R^n or are you familiar with general vector spaces

I'm actually working with R^3

ok, then it's true

when would it not be true?

and why?

when you're working with linear operators on more general vector spaces which have infinite "dimension"

[wintry steppe]

also, would this even be a valid argument for why it is?

[wintry steppe]

Like R^n?

no

R^n has finite dimension

n

[lucid glacier]

Cuz if not you could take say the embedding S $\mathbb R^2$ into $\mathbb R^3$, and then the map T that takes (x,y,z) to (x,y), then clearly $S\circ T $ is invertible as it's the identity in $\mathbb R^2$, but each transformation is not invertible

[stoic python]

ShiN

[wintry steppe]

lol tex bot is fucked up

[lucid glacier]

It's cuz I used a circumflex outside of the tex environment

[wintry steppe]

yeah that's a really annoying way to get an error lol

[lucid glacier]

If they map from the same space to itself then as tterra says this is true for finite dimensional spaces at least

[wintry steppe]

for infinite dimensional spaces i am thinking of the space of sequences of reals and the left and right shifts

but this isn't going to help them much i think

[lucid glacier]

Since if ToS is bijective, then S is injective and T is surjective, and between equidimensional spaces (At least if the dimension is finite) A linear transformation is injective iff it's surjective, so they are both bijective

[wintry steppe]

I think this is beyond the scope of my course

probably

if you haven't seen the abstract definition of a vector space then i wouldn't worry about it

[lucid glacier]

Tbh^

[wintry steppe]

Yeah

i have not

another proof: if TS = id, then 1 = det(TS) = det(T)det(S), so both det(T) and det(S) are non-zero (i.e., T and S are invertible)

[lucid glacier]

Yea this is pretty much the counterexample I gave, but since it's infinite dimensional then it's still an endomorphism but neither operator is bijective

Like it's analogous in inf. Dimension

[wintry steppe]

[glacial mango]

hi, im stuck on a markov problem. I have completed the entire question except PART D. the answer in the markescheme is:
20(2+(2/5)^n)

I'm not sure how to turn a transition matrix into a general expression. Can anyone teach me this?

[nocturne jewel]

Ok well $s_1=Ts_0$ right?

[stoic python]

Mosh

[nocturne jewel]

(the state vector after 1 week is the initial state after the transition has occured)

[glacial mango]

Yes

[glacial mango]

And Ok well $s_1=T^2s_0$ right?

[stoic python]

Nick Jojo

[nocturne jewel]

$s_2=Ts_1=T^2s_0$

[glacial mango]

Meant to write s2

[stoic python]

Mosh

[glacial mango]

My bad

[nocturne jewel]

so in general $s_n=T^ns_0$

[stoic python]

Mosh

[glacial mango]

Write, that’s the formula I derived. But the Mark scheme says 20(2+(2/5)^n)

[nocturne jewel]

so if you can find a nice expression for T^n, then you can find a function of n for the # of painting students

[glacial mango]

So how can I convert Sn into that form

That’s what I’m stuck on

[nocturne jewel]

you find T^n in general, then do the multiplication

then take the entry for painting students, that'll be your function of n

[silver heath]

I think this is false.. but not sure

doesnt it have to be an indexed set?

[nocturne jewel]

is A invertible?

[silver heath]

uh

[silver heath]

not sure... maybe? maybe not

[nocturne jewel]

well you can tell me from the info

[silver heath]

wait

no

because of the invertible matrix thoerlrme

thingine

missing a pivot column

[nocturne jewel]

missing a pivot means what about invertibility?

[silver heath]

its not invertible

[nocturne jewel]

right, and part of FTIM is independence and span of the rows and columns

[glacial mango]

How do you have a general expression for T^n?

[nocturne jewel]

if A is invertible, then the columns and rows span and are indep.

[nocturne jewel]

each entry would be a function of n

[silver heath]

yes i undertsand that

but why does it say V_k-1

wouldnt it be till v_n?

[nocturne jewel]

oh

[silver heath]

so would it be false then?

im pretty sure theres a theorem that goes exactly liek this but it requires A to be an indexed set

for whatever reason

[nocturne jewel]

I think it's still false cause they cant be independent with all n

[silver heath]

??

[nocturne jewel]

it's asking if {v1,...,vk} are an independent set

[silver heath]

no?

its just saying that v_k is a linear combination of v_1 -> v_k-1

[nocturne jewel]

yes, that's equivalent to independence

[silver heath]

mhm

Quick question

does a vector space have 1 basis?

or can it have mutliple

for exampe

R^2

[1,0] and [0,1]

is a basis

but so is [2,1] [0,3]

?

[glacial mango]

A vector space can have an infinite number of bases.

[nocturne jewel]

any set of vectors which span the space and is independent is a basis of the space

so yes, bases arent unique

[silver heath]

ok thanks!

[wintry steppe]

i thought it was

since

t_1(v-u)+t_2(u) = t_1v-t_1u+t_2u = t_1v+(-t_1+t_2)u

so its a linear combination of u and v

am i using the wrong logic or?

[teal grotto]

yes that’s correct. you show two way containment. if x is in span{u,v}, then
x = au + bv = au + bv - bu + bu = (a+b)u + b(v-u), so x is in span{v-u,u}.

you have shown that span{v-u,u} is a subset of span{v,u}.

i have shown that span{u,v} is a subset of span{v-u,u}

so the two are equal

[wintry steppe]

okay

what about this other one

span{u,v,2u-v}=span{u,v}

I said yes again since

au+bv+c(2u-v)=au+bv+2cu-cv=(a+2c)u+(b-c)v

conversely

uhhh

wait this one doesn't work

[teal grotto]

yes it does

[wintry steppe]

Oh

how did you think of the second step?

[teal grotto]

ah. i’m trying to think of a good way to word this sry

[wintry steppe]

its ok

[teal grotto]

this is fine. make sure to show the other direction of containment to complete the proof.

there is a more general statement that can be made for these types of problems. having trouble wording it rn, but it would treat this entire class of problems

[soft abyss]

Yeah you can get a very long and messy formula. I was asking if there is a nicer version

Anyway I found a paper with a nice clean formula

👍

https://projecteuclid.org/journals/missouri-journal-of-mathematical-sciences/volume-14/issue-2/The-Dimension-of-Intersection-of-k-Subspaces/10.35834/2002/1402092.full