linear-algebra

[stable kindle]

actually wait

no i think it works

can you have like. a set of all groups?

[winter harbor]

You can talk about monoidal categories

Which are categories endowed with a certain product, which we usually call a tensor product.

[stable kindle]

ah is that the generalised concept

[winter harbor]

And it satisfies properties which are analogous to associativity and having a unity element

[winter harbor]

Yeah

For instance

[stable kindle]

no, yeah

that's exactly what i mean

wrt. groups the identity is the trivial group

[winter harbor]

Vector spaces endowed with its usual tensor product are a monoidal category, in fact a symmetric monoidal category

[stable kindle]

and you can show associativity up to isomorphism

[winter harbor]

Sets endowed with Cartesian product too

The category of monoids endowed with direct sum is also a monoidal category

There are tons of examples

[gray dust]

no but we speak of Grp

[stable kindle]

oh

are there too many somehow

[winter harbor]

Yeah, Grp would be the category of groups

[stable kindle]

what about just finite groups

[gray dust]

category of groups with group homos as homos

[stable kindle]

i feel like you could definitely have a set of all finite groups

[winter harbor]

https://ncatlab.org/nlab/show/monoidal+category

[winter harbor]

It is not a set, it is a category tho

[stable kindle]

why is it not a set

[winter harbor]

It is the category Fin Set

At least that's how Mac Lane denotes it

[stable kindle]

all the elements are of finite order, how can you get any paradoxes

surely no recursion

[gray dust]

kai, distinguish class vs set

[stable kindle]

why

what is the problem lol

i understand not

[gray dust]

russell

[stable kindle]

there can't be russell for the set of finite groups

surely

the set of finite groups is infinite

but all of its elements are finite

so it cannot possibly be in itself

or anything like that

[left snow]

there are assumptions in category theory that aren't really like

the same

when you try to fit the metaphor of sets and classes

[winter harbor]

https://ncatlab.org/nlab/show/FinSet

[stable kindle]

none of this is explaining anything

even if you can't explain it

[left snow]

https://youtu.be/JOp7mH72Jlg
you can try this

[winter harbor]

Alright

[stable kindle]

can any of you say with 98% certainty that you could literally sit down and, in the specific case of defining a set of all finite groups, point to a contradiction

[left snow]

if you insist on calling it a set then yes it will result in problems

[stable kindle]

i know what russell's paradox is, btw

[left snow]

watch the video at 12:25

he addresses some solutions to this

[stable kindle]

and you also can't have a set of all sets bc the cardinality would be greater than the cardinality bc like 2^C < C and 2^C < C, using the power set

how do we extend that

[stable kindle]

yeah, i'm solving it with solution 1

bounding the size of the objects

[winter harbor]

Consider the proper class $\mathcal{U}$ of all sets. Consider the subclass:
$$
\text{Fin} := {S \in \mathcal{U} , \vert , \text{S is finite} }
$$
We will prove $\text{Fin}$ is a proper class. Indeed, suppose $\text{Fin}$ is a set, then for any $A \in \mathcal{U}$, we have that ${A}$ is a finite set, i.e ${A} \in \text{Fin}$. So if $\text{Fin}$ was a set, we would have that the union:
$$
\mathcal{C} = \bigcup {A} \subset \text{Fin}
$$
Is also a set, but $\bigcup{A} = \mathcal{U}$ is a proper class, a contradiction ; $\square$

God damnit

[stoic python]

MisterSystem

[stable kindle]

oh, fuck

it's hidden two layers down lol

[winter harbor]

Wdym?

[stable kindle]

like

the constraint that the elements are finite order isn't sufficient

since the elements of the elements can be infinite

so the elements can contain the entire thing

[winter harbor]

Idk how much this is a real thing

Like, idk much set theory.

[stable kindle]

no i think this is valid

[winter harbor]

But yeah, the problem is that if we take any set

[stable kindle]

it makes sense, whenever you have recursion you can get wild

[winter harbor]

Then {A} is finite

And oops

[stable kindle]

ok

[winter harbor]

But you see, we have to be careful dealing with categories.

[stable kindle]

what if we make the additional constraint that all the elements, and the potential elements of the elements, and the elements of the elements of the elements, and so on, all have finite orders

[winter harbor]

Uh, to be fair I don't even know how one would make this precise

Or if this can be made precise

So I can't answer :/

[stable kindle]

ok fair

many thanks tho

that contradiction blew my mind

smooth

[dusty pond]

not sure what to do here, could someone walk me through these questions? thanks so much!

[winter harbor]

Ofc

First

Notice the following

If we have a polynomial p in F[x] and I take p(T) for T some endomorphism of V

Then p(T) and T commute!

[dusty pond]

yeahh

[winter harbor]

Can you see this?

Nice

We will use this

We will prove (a) by induction. Let $k \in \mathbb{N}$, then $W$ is invariant under $T^{k}$. Indeed, for $k=1$ it follows by hypothesis.
\
\
Suppose now $W$ is invariant under $T^{k-1}$ for some $k \in \mathbb{N}$. Then, $\forall w \in W$ we have that $T^{k-1}(w) \in W$, since $W$ is $T$ invariant, this implies $T(T^{k-1}w) \in W$ and so $T^{k}(w) \in W$ and $W$ is invariant under $T^{k}, \forall k \in \mathhbb{N}$.
\
\
Now, suppose that a subspace $W$ is invariant under $T,U \in \text{End}{\mathbb{F}}(V)$. Then it is invariant under $T+U$.
\
\
Indeed, $\forall w \in W$, we have that $T(w) \in W$ and $U(w) \in W$, so $T(w) + U(w) \in W$ and $(T+U)(w) \in W$. So $W$ is invariant under $T+U$.
\
\
Finally, consider $p(T) = \sum\limits{k=0}^{m} \alpha_{k} T^{k}$. It is easy to see that if $W$ is invariant under a certain endomorphism $U$, then it is invariant under $\alpha U, \forall \alpha \in \mathbb{F}$.
\
\
Notice how $p(T)$ is a sum of terms of the form $\alpha T^{k}$ and we proved $W$ is invariant under each $T^{k}$, and thus is also invariant under each $\alpha_{k} T^{k}$. Since $p(T)$ is a sum of operators under which $W$ is invariant, then $W$ is invariant under $p(T)$ as well ; $\square$.

Here's how part (a) goes.

[stoic python]

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

[winter harbor]

There are prolly some typos I guess.

But this is the idea

You just need to make the proof cleaner.

[dusty pond]

omg this is super helpful - thank you so much!!!

[winter harbor]

Do you know how to apply part (a) to prove part (b)?

Think about this for a second

And if you get stuck

[dusty pond]

not sure rn, but looking over it i think i can

[winter harbor]

Try to use this

[dusty pond]

thanks!

[queen slate]

Hey Mathematical Homeslices, how much of linear algebra can I reasonably get from Khan Academy's course?

[winter harbor]

Some amount, yeah.

Keep in mind

It does everything under R^n

Doesn't go over a bunch of stuff like the formal definition of a vector space, thus doesn't go over stuff like dual vector space, quotients and other more abstract constructions

It also doesn't go over stuff related to invariant subspace decomposition and diagonalizable matrices.

So it doesn't talk about the spectral theorem for normal operators

Doesn't talk about the Cayley-Hamilton theorem

Doesn't talk about Jordan Canonical Form

It talks about more basic stuff

And is good at what it does

At least that's what I remember when I watched it 2/3 years ago

[queen slate]

Thank you, I just went ahead and snipped that comment and saved it as a PNG called "Rest Of Linear Algebra." I know that's not true either, but it's a start.

[winter harbor]

Good luck with your studies!

[sick sandal]

well i dont think you can ask to pay people
and if you have an issue with a certain question just send it and ask whats bothering you with it

[wintry steppe]

you know how you can visualize scalars as points on a line or 1D vectors and visualize vectors as arrows in space, is there a similar visualization/geometric intuition for rank 2 tensors?

[lavish jewel]

not really, maybe componentwise?

but a tensor is a transformation

it's related to some geometric concepts, but that's about it

you could try to characterize it through the action it performs

but it itself, not really

[tight grail]

thank god there's a linear algebra channel

[wintry steppe]

is the inner product a tensor?

[lavish jewel]

it also depends on what you call a tensor

unfortunately there are many definitions depending on who you ask

[winter harbor]

What do you mean by a rank 2 tensor? There's some confronting definitions in the literature. By a tensor of type $(p,q)$ on a finite dimensional vector space over a field $\mathbb{K}$ we mean a $(p+q)$ multilinear map:
$$
T : \underbrace{V \times \cdots \times V}{\text{p times}} \times \underbrace{V^{\ast} \times \cdots \times V^{\ast}}{\text{q times}} \rightarrow \mathbb{K}
$$
I've seen in the literature people referring me to a type $n$ tensor as a $(n,0)$ tensor, as a tensor of type $(0,n)$ or as a type $(p,q)$ tensor for which $p+q=n$.
\
\
In any case, since we are dealing with finite dimensional vector spaces, we can identify $V$ with its dual, and think about rank 2 maps as bilinear forms:
$$
T : V \times V \rightarrow \mathbb{K}
$$
Which can thus be identified with a matrix $A$ for which $T(u,v) = u^{t} A v$.

[stoic python]

MisterSystem

[wintry steppe]

don't multiple definitions of tensors define the same mathematical object?

[lavish jewel]

no

[winter harbor]

Sso sometimes in the literature, you will see rank 2 tensors being referred to as bilinear forms/matrices

[winter harbor]

If you are referring to a tensor by what I defined above

Then an inner product on a real vector space is indeed a tensor.

[wintry steppe]

then the inner product is a tensor right?

[winter harbor]

Yeah

But not on a complex vector space

Because hermitian inner products are sesquilinear

Not bilinear

A real inner product is defined as a symmetric, positive definite bilinear form on a vector space.

So it is a type (2,0) tensor

I.e, has contravariance 2 and covariance 0

[wintry steppe]

oh that kinda makes sense

[winter harbor]

In any case, this is the definition of a tensor that usually comes up in differential geometry and analysis

There's also something called the tensor product of two vector spaces, or more generally two modules over a commutative ring.

Which is a different concept

Ofc these are related

But tensor products ultimately generalize the concept of a tensor as a certain (p+q) multilinear map.

[wintry steppe]

wait what product is that? that X

[winter harbor]

Just cartesian product

[wintry steppe]

ah okay

[winter harbor]

In the usual sense

Sometimes it is referred to as direct product

In the context of algebra

[ionic laurel]

Hello

Can anyone help me clarify a question I have about subspaces?

If H is the solution set of the x vector in a linear system, and the parametric vector form of that is not parallel to the 0 vector, then it cannot be a subspace correct?

Like it is not passing through the origin*

Sorry if my question is confusing*

[winter harbor]

No, it is not.

You are correct, the solution set for Ax=b is usually going to be an affine space, but not a subspace.

For b≠0

[ionic laurel]

so its like

the solution set to my linear system has one free variable

but it also has scalars that are added onto it

if i had x1 x2 and x3 where x3 was free then x1 would be like x3 + 1 for example

that scalar at the end means that it doesnt run through the origin so it isnt a subspace correct?

[winter harbor]

Yup

That's exactly correct.

[ionic laurel]

👍

Thank you MisterSystem

[winter harbor]

Np

[dark brook]

What would be the easiest way of findind the inverse matrix of ?

[stable kindle]

I = I * I

[lavish jewel]

the hint would be to use the definition of nul(A) and linearity of A

for example, recall that A(x+y) = Ax + Ay

and that if a vector is in the null space of A, then Av = 0

[sour cloud]

im confused tho theres two matrices in Nul(A) ?

[lavish jewel]

what's the problem with having the null space be spanned by 2 vectors though?

this just means that any vector v = aw_1 + bw_2 satisfies Aw = aAw_1 + bAw_2 = 0, aAw_1 = 0, bAw_2 = 0

[sour cloud]

so v = 0 ?

[lavish jewel]

no, that's precisely the point

they're telling you that A[0,1,-1]^T = 0 and A[1,2,0]^T = 0

that's precisely what Nul(A) = span(...) means

[sour cloud]

by why are we using NUl(A) = span() to find the other three vectors to the Ax equation?

[lavish jewel]

because of linearity...

i'll give you an example, but you really will need to go review your definitions

you are told that A[0,3,0] = [-3,3,-6]

and also that A[0,1,-1] = 0 and A[1,2,0] = 0

by linearity, you can just add the 3 vectors together

A([0,3,0] + [0,1,-1] + [1,2,0]) = A[0,3,0] + A[0,1,-1] + A[1,2,0] = [-3,3,-6] + [0,0,0] + [0,0,0] = [-3,3,-6]

so [0,3,0] + [0,1,-1] + [1,2,0] = [1, 6, -1] is another solution to the equation

[sour cloud]

so your adding Ax with the two matrices in Nul A ?

[lavish jewel]

i added x with the two vectors in Null A

but any linear combination of the vectors spanning Null A works

for exactly the same reason: linearity and the definition of the null space

you need to go review those 2 definitions

[sour cloud]

ohhh ok i get it now

ok

thank you

so for the first solution i got [-2, 5, -7]

but since your adding the vectors i only got two solutions tho

[lavish jewel]

i cannot do anything more for you, i already explained how to find infinitely many solutions

you will need to go review what span means as well

[sour cloud]

first solution i did [-3, 3, -6] and second solution i used [0,3,0] since they are equal to each other

oh ok

[queen slate]

Hey @sour cloud , if you get this figured out, can you give me an idea of what it's asking? I'm new to linear algebra.

[sour cloud]

sure!

[queen slate]

Thanks, starting from the top, what is NulA?

[sour cloud]

Nul A is the Null Space of matrix A

also i got it @lavish jewel thank you!

[dark brook]

Can you find the inverse matrix of a matrix thats already on the RREF notation?

[lavish jewel]

sure, if after RREF you find the matrix has a full set of linearly independent columns

but note that the inverse of the RREFd matrix is not the same as that of the original

[dark brook]

Ahh okay. I've got the RREF(A) equals to I_n, but it only explains that there is existing an inverse matrix of A

[lavish jewel]

yep

[dark brook]

Seems like I need to start from scratch - seems like an easier options than finding / having the linearly independent columns

[lavish jewel]

what you can do is to deal with an augmented matrix

if you take your matrix A and augment it to [A I], where I is an identity matrix of the same size of A, then if you RREF the rows so that A becomes an identity matrix, the place where the identity matrix originally was turns into A^-1

[dark brook]

That sounds really fancy

We are quite new to this linear algebra stuff, so maybe I'll just stick to the book
And I don't quite seem to get how you even would do it like that (sorry)

[lavish jewel]

well

every time you do row operations on a matrix, this is equivalent to multiplying the matrix A by some matrix, let's call it R (for "row operation")

so something like RA

when you do one row operation after another, it's the same as composing several of these operations, which for matrices is the same as multiplying several matrices together

so Rn Rn-1 Rn-2 ... R2 R1 A

and these matrices (Rn Rn-1 Rn-2 ... R2 R1) multiply A to turn it into an identity matrix

meaning ( Rn Rn-1 Rn-2 ... R2 R1 ) = A^-1

by augmenting A into [A I], when you multiply by (Rn Rn-1 Rn-2 ... R2 R1), this is distributed to both blocks of the augmented matrix

so you get [(Rn Rn-1 Rn-2 ... R2 R1)A (Rn Rn-1 Rn-2 ... R2 R1) I]

and since (Rn Rn-1 Rn-2 ... R2 R1) = A^-1, this turns into [I A^-1]

just as we wanted

[sour cloud]

wait quick question how do you find a vector? if we dont have Nul span and it was just the matrix and T(v) (linear transformation) ?

[lavish jewel]

i need more info than that :x

a vector that satisfies some condition? or?

[sour cloud]

i dont know how to write matrix here lol

[lavish jewel]

invert the matrix directly or iteratively (if possible)

[sour cloud]

wdym ?

also it can be any matrix this is an example cuz i was curious on how to find a vector like this

[lavish jewel]

do gaussian elimination or invert the matrix, idk what else to say

[sour cloud]

but what do you do with T(v) then ?

the transformation is R^3 -> R^3

ok i found one like this:

[bold sun]

hi okay i need some help please if thats okay

i just sont get thiss

like what is it even asking me to do and how? its a bit wordy too

and where did they get the v from?

im just lost

[stable kindle]

any arbitrary v in R^3 i assume

[bold sun]

?? arbitary?

oh ok

so how do i do it? do you have any idea

do i just say how ab will be a 3x3 matrix so will ba and a^2 and b^2

lkke they all end up as 3x3 matrixes when multiplied?

[stable kindle]

the transformations, not the matrix products

[bold sun]

but how are we transforming it??

were multiplying it?

ngl this transformation thing is confusing me

basically i just read this example dont really get it but it looked like they did the transpose of it

it was in the lect notes but i didnt get the b one or c one ? like did they just transpose it like why are they sating r3 is r2

[bold sun]

??

oooooh okk multiplying is a type of transformation right?

[sick sandal]

linear transformation can be represented as a matrix , it basically does the job of T as it sends a vector from 1 space to another

[bold sun]

huh wdym linear transformation (sorry if its a dum question)

[sick sandal]

oh...my bad
i guess you're going through it backwards

[teal grotto]

what do you think T_{AB} would be? just take a guess

[sick sandal]

take is as such for now
if the matrix A is a mapping from R^2 to R^2
AB will also me a mapping that works similar to something familiar relating to functions( can you guess what it is?)

[bold sun]

idk a 9 by 9 matrix....i dont get this whole transformation thing

[teal grotto]

why do you say 9 x 9?

[bold sun]

oh 3x3 woops

[teal grotto]

right

so if AB is a 3 x 3 matrix

what would the corresponding linear transformation T be?

[bold sun]

r3 to r3

[teal grotto]

if Ax is T_A(x), then ABx is ...

[bold sun]

idk sorry

[teal grotto]

It should be T_{AB}(x) = ABx

[sick sandal]

treat them as functions :3

[torn stag]

The definition of matrix multiplication is $T_{AB} = T_{A}T_{B}$.

[stoic python]

IlIIllIIIlllIIIIllll

[bold sun]

okayyyy........

[teal grotto]

thats not really definitional

[torn stag]

almost a definition.

[gray dust]

it's a fact that follows from the def of matrix multiplication & matrix reps of linear maps

[teal grotto]

eh, it takes a bit of work to build that up, which is what 1u1 is trying to do

[bold sun]

im actually confused ngl

idk what im doing at all....

[sick sandal]

AB is similar to but not really a composite function starting with B and then A for every vector introduced on it

[gray dust]

if matrices $A,B$ represent linear maps $T_A,T_B$ respectively then the matrix product $AB$ represents the composition $T_A\circ T_B$

[sick sandal]

maybe that will help digest the idea

[bold sun]

whatt how?

so a and b are alredy transformed?

[stoic python]

RokabeJintaro

[gray dust]

you can prove this by def of matrix multiplication & matrix rep

[bold sun]

so am i trying to prove that??

i dont even get what its asking me to do woops

[gray dust]

i say you CAN prove it, as in it's not an 'immediately' given fact but it follows from definitions, not that you should take the time now to actually do it

[bold sun]

ooohh okk sorry

[gray dust]

the point is that the above fact gives intuition for viewing matrix multiplication

[dusky epoch]

roketto you need to remember help-seekers dont listen to anything but The Answer™️ /hj

[bold sun]

i actually do uko- i want to learn the process too to learn it an answer isnt always enough

or i will fail in the actual exams ...whenever that is ...if i dont get it.....

anyways so what am i trying to do im still lost....

[sick sandal]

have you taken linear transformation

or just matrices

[bold sun]

wdym ?

my courses are linear algebra calculus and analysis probabilty and business microeconomics (that was only optional module) and comp classes for this semester

so i think this matriceis is what we doing now part of lin al

not sure abt linear transformations?

i dont thinnk ive done it yet

dunno if i will

but im trying my best to keep up and understand everything thats taught cuz im finding this all really hard and overwhelming ngll

[sick sandal]

hmm correct me on anything i might explain poorly
so lets put it this way , each matrix corresponds to what you would call a linear transformation i wont define it or go into it but it works like a function and takes a given vector from 1 space say V to another say W in this case T: V-->W or T(v) just like you would work with a function
now this transformation can be represented as a matrix A or B that can take functions from say R^2 to R^3 in this case the number of colums would represent the initial dimension for reasons i wont dive into ,all you need to understand for now is that the 2 matrices A and B when multiplied work just like you would with 2 functions f and g ,only difference being we call them linear transformations and just like fg or f o g is a composite of functions same thing to A and B as each of them represent our so called "function" T namely TA and TB so T_AB would simply be represented by...well TATB and since A represents TA and B represents TB it would simply be T_AB=AB and works as T_AB(v)=ABv

[limber sierra]

holy run-on sentence

[orchid delta]

Hi I am struggling to learn linear algebra. We use Sheldon Axlers Linear algebra done right as our text book. I tend to go over the definitions and work my way through the chapter but then I come to the exercises and i do not understand how to even attempt the questions. I would highly appreciate any and all advise on how to approach studying this module

[limber sierra]

to summarize the point: theres a special class of functions called "linear transformations". these are what linear algebra is named after.

matrices are, in a sense, "alternate notation" for linear transformations, in that they exist in one-to-one correspondence with linear transformations. A matrix with m rows and n columns maps from an m-dimensional vector space to an n-dimensional vector space.

the point is that matrix multiplication and applying linear transformations are the same thing

this is why matrix multiplication is defined the way it is in the first place

[bold sun]

oh okk i got till the thid last line lol before getting little confused

i kind of get it

[bold sun]

ohh okk

so is a and b originally a transformation like what was it origionally if it is?

[limber sierra]

?

T_A and T_B are linear transformations

[sick sandal]

i mean if you are studying axler im assuming you are more interested that just using linear algebra
understand the theorems , try to prove them and make sure you know how you can use them

[limber sierra]

A and B are the matrices that represent those transformations

[bold sun]

so ta= a?

[limber sierra]

equality isnt quite accurate, T_A is a function while A is a matrix

but they represent the same thing

[bold sun]

oh....

[gleaming knot]

Honestly my professor never separated the notation T_A and A while learning linear algebra myself

[bold sun]

and a function is something that has been transformed?

[orchid delta]

Is axler enough to get this understanding or should I supplement my studying with other books?

[gleaming knot]

No, a function is a function, the same concept you know and love

[sick sandal]

i know axler is a 2nd pass that ignores determinants

[limber sierra]

to clarify

T_A = A is misleading since T_A and A are fundamentally different things

[sick sandal]

so might be a good idea to use a supplement if its your 1st time doing LA

[limber sierra]

(if you program, its kinda like comparing the int 2 and the float 2.0, i guess?)

but T_A(v) = Av is true for all vectors v

[nocturne jewel]

[T_A]=A is the notation I was shown with the input/output spaces bases on the brackets

[limber sierra]

the left hand side, T_A(v), means apply the function T_A to the vector v

the right hand side, Av, means multiply v by the matrix A

(on the left)

[orchid delta]

ahhh thank you. I was starting to feel hopeless

[limber sierra]

the point is that these are the same process

[sick sandal]

so its like f(x)=x^2?

[limber sierra]

thats why matrices are like

useful at all

[limber sierra]

thats not linear, but sure, same idea

you might have a function like:

[sick sandal]

ye just the overall idea

[bold sun]

ooh

[limber sierra]

[T\begin{pmatrix}v_1\v_2\v_3\end{pmatrix} = \begin{pmatrix}2v_1 - 3v_3 \ 0 \ v_1 + v_2 \ v_2/2\end{pmatrix}]

[stoic python]

Namington

[limber sierra]

T would be a linear transformation from ℝ³ to ℝ⁴

(or perhaps F³ to F⁴ for F an arbitrary field, but same idea)

so the matrix representation of T would have 3 rows and 4 columns

[bold sun]

okk yh

[limber sierra]

to determine the entries of those rows and columns, we typically look at what T does to a standard basis

[bold sun]

so do we only look at the rows for the no that goes on r

like what if both rows and colums have changed?

[limber sierra]

i dont understand the question.

[subtle walrus]

oh yeah, whats the mutliplicative inverse of 2 in a field of characteristic 2

[bold sun]

here you changed r3 to 4 by looking at the rows right

so i was saying the no that goes on top of r is that for the rows

[limber sierra]

the notation ℝ^n just means "the set of vectors containing n real entries"

[bold sun]

and also in your example the no of rows changed what if no of colums change or both numbers of rows and columns change

[limber sierra]

matrices are not members of ℝ^n (unless they're 1xn matrices, i.e. column vectors)

so the question is moot

its like saying "f(x) = x^2??? well, what if x is a hippopotamus!?"

[bold sun]

oh didnt know that i thought its what u used to describe the transformation

[limber sierra]

we're not talking about hippopotamuses

we're talking about numbers

and similarly, in this case, we're not talking about arbitrary matrices, we're talking about specifically column vectors with however many entries

i feel like youre missing some prerequisites here, try rereading earlier chapters

[bold sun]

i kinda did.....ive done all notes and qs whilst going along for practice so till like here so that included prev topics

but i got stuck here cuz it wasnt explained well at all- it was the 3 examples i showed thats what it sed then it moved on to other topics and propostions

[median ocean]

anyone know part 1?

[winter harbor]

Just compute the determinant

And find under which conditions it is zero.

[bold sun]

look its here about transformations

then few pages later they mentioned about the composites.....

i think those 2 are the only 2 things i needed for the q tho

[sick sandal]

if you understand it now ,works out fine :3 maybe he assumed you are familiar with vector spaces and linear transformation

which im assuming is coming soon

[bold sun]

vector spaces are the axioms right like v+0=v or v+(-v)=0 ? i think we did that but thats more to do with vectors not matrixs

not linear transformations- aint done that

[median ocean]

@winter harbor so I Just compute the determinant first?

[sick sandal]

vector spaces contains a field of scalars, a set of V objects called vectors and 2 rules of addition and scalar mulitplication with certain properties

[bold sun]

yh we did that with vectors like we did the propsitions

so uko would this be right then for the first bit this is what ive understood so far in how to do it t(ab) is r^3--->r^3 where tab(v) is avbv?

[sick sandal]

no

think back to composite of functions

[bold sun]

ok so ive misunderstood somewhere

[sick sandal]

if g=x^2 and f=x+1
you would not say
gof(3)=3^2*(3+1)
you would say its (3+1)^2 or g(f(x))
same thing applies here

[bold sun]

what dies the cirle stand for?

[sick sandal]

T_AB(v) would be ABv

applying B 1st followed by A

[sick sandal]

just a notation for composite

[bold sun]

so that would be a(bv)?

[bold sun]

oh ok

[sour cloud]

#linear-algebra message

anyone know how to solve that?

[sick sandal]

same thing as ABv yes :3

[median ocean]

so part 1 I just compute the determinant

first?

[bold sun]

oooh okkk so do i do the same thing with the rest so for example ba is r^3-->r^3 again but it would be tba(v)= bav this time

cuz we do a first

[sick sandal]

yep you got it

[bold sun]

ooooh okk im gnna make notes on this im my own words then try the rest of the q..... i thnk i get it now

thank you ever soooo much

[winter harbor]

Yeah, just calculate the determinant of C.

[bold sun]

ummm im soo sorry but now i dont get this its part of transformations...but diff

i manged to do a and b using what i just learnt like its eqaul to ax so i did that

but i dont get c

[left snow]

c is asking you to verify that the transformation is linear

you can let (x = \cvec{a \ b}) and (y = \cvec{c \ d}) and show that (T_A(x+y) = T_A(x) + T_A(y))

[stoic python]

ricey

[left snow]

show that it holds for all (a), (b), (c), and (d)

[stoic python]

ricey

[bold sun]

so i use a(x+y)= a(x)+a(y) right like do i use their value of a

[sick sandal]

ye just normal multiplication

you know what Ta(v) is

you're just using it

[left snow]

i think you have to show for all vectors x and y

[sick sandal]

ye

just take 2 arbitrary vectors x and y ( and show it holds for all x,y)

as mentioned

[bold sun]

so like this ignore the messy writing n background

[gleaming knot]

The writing of the logic needs work

[sick sandal]

could be more formal but yep pretty much it

[bold sun]

i dont do formal...something i need to work on

[gleaming knot]

I wouldn't say formal vs. not formal, I rather split it into makes sense vs. doesn't make sense

You did the right calculations but your proof does not make sense

[bold sun]

and then is the second bit saying a lambda x is equal to lmmbda ax

[left snow]

i think to be fair to him

the question asked him to verify

not prove

[gleaming knot]

A verification and a proof that two things are equal is the same thing

[bold sun]

like regardless of the place lambda is put

[left snow]

i think it would be acceptable to submit this, albeit it's not very rigourous

[left snow]

i was taught that verify means to check

so i wouldn't bother proving the linearity here

[gleaming knot]

Hmm?

To verify that 8 is a solution to x+3=11

and to prove that 8 is a solution to x+3=11

[left snow]

i substitute x = 8

[gleaming knot]

is exactly the same question

[left snow]

for the first one i would substitute x = 8 to check that it works

for the second one i would solve for x

[gleaming knot]

Nah if you solve for x you are actually proving the converse

that if x+3 = 11, then x = 8

[sick sandal]

why are they the same thing? im also used to verify and prove being seperate questions

[left snow]

https://math.stackexchange.com/questions/2636274/verify-vs-prove

[gleaming knot]

Look at "Verify Goldbachs' conjecture for all even integers less than 1 million"

[sic]

[vivid field]

How would I go about finding the domain and codomain ? Im confused

[gleaming knot]

Let's take off the 1 million part

Verify Goldbach's conjecture?

That's the same difficulty as proving (or disproving) Goldbach's conjecture hmm

is it not?

[sick sandal]

isnt that the whole PvsNP situation

[left snow]

verify that a sudoku solution is valid

prove that a sudoku solution is valid

[gleaming knot]

Same question yeah?

[left snow]

idts though verifying would be way easier compared to proving from the ground up

[sick sandal]

its more
prove you can solve this soduko

[left snow]

if you just dumped numbers into it

[sick sandal]

or

[bold sun]

okk i tried the same logic for that bit and it worked- Thank you bothhh

[sick sandal]

verify the solution

[gleaming knot]

have you all seen "the shortest paper ever"?

[left snow]

oh yeah that is indeed more apt

[gleaming knot]

for the proof that a conjecture proposed by someone was false

That's a full proof that Euler's conjecture is false, a proof that there exists a solution, etc etc etc

[left snow]

yeah but we aren't showing that it's false are we

[gleaming knot]

It's equivalent to proving that there exists a solution to the equation

Look, prove is not a word from another mysterious world

To prove something means to demonstrate it is true, logically

[left snow]

yeah but neither is verify

[gleaming knot]

If all it takes is a calculation, that's all it takes then

[left snow]

verify just means to check

[gleaming knot]

Check that it is true

[left snow]

and here he is checking that yes it works for all vectors x and y

so i don't see why his solution is not ok

[gleaming knot]

If you manipulate both sides of an equation and get 0=0 for example, is that a valid check?

If you assume a statement is true and you show that it implies a true statement, you haven't verified the statement at all

[left snow]

i mean yeah

if i show 0=0 i would consider it a valid check

[sick sandal]

i think the main idea is that verifying is a less rigorous version of proving

[left snow]

verify means
given these cases
test that the statement holds

and here the cases is all x and y with real entries

and he has shown that linearity does hold

[gleaming knot]

That's just a difference in the typical types of problems with "verify" and "prove" in them

But you can swap the word "verify" and "prove" and retain the same meaning

[sick sandal]

so here is what im thinking
does the tv work? verify it! i turn it on "yep it works "
prove it works i have to then explain in details why its working
if i press the power but i am then recieving a picture from the screen sound works and there seems to be no issue or assume the tv doesnt work which means it wont turn on or show anything but i turned it on then its indeed working!*
i am very open to know why this is wrong

[gleaming knot]

Turning it on also proves it works (at least at the time you turned it on) 100%

[left snow]

i feel like to prove that the tv works

i would have to show that the circuit diagram works

but to verify that the tv works

i just turn it on lmao

[gleaming knot]

Have you ever had to prove the existence of something?

[left snow]

yeah of course

[gleaming knot]

What's the proof like

In many cases it's just "Here's an example, it works, the end"

[left snow]

yeah

what does this have to do with verification

[gleaming knot]

How would you "verify" something exists?

You'd look for an example all the same

[left snow]

i mean it's just

in those cases sure you basically do the same thing

[limber sierra]

this just seems like a semantic misunderstanding

[left snow]

but i think in this matrix question

[sick sandal]

assume it doesnt and reach a contradiction

thats 1 way

[left snow]

verify works

[limber sierra]

"prove statement P" and "verify that statement P is true" mean the same thing

"prove statement P(x) is true for x = 7" and "verify that P(x) is true for x=7" also mean the same thing

but its more common to hear "prove" in the former case and "verify" in the latter

just... because

its just a language trend

[left snow]

yeah this is what i was taught

[gleaming knot]

And if your verification OR proof has a serious logical error, it fails to be a verification OR proof

[left snow]

and i don't think it's fair to say that his verification doesn't make sense because he has achieved the question's needs

[gleaming knot]

There's nothing that's a verification that's not a proof

[left snow]

where is the logical error in his proof though?

[gleaming knot]

He started by assuming A(x+y) = Ax + Ay

[left snow]

sure it doesn't look formal but i don't think it's fair to say that it "doesnt make sense"

[gleaming knot]

Already a fatal error

[shrewd thistle]

Anyone know how to solve this?

[left snow]

oh

[gleaming knot]

She* oops

[left snow]

ok

ok fair enough

she should've presented it as left = ... right = ...
because left = right then qed

[sick sandal]

prove
x^2-x has a solution of x=1
find x as if i dont know it
verify x=1 is one of the solution
substitute x

[gleaming knot]

Exactly

[left snow]

ok fair enough

[sick sandal]

what am i missing

[left snow]

didn't catch that

[limber sierra]

prove
x^2-x has a solution of x=1
find x as if i dont know it
this is unecessary

to prove that x=1 solves x² - x, it suffices to just plug in x=1

youre talking about deriving the solution

not proving it

[sick sandal]

fair enough

i guess its just the usage of language here

[limber sierra]

derivations work as proofs

(typically)

but theyre often overkill

[gleaming knot]

I only got worked up when I heard 2 people claim, what it seemed to me, that logical errors in a verification are okay [because we don't expect students to use correct logic in math unless they're taking a proof-based class]

[limber sierra]

(sometimes you do have to verify that your derivation actually produced a valid answer, e.g. you didnt introduce extrenuous solutions or whatever)

[bold sun]

okk lol ill change it to that 🙂

[limber sierra]

(but thats covered in grade school algebra so it should go without saying in #linear-algebra )

(though basically every manipulation in LA is invertible so... whatever)

[left snow]

i briefly skimmed through so my bad on that

lol

[gleaming knot]

Screams in multiplication by 0

[limber sierra]

how often are you multiplying by 0 in LA

[gleaming knot]

How about taking the derivative of a polynomial!

[limber sierra]

its not valid in any sort of row reduction/canonical form/whatever

[bold sun]

lol its okay thanks tho

[limber sierra]

eigenvalues explicitly disallow it

[gleaming knot]

What about multiplication by any matrix in $M_n(\bR)-\GL_n(\bR)$

[stoic python]

Icy001

[winter harbor]

[limber sierra]

that happens more often but a "typical" matrix in M_n(F) is invertible

with very high probability as n gets large

[gleaming knot]

Hmm to call that a manipulation though...

hmm...

What manipulations does one do in linear algebra besides row reduction and inverting a matrix and stuff

[winter harbor]

Depends on what you even mean by a manipulation.

It's not as it has a formal definition.

[limber sierra]

finding eigenvalues, canonical forms, that kinda thing

[winter harbor]

At least, in this sense.

[limber sierra]

change of basis

etc

[gleaming knot]

Linear algebra is fun

It's also the cornerstone of modern math because apparently mathematicians have not found an easier thing to reduce all their questions to than linear algebra

[winter harbor]

Finite dimensional vector spaces are just nice

[gleaming knot]

I have a lattice question that might be appropriate for this channel 👀

I'm stuck on proving that $L^{\vee\vee}\subseteq L$ for an integral lattice $L$

[stoic python]

Icy001

[gleaming knot]

where $L^\vee\coloneqq{v\in L\otimes\bQ:\langle v,x\rangle\in\bZ\text{ for all }x\in L}$

[stoic python]

Icy001

[gleaming knot]

An integral lattice $L$ is a free finite rank $\bZ$-module with an integer-valued symmetric bilinear form

[stoic python]

Icy001

[winter harbor]

Wait, so I suppose $\langle \cdot, \cdot \rangle : L \times L \rightarrow \mathbb{Z}$ is the symmetric bilinear form you are referring to, right?

[stoic python]

MisterSystem

[gleaming knot]

yep

[winter harbor]

Ok

And then we consider elements of the tensor product of L and Q

Which satisfy some property related to this bilinear form

Question

How exactly is this bilinear form defined for elements v in the tensor product of L and Q?

[gleaming knot]

It's extended by bilinearity

$\langle \frac 1nv,\frac 1mw\rangle=\frac 1{nm}\langle v,w\rangle$

[stoic python]

Icy001

[winter harbor]

Yes, and by proving that (L^vee)^vee is a subset of L I suppose actually means that we can embedd (L^vee)^vee in L, right?

[gleaming knot]

We can embed it in L tensor Q

[stable kindle]

14 times a day

[winter harbor]

Because taking (L^vee)^vee we would be dealing with elements in L tensor Q tensor Q

[stable kindle]

it's extremely satisfying

whenever i'm struggling to diagonalise some horrible matrix

just multiply by 0

boom

problem gone

[gleaming knot]

Yep, and Q tensor Q is Q

$\bQ\otimes_\bZ\bQ=\bQ$

[stoic python]

Icy001

[gleaming knot]

I just thought of the idea to choose a basis for both L and L^\vee

This could work

(integral basis of course)

[winter harbor]

We have to be a bit careful here

Because we can't really talk about (L^vee)^vee being a subset of L, as sets these are completely different.

So we want to construct an embedding of (L^vee)^vee into L

[gleaming knot]

They both live in L tensor Q canonically

[winter harbor]

Yes, after doing some identifications.

In any case, it is still just easier to construct an embedding between (L^vee)^vee and L

Them argue via some isomorphisms

[gleaming knot]

I think that's weaker than we want

for example, $\frac 12 L\hookrightarrow L$

[winter harbor]

That we have (L^vee)^vee as a subset of L if we instead view these as living inside another space or whatever.

[stoic python]

Icy001

[gleaming knot]

but $\frac 12L\nsubseteq L$

[stoic python]

Icy001

[winter harbor]

Yes

But we can identify 1/2 L via its isomorphic image

By the embedding

This is essentially the same thing you are trying to do

[gleaming knot]

The way I defined L^\vee

It's a lattice inside L tensor Q

L is also a lattice inside L tensor Q

that allows us to treat vectors in L and L^\vee on the same footing as vectors in L tensor Q

[winter harbor]

Yes, by considering it as its isomorphic image via the map q -> q tensor 1

So really doesn't matter

[gleaming knot]

yep

[winter harbor]

These are just technical details

The thing is that by dealing with embeddings

Is that dealing with linear maps is far more flexible

Than dealing with set theoretic relations

[gleaming knot]

It's flexible but...

I said that $\frac 12L$ can "embed" into $L$ via the multiplication by 2 map

[stoic python]

Icy001

[gleaming knot]

but for this question, if $L^{\vee\vee}$ turned out to be $\frac 12L$, then that would be a counterexample to the claim

[stoic python]

Icy001

[gleaming knot]

(via all the canonical identifications of the lattices with their images in L tensor Q of course)

[winter harbor]

I don't see how? But whatever.

[gleaming knot]

because 1/2 L has vectors not in L

so it would not be true that "v is in L^{\vee\vee} implies v is in L"

[winter harbor]

Is there a name for this construction?

Where is this problem from?

[gleaming knot]

Yeah, it's called the dual lattice

A paper I was reading defined $N$ to be minimal positive integer such that $\frac N2\langle x,x\rangle\in\bZ$ for all $x\in L^\vee$ and I wanted to check for myself that $N$ is the same as the absolute value of the determinant of the bilinear form matrix of $L$

and got stuck on the proof of the proposition I mentioned along the way

[winter harbor]

Wait, is this definition of N correct?

[gleaming knot]

Ahh oops

[winter harbor]

Yeah, it doesn't refer to N at all

[stoic python]

Icy001

[winter harbor]

Nice

Ok

The original problem seems nicer to solve lol

I will think about this for a bit

[gleaming knot]

Yea, let me know if you get anything! I also almost solved my problem, I just need to convince myself that you can always make a dual integral basis for L^\vee such that <ei, \eps_j> = \delta_{ij}

where e_i is an integral basis for L and \eps_j is the hypothetical dual integral basis for L^\vee

[brave lintel]

What even is the point of matrix multiplication

like

why does it exist

[fickle citrus]

What does that mean

[brave lintel]

Why is the multiplying matrix by matrix together useful

[stable kindle]

matrices correspond to linear transformations

multiplication of matrices corresponds to getting the transformation that's two transformations combined

[fickle citrus]

Do you mean $x^{T}y$ or you mean an inner product from vector space of matrices

[stoic python]

ShatteredSunlight

[stable kindle]

i think they mean AB

where A and B are matrices

lol

[brave lintel]

So in that sense its analogous to matrix composition

[stable kindle]

??

function composition, yes

[brave lintel]

yes thats what i meant

[stable kindle]

it's exactly the same, yes

it gives you it explicitly

the numbers

[brave lintel]

ok, ty. It just seemed unintuitive because it doesn't involve matrices of the same dimensions, like composition does

thanks!

[stable kindle]

no

[fickle citrus]

oh wait I thought there was a 'dot' word somewhere, I think I'm seeing words that weren't sent :(

[stable kindle]

function composition is very general

it applies to everything

[brave lintel]

i made a mistake, thats on me

[stable kindle]

even ones of not the same dimension

[stable kindle]

yeah they edited it

[fickle citrus]

_>

[stable kindle]

but composition doesn't have to be matrices of the same dimensions

to be clear

[brave lintel]

i was not aware

[gleaming knot]

Dah of course there is such a dual basis, just take it in $L\otimes \bQ$ for the given bilinear form....

[stoic python]

Icy001

[gleaming knot]

so that's my problem solved

[gleaming knot]

Have you ever solved a matrix vector equation before?

[nocturne jewel]

yes

T[v]=Av

it's stated in the screenshot that A is T's standard matrix

That question also wasn't directed at you.

It's not your question...?

It's not your question...

you're not roy347

[gleaming knot]

Some messages deleted...

[median ocean]

can someone help me part 2 part a I don't remember how to do it.

[sour cloud]

yes i have but idk why im getting this incorrect

do i just choose a vector from the matrix and set it equal to T(v) ?

[nocturne jewel]

You solve T(v)=[18,-26,-21], which is the system Av=[18,-26,-21]^T

@sour cloud

[nocturne jewel]

definition of basis

[sour cloud]

do i choose any row in the matrix?

[nocturne jewel]

??

have you solved systems of linear equations before?

[sour cloud]

i mean columns

yea i have

[nocturne jewel]

So.. solve the system you have

Av=[18,-26,-21]^T is a linear system

[sour cloud]

ohhhhh ok

wait sorry im confused how am i using matrix ?

do i do this equation for each row in the matrix ?

do i need reduced row echelon form ?

got it wrong 😢

i put [20, -30, -36]

and it was incorrect

[nocturne jewel]

how do you write Ax=b into an augmented matrix..?

[sour cloud]

i just wrote the matrix in reduced row echelon form

but ik thats wrong

[nocturne jewel]

Ok... but did you row reduce the augmented matrix?

or just the co-efficient matrix?

[sour cloud]

just the matrix that was provided to me lol

[nocturne jewel]

that's why

[sour cloud]

was i supppose to do something before?

[nocturne jewel]

You need to write the linear system as an augmented matrix then row reduce, like any other linear system

[sour cloud]

ok

[median ocean]

So do I just Solve the matrix:
1 1 1 3
1 -1 1 4
0 0 1 5

by RREF?

for 2 part b

[nocturne jewel]

yes

or you can reduce it to a 2 eqn 2 var system quite easily

[sour cloud]

so this is the augmented matrix : ```
[ 1 1 -4 18
-1 1 4 -26
0 4 1 -21]

[median ocean]

ill just do rref since im good at it

[nocturne jewel]

Ok, but do RREF on a 2x2 system instead, it's quicker

[nocturne jewel]

yes.

so row reduce that

[sour cloud]

ok

[median ocean]

I solved it and got
-3/2
1/2
5

[nocturne jewel]

$c_1+c_2=3-5 \ c_1-c_2=4-5$

[stoic python]

Mosh

[sour cloud]

i got [ 1 0 0 2 0 1 0 -4 0 0 1 -5 ]

[nocturne jewel]

minus sign is fucked up somewhere

[sour cloud]

?

[nocturne jewel]

2-4-5!=3

[sour cloud]

i put [2, -4, -5] and it was right

[nocturne jewel]

oh I was looking at Krusty's system, my bad

[sour cloud]

thank you sm!

[median ocean]

what about part

what about part c

[nocturne jewel]

convert [1,2,3] wrt B into a vector wrt the canonical basis

[median ocean]

So do I just Solve the matrix:
1 1 1 1
1 -1 1 2
0 0 1 3

then solve?

[nocturne jewel]

$[1,2,3]_{\mathscr{B}}^T=1b_1+2b_2+3b_3$

[stoic python]

Mosh

[nocturne jewel]

by definition of co-ordinate vectors

[median ocean]

@nocturne jewel so like this

[nocturne jewel]

yeah, not checking the algebra though

will check the fact you said 0+0+3=0

and 1+2+3=0

[median ocean]

So do I just Solve the matrix:
1 2 3 0
1 -2 3 0
0 0 3 0

[nocturne jewel]

????

[nocturne jewel]

.

[median ocean]

oh nvm then

[nocturne jewel]

just simplify this

properly

[median ocean]

huh

[nocturne jewel]

you said 0+0+3=0

and 1+2+3=0

both of which are false

[median ocean]

oh

it's supposed to equal 0 right?

[nocturne jewel]

no

idk where you got the idea =0 was required

[median ocean]

ok then no 0

[nocturne jewel]

just use definition of co-ordinate vectors, which I posted

[median ocean]

i was thinking something else then

Then I add those up right

[nocturne jewel]

yes

[median ocean]

so it would be
6
2
3

[nocturne jewel]

yes.

$[1,2,3]_{\mathscr{B}}^T=[6,2,3]^T$

[stoic python]

Mosh

[median ocean]

thats it, thats the answer for part c?

[nocturne jewel]

yes.

[median ocean]

for 3 do i write it like this
1 1
1 -1
2 0
1 4
0 0

[nocturne jewel]

transpose it, then you have the matrix.

[vivid field]

Hi, So ive gotten this far and I am trying to find the values of the parameter t for which the system has a unique solution. Can someone provide what direction I go from here. I used cramers rule.

[nocturne jewel]

Ax=b has a unique solution for all b iff A is invertible by FTIM

[median ocean]

@fringe archh if I transpose it then wouldn't it just turn to
1 1 2 1 0
1 -1 0 4 0

[sour cloud]

hey quick question how do you find T(v) if the vector isnt given to us?

i cant do reduced row echleon form since its only x1 and x2

[vivid field]

Not sure what you mean. I know that there is a unique solution for all b If A in not singular. Im confused on if that is all I need to do.

[nocturne jewel]

what do you know about det(A) if A is invertible?

[vivid field]

its not 0

[nocturne jewel]

yes, so you need det(A)!=0

[median ocean]

@nocturne jewel if I transpose it then wouldn't it just turn to
1 1 2 1 0
1 -1 0 4 0

[nocturne jewel]

yes.

[vivid field]

Okay! So in this case, A will have will unique solutions when t is equal to all real numbers excepts 2/3.

Because if t=2/3 than determinate is equal to 0

[nocturne jewel]

🤨 check your det(A) expression again

[vivid field]

?

Please elaborate ? Is that not what it would be than ?

[nocturne jewel]

$(3t)(5t)-(5)(12)$ is det(A)

[stoic python]

Mosh

[median ocean]

@fringe archh so this is the answer?
1 1 2 1 0
1 -1 0 4 0

@nocturne jewel

[nocturne jewel]

row reduce then write the basis.

[median ocean]

so i rref it?

[nocturne jewel]

.

Read what I said.

[vivid field]

So I would just set that to 0 and find the values ? I got t=2, and -2 ?

If that was all I had to do to find values of t did I just waste my time....bruh

[nocturne jewel]

yes, t=+-2 makes A non-invertible

[vivid field]

I really need to upgrade my understanding of linear algebra terminology. Sometimes, I just dont know whats being asked of me so I just do the ole relaible......determinant or RREF

[nocturne jewel]

if you can find FTIM show up, that's usually a good sign, yeah

[median ocean]

i row reduced and got
1 0 1 5/2 0
0 1 1 -3/2 0

[nocturne jewel]

you can reduce that more

[median ocean]

really i thought this was the lowest it can go

[vivid field]

If "S" is a set of all 2x2 singular matrices...

That means any 2x2 singular matrices that I can come up with will be elements of set "S" right ?

Just getting the terminology clear

And also...do 2x2 matrices contain an additive identity matrix such as the zero vector ? Or actually I should ask.. does "S" contain that ?

[nocturne jewel]

R^(2x2) is a vector space, you're asking if the space of singular matrices forms a subspace of it

[sour cloud]

@nocturne jewel do you know how to do mine i kept getting it incorrect.

[nocturne jewel]

Write v as a linear combination wrt the w vectors.

[median ocean]

aint no way i can reduce it anymore

[sour cloud]

how do you write v as a linear combination?

[vivid field]

Yes, so does S contain an additive identity matrix and can you describe it. What are the properties ?

[fringe arch]

Yeah man

[hollow finch]

its just a system of equations

if we suppose that $v=c_1w_1+c_2w_2+c_3w_3$ then thats a system of three equations in three unknowns.

[stoic python]

nix (@ me for the love of euler)

[teal grotto]

Really late but I finally have an answer just in case.

It suffices to count the number of $m$-dimensional sub-spaces of $\mathbb{F}_p^n$. We do this in two different ways.

Firstly, the number of $m$-dimensional sub-spaces of $\mathbb{F}_p^n$ is bounded above by $|\mathcal{P}(\mathbb{F}_p^n)|=2^{p^n}$, since the collection of all $m$-dimensional sub-spaces is a subset of $\mathcal{P}(\mathbb{F}_p^n)$.

Let $X$ represent the number of $m$-dimensional sub-spaces of $\mathbb{F}p^n$. Let $W$ be one such $m$-dimensional sub-space . Since $W$ is isomorphic to $\mathbb{F}p^m$, then there are a total of $p^m$ vectors in $W$, thus the number of ways to choose a basis for $W$ is given by $\prod{k=0}^{m-1}(p^m-p^k)$, since each collection of $m$ linearly independent vectors in $W$ generates the same sub-space $W$. It follows that $X\cdot\prod{k=0}^{m-1}(p^m-p^k)$ is the number of ways to choose a linearly independent set of size $m$ in $\mathbb{F}_p^n$.

Alternatively, we have that $\prod_{k=0}^{m-1}(p^n-p^k)$ also counts the number of ways to choose a linearly independent set of size $m$ in $\mathbb{F}_p^n$.

We obtain our result as
[\prod_{k=0}^{m-1}(p^n-p^k)=X\cdot\prod_{k=0}^{m-1}(p^m-p^k)]
so the number of sub-spaces of $\mathbb{F}p^n$ of dimension $m$ is given by
[\prod{k=0}^{m-1}\frac{(p^n-p^k)}{(p^m-p^k)}=\frac{(p^n-1)(p^m-p)\cdots(p^n-p^{m-1})}{(p^{m}-1)(p^m-p)\cdots(p^{m}-p^{m-1})}]
and we see that [\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{(p^n-p^k)}{(p^m-p^k)}] counts the number of sub-spaces of $\mathbb{F}_p^n$.

[stoic python]

c squared

[winter harbor]

[lavish jewel]

small typo in the second to last product?

[teal grotto]

yes

thanks for pointing that out

[teal grotto]

lol same

[wintry steppe]

hi, what’s algebra ?

[teal grotto]

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

[solemn lotus]

the space of all real polynomials is infinite-dimensional, right? the basis would be defined something like (x_1, x_2, ...)?

[teal grotto]

yea

its a hamel basis even, since each polynomial can be written as a linear combination of finitely many basis elements

[last girder]

Would anyone be able to tell me why this is not true?

[teal grotto]

what if n is even and dim S_1 = dim S_2 = n/2

or more generally, what happens if S_1 and S_2 intersect non-trivially

[last girder]

i see that now but if a vector space V has a basis of n vectors then every basis of V must have exactly n vectors. So why doesnt the union of basis 1 and basis 2 always guarantee a basis = n

[teal grotto]

you're missing the two key points of a basis. not every collection of n vectors is a basis (its not even guaranteed that B_1 U B_2 will have n elements)

they need to be both linearly independent and spanning

if my vector spaces intersect non-trivially, B_1 U B_2 either wont be spanning or linearly independent

[teal grotto]

an example: lets work in V = R^3 and let e_1 = (1,0,0), e_2 = (0,1,0) and e_3 = (0,0,1)

let S_1 = span(e_1, e_2) (this is the x-y plane)
S_2 = span(e_1) (this is the x-axis)

Then dim(S_1) = 2, dim(S_2) = 1 and 2 + 1 = 3 = dim(R^3) and
B_1 = {e_1, e_2} and B_2 = {e_1} are bases for S_1 and S_2 respectively, but B_1 U B_2 = {e_1, e_2} is not a basis for R^3

[last girder]

ahhh

i see it now

i appreciate the help 🙂

but i have one question about what you said earlier

[teal grotto]

sure

[last girder]

what does it mean when two subspaces intersect "non-trivially?" The point at which they intersect is the non trivial solution?

[teal grotto]

two vector subspace S_1 and S_2 intersect trivially if S_1 intersect S_2 = {0}

phrased differently, the zero vector is the only vector that S_1 and S_2 share in common

[last girder]

so if the zero vector is one off n vectors that S1 and S2 share in common, they intersect non trivially

[teal grotto]

S_1 and S_2 could share infinitely many vectors in common

but S_1 and S_2 intersect non-trivially if there is at least one non-zero (or non-trivial) vector that S_1 and S_2 share in common

[last girder]

got it

appreciate the help!

[median ocean]

3 please

[teal grotto]

what have you tried

[solemn lotus]

if you have $A$, $B \in \bR^2$ such that $\cancel{\exists} c \in \bR\ cA = B$, how do you prove that ${A, B}$ spans $\bR^2$?

[stoic python]

CoolShot

[median ocean]

@solemn lotus Do I just write it like this
(1, 1, 2, 1)
(1, -1, 0, 1)

Then row reduce?

[solemn lotus]

?

[median ocean]

Oops

I meant @teal grotto

[wintry steppe]

it's not true. take A = (0, 0) and B = (1, 0).

[solemn lotus]

sorry forgot to mention A, B ≠ (0, 0)

[winter harbor]

Let $A = (a,b)$ and $B = (a',b')$ and consider the matrix:
$$
M

\begin{bmatrix}
a & a' \
b & b'
\end{bmatrix}
$$
Notice that since the columns are linearly independent, then $M$ is non singular which implies that $\forall y \in \mathbb{R}^{2}$, $\exists ! x \in \mathbb{R}^{2}$ for which $Mx = y$.

[stoic python]

MisterSystem

[winter harbor]

That's one way to do it.

Another way to do it

Is just using the fact that R^2 has dimension 2

[winter harbor]

And you requirements imply that A and B are linearly independent.

And since a basis is a maximal linearly independent set and we know the dimension of R^2 beforehand.

It must the base that A and B form a basis

thus, equivalently, they are a linearly independent set that span R^2

[solemn lotus]

gotcha, ty

[wintry steppe]

Just to clarify: If $U$, $W$ are subspaces of $V$ and $U$ and $U\oplus W = V$, then every element/vector $v\in V$, $w \in W$ is linearly independent, right?

[stoic python]

justini

[gleaming knot]

I think so, using plus with circle is also asserting that their intersection is zero

Implicitly

[bold sun]

hey i kinda need a little bit of help basically im combining the printed lecture notes with the ones i made like in the lectures but i dont get this how is 0 -0.5 the same as 2 1

i put a star on it

i tried some working in pencil but its not giving me the same answer like i dont get why he did that step saying 0 -0.5 + lambda 4 3 is equal to 2 1 plus lambda 3

can some one explain?

[gleaming knot]

I can confirm they are in fact different things

No wait

[bold sun]

oooh he sed they were the same

[gleaming knot]

I can confirm that (0, -1/2) is different from (2,1), HOWEVER

The line (0, -1/2) + lambda*(4,3) is the same as the line (2,1) + lambda*(4,3)

[dusky epoch]

(0, -1/2) and (2, 1) differ by a multiple of (4, 3)

[bold sun]

in the lect notes it just has 2 1 plus labda 4 3 as the answer

[dusky epoch]

...you really should use parentheses

(2, 1) + λ*(4, 3)

[bold sun]

yh lol so i dont really get it

[dusky epoch]

you cannot, cannot do away with the parentheses here.

anyway

[bold sun]

would both give the marks?

he wrote both and put eqal signs on the borad

[dusky epoch]

any vector in the form (2, 1) + λ * (4, 3) can be written as (0, -1/2) + λ' * (4, 3) as well

and vice versa

[dusky epoch]

assuming your teacher isn't an incompetent piece of garbage, yes.

[bold sun]

but how did they get there?>

[gleaming knot]

https://www.desmos.com/calculator/x9pf31svxh

I made this cute thing

[dusky epoch]

sounds like they might have just gone for the y-intercept as their base point

[bold sun]

aah okk

[bold sun]

like i dont get what the multiple is for same line?

like i cant multiply cuz anything time 0 is 0 and if i add 2 then id get (2, -0.5+2) which is which is (2, 1.5) not (2,1) if ugm

[gleaming knot]

Did you play around with my cute thing?

You can the numbers in the range for t in both things on the left

[bold sun]

oh okk lemme try that

they just show how they both the same line ??

[gleaming knot]

Did you figure out how it works?

[bold sun]

wdym? i just substituted nos in

[gleaming knot]

Did you figure out how what it's showing on the right corresponds to what you type in on the left

[bold sun]

yh it extends the line or shrinks it

[gleaming knot]

Ya

[bold sun]

the lager the values the longer the line?

[gleaming knot]

Try changing the equation now

[bold sun]

to?

[gleaming knot]

change (0, -1/2) to something else

anything you want

[bold sun]

kk

[gleaming knot]

Play around with it and have make observations and try to figure out how it works

[strange delta]

why is it not onto and one to one?

[bold sun]

the line is parallel if i change that

[gleaming knot]

Ya

How can you predict where exactly the line ends up

based on what you change (0, -1/2) to

also if you change (0, -1/2) to (2,1) what happens

can you find other vectors besides (2,1) where the same thing happens

[bold sun]

thats the same line

[bold sun]

bigger values the more the parallell

[bold sun]

where its the same as (0 -1/2)

thats scale factors of the numbers right

[gleaming knot]

Test your hypotheses

[bold sun]

omg it isnt scalar factors that mean on same linee

[gleaming knot]

:)

[bold sun]

that means scalar factors are parallel

so how do we know if same lineeee

[gleaming knot]

Keep playing around until you find something

[bold sun]

like keep changing the (0, -0.5)

[gleaming knot]

yes

[bold sun]

nothing is workinggggg

[gleaming knot]

Keep trying

Look at which attempts got close

[bold sun]

okk i got 4 and 2.5

[gleaming knot]

Nice

[bold sun]

dunno how it realtes tho?

[gleaming knot]

Well keep playing then

[bold sun]

cuz one is times 2 another is times 2.5

[gleaming knot]

Until you can reliably make another one

[bold sun]

im just gnna end up with similar logic uko

[gleaming knot]

It's not the right logic yet

[bold sun]

6 and 4 worked

thats times by 3 is 6 and times 3 add 1 is 4

[gleaming knot]

Think of other possible ways to get (6,4)

I think the (4,3) is important here