linear-algebra

[winter harbor]

And another function g (say sin(x)) or whatever

And we want to define what f+g does to an element x

We just define it pointwise

[sonic beacon]

f+g = f(x) + g(x)

(f+g)(x) = f(x) + g(x)

[winter harbor]

Exactly

In the case of linear functions

It is the same thing

If I have two linear maps

T,T' : V -> W

[sonic beacon]

(T+T')(x) = T(x) + T'(x)

[winter harbor]

Yup

Notice that

In this case

[sonic beacon]

whats the type signature of this sum

[winter harbor]

We must also prove that the sum of two linear maps is linear

For this sum to be well defined

But is quite trivial

I would recommend doing it as an exercise

[sonic beacon]

ok

[winter harbor]

In general, as Luna has stated before

Notice the following

I could have any set X

And another vector space W fixed

I can then define

$\mathcal{F}(X,W) := {f : X \rightarrow W}$ the set of all functions between X and W

[stoic python]

MisterSystem

[winter harbor]

This is also a vector space as Luna has stated

And we can take the domain to be anything

[sonic beacon]

let me organize some facts to see if i understand: the set of all linear maps between two vector spaces is a vector space (what i want to prove) and what you're saying is that this set under addition and scalar multiplication is also a linear map?

(ill prove that too)

[winter harbor]

Nope

The thing is this

We have the set of all linear maps between V and W

And we have already defined a sum on it

Which is great

But

A priori

We don't know

If this sum is well defined

Meaning that

If I take two linear maps T,T'

Their sum is T+T' is still a linear map

Or that λ T is still linear

All we know so far is that these are functions

But we have also to prove that they are linear

And you use the same definition as before

To prove T+T' is linear

We would go as follows

$\forall u,v \in V$, we have that:
$$
(T+T')(u+v) = T(u+v)+T'(u+v)
$$
But $T$ and $W$ are linear, so we have that
$$
T(u+v)+T'(u+v) = T(u)+T(v)+T'(u)+T'(v)
$$
Using commutativity and associativity, we can rearrange this sum as
$$
(T(u)+T'(u))+(T(v)+T'(v))
$$
But notice that $T(u)+T'(u) = (T+T')(u)$ and $T(v)+T'(v) = (T+T')(v)$
This means that
$$
(T+T')(u+v) = (T+T')(u)+(T+T')(v)
$$

[stoic python]

MisterSystem

[winter harbor]

Here, that's part of the proof of how to prove that the sum of two linear maps is also linear

You'd also need to prove what?

[sonic beacon]

scalar part of linear map condition

[winter harbor]

Exactly

[sonic beacon]

c * T(u+v) = T(c*u+v) ?

^ need to prove that ?

[winter harbor]

You would have to prove that (T+T')(λu) = λ(T+T')(u)

So we can put scalars to the front

And there we go

With this we prove that the sum of two linear maps is also linear

You would have also to prove that multiplication of a linear map by a scalar is also linear

These things are quite conceptual, right?

[sonic beacon]

yeah!

[winter harbor]

They are not intrinsically hard

Just like

Formalism

These things are so intuitive that you can get confused

Try to do it on your time and you will know exactly what you have to do

[sonic beacon]

i will

thank you

i dont have a lot of time to work through it now, but have enough to start and will come back to this

you helped me gretaly

thank you

you gave me a lot so i can start y own proof

[marble lance]

You also need to prove the same things for λT to get that it is linear.

[winter harbor]

Yeah, I stated this here.

I just went over the sum part

Because I thought it would be like

Less confusing

Like, proving that λT is also linear is so trivial that I think can be like

Even a bit confusing pedagogically lmao

[marble lance]

[sonic beacon]

yeah, you guys gave me enough so i can start the proofs. i will come back when i get the time and try the proofs, but i got enough lego pieces to start which i didnt when i asked initially

so thanks again!!!

[winter harbor]

Np

[glad acorn]

One of my answer is wrong and I cannot find out which one is it

[hot swallow]

Can someone explain to me what it means when vector spaces are finitley generated?

[limber sierra]

it means any basis has finitely many elements

i.e. finite-dimensional

[hot swallow]

Ok, so if a question ask which of the following vector spaces are finitley generated? Do i just check if it has finitley many elements?

So if the vector space maps from R to R then it is not finitley generated, right? Cause there are infinite elements

[lavish jewel]

you have a bunch of issues there

something mapping from R to R is a... map, not a vector space. unless you mean some sort of vector space of functions from R to R

[hot swallow]

Ye i meant vector space of funtions from R to R

M bad

[limber sierra]

Do i just check if it has finitley many elements?
no, this doesnt work

ℝ² is finitely generated, say by the vectors (1, 0) and (0, 1)

but it has infinitely many elements

show that any basis of your vector space must have infinitely many elements

[hot swallow]

Okay i think im starting to get it, thanks

[limber sierra]

starter hint: note that the polynomials form a subspace

the space cant be finite dimensional if it has an infinite dim subspace

[glad acorn]

Why 4 is false? I have checked the answer

Are there counter examples?

[slow scroll]

Define equivalent system?

[glad acorn]

Same solution

[nocturne jewel]

Systems are equivalent if they have the same solution space

[winter harbor]

Are you familiar with linear transformations?@glad acorn

At least for me it is easier to see this way

[glad acorn]

How? Can you show it to me?

[winter harbor]

Ok so

We can associate to the system of equations (S)

A linear map

Such that the system of equations may be rewritten as S(x) = b

Where x = (x_1,x_2,x_3, x_4,x_5) and b=(b_1,b_2,b_3)

We can a analogously associate to (T) a linear transformation T and rewrite it as T(x) = d

Notice that

The systems (S) and (T) are row equivalent iff S and T have the same kernel (null space)

And the systems (S) and (T) are equivalent

Iff the solution set S(x) = b and T(x) = d are the same, i.e:
$$
S^{-1}(b) = T^{-1}(d)
$$

[stoic python]

MisterSystem

[winter harbor]

So what we want to try to construct

Are two linear maps

That have different kernels

[wintry steppe]

I'm looking for a rigorous book on linear algebra which I can find on pdf by free, like just first or second undergrad year linear algebra topics. Currently I'm using a book meant for engineers, because is the one I found lol, it's good, but not too rigorous on everything. I want to get the fundamentals right. Thanks

[winter harbor]

But that have the same solution set for two (potentially distinct) vectors b and d

And like, we can construct such maps artificially in some sense lol

For instance

If we fix a vector x \in R^5 in this case

We can extend it to a basis of R^5

Meaning that there are vector y_1, y_2, y_3, y_4 such that {x,y_1,y_2,y_3,y_4} is a basis for R^5

Now

Let b,d,p \in R^5 be distinct vectors non zero vectors in R^5

Define the map S(x) = b, S(y_1) = 0, S(y_2) = 0, S(y_3) = 0, S(y_4)=0

Since we have taken a basis for R^5

This uniquely defines a linear map

Analogously

We can define the linear map T(x) = d, T(y_1) = p, p≠ 0, T(y_2) = 0, T(y_3) = 0, T(y_4) = 0

And there we go

Notice that S and T have different kernels

Since I picked p ≠ 0

But the systems S(x) = b and T(x) = d have the same solution sets (i.e (S) and (T) are equivalent systems)

So they define equivalent system of equations

Whose augmented matrices are not row equivalent

[sonic beacon]

@winter harbor going back to the earlier discussion of the set of all linear maps between two vector spaces.. if i want to define what the sum and scalar multiplication means between elements in this set then am i understanding you correctly by defining it in the following way: T,T': V->W where v in V, and T(v), T'(v) in W and addition is defined by the function signature (T+T'): V->W, where the function is defined as (T+T')(v) = T(v) + T'(v) and scalar mult is defined by the function signature cT: V ->W where the function is defined as (c * T)(v) = c*T( v) ?

[winter harbor]

Yup, that's it

(cT)(v) := c*T(v)

Exactly

What would be a function signature tho?

Ah

Function assignment I guess

[sonic beacon]

what was throwing me off is typically linear maps are defined by one function signature: T: V->W for both sum and scalar mult, but in the above i have two different function signatures for add and scalar?

is that because im taking into account 2 linear maps

and what sum and scalar mult means for them

i usually refer to function signatures as type signatures (due to programming). not sure what to stick with but type signatures make sense to me since you're mapping from type V to type W

[winter harbor]

Damn, I am not so used to type theory lmao

So I might not be of much help

[sonic beacon]

nah dont worry

it's just how i translate things in my head

it's the same thing im doing math not cs

stick with standard LA for the sake of this haha, just a personal translation going on in my head

[winter harbor]

Yeah so, we define the sum to be like this
\begin{align*}

• : \text{Hom}(V,W) \times \text{Hom}(V,W)& \mapsto \text{Hom}(V,W) \
  (T,T') \mapsto T+T'
  \end{align*}
  Where T+T' is the linear map defined by
  \begin{align*}
  T+T' : V& \rightarrow W \
  v \mapsto T(v) + T(v')
  \end{align*}

[stoic python]

MisterSystem

[winter harbor]

So like

Sum of functions is a map that takes two functions

and assigns to it another function

and same goes to multiplication of a function by a scalar

[sonic beacon]

what does the v |-> T(v) + T'(v) notation mean

[winter harbor]

it means that (T+T')(v) = T(v)+T'(v)

[sonic beacon]

whats the function signature/assignment for scalar mult

[winter harbor]

so the function T+T' applied to v is equal to T(v)+T'(v)

\begin{align*}
\cdot : \mathbb{R} \times \text{Hom}(V,W) \rightarrow \text{Hom}(V,W) \
(\lambda, T) \mapsto \lambda T
\end{align*}
Where
\begin{align*}
\lambda T &: V \rightarrow W \
v \mapsto \lambda T(v)
\end{align*}

[stoic python]

MisterSystem

[winter harbor]

So yeah

We are dealing with functions

that take another functions as arguments

and that's how we define sum and scalar multiplication of functions and give Hom(V,W) a vector space structure

[sonic beacon]

the way we define addition/scalar mult across linear maps in this set is the definition of a linear map itself, so do we need to prove that this is the case still?

[winter harbor]

Idk what you mean by that exactly

Like

We have to prove that + is well defined

Meaning that it in fact T+T' is a linear map

[sonic beacon]

(why do we need to prove + is well-defined?)

[winter harbor]

And λ T is also linear

[sonic beacon]

(what does well-defined mean here)

(and why do we need to prove lambda is also linear)

[winter harbor]

Well defined is a math jargon

It basically means like

If we define a function f : X -> Y

We have to make sure that for every x in X

There exists a unique y in Y with y = f(x)

In this case

Uniqueness is a not a problem

The problem is that like

• : Hom(V,W) × Hom(V,W) -> Hom(V,W) in fact maps two linear maps (T,T') to another linear map T+T'

T+T' is a function sure

But is it linear?

It indeed is

But we have to verify

[sonic beacon]

what i meant was X={T, T', ....}, where X is the set of all linear maps between vector spaces V, W. we want to define what it means to add any two vector spaces in X T and T' together and what it means to take a c in F and scalar mult it with T, ie, c * T -- the way we did that is by using the linear map definition on what add and scalar mult means --- is it self-evident from here that this is linear or do we need to verify it? if it isn't self-evident why isnt that the case

you're helping a lot btw, ty for all of this

[winter harbor]

The thing is that we didn't use the fact that T and T' are linear in the definition at all

Like

We can sun two functions

Even if they are not linear

Say sin(x)+x^2

This is a sum of functions

The nice thing about linear functions

Is that if you sum two linear functions the way I had defined it before

We still get a linear function

Get it?

And you can prove this

Using the properties of linear maps

And here we explicitly have to use the fact that these maps are linear

Not necessarily to define the sum

Because as I mentioned before, you can sum even non linear functions.

But linear functions have the property that if we sum two linear functions, the sum is still linear.

[sonic beacon]

ahhh i see why we need to prove this now

ok ty

ok im going to work more on this, ty again for the help

[azure bridge]

can someone help w this question? im not sure what to even do to start

[still lodge]

is there a visual representation of the wronskian

[azure bridge]

umm i just googled that I never heard of what that is

but what i sent is everything i see for the question

[nocturne jewel]

Yeah the Wronskian has nothing to do with your question

[stable urchin]

I think they were just asking their own question

this is just for me, just wondering if the little proof on the bottom right is sound logic with regard to the preservation of linear relations in the row reduction of a set of vectors

[frail trail]

The key concept is to
represent the current orientation of the turtle in space by three vectors
H,L, U , indicating the turtle’s heading, the direction to the left, and the
direction up. These vectors have unit length, are perpendicular to each
other, and satisfy the equation H ×L = U

What is "unit length"

[stable urchin]

this uhm

isnt

linear algebra

[wintry steppe]

When you are doing an LU factorization like this, how do you decide what the pivot will be for the rightmost column? I’ve done two of these and the pivot of the rightmost column is always the wrong part in my answer

Here is an example where my L is correct but my U is not because of the 3rd pivot.

[jagged forum]

Are the blank spots meant to represent 0 in the entries?

[dawn fractal]

funny

[jagged forum]

huh

Did it not upload?

Yes sorry about that, discord was being weird

[dusky epoch]

yeah blanks stand for zero

[jagged forum]

Ok thanks 🙂

[ripe reef]

Can i get some help...

[zinc timber]

it is

I don't see anything wrong here tho

[jagged forum]

Is the answer key wrong? It is claiming (B). However note A^2-2A=0 implies that the minimal polynomial is an unit multiple of P(X)=X^2-2X=X(X-2)=0. By cayley hamilton, the characteristic polynomial is a multiple of the minimal polynomial of any matrix. This implies that 0 is an eigenvalue of A.

[zinc timber]

III is also correct

0 indeed is a eigen value of A

[jagged forum]

Yeah that's what I thought. Obvious I is false. How do we get that A is diagonalizable from A^2=2A though?

[zinc timber]

A has eigen values 0 and 2

as 2 are distinct, there are 2 LI eigen vectors

so...

[jagged forum]

And it could be an n by n matrix

For n=3 for example

Then you'd be missing an eigenvector

So just knowing our two isn't sufficient

[zinc timber]

oh it's nxn, wait let me think

actually yeah

A^2=2A is the minimal polynomial and it's splits into linear factors

with power 1

so it'll be diagonalizable

you get non-diagonalizability if the minimal poly is of the form $(x-a)^n \cdots p(x) = 0$

[stoic python]

Ryuzaki

[zinc timber]

where n>1

[jagged forum]

interesting, I will check that out

[random axle]

What's the typical operating procedure when you calculate the eigenvalues of a 3x3 matrix and one of the roots of the characteristic equation are repeated? In your eigenvalue matrix (diagonalized) matrix do you list just 2 eigenvalues or do you list the repeated solutions in the diagonal matrix?

Just listing two eigenvalues in the diagonal matrix will produce a vector which isn't conformable and hence you wouldn't be able to find a corresponding eigenvector matrix for the diagonalized matrix

[zinc timber]

you are already given that 4 is an eigen value

then why don't u just calc the vector with it?

[random axle]

Right, and I calculated further that -2 is another eigenvalue, but it's repeated

I can calculate the vector for the 4 eigenvalue

[zinc timber]

what's the issue if it is repeated?

[random axle]

The diagonal matrix

Do i put two eigenvalues in the diagonal matrix

[zinc timber]

check the nullity of A-(-2)I

[random axle]

or 3

[zinc timber]

if it's 2 then you have yr diagonal, otherwise you won't

no you can't put the EV's in the diagonal if they are repeated

[jagged forum]

Yeah you were right ryuzaki

https://kconrad.math.uconn.edu/blurbs/linmultialg/minpolyandappns.pdf

[zinc timber]

unless the nullity of A-mI matches with the power of (x-m)

[zinc timber]

yeah I know I am right

[random axle]

sorry, im doing basic high school linear algebra course. i have no clue what this means

[zinc timber]

oh

find the other other eigen vectors

that's what I meant to say

[random axle]

Should the diagonal matrix be like this

$$\begin{bmatrix}
4& 0 & \
0 & -2 & \
\end{bmatrix}$$

[stoic python]

azeem321

[zinc timber]

NO you have 3x3 matrix not 2x2

you said -2 is repeated

[random axle]

yes

[zinc timber]

then?

[random axle]

Ok, I see. But for the eigenvectors

should i have two columns

[zinc timber]

no

[random axle]

with the same eigenvectors

[zinc timber]

3

[random axle]

sorry yes i meant 3

[zinc timber]

take example the identity matrix of order 2

any vector in R2 is an eigen vector

you just have to choose any two of them

such that they are not dependent

do something similar here

[random axle]

What do you mean by dependence

[zinc timber]

one is not a constant multiple of the other

[topaz linden]

Let A be a matrix such that $A^TA=I$ and $Y$ be a symmetric matrix. Find a matrix $X$ such that $Y = A^TX+X^T*A$

[stoic python]

Finitely Many Bananas

[topaz linden]

I need this result for my topology hw

Can someone tell me how to choose X?

[zinc timber]

I can guide you through

remember A'A = I

[topaz linden]

Is this true?

[zinc timber]

try multiplying A and A' in some way with yr original eqtn

[topaz linden]

Just need to verify that first

[zinc timber]

true what?

[topaz linden]

Does such an X exist?

[zinc timber]

you are asked to find one X s.t. its true

[topaz linden]

Well that isn't what the problem is

I'm trying to prove something

[winter harbor]

Bananas

[zinc timber]

you won't know unless you try will you?

[winter harbor]

Have you read the bible?

You would have known...

It's in the bible

[topaz linden]

I see

I have not read the bible

[zinc timber]

lol

[random axle]

The system of equations formed when finding the set of eigenvectors which satisfy the corresponding eigenvalue all lie on 1 line. So how can you find an eigenvector for that eigenvalue which is not a scalar multiple? Unless you just use (0,0,0) as an eigenvector

[zinc timber]

show you work

corresponding to -2?

[random axle]

nevermind, brain wasn't working as usual. thx for the help

[topaz linden]

Won't A^T^{-1} just be A?

[winter harbor]

Yeah, that's the indeed the case since
$
A \in \text{O}(n) \subset \text{GL}(n, \mathbb{R}) \implies AA^{T} = I \implies (A^{T})^{-1} A^{-1} = I \implies (A^{T})^{-1} = A
$

[topaz linden]

$A \in \text{O}(n) \subset \text{GL}(n, \mathbb{R}) \implies AA^{T} = I \implies (A^{T})^{-1} A^{-1} = I \implies (A^{T})^{-1} = A$

[stoic python]

Finitely Many Bananas

[winter harbor]

But like, notice that in this proof we are not using much about O(n) at all

So in general

We could take X = 1/2(A^T)^(-1) B for a general matrix A \in GL(n,R)

So the argument is more general

[upbeat plover]

Say I have an integral with respect to x multiplied by some other function, g(y): g(y) * integral(f(x) dx)
In this case, I can move g(y) inside of the integral because it's not dependent at all on x. But what if g(y) and f(x) don't commute? Do I move g(y) to the left hand side of f(x), or the right hand side of f(x)? That is, which of the following would I do?
g(y) * integral(f(x) dx) = integral( g(y) * f(x) dx)
or g(y) * integral(f(x) dx) = integral( f(x) * g(y) dx)

[zinc timber]

don't commute?

bruh, it's just multiplication

[upbeat plover]

so fg =/= gf

[zinc timber]

what made you think they won't commute

[upbeat plover]

That's what I'm asking though, in a scenario where they don't commute, how would I do this

[zinc timber]

there won't be any scenario when they don't commute

f(x) and f(y) are real(cpx) numbers

[upbeat plover]

ok, I should be more specific - I was trying to be more general but I might have generalised too much

In the scenario I'm actually trying to solve, it's operators in place of those functions

[zinc timber]

you if are working with that general field, then redefine the integral and come back

or just write the integral in terms of infinite sum and see what the order should be

[upbeat plover]

ah good idea

[zinc timber]

that was a joke btw

LMAO

[upbeat plover]

why wouldn't approximating the integral in terms of a sum work though? That implies that g should be on the left hand side of f, what's wrong with that?

[zinc timber]

that will work

but again

they always commute so

[upbeat plover]

I must not be explaining it well, but I'm specifically referring to a scenario where they don't

maybe referring to them as functions at the beginning was a bad way to try to generalise it

[winter harbor]

The approach he mentioned before doesn't make sense in a general field because you need it to have some topology in order to define convergence of series. Also, even working over a general topological field it may still not be enough in order to define a nice notion of riemann-stieljes integration.

[upbeat plover]

ah right

thanks

but yeah, I'm referring to two objects, G, which is not dependent on x, and F, which is partially dependant on x, and I know they don't commute

so I'm still unsure on whether G int(F dx) = int(G F dx) or int (F G dx)

[zinc timber]

so you want to define Integration on a object that doesn't commute?

whatever it's G*F

[upbeat plover]

why is that though? Is there some clear proof I'm missing?

[lavish jewel]

i think ryuzaki doesn't understand your question, and you might be better off asking in the analysis channel

[zinc timber]

No I do, I just don't see the point of defining the integration that way

sure you can, may be u have figured out to define the integration, $g\sum f_i = \sum gf_i = \int gf$

[stoic python]

Ryuzaki

[lavish jewel]

if the functions are matrices, for example, they won't commute

[zinc timber]

integration over matrix?

[lavish jewel]

anywho

depending on the type of integration and the outputs you get

the G would go on the left or right

we can't say without knowing what the transformations are

[upbeat plover]

yeah, specifically I'm working with spinor components

right, ok

[lavish jewel]

if you can treat integral f(x) dx as a single operator, and then you want to use g after that, then g goes on the left

it might also be that g(y) f(x) exists and f(x) g(y) is undefined

i don't really know about spinors to help you out more than this

[upbeat plover]

ok, thanks anyway!

[lavish jewel]

ah but

since they're tensors, you can represent them with a matrix and use usual matrix-vector calculus and such

so if you originally had g on the left, it should stay on the left when it goes inside the integral

since composition of matrices behaves that way too

[upbeat plover]

Ahhh, gotcha! Thank you!

[dusky epoch]

<@&268886789983436800> advertising

[limber sierra]

ty

[undone hamlet]

Are there any good resources for learning the linear algebra proofs?

or the abstractness?

[lavish jewel]

axler's and gilbert strang's books

[undone hamlet]

ty

[little scarab]

Ex: In each of the following, find (if possible) conditions on a and b such that the system has no solution, one solution, and infinitely many solutions.
x+by= −1
ax+2y= 5

not entirely sure what to do here. i converted it to matrix form to try get something in row echelon form maybe, which allows reading off the no solution case I suppose, but not sure about the other two cases

[quasi vale]

@little scarab For unique solution, you need the rank of the matrix of coefficients = rank of augmented matrix = no. of variables

for infinitely many solutions, you need rank of the matrix of coefficients = rank of augmented matrix but not equal to no. of variables

[sturdy portal]

Can you give me a hint (just a direction) how to efficiently solve such ($a_{ii}=0$, $AC=C$) matrix equations in general?

[stoic python]

JohnDark

[hollow void]

What is caliey hamilton theory. I actually can't understand that.

[zinc timber]

u mean the theorem?

[limber sierra]

what part dont you understand?

do you know what a characteristic polynomial is?

[hollow void]

No

[limber sierra]

given a matrix $A$, the characteristic polynomial is $\mathrm{det}(tI - A)$. $t$ is a variable, so the resulting determinant will be a polynomial in $t$

[stoic python]

Namington

[hollow void]

Ohh charesteristic equation

Got it

But it's wrong

[limber sierra]

for example, given the matrix $\begin{pmatrix}4&2\3&1\end{pmatrix}$, the characteristic polynomial is[\mathrm{det}\begin{pmatrix}t - 4 & -2 \ -3 & t - 1\end{pmatrix}]

[stoic python]

Namington

[hollow void]

| A - Xl | = 0

[limber sierra]

which we can compute by hand to be $(t-4)(t-1) - (-2)(-3) = t^2 - 5t - 2$

order doesnt matter

[hollow void]

Ohh ok

[stoic python]

Namington

[limber sierra]

anyway, the statement of cayley-hamilton is that each matrix satisfies its own characteristic equation

[hollow void]

Yes

[limber sierra]

so if we replace $t$ in the above with the matrix $A$

[stoic python]

Namington

[hollow void]

Then

[limber sierra]

\begin{align*}A^2 - 5A - 2 &= \begin{pmatrix}4&2\3&1\end{pmatrix}^2 - 5\begin{pmatrix}4&2\3&1\end{pmatrix} - 2I \&= \begin{pmatrix}22&10\15&7\end{pmatrix} - \begin{pmatrix}20&10\15&5\end{pmatrix} - \begin{pmatrix}2&0\0&2\end{pmatrix} \&= \begin{pmatrix}0&0\0&0\end{pmatrix}\end{align*}

bleh

one sec

[hollow void]

A² - 5A - 2 = 0?

Ohh

[stoic python]

Namington

[limber sierra]

right

it equals 0

[hollow void]

Thanks brother. I actually hates bookish language.

So book is very hard for me

[limber sierra]

and this is the case for the characteristic polynomial of any square matrix.

(over a commutative ring with identity)

[hollow void]

Thank you @limber sierra

[sonic beacon]

i'm trying to prove the set of all linear maps between vector spaces form a vector space and am trying to show that associativity is satisfied but stuck, where (T+T'): V->W, (T+T')(v) = T(v) + T'(v), i have ((T+T')(u+v)) + (T+T')(w) = ((T(u+v) + T'(u+v)) +T(w) + T'(w) -- and then stuck from there, how can I proceed? Can I then say ((T(u+v) + T'(u+v)) +T(w) + T'(w) = (T(u) + T(v) + T'(u) + T'(v) ) + T(w) +T'(w) (due to T, T' being linear maps) = T(u) + T'(u) + T(v) + T'(v) ) + T(w) +T'(w) (due to T,T' evaluating its inputs to fields and fields being assoc) = (T+T')(u) + ((T+ T')(v+w) ) ?

[tranquil steeple]

just start multiplying, from the first row you see a_13 must be 1 for example

[sturdy portal]

@tranquil steeple You've missed the "in general" part but thanks for the reply. I appreciate it

[tranquil steeple]

that would be how you do it in general.

obviously easiest using just a symbolic solver

[sturdy portal]

Alright, thank you for your advice

[tranquil steeple]

if C can be of any form then I can not see any other way

[hollow void]

What is singular matrix?

[sonic beacon]

can anyone look over my proof?

[hollow void]

Ohh ok

😳

[limber sierra]

so a singular matrix is a noninvertible matrix.

[hollow void]

Very few times google shows direct answer

[hollow void]

Yeah. Because determinants = 0
Product of Eigen value also zero

[sonic beacon]

?

[zinc timber]

@sonic beacon ?

[sonic beacon]

@zinc timber :i'm trying to prove the set of all linear maps between vector spaces form a vector space and am trying to show that associativity is satisfied but stuck, where (T+T'): V->W, (T+T')(v) = T(v) + T'(v), i have ((T+T')(u+v)) + (T+T')(w) = ((T(u+v) + T'(u+v)) +T(w) + T'(w) -- and then stuck from there, how can I proceed? Can I then say ((T(u+v) + T'(u+v)) +T(w) + T'(w) = (T(u) + T(v) + T'(u) + T'(v) ) + T(w) +T'(w) (due to T, T' being linear maps) = T(u) + T'(u) + T(v) + T'(v) ) + T(w) +T'(w) (due to T,T' evaluating its inputs to fields and fields being assoc) = (T+T')(u) + ((T+ T')(v+w) ) ?

[stray meteor]

the real coordinate space of n dimensions is basically that if u take the n components of ur vector, they all shud be real numbers

am i right or is this definition flawed?

[zinc timber]

yeah why not

they are linear from VS to VS and the addition you are doing, i.e. T'(a)+T(b) is actually a sum in V' and + is associative in a VS so

yeah

[sonic beacon]

okay cool, ty for checking, does the proof overall look good

[zinc timber]

the proof is simple, u can't go wrong unless u horribly mess something up

[stray meteor]

?

[sonic beacon]

would the last line be something like 'therefore hom(V,W) satisfies associativity"

[zinc timber]

no

[sonic beacon]

ah

how should i end that proof then

[zinc timber]

• is associative in L(U,V)

[stray meteor]

??

[zinc timber]

@stray meteor yeah

[stray meteor]

oh alr then ty

[sonic beacon]

@zinc timber i also proved commutativity and distributivity for hom(V,W), how should i word the last line of those respective proofs? like 'therefore hom(V,W) satisfies commutativity/distributivity" or should i word it differently

[zinc timber]

no

• is commutative in L(U,W)

[sonic beacon]

oooh

[zinc timber]

same language

[sonic beacon]

ok i see

[zinc timber]

a space can only be associative/ comm only under some kind of operation

a set cannot be comm/asso itself

[sonic beacon]

is L(U,V) the same thing as hom(V,W)

ohh i see

[zinc timber]

ya

[sonic beacon]

so i'd say 'therefore + and * is distributive in hom(V,W)' for the other one

[zinc timber]

*?

[sonic beacon]

• for scalar multiplication

[zinc timber]

oh by a scalar

yeah

[sonic beacon]

kk

[fast comet]

I'm a little confused about a homework problem regarding free variables. In a 3x4 matrix, using RREF I get rows 1 and 2 having values, and rows 3 and 4 having all zeros. Normally this means using x3 as a free variable. The question wants me to use x1 as the free variable and I'm confused about how to go about that. Am I missing something, or is it as simple as reordering the columns in the original matrix - swapping C1 and C3?

[lavish jewel]

i'm thrown off right from where you say 3x4 matrix but you say it has 4 rows

[sonic beacon]

@zinc timber i was told i proved assoc wrong, that i needed to prove: (T + T’) + T’’ = T + (T’ + T’’) --- why isn't what i wrote correct and what do i need to prove

[sonic beacon]

in reference to this

[sonic beacon]

i dont understand how to prove (T+T') + T'' = T + (T' + T'')

[gray dust]

for maps T,S:V->W we say T=S if T(x)=S(x) for all x in V. so we must show ((T+T') + T'')(x) = (T + (T' + T''))(x) for all x in V

[sonic beacon]

ok

let me use that to re-attempt

@gray dust will i still use the fact (T+S)(x) = T(x) + S(x)

[gray dust]

yes

[sonic beacon]

ok i will re-try thank you

i was confused because the way i saw it written was like ((T + T') + T'') = T + (T' + T'') without a function input

and was very confused

is it implicitly understood when written in the form of: ((T + T') + T'') = T + (T' + T'') that this means ((T+T') + T'')(x) = (T + (T' + T''))(x)

[gray dust]

right we must distinguish equality of vectors vs equality of maps

[sonic beacon]

i wanted to prove commutativity too which is T + T' = T' + T, so what is the explicit formulation of that? is it T(x) + T'(x) = T'(x) + T(x) ?

[little scarab]

(hoping I can ask a y/n Q: if you have an augmented matrix with a row [0 0 0 1], is that classed as a 'leading 1' for the purpose of determining the rank of the matrix?)

[gray dust]

(T+T')(x)=(T'+T)(x) for all x in V

[little scarab]

(my thinking is no but my textbook isnt really clear)

[gray dust]

that's the definition of T+T'=T'+T

[gray dust]

the 1 is leading bc all entries before it are 0

[sonic beacon]

and lastly i was trying to formulate c * (T + T') = c T + c T' correctly with the (x), is that c * (T(x) + T'(x) ) = c * T(x) + c * T'(x) ?

[gray dust]

the formulating IMMEDIATELY after applying the definition isn't that

that's what you get after also using the definition of T+T'

[sonic beacon]

i wanted to write 'prove expression 1 = expression 2' correctly

so i prove the correct thing

[gray dust]

so recall the def of equality of maps

for maps T,S:V->W we say T=S if T(x)=S(x) for all x in V.

[sonic beacon]

im not sure how to correctly translate c* (T + T') = c * T + c * T'

with the input (x)

[gray dust]

each side is a map

we plug x into each map

[sonic beacon]

c* (T + T')(x) = (c * T + c * T')(x)

like that ^ ?

[gray dust]

rhs yes, lhs should be (c(T+T'))(x)

[sonic beacon]

oh i see

[gray dust]

THEN applying the definition of scalar*map turns that into c(T+T')(x)

[sonic beacon]

ok i have a correct formulation of what to prove for the 3 axioms i wanted to show. i'll work on them to see where i get

[zinc timber]

@sonic beaconu only asked if you can write T(x)+T(y) = T(y)+T(x) or not which it is and that's what I said

also associativity means a(bc) = (ab)c

in case yr operation is + so it'll be T+(T'+T") = (T+T')+T"

[sonic beacon]

cool

ty both

i have enough to re-try the proofs

[chilly wing]

What do the concepts of orthogonal and symmetric matrices mean exactly?

And how are they both related to each other?

[wild fulcrum]

orthogonal matrix has columns form a set of orthonormal vectors

symmetric means when you transpose you get the same matrix (square matrix)

[chilly wing]

What are orthonormal vectors exactly

[wild fulcrum]

• they're orthogonal to each other
• unit length

[chilly wing]

Oh right

Ok thanks

[wild fulcrum]

$\mathbf{Q}^T \mathbf{Q} = \mathbf{I}$ is a direct effect, so whenever you see this, you're dealing with orthogonal matrix

[stoic python]

Vee

[hard drum]

That's how I normally see orthogonal defined (along w Q being square)

[sonic beacon]

is this a valid proof of assoc for Hom(V,W) under +: ((T+T') + T'')(x) = (T + T')(x) + T"(x) = T(x) + T'(x) + T"(x) = (T(x) + (T'(x) + T"(x))) = (T(x) + ((T+T')(x) )= (T + (T' + T")(x)

@zinc timber or @gray dust

[gray dust]

yes

[sonic beacon]

ty for checking

[golden monolith]

If anyone is free for helping on Python, relating to matrices, you can join vc

I really need an exterior point of view on the question asked right now. It's mostly about understanding it

It doesn't seem to make sense

[undone hamlet]

What is the difference between a trivial and nontrivial solution? The book makes it confusing

[golden monolith]

It's not a clear thing, it's just when the reasoning behind finding the solution is easy enough

So it doesn't have to be explicitly written

[nocturne jewel]

like for example, Ax=0 has the trivial solution x=0 cause "no shit x=0 is a solution"

however if Ax=0 also had say x=[1,2,3] as a solution, this would be a nontrivial solution

cause... it's not trivial

[winter harbor]

It is just function notation

If we have a function f : X -> Y between two sets

We denote it in this way

Where X is the domain

And Y is the codomain

If you have two functions f : X -> Y and g : Y->Z where the codomain of one function equals the domain of the other

You can compose them

To get a function g ° f : X -> Z

This question is just asking you to like

Write down the domain and codomain of the composition of these two functions...

g°f : X -> Z is the function such that for all x in X we have (g°f)(x) = g(f(x)).

Like, you have never seen this before?

That should prolly be a prerequisite for linear algebra...

You can search it up on the internet

Function composition or whatever

[silver heath]

Any suggestions for how to do this problem?

I am not quite sure where to start

there are 3 non singular values so

i guess we cant have any variablility

in U and sigma

and V?

idk but i am prolly missing something

[stable swift]

Factor the matrix
$$M=\left[\begin{array}{cc}
3 & 4 \
8 & 11 \
\end{array}\right].$$
as a product of elementary matrices.

[stoic python]

burga

[stable swift]

looking for help youtube isnt doing well

D:

is there an easy way to do it with code?

Im looking to learn that problem, or this one

Demonstrate the use of Gauss-Jordan elimination to compute $M^{-1}$, where
$$M=\left[\begin{array}{ccc}
1 & 4 & -2 \
4 & 15 & 6 \
0 & 0 & 2 \
\end{array}\right].$$
(Recall that $\mathrm{rref}\left[M|I\right]=\left[I|M^{-1}\right]$ whenever $M$ is invertible.

[stoic python]

burga

[silver heath]

Uh

[stable swift]

I think im going with the second problem

First, I form the block matrix M = [A,I] and row reduce M to an echelon form:

[nocturne jewel]

yeah you just augment A with I, then perform Gauss Jordan

[stable swift]

I'm having a hard time writing and understanding it in latex as this example does

[silver heath]

??????

what does that mean, this workbook isn't in latex?

[stable swift]

sorry, Im new to latex and linear algebra, its hard for me, ill ask in the proper channel for latex help

its those combined matrix

[silver heath]

wait

no

thats not latex

that is the propper notation

[stable swift]

yes sorry

[silver heath]

whats confusing about the notation

[stable swift]

well when I tried to copy the format and write that double matrix (the one with M, and the 3x3 identity

i get jumbo #latex-help

[silver heath]

???

are you trying to type it out in latex or write it on paper

[stable swift]

im trying to write in latex, what the book is showing

[silver heath]

Oh

yes then go to the #latex-help

[stable swift]

the block matrix

$$M=\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
4 & 15 & 6 & 0& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right].~$$

[stoic python]

burga

[stable swift]

that seems right so far, can anyone help with the next step

https://cdn.discordapp.com/attachments/540211747613704221/892567438145761280/212329986680422400.png

[silver heath]

do you know how to

put something into reduced row echeclon form?

[stable swift]

well im trying to figure this problem going by the books example so hopefully it will teach me better

im pretty bad at it

[silver heath]

Dude

gauss-jordan is a way to do that

[stable swift]

yes is that the way the book is doing it?

in the example i posted?

[silver heath]

yes

[stable swift]

https://cdn.discordapp.com/attachments/540211747613704221/892569281378799616/Snapchat-1926426597.jpg

so it looks like its just row idenitys or something

but im still a bit lost

[silver heath]

have you even read the textbook?

[nocturne jewel]

ERO's yeah

[silver heath]

or did you skip

a lot of material

[stable swift]

im trying we all learn different sorry man

feeling really sad on these

[silver heath]

maybe watch a yt vid of gauss-jordan elim

[stable swift]

i watched like 3

they lost me a bit with some row thing

[silver heath]

do you know what an elementary row operation is

[stable swift]

something simple like multiplying maybe?

i saw the video explanation on inversing after figuring some stuff out

[silver heath]

I suggest

going back to the basics and don't start learning about inverses till your comfortable with that

stuff

My question got buried :/

how do i aproach this problem?

[stable swift]

write a system of equations maybe who knows not me

[silver heath]

<@&286206848099549185> ?

[stable swift]

help me 2

https://cdn.discordapp.com/attachments/540211747613704221/892569281378799616/Snapchat-1926426597.jpg how did we get from the first matrix to the second

they keep the top the row the same

the second row ___

[stable swift]

Demonstrate the use of Gauss-Jordan elimination to compute $M^{-1}$, where
$$M=\left[\begin{array}{ccc}
1 & 4 & -2 \
4 & 15 & 6 \
0 & 0 & 2 \
\end{array}\right].$$
Known that $\mathrm{rref}\left[M|I\right]=\left[I|M^{-1}\right]$ whenever $M$ is invertible.

First set up the matrix with the 3x3 identity, then add -4 times the 1st row to the 2nd row
$$M=\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
4 & 15 & 6 & 0& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right]~.\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
4 & 15 & 6 & 0& -1 &14\
0 & 0 & 2 & 0& 0& 2 \
\end{array}\right]$$

[stoic python]

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

[stable swift]

how am I doing>?

I dont know why my ~ needs a . to work but thats for another ch

[silver heath]

bro

you have to do it for both sides

[silver heath]

can someone help

[stable swift]

(b) Demonstrate the use of Gauss-Jordan elimination to compute $M^{-1}$, where
$$M=\left[\begin{array}{ccc}
1 & 4 & -2 \
4 & 15 & 6 \
0 & 0 & 2 \
\end{array}\right].$$
Known that $\mathrm{rref}\left[M|I\right]=\left[I|M^{-1}\right]$ whenever $M$ is invertible.

First write the augmented matrix with the 3x3 identity, then RO1: add -4 times the 1st row to the 2nd row. eliminate each column into identity matrices.
$$M=\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
4 & 15 & 6 & 0& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right]~.\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
0 & -1 & 14 & -4& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right] ~.$$

$$!~.\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
0 & -1 & 14 & 0& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right]~.\left[\begin{array}{ccc|ccc}
1 & 0 & 54 & -15 & 4 & 0\
0 & 1 & -14 & 4& -1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right]$$
$$!~.\left[\begin{array}{ccc|ccc}
1 & 4 & -2 & 1 & 0 & 0\
0 & -1 & 14 & 0& 1 &0\
0 & 0 & 2 & 0& 0& 1 \
\end{array}\right]~.\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & -15 & 4 & -27\
0 & 1 & 0 & 4& -1 &7\
0 & 0 & 1 & 0& 0& 1/2 \
\end{array}\right]$$
After eliminating the 3rd column, we arrive at the inverse matrix of \left[\begin{array}{ccc}
-15 & 4 & -27\
4& -1 &7\
0& 0& 1/2 \
\end{array}\right] $$

[stoic python]

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

[stable swift]

can someone help me polish this up

or check it

[winter harbor]

That looks good, yeah.

,w calc inverse of {{1,4,-2},{4,15,6},{0,0,2}}

[stoic python]

Results provided by WolframAlpha

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

Requested by MisterSystem#1236

[stable swift]

Suppose $A,B\in\mathbb{R}^{n\times n}$ are matrices. Say that {\it $A$ is similar to $B$}
and write $A\sim B$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.
Prove that this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
(This is comparable to exercise 3.76 on page 109 of the text. This is also foreshadowing
Chapter 9.)
To prove this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
by checking 3 steps;

1. A \approx A

A=IAI

2)If A \approx B , then B \approx A

If A = PBQ, then B = P$^{-1}$AQ$^{-1}$

3. If A = PBQ and B = P$^{'}$CQ$^{'}$, then A=(PP$^{'}$)C(Q$^{'}$Q)

[stoic python]

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

[stable swift]

does this make any sense at all, I pretty much just copied the example

[sonic osprey]

1 is fine

2 and 3 are not. Read the definition of similar in the second sentence again

[stable swift]

D:

can you point me toward some help @sonic osprey ❤️

[sonic osprey]

Read the definition again

There's only P, there's no P and Q

[stable swift]

O darn

So my answer would be if A=PB, then B = P^-1(a)?

[sonic osprey]

?

Read the definition again

[stable swift]

😵

over my head like woosh

[sonic osprey]

When is A similar to B?

[stable swift]

When B is similar to A

o_O

because uhh

communitative property of equivlence or what

[sonic osprey]

I'm just asking you to read the definition in the second sentence

When is A similar to B?

[stable swift]

Say that {\it $A$ is similar to $B$}
and write $A\sim B$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.

[stoic python]

burga

[sonic osprey]

So can you show that if A is similar to B, then B is similar to A using this definition?

[stable swift]

Suppose $A,B\in\mathbb{R}^{n\times n}$ are matrices. Say that {\it $A$ is similar to $B$}
and write $A\sim B$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.
Prove that this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
(This is comparable to exercise 3.76 on page 109 of the text. This is also foreshadowing
Chapter 9.)
To prove this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
by checking 3 steps;

1. A \approx A

A=IAI

2)If A \approx B , then B \approx A

If A = PB, then B = P$^{-1}$AP

3. If A = PBQ and B = P$^{'}$CQ$^{'}$, then A=(PP$^{'}$)C(Q$^{'}$Q)

[stoic python]

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

[tall rock]

maybe try looking up about matrix multiplication rules first, especially about dimensions

[stable swift]

Does that make more sense for 2?

[sonic osprey]

No?

Why does A = PB make sense? Read the definition of similarity again

[stable swift]

because there exists an invertible matrix P that when mutliplied by B is A

[sonic osprey]

Is that what the definition of similarity says

[stable swift]

yessir

[sonic osprey]

Uh

Say that {\it $A$ is similar to $B$}
and write $A\sim B$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.

[stoic python]

Zopherus

[sonic osprey]

Is this the same as what you said?

[stable swift]

guess not

If there exist an invertible matrix $P$ such that $B=P^{-1}AP$, we can say A ~ B

[stoic python]

burga

[winter harbor]

Yup, it is.

Since reflection is a linear transformation, and we are considering matrices in the standard basis

We only need to check where (1,0) and (0,1) are mapped under reflection on the x axis

We can see that (1,0) is mapped to (1,0) ofc

And (0,1) is mapped to (0,-1)

I think you can see that intuitively

Np

[stable swift]

Im still trying to fix B)

3. [10] Suppose $A,B\in\mathbb{R}^{n\times n}$ are matrices. Say that {\it $A$ is similar to $B$}
   and write $A\sim B$ if there exists an invertible matrix $P$ such that $B=P^{-1}AP$.
   Prove that this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
   (This is comparable to exercise 3.76 on page 109 of the text. This is also foreshadowing
   Chapter 9.)
   To prove this defines an equivalence relation on $\mathbb{R}^{n\times n}$.
   by checking 3 steps;

1. A \approx A

A=IAI

2)If A \approx B , then B \approx A

If A = PB, then B = P$^{-1}$AP

3. If A = PBQ and B = P$^{'}$CQ$^{'}$, then A=(PP$^{'}$)C(Q$^{'}$Q)

[stoic python]

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

[stable swift]

I mean im still trying to fix 2)

[winter harbor]

Just write down a 2×2 matrices with entries a,b,c,d

Then solve a linear system of equations

[tall rock]

you can also answer it with trivial values tbh

[winter harbor]

Yeah so

[tall rock]

@charred root #linear-algebra message

[stable swift]

;x

what operation do we do

with the trivial (A B \ C D)

[winter harbor]

Suppose that $A$ is similar to $B$, then $\exists P \in \text{GL}(n,\mathbb{R})$ an invertible matrix with
$$
B = P^{-1}AP
$$
Then, we have the following chain of equivalences
\
\
$
B = P^{-1}A P \iff P B = P(P^{-1}AP) \iff PB = AP \iff (PB)P^{-1} = (AP)P^{-1} \iff A = PBP^{-1}
$
\
\
Now, notice that this implies
\
\
$
A = (P^{-1})^{-1} B P^{-1}
$
And $P^{-1} \in \text{GL}(n,\mathbb{R})$.
\
\
So $B$ is equivalent to $A$.

[stable swift]

Hmm okay thats a big one

its explained to be understood though

[stoic python]

MisterSystem

[winter harbor]

I will let you prove transitivity

But really

There's not much to it

Just use the definition

[stable swift]

transitive property would be something like if a =b , and b = c, then a =c because transitive property

but there is only A and B?

or do we throw in P there

[winter harbor]

We know that A ~ B (use the definition)

And B ~ C (B and C are similar)

And then you apply the definition

There are P \in GL(n,R) with B = P^-1AP

And S \in GL(n,R)

With C = S^-1AS

And you want to find another invertible matrix

Say H

With C = H^-1 A H

Think about what operations you can do

With invertible matrices

That still yield another invertible matrix

And use it to construct such H

[zinc timber]

that's the defination

[sudden heart]

Hi, I have a question about finding the general solution to a matrix. My textbook mentions that the general solution = the particular + homogeneous. Do I simply take the matrix I was given and set the vectors equal to 0 to find the general solution? And then use the original matrix that is equal to other constants to find the particular solution?

Sorry I am mad confused about it

[lavish jewel]

pretty much

although you can just RREF in the original form and see what lies in the null space

[limber sierra]

set the vectors equal to 0
what vectors?

[sudden heart]

The vectors inside the matrix, or are those just equations?

[lavish jewel]

if you have a system Ax = b, with a matrix A and vectors x and b, you'd have to set b = 0

[limber sierra]

setting the vectors in a matrix to 0 makes me think youre zeroing out the matrix

you mean to set the constant vector equal to 0

which is correct

[sudden heart]

Okay yes, I was thinking of setting the b =0.

[wintry thorn]

dont know if this is regarded as linear algebra

but have some problems regarding mathematical induction

so the first line below for m=1 in the solution section
i dont get how it is derived from the question
please help

[zinc timber]

@wintry thorn#proofs-and-logic

[acoustic path]

x+3=9

[silver heath]

This is a problem for #prealg-and-algebra

[delicate cairn]

Hi all I was finding inverses of matrices today and I was thinking why is the determinant a thing like why does it work where does it come from?

[silver heath]

Do you mean geometric interpretation or ...?

[delicate cairn]

Yeah

And more I just wanna understand it deeply

[silver heath]

So the determinant is the change of area of a linear transformation

for example T(x) = Ax

[delicate cairn]

Yeah we was doing that today

Det of a 3x3 is the change in vol

[silver heath]

yeah

[delicate cairn]

But still like is there a proof of determinant or something ?

[silver heath]

Um.. I don't think there is a proof but if you look up "why does the determinant work" there are some good resources

[delicate cairn]

Alright thanks

[lavish jewel]

the easiest way to understand it is with the eigenvalues, if you've already covered that topic

[silver heath]

he probably hasn't noting that he just learned the determinant, but when you do learn eigenvalues it might be helpful to relook over this

[delicate cairn]

Eigenvalues is next lesson

So I am probably a bit early lmao

We did determinant of 2x2 last year

[wild fulcrum]

maybe watch some visualisation for some intuition

[limber sierra]

its kind of abstract, but the conceptual motivation for the determinant in a mathematician's eye is that it's a scalar value that carries multiplicative info about the matrix

mathematicians like to understand things by looking at simpler things

scalars are simpler than matrices

and the property det(AB) = det(A)det(B) tells us that determinants preserve the multiplicative structure of matrices

this is where things like "invertible iff nonzero determinant" come in

every real number has a multiplicative inverse... except 0

this explanation might not connect with you right away, and thats fine

its an abstract motivation

but its a very handy one to know

[delicate cairn]

It’s a very good explanation I like the way u said it carries info about the transformation

My brain feels dead after doing inverses of 3x3 all day lmao especially when it’s algebraic and no 0 terms

[silver heath]

why would your teacher make you do that

[delicate cairn]

Nah becuase it was in many different contexts like we was just doing the exercises from the text book so some were just find determinant and some were find inverse or solve an equation or find the image of points after a transformation but they all involved calculating Inverse

And when I say all day I mean like 3 hours and probably only like 1 hour total doing the questions

He is a good teacher

[silver heath]

Were you using the 1/det * conj A = A^-1

[delicate cairn]

Yeah

[silver heath]

cuz that one is pure pain

[delicate cairn]

That one

It was easy enough just kinda long

[silver heath]

idk why it exists, because its soo much easier just to do the (A | I) --> (I | A^-1)

[delicate cairn]

Really I thought that way was so much longer

We have only done it using det tho

[silver heath]

rlly, for 3x3+ matrices doing the determinant sucks

but 3x3 rref is much easier

[delicate cairn]

I’ll have a look into it , I mean it’s not that bad when you get used to it because once you have the minor matrix you just transpose and apply the sign matrix in one go

And then det is easy since it’s just the row 1 x row 1 but remembering the +-+

[silver heath]

what if u have a 4x4 or 5x5 matrix?

[delicate cairn]

We won’t on our spec becuase it’s just pointless

I am further maths a level btw going uni within a year

Uk

[silver heath]

learning lin alg before uni?

[summer sinew]

Hi, does anyone have a good simple guide on changing standard basis to a basis consisting of orthogonal vectors given 3 vectors?

v, b1, b2 and b1 and b2 are orthogonal to each other

so im trying to find b in the basis defined by b1 and b2

[winter harbor]

What do you mean exactly?

That we have a certain basis B = {v,b_1, b_2} of R^3

[summer sinew]

I can give like an example

[winter harbor]

and we want to find an orthogonal basis that is induced by this one?

[summer sinew]

im just confused about how to incorporate the dot product

to find the answer

yea i already checked the CosTheta to check if it is orthogonal

so im aware of that

[winter harbor]

If that's what you are looking for

There's something called the Gram-schmidt process

[summer sinew]

so like given vectors v = [5, -1], b1 = [1,1] , b2 = [1, -1]

im trying to find v in the basis defined by b1 and b2

theyre all standard basis

and b1 and b2 i checked are orthogonal

I watched the khan academy video on it but they were using more variables than I was

[winter harbor]

ok, we have v,b_1, b_2 all written in the standard basis of R^2

and we want to write down

[summer sinew]

so It was hard to apply the theory

[winter harbor]

v

written in the basis of b_1 and b_2?

I get you!

[summer sinew]

Thats another interesting principle, my professors recording talked about an R

[winter harbor]

Yeah so

[summer sinew]

but he just defined R in a previous example with no explanation

so IM not sure what R even is, since its not given

I just am looking for essentially Vb

if im not mistaken thats the format

[winter harbor]

Since $\mathcal{B} = {(1,1), (1,-1)} \subset \mathbb{R}^{2}$ is a basis, that means that $\exists a,b \in \mathbb{R}$ such that
$$
(5,-1) = a \cdot (1,1) + b \cdot (1,-1)
$$
Notice that this defines a linear system of equations
\
\
\begin{cases}
5 = a + b \
-1 = a - b
\end{cases}
\
\
So in order to write down $v$ in the basis $\mathcal{B}$ we have to solve this system of linear equations.

[stoic python]

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

[winter harbor]

And notice that the solution to this system of linear equations is a = 2 and b = 3

so v in the basis of b_1 and b_2 has coordinates (2,3)

[summer sinew]

Interesting. Im going to have to break this down.

So what is that sign before R^2

and what does R^2 mean?

[winter harbor]

R^2 is the plane

[summer sinew]

so the size of the vectors right?

so if there were 3 numbers in the vector it would be r^3

[winter harbor]

i.e, $\mathbb{R}^{2} = {(x,y) , \vert , x,y \in \mathbb{R} }$

[stoic python]

MisterSystem

[winter harbor]

yup

in general, R^n denotes the vector space of n tuples of real numbers

the symbol before R^2 means that the basis B is a subset of R^2

[summer sinew]

Ok so is just defining it in linear terms

[vestal cobalt]

help

[summer sinew]

nice,

so where did the 5 come from?

[winter harbor]

v = (5,-1)

[summer sinew]

ah rght right

[winter harbor]

and we want to write it down

as a linear combination of b_1 and b_2

[summer sinew]

so essentially Vb = A(b1) + b(b2)

[winter harbor]

yup

[summer sinew]

now how you transformed that into a linear system of equations

how did you do that

[winter harbor]

v written the basis B is ab_1+b b_2

[winter harbor]

just look at the pairs

(5,-1) = a(1,1) + b(1,-1) = (a+b,a-b)

but like

remember equality of pairs

this means that 5 = a+b and -1 = a-b

[summer sinew]

so is there a more simpler way to get the solution with the dot products?

that was what I was originally trying to go for after the lesson taught it that way

[winter harbor]

There indeed is

and we can use that b_1 and b_2 are orthogonal in order to find that

notice this

[summer sinew]

Thank you so much btw for your time, I hope this isnt annoying man.

I really like linear algebra and like our professor just jumped into the how rather than the why so Im confused.

If im projecting this correctly, B1 and B2 are right angle vectors with V coming straight through the middle, but I dont understand the visual of what the question means im kind of just trying to under stand how to solve]

but I feel like if I KNEW why more it would be so simple

[winter harbor]

Suppose that
$$
v = a b_{1} + b b_{2}
$$
For some scalars $a,b \in \mathbb{R}$, we can then take the dot product on both sides with respect to $b_{1}$
$$
v \cdot b_{1} = a (b_{1} \cdot b_{1}) + b (b_{2} \cdot b_{1})
$$
Notice that since $b_{2} \cdot b_{1} = 0$, we have that
$$
v \cdot b_{1} = a (b_{1} \cdot b_{1})
$$
This then implies that
$$
a = \dfrac{v \cdot b_{1}}{b_{1} \cdot b_{1}}
$$
So we can plug now the numbers in and get that
$$
a = \dfrac{(5,-1) \cdot (1,1)}{(1,1) \cdot (1,1)} = \dfrac{4}{2} = 2
$$
Similarly, we can take the dot product on both sides with respect to $b_{2}$ and get
$$
b = \dfrac{v \cdot b_{2}}{b_{2} \cdot b_{2}} = \dfrac{6}{2} = 3
$$

[stoic python]

MisterSystem

[winter harbor]

And there we go

we got the same answer as before

[summer sinew]

let me look

Thank you so much man

I hope oneday I can be able to understand the theory behind it

because you really summed it up exactly How i wanted

[winter harbor]

Np man

I wanted to show the first method

because it is more general

it works for any sort of basis

this one works for orthogonal basis only

you can try as an exercise

choosing another orthogonal basis

and another vector v

and trying to find the coordinates of v with respect to this basis

Using the two methods I have presentend

[summer sinew]

I am going to thanks.

So when adding up vectors

on the step where it says a = ....

how are you getting 4 / 2

which then = 2

[winter harbor]

I calculated both dot products

[summer sinew]

I know how

Im blanking out

Oh nvm I remember

you multiply the first iteration by the other first iteration

and so on

[winter harbor]

(5,-1) * (1,1) = 5*1 -1 * 1 = 4

yeah

[winter harbor]

#prealg-and-algebra or #❓how-to-get-help

that's not linear algebra

[summer sinew]

so in the case im given more vectors

would I just use v = ab1 + bb2 + cb3 + db4 ?

so c = v.b3/b3.b3

d = v.b4/b4.b4

and so on

[winter harbor]

yup

but why?

The same argument applies

given that b_1, ..., b_4 are orthogonal

[summer sinew]

just how im plotting it out on my mind

yea they're all pairwise

orthogonal

oh so your saying that since theyre orthogonal

the last part of that beginning theorem is to be ignored

since it = to 0

and pretty much the same formula for a and b is the same all around

for all variables

[summer sinew]

whats the fastest way to find whether 2 vectors are linearly independent or dependent

I know the vectors a, b and c are dependent if a = q1b+q2c where q1 and q2 are scalars

but if im only given two vectors A and B

what the hell is C and what does q1 and q2 even mean

[nocturne jewel]

well 2 vectors are dependent iff you can non-trivially take a linear combination and get the 0 vector

[summer sinew]

Nice

[nocturne jewel]

fixed it*

so $av_1+bv_2=0$ and at least one of a and b aren't 0, then $v_1$ and $v_2$ are dependent

[stoic python]

Mosh

[nocturne jewel]

It's strange that you weren't given the general definition for linear independence.

[summer sinew]

I was

its just hard to apply the terminology

like i take notes and my notes say that basis is a set of n vectors that are not linear combination, span the space, and that the space is n-dimensions

but when it comes to applying that it gets interesting so

what does q1 and q2 mean

in the case of a = q1b+q2c

[nocturne jewel]

basis is a linearly independent spanning set, to which the cardinality of the basis (in the finite sense) is the dimension of the space

linear independence is characterized by the ability for the set of vectors to "reach" 0 in a sense.

They're independent if the only way to reach 0 is to set all scalars in the linear combination to 0 and dependent if there's a non-trivial way to reach 0

[summer sinew]

heres a nice image

[nocturne jewel]

so for ${v_1,v_2,v_3}$, they're independent if $$q_1v_1+q_2v_2+q_3v_3=0\implies q_1=q_2=q_3=0$$

[stoic python]

Mosh

[summer sinew]

makes sense

anything other than 0 it is dependent and must be apart of the same dimensional plane

[nocturne jewel]

sure

[summer sinew]

if I was trying to find a = q1b + q2c

is that entailing

vector a = b and c

and then use that to find the q1 and q2?

i am confused on what q1 and q2 imply

[nocturne jewel]

q1 and q2 are just the scalars

[summer sinew]

interesting interesting ok

What if I wanted to find out the scalars with respect to b and c

[nocturne jewel]

however, you're still using a corollary of dependent vectors

instead of the definition of independence/dependence

[summer sinew]

ahh

[nocturne jewel]

testing independence by definition results in just solving a linear system, which is algorithmic

[summer sinew]

got you

thanks!

but in the case of the image above

how would i find q1 and q2

[nocturne jewel]

solve the linear system

[summer sinew]

ah so

[nocturne jewel]

visually for that image, you can just note that 1b+3c=a

[summer sinew]

q1 = 1

and q2 = 0

right?

[nocturne jewel]

no, cause clearly a isnt b

[summer sinew]

i thought q1 is the first iteration of the vector

i must be mistaken then

[nocturne jewel]

a=[2,2]
b=[1,-2]
c=[-1,0]

[summer sinew]

correct

so q1 of b is 1

and q2 of c is 0?