linear-algebra

[humble oak]

{[1 2], [3 5], [3 2]}

[gray dust]

actually i don't wanna use that, and before examples, do you get everything this says?

every vector in the space is a linear combo of the vectors in the given set

[humble oak]

in my mind

i read it as every vector in the set can be made in the vector space

[gray dust]

wdym 'made in the vector space'

[humble oak]

every vector in the set can be made from the vectors in the vector space using linear combinations

[gray dust]

no you flipped it

every vector in the vector space can be obtained by a linear combo of the vectors in the given set

[humble oak]

oh

does it matter how many vectors we can use for the linear combination from the given set?

nvm i just thought about it

[bold thicket]

Why is it called linear algebra?

And oh shit we got to be good with vectors

[gray dust]

here are some easy exercises. show span{(1,0),(0,1)}=R^2, span{(1,0),(0,1),(0,0)}=R^2, span{(7,0)}=x axis

[humble oak]

okay, i'll see what i can do o:

the vectors are already made into RREF, so is it standard that the number of pivots = dimension of span?

from
[1 0
0 1]
can we conclude that it spans R^2?

[gray dust]

no real need for rref. use the defn of span directly

take an arbitrary vector in R^2; it has a form (x,y) where x,y are in R. write (x,y) as a linear combo of (1,0) & (0,1)

[humble oak]

so could i just say something like, let x, y in R.
x[1 0] + y[0 1] = [x y]
therefore the set spans R^2

[gray dust]

we start with 'take an arbitrary vector in the vector space. depending on the vector space this vector will have some form ...' THEN show this vector is a linear combo of the vectors in the given set

note what we're really doing is showing the vector space is a subset of the span of the set (subset going the other way is given almost for free so we don't worry about it)

[humble oak]

okay, i think i get it

the second question would be the same as the first question because the zero vector doesn't contribute to the dimension of the span right?

we can always make that [x y] vector through linear combination

[gray dust]

there's a certain sense in which (0,0) doesn't 'contribute more' to the span. but you should go through it with the same process. directly use the definition of span

[humble oak]

alrighty okay, thank you very very much i will continue to work 😄

[gray dust]

you're welcome

[topaz linden]

So I had a problem that I'm trying to do and it essentially boils down to proving the following:

1. Given $n$ unknowns and $n-1$ homogeneous equations, is the solution set exactly equal to a line?

[stoic python]

Have a Banana, Bitch

[topaz linden]

2. Say W is an $n-1$ dimensional subspace of $k^n$. Given a basis $S=w_1,\cdots, w_{n-1}$ define the matrix $M(S)$ as the matrix having $w_1$ for the first row (in standard basis representation of $w_1$), $w_2$ as second,..., $w_{n-1}$ as the $n-1^{th}$.

[stoic python]

Have a Banana, Bitch

[topaz linden]

Is it true that if we look at the solution set of M(S)*x=0 then it will be the same as the solution set of M(S')*x=0 for some other basis S' of W?

[pure tangle]

If i solve and find x=2, y=1, and z=1/2 would the vector form just be [x,y,z] = [0,0,2] + t*[2,1,1/2]?

[covert trellis]

How do you determine if a matrix is consistent or inconsistent without RREF?

[acoustic path]

lmao can yall not ask questions while another one is posted

[pure tangle]

<@&286206848099549185>

[slow scroll]

@pure tangle wdym by vector form? I would interpret "the vector form of a solution" as (2,1,1/2) personally.

[tame mural]

@covert trellis one way is by the number of rows and columns

[covert trellis]

How do you do that?

[tame mural]

a consistent system has equal rows and columns

but I'm not sure how you can get all the information you'd get from RREF without doing it

because you still need to know more

[hybrid charm]

is there a difference between a matrix in row reduced echlon form and just echlon form?

[nocturne jewel]

Echelon form isnt reduced

[gray dust]

@hybrid charm rref requires being in ref. it further requires that each leading entry must be the only nonzero entry in its own column, and each leading entry must be 1. ref doesn't require these.

[still elbow]

is the powerset of a set a vector space over F2?

under exclusive or

[blissful vault]

how do i prove that this is closed under scalar multiplication?

i already determined that it is not closed under addition

[limber sierra]

@still elbow i dont really follow how youre trying to construct this; would your addition be unions? whats your multiplication?

[still elbow]

the operation is symmetric difference

[limber sierra]

ah

[still elbow]

or exclusive or

[limber sierra]

then whats scalar multiplication

from F_2

like 0S = {} and 1S = S for all sets?

[still elbow]

yeah im not exactly sure about that part

[limber sierra]

if so i think that actually is a vector space

since distributivity holds

(Powerset(S), symmetricdiff) is definitely an abelian group

[still elbow]

Yeah, definitely

Powerset(S), union and powerset(s), intersection aren't abelians though

right?

[limber sierra]

union and intersection arent invertible

so theyre not groups

they are commutative however

but yeah, any abelian group forms a vector space over F_2

by letting 0 * v = the zero element of the group

and 1 * v = v

[still elbow]

Ty so much

[limber sierra]

its a bit of a weird construction though

[still elbow]

Definitely

[still elbow]

wait a bit

what about lets say

F8

F_8

[limber sierra]

im not sure how youd define multiplication between the field of 8 elements and power sets of a set

[still elbow]

nonno

[limber sierra]

there might be a way to do it sensibly but i doubt it

[still elbow]

i mean like you said that any abelian over F_2 forms a v space

but take F_8 over F_2 (usual operations)

[limber sierra]

oh like

the additive group of F_8 over F_2?

yes thats a vector space

[still elbow]

Uhmm

what about 5(1+1)

[limber sierra]

what about it

[still elbow]

Like from one side 1+1=0 so 5*(1+1)=0

but then

[limber sierra]

whats 5

[still elbow]

5(1+1)=5+5=2

5 is an element in F_8

[limber sierra]

F_8 is not Z/8Z

Z/8Z is not a field

[still elbow]

oh yeah, right , i meant Z/8z

but its an abelian though

[limber sierra]

Z/8Z is an abelian group yes

[still elbow]

But is Z/8Z a vector space over F2?

[limber sierra]

yes

i dont see whats wrong with having 5(1+1) = 2

[still elbow]

Isn't 1+1=0 in Z/2Z?

[limber sierra]

okay maybe this confusion would be cleared up by clarifying whether each element is from Z/2Z or Z/8Z

since the "1" in Z/2Z is different from the "1" in Z/8Z

[still elbow]

I see that

[limber sierra]

oh i think i see what youre taking objection to

(1+1)5 where 1 is from the scalar field and 5 is a vector

hm

actually yeah youre right

my bad

i forgot that you need nilpotence

let me fix my statement:

every abelian group in which x + x = 0 for all x forms a vector space over F_2

@still elbow sorry, hopefully that clears this up

in any case, that does still work for your example

of (powerset, symmetricdifference) over F_2

since the symmetric difference of any set with itself is {}

[still elbow]

Now this totally makes sense, thank you!

[tame mural]

How annoying that symmetric difference and direct sum of spaces uses the same symbol

[tame mural]

Yes, some people use ⊕ for symmetric difference / XOR

[digital bough]

#❓how-to-get-help

[viscid kernel]

@pure tangle as you see youve got 3 equations in the form of z = f(x,y) those are plains in 3d. The intersection of those 3 plains is the solutionand since there are all constants at the right hand side of the equations they dont go through the origin which means that at least 2 plains cant lay on each other, if that was the the case there would be no solutions. So the functions are not the same. Which means that it has one solution because each is a different function. To check if it really doesnt have one solution or non, you could check if the matrix A has a determinant of 0 and if the you consider the right hand side as a columnvector. If the columnvector is linear combination or of the columnvectors in A and the deterinant is 0 the sollytion would be a vector + nullspace ( which is an affine subspace ) which means that there are infinitely many solutions. But luckily in this case the determinant is not 0 which means theres only one solution.

[charred star]

go to #prealg-and-algebra, linear algebra is a whole other story

[dark valley]

guys whats the requeriments to a A U B subspaces to be a vectorial subspace?

[pallid rampart]

if I remember correctly the union of two subspace is a subspace iff A is a subset of B or B is a subset of A

[dark valley]

u sure? not trying to be mean but rly didnt heard about that rule

[pallid rampart]

it's not really a rule

it's a fairly standard exercise

Ok here

Let's say $A\cup B$ is a subspace and $A\not\subseteq B$ and $B\not\subseteq A$, then we can choose $x\in A-B$ and $y\in B-A$. Now since $A\cup B$ is a vector space $x+y\in A\cup B$. If $x+y\in A$, say $x+y=a\in A$, then $y=a-x\in A$ contrary to our choice of $y$. If $x+y\in B$, say $x+y=b$, then $x=b-y\in B$ contrary to our choice of $x$

[stoic python]

Whoever

[pallid rampart]

so we arrive at a contradiction either way

This shows that the union of two subspaces is a subspace only if one is a subset of the other

The other direction is obvious

[dark valley]

okok

ty

but if it is a intersetion is it A is a subset of B AND B is a subset of A

[pallid rampart]

well the intersection of two subspaces is always a subspace

[dark valley]

ah yes

right

[wintry steppe]

#help-0 message

[dire thunder]

!r3

3. Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[dark valley]

Using the formula: a + bx + cx^2 + dx^3, for (a,b,c,d)

the 5 + x means (5,x ...) or (0,5+x ...)?

[pallid rampart]

It means (5,1,0,0)

Since $a+bx+cx^2+dx^3$ is $(a,b,c,d)$, then $5+x=5+1\cdot x+0\cdot x^2+0\cdot x^3$ is $(5,1,0,0)$

[stoic python]

Whoever

[pallid rampart]

@dark valley

[dark valley]

k, ty again

[dawn remnant]

so you have a linear operator and a set of vectors such that the image of that set is independent. You need to prove that that means that the original set is also independent.

Prove by contradiction. Write down what it means for the original set to have a dependency, and prove that then the transformed set must also have one.

[acoustic path]

ohh

[pallid rampart]

Also if you want to be direct you can just use proof by contrapositive

[acoustic path]

if the set is linearly dependent its gotta do that one of these v1.. vm can be written as a linear combination of the others, however that cant be because Tv1.. Tvm is linearly independent?

[acoustic path]

how would i go about doing that?

[pallid rampart]

So the contrapositive statement of the problem is if $v_1,\dots,v_m$ is linearly dependent then $Tv_1,\dots,Tv_m$ is linearly dependent, then you use exactly what ConfusedReptile said

[stoic python]

Whoever

[acoustic path]

ahh

ok yall thx again

[blissful vault]

no one asnwered :(

i need help with (ii)

how to prove it's closed uner scalar mult

i understand the concept

like if it's a scalar

it won't turn a real polynomial imaginary

but how to prove it

[dawn remnant]

All the roots of $c f$ are the roots of $f$ if c is nonzero. If it's zero, that's a zero polynomial, which is still in the set.

[stoic python]

ConfusedReptile

[dawn remnant]

that's pretty much it

[quartz compass]

what have you tried?

[acoustic path]

well ive tried thinking of a function that gets reduced to R1 but we are still able to factor out a scalar yet additivity doesnt stand.. i cant rlly think of any

i wanted to say the derivative would be good but additivity would stand

[quartz compass]

here's the kind of trick I have in mind to make one, use the fact that x/y = (ax)/(ay)

[acoustic path]

a=1?

[quartz compass]

a is arbitrary

[acoustic path]

oh my b

its the function we gotta give an example of

[quartz compass]

yeah and since $(x,y) \in \bR^2$ what I showed is like plugging in for $(ax,ay)$ but ending up with the same thing, see what I mean?

[stoic python]

Merosity

[acoustic path]

ohh

so a function that kills off the y-dimension is enough?

because we're supposed to end up in R1

[quartz compass]

not sure what you mean by that

phi(x,y) = x won't work because it's linear for instance

a projection is linear

[acoustic path]

well with the example you used then how is addivity not defined?

[quartz compass]

I didn't give an example, I gave a hint

as a way to help you try to construct a function

$f(x,y) = \frac{x}{y}$ satisfies $f(ax,ay) = f(x,y)$

[stoic python]

Merosity

[quartz compass]

that's not quite what you want

you want f(ax,ay) = af(x,y)

[acoustic path]

ok i got it

i think sqrt(x^2+y^2) works

[quartz compass]

that looks good to me

I think what I had in mind was a bit weirder

phi(x,y) = y sin(x/y) when y !=0 and =0 when y=0

[acoustic path]

wait youd get sin(x/0) tho

[quartz compass]

yeah that's what the second part is

that's when y !=0 , when y=0 it's just 0

it's piecewise

[acoustic path]

ah

and where does the "a" scalar work in that function?

[quartz compass]

just plug it in and work it out, show me where you get stuck

your answer works just fine though so don't sweat it

[verbal chasm]

x^2/y seems like it should work too (?)

[quartz compass]

yeah any function where you do yf(x/y) then redefine at y=0

will work

[blissful shell]

sorry for interrupting , i have a simple question. Let A nxn matrix, if I find det(A) using the Laplace expansion (cofactor expansion? idk how you call it) for the 1st row it should be the same for the 2nd,3rd...i-th row too ? And what about the j-th column ?

[nocturne jewel]

Laplace/Cofactor expansion works for any row / any column

[blissful shell]

and should give the same result for all of them ?

[nocturne jewel]

yep

[blissful shell]

thanks

[wintry steppe]

What are some interesting Linear Algebra project ideas?

Preferably either in the areas of pure mathematics or economics.

[bright vapor]

depends on your level but I would say SVD decomposition for image compression is an personal favorite

there's a book called "Linear Algebra in Action" which gives a lot of applications of linear algebra

discrete fourier transform also

[chrome current]

Hey! I'm really struggling with this question, how am i supposed to determine the vector AT with the base AB, AC?

[misty storm]

"Classify the point conglomerate in R² whose sum of distances tween A(3,0) and B(-3,0) equals 7"

[misty storm]

oh yeah

forgot to explain this further

how am I supposed to calculate the distances between a conglomerate of points and a single point?

do I calculate from an average or the closest point?

and by "classify" the exercise asks if its a circumference/hiperbole/whathaveyou

[quartz compass]

this doesn't work because a could be negative

[acoustic path]

but we would have a^2 on the inside tho

a*sqrt(x^2+y^2)=sqrt(a^2 x^2 + a^2 y^2)

right?

[quartz compass]

well do it with a=-1

phi(-v)=-phi(v)

left side is always positive

so right side must be too, but -phi(v) is negative

$-\sqrt{x^2+y^2} \ne \sqrt{(-1)^2 x^2 + (-1)^2 y^2}$

[stoic python]

Merosity

[acoustic path]

yea

i get what u mean

i got no fckn idea then

[quartz compass]

well you still have the solution I offered earlier at least

[acoustic path]

i didnt understand it rlly

can you just divide x/y like that?

[quartz compass]

pick any function f(z)

that's defined for all z in R

then you can define $$\phi(x,y) = \begin{cases}y f(x/y) & y \ne 0 \ 0 & y = 0 \end{cases}$$

[stoic python]

Merosity

[quartz compass]

I picked f(z) = sin(z) cause I liked it

[acoustic path]

how is scalar multiplication defined there tho? like how is y*sin(a(x/y))=a ysin(x/y)

[quartz compass]

a(x,y) = (ax,ay)

[acoustic path]

ohhhh

so it cancels out to be 1?

[quartz compass]

yeah

[acoustic path]

oh thats smart

[quartz compass]

cool 👍

[acoustic path]

thx again dude

esp coming back to tell me im wrong

appreciated

[quartz compass]

yup you're welcome lol just randomly about to throw some paper away and just flashed in my mind about your problem earlier lmao

[acoustic path]

lmao

[quartz compass]

sort of part of a larger topic called homogeneous equations

[blissful vault]

thanks for the reply earlier, i have another problem

how do i start this question? a hint would be enough, as i still want to do it myself

[quartz compass]

take some elements from W_c and try to do stuff they should be able to do in a subspace with them

[blissful vault]

so i just invent some values and play with them?

[quartz compass]

well I would write f(x) and g(x) and say they're both in W_c

now like, add them, do you get another element in W_c?

[blissful vault]

ok, let me try

[misty storm]

How do I know what kind of conical shape it is based only on the info of the distances between the shape and certain points?

Circumference/Ellipse/Hyperbole/Parable/degenerate conic

[tame mural]

Has anyone encountered notation for the dual space like $A^{#}$?

[stoic python]

meow

[misty storm]

"Classify the shape in R² whose sum of distances to the point (3,0) and (-3,0) equals 7

I'm pretty sure that it's an ellipse

but how do I formally prove it?

I know that the sum of distances to 2 focal points in an ellipse equals 2a

but who's to say that this isn't, say, an hyperbole?

[wintry steppe]

in a matrix, does changing the column order change the matrix?

like if I was told to make a matrix out of vectors a,b,c

and I made a matrix [a,b,c]

is it different from doing [b,c,a]

[hollow finch]

besides the obvious fact that the matrix is literally different, yes that has a noticeable effect

[wintry steppe]

how?

[hollow finch]

so say A=[a b c] and B=[b c a]
the columns of A tell us that e1 gets mapped to a, e2 to b, and e3 to c.
the columns of B tell us that e1 gets mapped to b, e2 to c, and e3 to a. which is completely different

[wintry steppe]

ok yeah

[hollow finch]

the way things are getting mapped is swapped

[wintry steppe]

but would it make a difference for other stuff

[hollow finch]

...like?

[wintry steppe]

inverse, independence, null space, Ax=0 homogenous stuff etc.........

[tame mural]

Swapping columns is a permutation

So anything you'd expected to be preserved under permutation

[hollow finch]

well yeah all of those things are different

the inverse wont be very different but still different

[tame mural]

Independence won't change with permutation

[hollow finch]

^

[tame mural]

thus rank and nullity won't change

[hollow finch]

the determinant will only change by a sign

if at all. its possible it will be the same

[wintry steppe]

i see ok

thx

[stoic python]

meow

[pallid rampart]

Probably the field in which the vector spaces V and W are over

[safe jay]

hey guys I don't really get what is the purpose of non-trivial solution ?Is it like an infinity solution?
Because in high school, I've study that a system of solution that have n variable and m solution, such that n > m, then there will have an infinite solution

[limber sierra]

"non trivial" just means "not equal to 0"

if you take a homogenous system, say Av = 0

its obvious that the zero vector always solves this

thats not a very interesting observation

so one often asks for nontrivial solutions

we already know the zero vector solves it, but does it have any other solutions?

[pure tangle]

for this matrix why can't the bottom 2 rows (y and z) cancel, become free variables and end up with x = 1-s+t, y = s, z = t

is it because if I use combinations to get one row to zero the other wont be able to make the same operation because it's now a row of zeros?

[safe jay]

I think each operation can be do once for each row
ex : if u subtract second row from third row you will get
(1 1 -1 1)
(0 -3 5 2)
(0 0 0 0)

[pure tangle]

thanks for the response. thats what my convulted last sentance was getting at. Thanks so much

[nocturne jewel]

Need a hand with this, not entirely sure how to start

[hoary osprey]

try writing out a basis for kerT, expanding it to a basis of KerST and then finding a basis for KerS from that and comparing them the dimensions

[nocturne jewel]

How do you write a basis for a general subspace?

just let {v1,v2,...vn} be a basis?

and not quite sure i understand what you mean by expanding a basis

[hoary osprey]

yea just write it out

[hoary osprey]

kerS is a subspace of kerTS so u can expand its basis to a basis for KerTS

[nocturne jewel]

ok you lost me lol

kerS is the set of elements in U which map to 0, how is that a subspace of the elements of kerTS?

is T(S(u)) always 0 if S(u) = 0?

[hoary osprey]

well yea

T and S are linear transformations

[nocturne jewel]

Ok right

[nocturne jewel]

@hoary osprey would you be able to give a hint for the image inequality?

[reef sleet]

For this, I would solve for all of the values of a_n with the coordinates (1, 0), (2, 0) etc right?

[wintry steppe]

yes. since this is the linear algebra channel, i'm assuming you're supposed to do this using lagrange interpolation

this is not linear algebra and is more suited for #prealg-and-algebra or #precalculus

[reef sleet]

@wintry steppe I don't know what Lagrange interpolation is 😅 this is a question about curve fitting from my LA textbook

Solving for a_n with Gaussian elimination

[frosty vapor]

you could gauss jordan but that's pain

[wintry steppe]

that's another way to do it lol

[reef sleet]

yeah I hate g-j lol

[frosty vapor]

assuming tterra's method is nicer

[wintry steppe]

lagrange interpolation is usually very nice

let me grab a screenshot

[reef sleet]

What is it? Is there any way you could explain it to someone new to LA or is it too complicated?

I'm really enjoying this subject so far, feels like doing a puzzle 😛

[wintry steppe]

it's a way of finding the unique n-th degree polynomial taking on specified values at n + 1 distinct points

all you need to understand it is to know what a basis is

[reef sleet]

Is n here 3, then, because the amount of given points is 4?

[wintry steppe]

yeah

well

[reef sleet]

I don't know what a basis is 😛

[wintry steppe]

the wikipedia page is good imo https://en.wikipedia.org/wiki/Lagrange_polynomial

it has a worked example too

two actually!

meh you don't even need to know what a basis is lol

[reef sleet]

So to perform Lagrange interpolation, none of the x coordinates can be the same?

[wintry steppe]

well, you can't have a polynomial taking on two different values at one point

so yeah

[reef sleet]

oh yeah

LOL

What's the big pi-esque symbol mean ?

[wintry steppe]

product

[limber sierra]

product

capital sigma stands for sum

capital pi stands for product

[reef sleet]

Trying to wrap my head around this

[limber sierra]

side tangent: informally speaking, a basis is sort of the "least possible information you need" to entirely determine what linear functions to do a space

for example, to determine what a linear transformation in R^2 (the x-y plane) does, you only need to look at how it transforms the x axis and y axis

so the vectors (1 0) and (0 1)

so {(1 0), (0 1)} is a basis for R^2 [the "standard basis", and the formal interpretation of cartesian coordinates]

but it has many other bases; for example, {(4 2), (-1 -1)} would also be a basis

there's of course a more formal definition than that, but this property is why we "care" about bases

there are some cool facts about bases, such as the fact that every basis of a given space has the same size

which we call the space's "dimension"

R^2 is called 2-dimensional (or 2D) because all its bases have 2 vectors

[reef sleet]

Do I learn this later in the course too?

[limber sierra]

you should, yes

theyre one of the more important parts of linear algebra

in fact, a lot of mathematics in general can be boiled down to "in order to study [x transformation], we should look at what it does to [some special "part" of the structure]"

e.g. to determine group homomorphisms it suffices to look at a group's generators [assuming you know the map is a homomorphism in the first place]

obviously that's beyond the scope of linear algebra

but the point is that it's a common theme

[reef sleet]

and my brain 😝 what is "structure"?

Like R^2?

[limber sierra]

R^2 is one structure - in fact, it's a specific type of structure, called a "vector space"

a "structure" is essentially something we can do mathematics on

slightly more formally, it's a set together with some operations

in the case of vector spaces, your set consists of vectors

[reef sleet]

ah yeah, linear algebra is the study of vector spaces right?

[limber sierra]

and your operations are being able to add vectors

and multiply vectors by scalars

right

[reef sleet]

how about products? can't you do those with vectors?

oh isn't thjat just addition?

[limber sierra]

well you cant multiply vectors "naively" but you can define cross and dot products

when you do so, youre adding additional "structure" because youre creating a new operation

[reef sleet]

what are other types of structures?

besides like R^n

[limber sierra]

well other types of vector spaces specifically might simply be where you take your elements from a different base field

so rather than working in R^n, you work in Q^n

the space of vectors consistng of rational numbers

(this is what computers do in practice)

or in C^n, the space of vectors consisting of complex numbers

you can get more... interesting than this though

are you familiar with modular arithmetic?

[reef sleet]

And Q^n has different properties than R^n

uhhh I know that 7 mod 3 is 1 😝 that's about it

(right? LOL)

[limber sierra]

yes

[nocturne jewel]

Lagrange Polynomials was a pain to understand (at least the formula cause my prof had it not in big Pi notation at first)

[limber sierra]

you just need to have a basic notion of what "mod" means

essentially if you imagine that we take the integers

but then take them "modulo" the same number

we get a new structure, which we can then make vector spaces out of

let me be a bit more concrete

and say we're considering the field Z/3Z, the integers modulo 3

(the notation might seem weird, theres good reasons for it but you dont need to worry about it)

this structure has 3 elements

{0, 1, 2}

and addition and multiplication behave kinda like you'd expect

except if we "overflow", for example 2 + 2

we "loop back around"

so 2 + 2 = 4 = 1

since 4 = 1 mod 3

similarly, 2 + 1 = 3 = 0

besides that its not that weird of an object though

and then we can consider (Z/nZ)^k as a vector space

where the vectors consist of k entries from Z/nZ

[in my example, from Z/3Z]

this vector space, unlike examples such as R^n or C^n

is finite

there are only finitely many vectors in it

[frosty vapor]

ooh.. finite vector space...

[limber sierra]

this gives it some weird behaviours

[reef sleet]

like the overflow thing?

[limber sierra]

for example, if youve worked with finding solutions to linear systems of equations in the past

[frosty vapor]

i recently learned about the Z/nZ group but never considered it as a vector space that is very cool

[limber sierra]

you might know theres situations where you'll have infinitely many solutions to a system

[reef sleet]

Yes

[limber sierra]

but of course, that only makes sense if your vector space has infinitely many vectors!

in the case of finite vector spaces, we just end up with finitely many solutions

but with more than 1

[reef sleet]

so a vector isn't just a line in space? 😝

[limber sierra]

vectors in R^n are lines in space!

and this is useful intuition to have

but the notion of "vector" is more general than that

not all courses cover this in all its nuances

but its a very handy construction

another example of a vector space would be where your vectors are the real numbers R, but your scalars are the rational numbers Q

in other words, you can add real numbers together, but you can only multiply real numbers by rational numbers

this might seem like a small restriction, but it actually makes our space behave very... chaotically

i was talking about a "basis" earlier, but it turns out that the basis of R with scalars Q isn't very nice

in fact:
(1) it has infinitely many entries
(2) there's no way to make an algorithm that describes all those entries

we call R over Q "infinite dimensional" as a result

[reef sleet]

R over Q?

like the vector space created by R and Q?

[limber sierra]

the vector space created by letting your vectors be real numbers and your scalars be rational numbers

obviously "normally" when we work with R

we take our scalars to also be from R

in which case you have no issues

and get a nice, simple, one-dimensional space

the "real numbers"

but if you "restrict" multiplication so that you cant multiply real numbers by real numbers, you get something much weirder

[reef sleet]

So I can do 6pi in R, but in R over Q I can't because pi is irrational?

wait

[limber sierra]

no, you can do 6pi since 6 is a scalar and pi is a vector

but you couldnt do say

pi * pi

anyway, this is a massive side tangent from the original question

R over Q specifically... isnt a very useful structure to work with

honestly

its mostly discussed as a way to sort of reinforce

"we take a lot of the 'niceness' of R^n for granted"

"if we remove something simple, we get something much weirder and harder to describe"

i think some irrationality arguments look at R over Q, like the proof that sin(x) is irrational whenever x is rational and not equal to 0

[taking sine to be from radians]

but in general its certainly not as useful as R^n or C^n

which are, like, the foundations of modern physics + computing

[reef sleet]

What courses do I have to take to learn all of this? It's so interesting but I feel like I'm just limited to ever learning it because I've ignored math most my life lol, maintaining formality and defining everything is so difficult for me :( feels like if I tried to learn all of this I'd fail anyways

[limber sierra]

well sometimes a linear algebra course will cover it

some do, some dont

failing that, an abstract algebra course or two

(which of course will require proficiency with proofs as well)

[reef sleet]

what do you learn in abstract algebra? all I know are groups (group theory), rings, and fields rofl

and idek what those mean

something to do with defining addition and multiplication

[limber sierra]

a first abstract algebra course will typically introduce the theory of finite groups

and maybe a bit of ring/field theory

[reef sleet]

is a finite vector space a type of finite group?

[limber sierra]

well kinda

if you take a vector space and only look at, say, vector addition

you get a group

(specifically an abelian group)

or you can go the other way, "extending" abelian groups into vector spaces by picking a field and defining multiplication

(the latter technique is a very common proof technique in group theory, actually, since vector spaces/linear algebra is so nice and well-understood)

[reef sleet]

but because vector multiplication isn't as "simple" (?? dunno the word) you can't call it a group with multiplication?

[limber sierra]

well formally groups only have one operation

[reef sleet]

oh

[limber sierra]

they have a set and a single operation on that set

if you start adding more operations, you get different structures

(most immediately rings, but eventually vector spaces if you pair it with a scalar field)

[reef sleet]

and this is why it's called abstract? because you can introduce weird things like only being able to multiply by rational scalars?

that was a vector space example but I just mean it like

defining things in ways that "normal" (!??!?!?!?!) math doesn't have

I don't even know how to say what I'm trying to say

LOL

I'm sorry

[limber sierra]

no i getcha

and yeah its called "abstract" since the constructions it looks at usually arent very... you know, concrete

like theyre often difficult or impossible to visualize

and the connections to "practical applications", if there are any, are usually a bit more contrived

[reef sleet]

oh I don't care about practicality, this is so interesting in its own right

[limber sierra]

e.g. its not immediately obvious how elliptic curve groups, for example, could possibly relate to encryption and decryption

since theyre just groups of points on a funky curve with weird addition

but elliptic curve cryptography is a thing

[reef sleet]

can you give an example of weird addition?

[limber sierra]

well the modular arithmetic structure i defined earlier is one example

[nocturne jewel]

One of my vector space questions from a problem set had the min function as addition

(also when you guys are done lmk cause i have a question lol)

[reef sleet]

and that structure was a group with elements {0, 1, 2}?

2 + 5 would be 1?

and the only defined operation is addition?

[limber sierra]

right

[reef sleet]

that's awesome

[limber sierra]

or you could give it more elements if you want

{0, 1, 2, 3, 4, 5, 6, 7} for example

makes total sense

and works the same way

except you "overflow" at 8 instead of 3

[reef sleet]

and if you extend the operations to addition, what does the group turn into?

[limber sierra]

thought experiment: what if we take {0, 1, 2, 3...} and have this set include all positive integers?

||you just get the standard natural numbersN with regular old addition||

not sure what you mean by that feather

[reef sleet]

uhh

wait

I said addition

LMAO

I meant multiplication

if you add multiplication how does that transform the structure?

[limber sierra]

oh, then you get the ring Z/nZ

[reef sleet]

or how we refer to it?

[limber sierra]

its a very nicely-behaved ring all things considered

[reef sleet]

and in that ring, 2*4 is defined to be 1 as well?

[limber sierra]

uh

in Z/3Z you mean?

2 * 4 = 8 = 2 modulo 3

so 2*4=2

or alternatively, 4 = 1, so 2*4 = 2*1 = 2

both of these get you the same result

[reef sleet]

uhh I meant in the set {0, 1, 2, 3, 4, 5, 6, 7}

[limber sierra]

ah

then 2 * 4 = 8 = 0

not 1

[reef sleet]

OH

because 8 mod 8 is 0

and that ring would be Z/8Z>

?*

[limber sierra]

yes

in fact, theres a stronger claim: a * b never equals 1 in this group of integers mod 8, unless a and b are both 1

[reef sleet]

hmm

[limber sierra]

er wait

i stated that wrong

LMAO

since ofc 3 * 3 = 1

[reef sleet]

2 * 5 = 10; 10 mod 8 = 2?

[limber sierra]

but there ARE elements a such that you cant multiply them by anything to get 1

is what i was trying to say

for example, no matter what you multiply 2 by

2 * b will never equal 1

in Z/8Z

[reef sleet]

because 9 is odd

[limber sierra]

this is a fairly important fact since it means Z/8Z is not a "field"

[reef sleet]

and our set is only composed of integers

is that right?

[limber sierra]

i.e. it doesnt always have multiplicative inverses

meaning division doesnt make sense in Z/8Z

i mean you can check this fact manually

[reef sleet]

that's the abstract part? it's hard to imagine not being able to divide

[limber sierra]

2 * 0 = 0
2 * 1 = 2
2 * 2 = 4
2 * 3 = 6
2 * 4 = 8 = 0
2 * 5 = 10 = 2
2 * 6 = 12 = 4
2 * 7 = 14 = 6

[reef sleet]

wow

LOL

[limber sierra]

and as you might be able to expect

this pattern repeats

and yeah, as you correctly observed

its because anything congruent to 1 modulo 8

must be odd

1, 9, 17, 25

so on

all must be odd

in fact, theres an even stronger statement:

Z/nZ is only a field (i.e. division only makes sense) if n is prime

[reef sleet]

why?

what does it mean for division to make sense?

[limber sierra]

well when we talk about "division"

[reef sleet]

thanks for spending the time to explain, by the way

[limber sierra]

this means being able to "undo" multiplication

right?

[reef sleet]

yes

[limber sierra]

for example, in the real numbers, to divide by 5

you multiply by 1/5

[reef sleet]

like how multiplcation can be thought of as repeated addition, division can be thought of as repeated subtraction

[limber sierra]

and this "undoes" multiplication by 5

the way we express this is that

if a is in our field and not equal to 0

then we define a^-1 to be the number such that a * a^-1 = 1

so for example

in Z/5Z

this is a field, and division makes sense

to divide by 1, you just multiply by 1

since 1 * 1 = 1

but how do you divide by 2?

well, 2 * 3 = 6 = 1

sine we're working modulo 5

and 6 =1 mod 5

so division by 2 is the same thing as multiplying by 3

in Z/5Z

in other words, 2^-1 = 3

similarly 3^-1 = 2

[reef sleet]

brain going ouchie

[limber sierra]

and 4 * 4 = 16 = 1 modulo 5

so 4 is its own multiplicative inverse

ie dividing by 4 is the same thing as multiplying by 4

(this is comparable to -1 in the real numbers btw)

(and this should make sense; -1 = 4 modulo 5, after all!)

now lets say we took Z/nZ and assume n is not a prime number

if n isnt prime, that means n = a*b for some integers a, b in your ring Z/nZ

[reef sleet]

yes

[limber sierra]

and a and b are both > 1

(">" doesnt really make sense here but you know what i mean; theyre neither 1 nor 0)

[reef sleet]

the notation Z/nZ makes no sense to me because I keep wanting to cancel the Z LOL

[limber sierra]

yeah its weird notation, theres ring theoretic reasons for it

[reef sleet]

still wrapping my head around the "dividing by 2 by multiplying by 3" thing

[limber sierra]

but its kind of random until you see those

anyway

to finish my argument above

since n = a*b

and n = 0 in Z/nZ

this means 0 = a*b

so we should be able to "divide both sides by b"

that is, multiply by b^-1

to get 0 = a

but... thats a contradiction since we assumed a is not 0 or 1

[reef sleet]

yeah that's what I was about to say

[limber sierra]

so Z/nZ cant be a field (ie division cant make sense)

if n is composite

but it does make sense if n is prime!

[limber sierra]

one thing i can say that might clear this up a bit

when we study algebraic structures like groups, rings, fields, vector spaces, etc.

while of course the way we write things is important for communication

what we're really studying is how the operations behave

in the case of Z/nZ, we're studying how modular arithmetic behaves

not really how "2" and "3" behave

since the "2" and "3" in this structure are different from the regular ol' "2" and "3" in the real numbers

if you're familiar with boolean logic, for example

this is a system with two truth values, TRUE and FALSE, and the operations AND and XOR

and they follow these rules:

TRUE AND TRUE = TRUE
TRUE AND FALSE = FALSE
FALSE AND TRUE = FALSE
FALSE AND FALSE = FALSE

TRUE XOR TRUE = FALSE
TRUE XOR FALSE = TRUE
FALSE XOR TRUE = TRUE
FALSE XOR FALSE = FALSE

but what if i instead relabel this

so that "TRUE" is "1" and "FALSE" is "0"

and "AND" is *, "XOR" is +?

then we get:

[reef sleet]

I think I see it

[limber sierra]

1 * 1 = 1
1 * 0 = 0
0 * 1 = 0
0 * 0 = 0

1 + 1 = 0
1 + 0 = 1
0 + 1 = 1
0 + 0 = 0

hey

that's just Z/2Z!

[reef sleet]

1 + 1 = 0 :P

oh wait

[limber sierra]

oh actually

[reef sleet]

yeah that is z/2z

[limber sierra]

id idnt state that

quite right

let me use XOR instead

that makes more sense haha

[reef sleet]

I've seen XOR but idk what it is

[limber sierra]

XOR is just addition!

no seriously speaking

XOR is like

"is ONLY one of these true?"

so true xor true is false

since BOTH are true

[reef sleet]

ah

[limber sierra]

anyway the point is that

the way we "label" the elements of Z/2Z

or of boolean algebra or whatever

doesnt matter

what matters is how the operation behaves

so saying "division by 2 is the same thing as multiplying by 3" might not make sense

but what matters is that we're looking at how the operations behave

not the numbers

we could replace that with "division by Wombat is the same thing as multiplying by Germany" if we wanted

this... still wouldnt make much sense

[reef sleet]

LOL

[limber sierra]

but thats fine since we're looking at the structure of the operations

(in this case modular arithmetic)

rather than the elements themselves

this focus on the "operations" is expressed by the notion of an "isomorphism"

which is basically when we "relabel" elements of a group/ring/field/vector space/whatever

[reef sleet]

I've seen that word 😛

I thought it was a topology thing kekw

[limber sierra]

it's also a topology thing!

isomorphism is a nice concept since its a way of saying the structures are "the same" in how they "behave"

just "labelled" differently

for example, R^2 and C are isomorphic as vector spaces

(where here C has real number scalars)

this fact is what justifies us writing the complex number plane as a 2-dimensional x-y plane

where the "x axis" is the real line and the "y axis" is the imaginary line

of course, C can be given a bit more structure than R^2 - in particular, we have a standardized way to multiply complex numbers

whereas there isnt quite such a nice thing for vectors from R^2

(there's the dot product but it gives us a scalar, not a vector)

(so it's not as nice)

but if you look at C only as a vector space, it behaves like R^2

just with 4+2i relabelled as the vector (4 2)

or -3 + 7i relabelled as (-3 7)

or whatever

[also, if youre wondering: our choice to label a+bi as the vector (a b) is totally arbitrary]

[we could do (b a) instead and it wouldnt change anything - (a b) is just more common]

anyway

if you want a really whacky example of "addition"

which doesnt really feel like... addition

let me bring up my elliptic curve example from before

elliptic curves are a special type of curve; here's one example

you dont need to worry too much about the technical details, essentially theyre a specific class of 2-dimensional cubics with nice number theory properties

let's label two points on this elliptic curve, the green and purple

how do we "add" them?

well its a bit of a weird process:

[reef sleet]

how do you add points in general? add x coordinates and y coordinates?

[limber sierra]

well for common spaces

like R^2

yes

thats how you add vectorS!

but in this context

we're looking for a different group structure than that

so our addition is a bit weird

start by drawing a line, and label the point where it crosses the curve, here the light blue

then draw ANOTHER line, this one parallel to the y axis

[reef sleet]

.

[limber sierra]

the point where it crosses - the pink point here - is the sum of the green and the purple

[reef sleet]

that is weird

[limber sierra]

(now you might have concerns like 'what if the line between them never intersects the curve again?' for that reason our elliptic curve group has an additional "point at infinity")

anyway

as i said this may seem like a totally random and weird process

but it turns out that, despite the abstractness and bizarre behaviours

this actually has very deep number theoretic applications

and, in particular, that number theory can be applied to cryptography

a full explanation of this is beyond the scope of a discord discussion, i'm afraid

but hopefully this demonstrates that our group operations may not necessarily look like "regular ol'" addition or multiplication

[reef sleet]

yeah, I see it

abstract algebra seems like something I'd like

what courses do you recommend someone take before taking it?

besides a proofing course

[limber sierra]

proofs + linear algebra are enough honestly

[reef sleet]

oh sweet

[limber sierra]

examples from basic combinatorics sometimes come up

but like

those arent hard to learn on the fly TBH

(aside: if youre wondering why these curves are called "elliptic curves", theyre not actually related to ellipses)

[reef sleet]

I figured that

LOL

[limber sierra]

(the terminology dates back to a certain type of integral that was originally used to study ellipses)

(so the integral was named after ellipses, then elliptic curves were named after the integral)

(but by that point we're pretty far from the original ellipse inspiration)

(a lot of names in mathematics have some... unfortunate overlappings)

https://en.wikipedia.org/wiki/Normal#Mathematics

[reef sleet]

LOL

[limber sierra]

(it gets even worse when you try and translate from other languages - famously "variete" in French SOMETIMEs means "variety" but SOMETIMES means "manifold")

(and you have to figure it out from context, which is of course trickier for english speakers not used to french nuances)

[reef sleet]

I've got to get back to work but thank you very much for your time :) the abstract algebra group/ring thing was so cool LOL

[nocturne jewel]

part 2

[hoary osprey]

you've shown that the dim of a subspace is <= that of the parent space right?

[nocturne jewel]

I think so?

[hoary osprey]

alright well you can use that to show that dim(imTS)<= dim(T)

[nocturne jewel]

Like ik i need to get to dimImTS < dimImT or < dimImS

[hoary osprey]

you need to show both for the result to hold

[nocturne jewel]

Yeah

[hoary osprey]

dim(imTS)<= dim(T) is pretty straightforwaad, since im(TS) is clearly a subset of imT, and theyre both subspaces of W so the result follows

[zealous junco]

i got a simple q

so in the analysis class they said

[wintry steppe]

Ok

[zealous junco]

|x| ≥ 0, |x| = 0 if x = 0

|x+y| ≤ |x|+|y|

for real number x, y

this shows that this particular | | is a norm right

and R with |•| is a normed space

[wintry steppe]

Yes

[zealous junco]

And when they also said 4) |x•y| = |x||y| only works for R, are they really saying that R is also an inner product space?

wayt no

[wintry steppe]

No it's symplictic

[zealous junco]

i think inner produc space is |x•y| ≤ |x||y|

right

[wintry steppe]

yeah, that's cauchy-schwarz

[zealous junco]

symplictic space? It seems if I define <x,y> = xy then R is an inner product space, since it satisfies everything

[wintry steppe]

in R, all pairs of vectors are linearly dependent, so equality holds here

[zealous junco]

ok yea thanks

[wintry steppe]

(that the equality case for cauchy schwarz, i.e. |x dot y| <= |x||y| if and only if x and y are proportional)

[wintry steppe]

it's not antisymmetric

le antisymmetric inner product has arrived

[digital bough]

Thanks for the great explanation to him! Very interesting, gives me more incentives to read this channel

[novel hamlet]

I got set E and P(x) is all functions f:E->R, then defined p ∈ E for χp : E → R such as x-> 1 if x=p and otherwise 0. I need to show that (Xp1...Xpn) is basis for space P(x) if E has n elements and {p1...pn}=E. I have gotten to point where i have f∈ P(x) then i have f=a1X1+...anXn if and only if aj=f(pj), but im not sure on how to proceed from there.

[novel hamlet]

Thank you altought im not sure what notation := means here and that lambada

[spiral star]

:= means defined to be equal

i define f(x) to be equal to whatever i wrote on the right hand side

[novel hamlet]

iirc lambada had something to do with eigenvalues?

[spiral star]

lambda is just an arbitrary scalar here

i applied the definition of linear independence

i showed that if i take a linear combination of the vectors x_p1 ... x_pn that results in the 0 vector

then all scalars in that combination were 0

so lambda_1, lambda_2 and so on are just scalars from R

[novel hamlet]

i think i understood that

[spiral star]

and to show that the set is independent, i have to prove that all lambdas were 0

but thats easy since f is the zero function

[novel hamlet]

yeah that seems logical

[paper pebble]

How can you make a non singular matrix into a singular matrix?

[wintry steppe]

subtract it from itself

[frosty vapor]

lmfao tterra

bahahah

[north sierra]

no clue where to start

[native rampart]

What should p be if T(p)=0

[north sierra]

no clue

it should be in null space right

[native rampart]

Yee

What is 0 in R^2?

[north sierra]

(0,0) ?

[native rampart]

Yes

And what kind of p does T map to (0,0)?

Hint:||p such that p(0)=0||

[north sierra]

p that is a subspace of P_2?

[native rampart]

The set of such p forms s subspace

[north sierra]

so is that it?

also, what does the subscript denote in P_2

the 2

[native rampart]

P_2 means Space of Polynomials of degree at most 2

[north sierra]

oh i see

so how do i find a polynoimal?

that would span the kernel of T

[native rampart]

An element of P_2,p will be of the form ax^2+bx+c

[north sierra]

i see ok

i think i get it now

thankss

[novel hamlet]

i have managed to get to dim(w+w¨)=dim v, can is just take dim out of each side to get w+w'=v?

[native rampart]

If w+w' is a subspace of V,yes

[novel hamlet]

yes, but im not sure how to justify it

dim W ≥ (3/4) dim V , dim W0 ≥ (3/4) dim V and dim(W ∩ W0
) ≤ (1/2) dim V .

i got them defined like this

[native rampart]

A basis of V consists of dim V elements

[novel hamlet]

just slapped them to dimension rule and got that dim(w+w¨)=dim v

was not sure if there exist some dim^-1(x)

ie. counter operator for dim

[native rampart]

And any LI set of dim V elements is a basis

Implying a basis of W is a basis of V

Implying V =W

[novel hamlet]

yeah i did it like that

but i changed it to w+w0=b and then showed that if b is subspace of V then dim b = dim V= B=v

so kinda sme but with extra steps

[native rampart]

You are talking about lagrange interpolation?

[novel hamlet]

whats that?

[pallid pumice]

Hey everyone, I currently have this question on a homework assignment. I created a matrix and reduced it to (1 0 -3 2), (0 h 6 k-4). I'm a little confused on where to go from there. Since x3 is a free variable shouldnt the equations only have infinitely many solutions?

[wintry sphinx]

basically

only if the equations can be made inconsistent by adding them up will you get no solution

but there will never be a unique solution

the x_3 could be a typo

[pallid pumice]

yeah thats what i was thinking to

the matrix i calculated from that is correct tho right?

[gritty frigate]

Two vectors in Rn are always l.i if none of them is zero and they are not proportional ?

I know that this is a fact for R1, R2, R3, but... Can this be generalised for Rn ?

Oh this has to be true.. and can be proven..

[brisk fractal]

a set ${v_i}$ of vectors is said to be linearly dependent if there exists a non-trivial linear combination $\sum_i c_i v_i = 0$ where not all $c_i = 0$

[gritty frigate]

Ignore my msg

[stoic python]

bacono

[brisk fractal]

sorry I meant linearly dependent

[gritty frigate]

Oh right hahaha

[brisk fractal]

in other words, no vector in the set is in the span of any sub collection of other vectors (if they are linearly independent)

[gritty frigate]

Yeah, if there is a solution, for a set of n vectors and that solution is {s1,s2...,sn}

[brisk fractal]

and if all s_n are 0, then they are LI

that's the natural generalization not just for R^n, but any vector space

imo that's the most straightforward definition for proving things related to dimension

[north sierra]

is this true?

not sure

i say false

because if you take a subset then it couldn't span a vector space ?

[wintry sphinx]

the subset doesn't have to be proper

the subset could very well be S

it's saying that there exists some subset that is a basis, not that a particular subset is

[north sierra]

oh i see

makes sense

[acoustic path]

so my thinking is that when Tv=0 we have two cases that either v=0 or T=0. however v cant be equal to 0 because v isnt in subspace U which naturally must have the 0 vector for it to be a subspace. so that means that T=0. and then im not sure what to do

oh if Tv=Sv it says that v is in subspace U

so then T=S

and so for both cases T is just linear map S but S isnt in the map of V so neither is T??

[hoary osprey]

uhh

not sure what u mean by T=0

[acoustic path]

well in the Tv=0 case

T must be equal to 0 right

cuz v cant be equal to zero

[hoary osprey]

why not

[acoustic path]

cuz v cant be in subspace U

and since its a subspace it must have the zero vector

am i just wrong about this

[nocturne jewel]

Sorry for ping but is this what you meant? (Only getting around to finishing the question now lol)

[brisk fractal]

well that's why you would be able to invoke that argument

in general any subspace of a vector space has dimension less than or equal to the parent space

[nocturne jewel]

right, but how is Im(TS) a subspace of Im(S) or Im(T) is what im more confused about

wouldnt Im(TS) be bigger?

[brisk fractal]

TSv = T(Sv), and so im(TS) subset im(T)

[nocturne jewel]

wait no, cause Im(TS) specifically needs S(u) vectors, where as Im(T) can be any vector from the input space

[brisk fractal]

let me look at this a bit more carefully one sec

[nocturne jewel]

kk

[brisk fractal]

ok

Let $v \in Im(TS)$. Then $v = TSx = T(Sx)$ for some $x \in U$. Since $Sx \in dom T = V$, then $v \in Im(T)$. Therefore $Im(TS) \subseteq Im(T)$.

[stoic python]

bacono

[nocturne jewel]

what's domT?

[brisk fractal]

domain of T

[nocturne jewel]

v in Im(T) is just from the def of image right?

[brisk fractal]

I was just assuming that for the sake of proof

if you want to show that they are subspaces, you'll need to further show that they fulfill the vector space axioms

[nocturne jewel]

Ok the jump from Sx in V to v in ImT isnt clicking for me

Oh wait

[brisk fractal]

okay so TS is a composition of linear maps right, and so it's the same as multiplying their representative matrices

matrix multiplication is associative

[nocturne jewel]

I think it clicked but cant put it to words lol

[brisk fractal]

you can view TSx as both (TS)x or T(Sx) = Tv

[nocturne jewel]

Im(TS) and Im(T) are both sets of vectors in W

[brisk fractal]

yes

really, vector spaces but you get the point

[nocturne jewel]

yeah, so does a similar argument hold for Im(TS) subset Im(S)

[brisk fractal]

let me think about that

do you have rank-nullity yet?

[nocturne jewel]

yeah and dimension formula

[brisk fractal]

okay that makes things easier

it turns out I am extremely rusty on basic linear algebra

I can't help you much here

[nocturne jewel]

rip no worries

[wintry steppe]

helping with LA in this channel is the best way to learn LA

[nocturne jewel]

want to learn LA then Terra? Lol

wait i think i got it

$dim(U) = dim(Ker(TS)) + dim(Im(TS)) = dim(Ker(S)) + dim(Im(S)) \ dim(Ker(TS)) = -dim(Im(TS)) + dim(Ker(S)) + dim(Im(S))$

[stoic python]

moshill1

[nocturne jewel]

Got to: $$dim(Im(TS)) \geq dim(Im(S)) - dim(Ker(T))$$

[stoic python]

moshill1

[nocturne jewel]

but then dk how to deal with the dim(ker(T)) part

[brisk fractal]

well I've noticed a neat pair of relations

$S : \ker (TS) \to \ker (T)$

[stoic python]

bacono

[brisk fractal]

$T : \Im (S) \to \Im (TS)$

[stoic python]

bacono

[brisk fractal]

now, what information can we gain from this using rank-nullity?

(which for arbitrary linear maps is written)

[nocturne jewel]

dim(ker(TS)) = dim(Ker(S)) + dim(Im(S)) <= dim(Ker(S)) + dim(Ker(T))

[brisk fractal]

badabing

you got it

I'm not sure if that's the intended way of doing it but I thought it was a neat approach

did I say something incorrect

[nocturne jewel]

ii

specifically the Im(TS) is a subset of Im(S)

[brisk fractal]

well when you do this it's just a couple of additional terms

[nocturne jewel]

Ok well I only know dim(Im(TS)) <= dim(Im(T)), so how am i suppose to relate dim(Im(TS)) with dim(Im(S))?

Ok, I tried dimension formula and you said it's wrong so....

[brisk fractal]

slim, what is fishy about the argument above?

I see that it seems a bit off but I don't know how to correct it

[nocturne jewel]

ii

[brisk fractal]

wait I see the problem

it's not null(S) + null(T), it's null(S restricted to ker(TS)) + null(T)

[nocturne jewel]

It's called trying to solve the question...?

Dont ?? me

[brisk fractal]

but null(S restriction) is less than null(S) so we get the inequality

[nocturne jewel]

Ok well im trying to solve ii

[brisk fractal]

okay I have a solution

use the second relation to get

$\mathrm{null} (T \mid_{\Im(S)}) + \rank (TS) = \rank (S)$

[stoic python]

bacono

[brisk fractal]

then rank(TS) <= rank(S)

[nocturne jewel]

what's null and rank compared to im and ker?

[brisk fractal]

and you get rank(TS) <= rank(T) from the subspace relation

[brisk fractal]

null(T) = dim(kerT)

rank(T) = dim(Im T) = dim(Col T) = dim(ran T) = dim(Row T)

[nocturne jewel]

ok and what's the 2nd relation lol

[brisk fractal]

or rather, T restricted to Im(S)

[nocturne jewel]

Ok so define a map from ImS to ImTS?

[brisk fractal]

no, just take T and restrict it to Im(S), and then consider rank-nullity on this map

[nocturne jewel]

ok but what do you mean restrict a map to a set?

[brisk fractal]

this relation comes from the fact that if v in Im(S), then v = Sx for some x in dom(S), and so Tv = TSx, and Tv is in Im(TS)

it's literally just setting the domain to a subset

[nocturne jewel]

Right, so cant I just define a new map that does that?

[brisk fractal]

oh that's what you meant, yes

sorry my bad

[nocturne jewel]

no worries lol

$A: Im(S) \to Im(TS) \ dim(Im(S)) = dim(Ker(A)) + dim(Im(A))$ ?