linear-algebra

[limber sierra]

so your vectors are all real polynomials

[radiant yarrow]

Are there prereqs I need to study before getting into LA?

[limber sierra]

and your scalars are specifically constant polynomials

high school algebra

[native rampart]

You need to be able to solve systems of linear equations

[radiant yarrow]

I did that. I studied till calc 1

[limber sierra]

well its just you said you knew what fields were

so i assumed youd know of common examples of rings

but i guess not

[radiant yarrow]

I read about fields in the appendix

I just know it's a set of elements with + and * operations which give out a result in the same set

[limber sierra]

okay then the important takeaway is

when we talk about a vector space V over a field F

we mean that your scalars come from F

and your vectors from V

when we write something like ℝ⁴ with no further information, this typically means the vector space ℝ⁴ over the field ℝ

so your scalars are any real number

and your vectors are composed of 4 real numbers

but if we said "the space ℝ⁴ over ℚ" instead, here our vectors are unchanged

they still consist of 4 real numbers

but now our scalars have to be rational

so sqrt(3) would be a scalar in ℝ⁴ over ℝ, but not in ℝ⁴ over ℚ

(hence you can multiply by sqrt(3) in the former, but not the latter)

[radiant yarrow]

Ohhh

Damn

[limber sierra]

this actually gives rise to radically different vector space structures

[radiant yarrow]

Ohh

[limber sierra]

for example, i can express ANY vector in ℝ² (over ℝ) by taking the following equation:[
a\begin{pmatrix}1\0\end{pmatrix} + b\begin{pmatrix}0\1\end{pmatrix}
] and setting $a$ and $b$ to appropriate values (this is called a ``linear combination")

[stoic python]

Namington

[limber sierra]

but i would NOT be able to do that in ℝ² over ℚ

for example, $\begin{pmatrix}\sqrt{2}\0\end{pmatrix}$ would be impossible to express

[stoic python]

Namington

[limber sierra]

since $\sqrt{2}$ is not a scalar and so i cannot set $a$ to it

[stoic python]

Namington

[native rampart]

Well do 1(√2,0)

[limber sierra]

?

thats a different equation

[native rampart]

Take that as a basis element

[limber sierra]

yes, im trying to show that ℝ² over ℚ has strictly greater dimension

sigh

the point is that ℝ² over ℚ has a fundamentally different structure

since we cant express all its vectors by just taking "linear combinations" of 2 vectors

[radiant yarrow]

okayy...

[limber sierra]

in fact, if you do the algebra, it turns out you need infinitely many vectors

(this is what i meant by "infinite dimensional" earlier)

the axiom of choice asserts the existence of such a set for any vector space, but its typically unconstructible

[radiant yarrow]

Oh so the set of vectors isn't bounded?

[limber sierra]

its size isnt bounded by a finite number

[radiant yarrow]

Ohh okayy

[limber sierra]

but again, this is different from "standard" ℝ² (over ℝ)

in ℝ² over ℝ, we only need 2 vectors

(this is known as the "standard basis"; there are many other choices of basis, but this is the simplest.)

this is mathematically important since it turns out

you can figure out EVERYTHING a linear transformation does

JUST by looking at what it does to these "basis vectors"

this makes linear transformations in ℝ² over ℝ "simple to study" comparatively

since we only need to look at what they do to 2 [linearly independent] vectors

whereas in ℝ² over ℚ, we would need to look at what they do to infinitely many vectors to fully determine them

which, as one can imagine, isnt really feasible

(the study of ℝ over ℚ is closely related to galois theory.)

(the space is so messy that you cant use the "nice" techniques of linear algebra as readily)

aside: this idea of studying transformations of a space by looking at what they do to some "generating element" (i.e. some elements you can recover the entire space from, like (1 0) (0 1) in the above examples) comes up many times in algebra

we say "homomorphisms are determined by how they act on generators"

thats beyond the level of #linear-algebra , but its important to mention

since it emphasizes that this idea is actually very deep and useful

[radiant yarrow]

Okay I read through it twice slowly now, and I think I get it

[digital bobcat]

Can anyone help me with developing an intuitive understanding of the bra and ket in the dirac notation?

Or maybe suggest a resource which does that.

[lavish jewel]

you can think of bras as being in the dual space of a vector space

and the kets are vectors in that vector space

a simple example in R^n would be to consider kets as vectors in R^n, while bras are transposed vectors in R^n

noting that the scalar product of 2 vectors v and w, v dot w, can be written also as v^T w, you could say v^T is a bra, while w is a ket

since v^T is a linear form that takes a vector w in R^n and maps it to the field R

for inner products of square integrable functions, bras would be e.g. integration of some kernel, and bras would be square integrable functions

[dusky epoch]

it's $v^*$ rather than $v^T$

tho i guess you would need the base field to be C

[stoic python]

Ann

[lavish jewel]

i took the field as R and decided to tell no one about it

3:

[stoic python]

!superficialsicko

!superficialsicko

[quartz compass]

you don't as far as I know

but maybe there's some other proof they have in mind

[dawn fractal]

not sure if i even need that

typo.

[stoic python]

!superficialsicko

[dawn fractal]

there

[native rampart]

That should be A*

[dawn fractal]

yeah

[quartz compass]

I assume you just wrote A because it's hermitian

[dawn fractal]

so close to being Hermitian though, idk

[native rampart]

Anyway you know (A-\lambda I)v=0 implies (A-\lambda*)v=0

[dawn fractal]

A is Hermitian as in the given

[native rampart]

Which means (\lambda-\lambda*)v=0

(c-c*)=0 implies c is real

[dawn fractal]

\lambda *?

do u mean $\overline{\lambda}$?

[stoic python]

!superficialsicko

[native rampart]

Yes

[dawn fractal]

right, different notations

[stoic python]

!superficialsicko

[dawn fractal]

i dont follow

[native rampart]

Yes

Ok, That's not straightforward

mb

[lavish jewel]

an alternative way would be to consider x* A x and use associativity

[native rampart]

You know <Av,Av>=<A*v,A*v> because A is hermitian

[lavish jewel]

that's a really messed up path, interesting

[native rampart]

Then just write <Av-cv,Av-cv> and <A*v-c*v,A*v-c*v> and compare both

[lavish jewel]

wouldn't <v, Av> = <Av, v> be easier?

[native rampart]

Probably

[lavish jewel]

cuz anyway the one you wrote will show lambda squared are real

[native rampart]

But,That gives you a useful result for normal operators in general

[marble lance]

My dumbass almost wrote but if lambda^2 is real then so is lambda.

[lavish jewel]

epic i^2 moment

[marble lance]

[native rampart]

Ok, That just gives you the answer

[broken sun]

Hello. How can I show that the double orthogonal complement of a finite-dimensional subspace of an inner product space is the subspace?

[dusky epoch]

are you sure this is true as stated?

are you sure you want only the subspace to be findim, and not the ambient space?

[broken sun]

I think so.

Yes.

[dusky epoch]

let me see if i can find my functional analysis notes from last year...

showing $W \subseteq (W^{\perp})^{\perp}$ is comparatively easy if you just write out the definitions of everything (though i suspect you in particular would find this hard too)

[stoic python]

Ann

!superficialsicko

[dawn fractal]

not sure if this is what you meant by lambda squared being real

i got the first equality by writing it out as the Euclidean inner product

[lavish jewel]

what cursed notation. by v_i^2 i'm guessing you meant |v_i|^2, because the v_i are complex

[dawn fractal]

oh i meant the complex conjugate of all the v_j's

and yes, v_j not v_i

[native rampart]

<v,Av>=c*<v,v>=<Av,v>=c<v,v>

implying (c-c*)<v,v>=0

[lavish jewel]

this, what buncho wrote, gives you the solution directly. c* = c for real c

in what you have, even after correcting the missing conjugates, you get that lambda^2 is real, which doesn't directly help you

you then have to study the relationship between lambda^2, the eigvals of A^2, and lambda, the eigvals of A

[dawn fractal]

yeah, lambda could've been complex while having a real square

[lavish jewel]

indeed

[dusky epoch]

okay so apparently the real thing to prove here is that double orthogonal complement equals the original if your subspace is closed

from which your result will follow since all finite-dimensional subspaces of an inner product space are closed

("closed" here meaning closed in the topological sense)

[broken sun]

Yes. Although I may not be smart enough, in my opinion, that direction is obvious, isn't it?

[dusky epoch]

oh you deem it obvious

mkay

[broken sun]

What does "mkay" stand for?

[lavish jewel]

it's a stylized "ok"

[radiant yarrow]

ok

[broken sun]

Is it possible that we do not use topological concept here?

[dusky epoch]

don't think so

[broken sun]

Why?

I saw that statement in an elementary linear algebra text.

[dusky epoch]

if it was elementary linear algebra then theyre probably assuming the ambient space to be findim too, but you did not

[broken sun]

No, it was not assumed there.

[dusky epoch]

really?

can you send me the textbook

[broken sun]

Yes.

[dusky epoch]

because now i'm curious

does it provide a proof

or does it just assert that

[broken sun]

No.

Friedberg

[broken sun]

Its ebook?

[dusky epoch]

i'd prefer a pdf myself

and also can you

[dusky epoch]

please

[dusky epoch]

stop

[dusky epoch]

reply

[dusky epoch]

pinging

[dusky epoch]

me

[dusky epoch]

every

[dusky epoch]

time

seriously it's getting really annoying

[broken sun]

Sorry for last pinging.

[dusky epoch]

okay im done here

[broken sun]

I did not want to ping you in the last message; I mistakenly edited it.

If I must not ping you, then how can I reply to a specific message of yours?

[dusky epoch]

you can turn off the ping in a reply.

[broken sun]

How?

[broken sun]

Let me check.

[north hedge]

lmao

[wintry steppe]

Yes, they're independent.

but I have to show that for every sin(x) and for every cos(x).

since they're not constant.

any ideas?

[lavish jewel]

my google fu says that if $a \cdot \sin(x) + b \cdot \cos(x) = 0$, then $a \cdot \sin(0) + b \cdot \cos(0) = 0$ and also $a \cdot \sin(\pi/2) + b \cdot \cos(\pi/2) = 0$

[stoic python]

Eρρa

[lavish jewel]

since the linear combination should yield the 0 function, yeah?

but from those two special cases, you can see this implies a = b = 0

[wintry steppe]

fu?

[lavish jewel]

kung fu, but replace kung with google

or google-jutsu, if you prefer

[wintry steppe]

Yeah, yeah.

placing special numbers to it.

I did it too.

But didn't convince me.

you know.

Just placing special numbers on it.

[dusky epoch]

0 means the zero function

[lavish jewel]

the definition of the 0 function is that it is 0 for any value of x

[dusky epoch]

meaning a and b have to be such that a sin(x) + b cos(x) = 0 for all real x, all at once

and in particular it must be true for x=0 and for x=pi/2

[lavish jewel]

what we did was show that you only get the 0 function if a and b are 0; otherwise, you get a nonzero value at at least 1 point, and then it is not a zero func

[wintry steppe]

next time, @lavish jewel will throw me a big giant zero function if I'd ask sth. in terms of zero function again.

[wintry steppe]

okay, but don't we have to go through all possibilities?

Okay, in that case we got. 0, 0.

[lavish jewel]

that's already enough

[dusky epoch]

no, nobody says we "have to" do anything

i mean, if you want to spend literal forever examining continuum-many equations that's on you

[lavish jewel]

those two cases show it's impossible to get 0 from sin and cos at x = 0 and x = pi/2 if a and b are not both 0

we only needed 2 points to show that a and b have to be zero for the result to be 0 at those specific points

if a and b are not both 0, then you have a counter example to a sin (x) + b cos (x) = 0, and none of the following analysis matters

[wintry steppe]

okay. got it.

[lavish jewel]

cuz you couldn't even apply the definition of linear independence to this in that case

[wintry steppe]

yeah.

[lavish jewel]

and everything is fucked

[wintry steppe]

fuked

:D

Thanks @lavish jewel @dusky epoch

[lavish jewel]

tteppann

[dawn fractal]

if an n-by-n matrix A is Hermitian, then how do i prove that eigvectors corresponding to distinct eigvals are orthogonal relative to the Euclidean inner product?

say x1,...,xn and y1,...,yn are the respective coordinates of arbitrary eigvectors x and y corresponding to distinct eigvals

[quartz compass]

take two eigenvectors u, v with different eigenvalues and look at $u^\dagger A v$

[stoic python]

Meρρa

[dawn fractal]

i came up with <x,y> = sum_{k=1}^n x_k\overline{y_k} at the end

[quartz compass]

you can evaluate it in two different ways

[dawn fractal]

hm, idk what $u^\dagger A v$ is

[stoic python]

!superficialsicko

[quartz compass]

conjugate transpose

[dawn fractal]

i did it using the basis of C^{nx1} though

[quartz compass]

take the complex conjugate of the entries and transpose it, just one of many notations to represent this, it's not too serious

u is a column vector

it's an eigenvector of A

[dawn fractal]

for any two distinct vectors in the basis of C^{nx1} = V, the image under f (Euclidean inner product) is always 0

but <v,v> for any v in the basis of V is always equal to 1

[quartz compass]

sure

that's what you're trying to prove, that it has orthogonal eigenvectors

[dawn fractal]

i meant for any two distinct vectors in the basis, not for any two eigenvectors corresponding to distinct eigvals

[quartz compass]

don't think about the components of the eigenvectors, it's not going to be useful for this proof

[dawn fractal]

oh

[dawn fractal]

so i would have $$\begin{bmatrix}u_1\\vdots \u_n\end{bmatrix}^* A\begin{bmatrix}v_1\\vdots \v_n\end{bmatrix}$$

[lavish jewel]

you can use vdots

[dawn fractal]

figured

[stoic python]

!superficialsicko

[quartz compass]

like I said don't waste your time

just write $u^* A v$

[stoic python]

Meρρa

[quartz compass]

now you can evaluate it two ways by making A act on u or v to get different eigenvalues

[dawn fractal]

$u^* Av = u^* (\lambda v)$
\$u^* Av = (u^* A^)v = (Au)^ v = (\gamma u)^* v = \overline{\gamma} u^* v$

here

[quartz compass]

close

u and v have different eigenvalues

[dawn fractal]

right

[quartz compass]

also the overline is unnecessary over the eigenvalue

[stoic python]

!superficialsicko

[dawn fractal]

yes, because they are both reals

[quartz compass]

yeah, good so now you're nearly done with the proof

[dawn fractal]

hm

$u^* (\lambda v) = \lambda (u^* \ v) = u^* Av = \gamma u^* \ v?$\
but $\lambda$ is distinct from $\gamma$

[stoic python]

!superficialsicko

[quartz compass]

in particular focus on this: $u^* (\lambda v) = \gamma u^* v$

[stoic python]

Meρρa

[quartz compass]

as a hint, you can think of $u^*v$ as just being a scalar

[stoic python]

Meρρa

[dawn fractal]

because it's a 1x1 matrix

let me think

the definition of orthogonality for vectors x,y is that <x,y> = 0

[quartz compass]

nah just look at this, you don't have to think about any linear algebra

it's just an equation with scalars in it effectively now

[dawn fractal]

i mean isn't that a property of multiplying matrices by scalars?

jk

i misread

[quartz compass]

yes, that's what it means for vectors to be orthogonal, <x,y>=0

all good

[dawn fractal]

all i'm seeing is some sort of 'associativity' and 'commutativity' going on

[quartz compass]

$u^* (\lambda v) = \gamma u^* v$

[stoic python]

Meρρa

[quartz compass]

write the left side in terms of $u^*v$

[stoic python]

Meρρa

[quartz compass]

like right now lambda is in between there

[stoic python]

!superficialsicko

[dawn fractal]

in here

[quartz compass]

ok so divide by $u^*v$ since it's a scalar

[stoic python]

Meρρa

[dawn fractal]

isn't that a contradiction though

[quartz compass]

👏

exactly good job

[dawn fractal]

now to connect this with orthogonality

[quartz compass]

when can you not divide by a number?

what does that number have to be

[dawn fractal]

nonzero

[quartz compass]

you can't divide when it's zero

[dawn fractal]

i answered ur 2nd question lmao

[quartz compass]

lol

so... what's u*v

[dawn fractal]

nonzero

[quartz compass]

if it's nonzero then you have a contradiction though

[dawn fractal]

..yes..

[quartz compass]

so it must be 0

[dawn fractal]

but we didn't define vector division though

we don't have that

but even if u didn't ask me to divide

[lavish jewel]

[dawn fractal]

i would've still concluded they were the same eigenvalues

[quartz compass]

what are you talking about

it's a scalar

[dawn fractal]

i mean

[quartz compass]

?

[dawn fractal]

how do i even explain that in my proof

[quartz compass]

there's another way to say it but the fact that you are having trouble with it being explained this way doesn't make me want to say it

you could subtract it to one side and factor it out

$(\lambda - \gamma)u^*v = 0$

[stoic python]

Meρρa

[quartz compass]

now you know $\lambda - \gamma \ne 0$ so you can divide it and get $u^*v = 0$

[stoic python]

Meρρa

[quartz compass]

but this is still a scalar

[lavish jewel]

btw watch out for the case of equal eigenvalues there

[quartz compass]

we assumed from the start they were distinct

[lavish jewel]

ah

[dawn fractal]

Alright, my thanks

[quartz compass]

compute some more examples with this stuff

[dawn fractal]

i was expecting it to be similar to the proof of the problem i asked about earlier here

and it was

[quartz compass]

you seem to be unsure about what a lot of this stuff actually is, like when you were talking about vector division

[quartz compass]

yeah good, it is pretty similar luckily

[dawn fractal]

i just didn't think of it more creatively i suppose

[quartz compass]

haha all good, but now you know, pretty fun proof really

[swift agate]

Hey can anyone offer help with this?

https://cdn.discordapp.com/attachments/373217754666500097/846401104782557244/unknown.png

I'm pretty stuck on it

[dusky epoch]

this looks more like multivariable calculus than linear algebra

but other than this it's just a matter of finding all the partial derivatives & arranging them into the jacobian matrix, & then evaluating it all at (-2,1)

[swift agate]

Okay so the first step should be to find the partial derivatives of the 4 different terms in F(x)?

[lavish jewel]

pretty much

[verbal vessel]

Can anyone help me? I have 3 LA questions

Show if R_i < |a_ii| for all i, then A is invertible

[dusky epoch]

something something eigenvalues?

https://www.wikiwand.com/en/Gershgorin_circle_theorem

[lavish jewel]

can also look up diagonally dominant mats

[zealous junco]

yea u basically r show all matrices are invertible

diag dom matrice*

[native rampart]

What have you tried?

[glass cedar]

i need some help on recurrent sequences.

Let's take the fibonacci sequence first
using this matrix formula

https://cdn.discordapp.com/attachments/807226904566038578/846453229029031976/unknown.png

then

https://cdn.discordapp.com/attachments/807226904566038578/846454642665324574/477215470907424807.png

which is very useful in computer science, as raising to a power can be done in logarithmic time complexity
I have noticed that more recurrent sequences have this type of formula for the nth term (multiplying the first few terms by a certain constant matrix)
now the questions i have are

1. do all recurrent sequences have this type of formula
2. how do you find the formula

ping if answered or if you think i should have posted this question in another channel (maybe a more advanced one?)

[quartz compass]

if they're linear recurrent sequences, yes

[glass cedar]

thought so

how would you find the formula then?

[quartz compass]

probably best to think about how the formula for fibonacci sequence is made

it's the exact same idea really

multiply the matrix out and look at why corresponding terms are equal

[glass cedar]

i'm gonna ask a dumb question

how is the one for fibonacci made

i don't have a path for finding it, just the formula

[quartz compass]

do this

[glass cedar]

oh oh alright

[native rampart]

F_n=(1)F_(n-1)+(1)F_(n-2)
F_(n-1)=(1)F_(n-1)+(0)F_(n-2)

[stoic python]

Buncho Dragons

[native rampart]

This is the easier form to digest

[glass cedar]

and we chose a vector with 2 elements because any fibo number is written in terms of the previous 2 elements?

[native rampart]

Yes

[glass cedar]

wow alright that's awesome

makes sense

so now to make sure i got this right

if we have the seq $$A_n=A_{n-1}+A_{n-2}+A_{n-3}$$
then it's safe to say
$$\begin{pmatrix} A_n \ A_{n-1} \ A_{n-2} \end{pmatrix}=\begin{pmatrix} 1 & 1 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix}A_{n-1} \ A_{n-2} \ A{n-3} \end{pmatrix}$$

[stoic python]

ℨєɟɟ̶ค𝔯

[native rampart]

Yes

[glass cedar]

thanks!

[wintry steppe]

i dont understand this statement: "if we have $ax=b$ then if $a=0$ and $b \neq 0$ then $0x = 0 = b$ which is false for any value x, and so there is no solutions"

[stoic python]

zeffs

[stable kindle]

if a is 0, then ax is 0 for all x

so if b isn't 0, there's no b such that ax = b

[wintry steppe]

so if b is 0 then there is no solutions because a would also be 0

[short coral]

Consider a 2D XY plane. Is rotating XY plane by 90 degrees a Linear transformation?

Consider a 3D XYZ plane. Is rotating XY plane by 90 degrees but keeping Z axis constant a Linear transformation?

[wintry steppe]

idk you tell me

can you write out these functions?

is it at least intuitively clear? do your rotations r satisfy r(x + y) = r(x) + r(y) and r(scalar*x) = scalar*r(x)?

[wintry steppe]

can an equation with something on left and right side of equal sign ever have one solution?

[stable kindle]

watch this

x = 1

there's something on the left side and the right side

[wintry steppe]

can i have an example o this?

[stable kindle]

0 x 1 = 0

[wintry steppe]

yes but $b \neq 0$

[stoic python]

zeffs

[stable kindle]

exactly

so there are no solutions

[wintry steppe]

but then $"if we have $ax=b$ then if $a=0$ and $b \neq 0$ then $0x = 0 = b$ which is false for any value x, and so there is no solutions"$ doesnt make sense to me

[stoic python]

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

[wintry steppe]

because we cant have a = 0 and b not equal to 0

[stable kindle]

exactly

there's no values of x that work

[wintry steppe]

but why would a linear system ever have that?

[stable kindle]

a system doesn't need to have solutions

[wintry steppe]

if $a \neq 0$ then it has the unique solution x = b/a for any value of b

[stoic python]

zeffs

[stable kindle]

sure, why not

[minor bluff]

Someone help on q7 pls

I can't find the linear transformation its really frustrating

Okay so I figured out the answer but idk how on earth it worked

[nocturne jewel]

$[5,-3]=5e_1-3e_2$

[stoic python]

Mosh

[minor bluff]

Yeah so then you do 5y_1 -3y_2 etc

But it took me so long to notice that

[nocturne jewel]

yeah cause T is linear

[minor bluff]

The matrix that maps e1 to y1 is not the same as the one that maps e2 to y2

Is it

[nocturne jewel]

T does that, yes

T maps e1 to y1 and e2 to y2

[minor bluff]

A linear transformation can be represented by a matrix right

[nocturne jewel]

yes

[minor bluff]

Why can't we just find that matrix and apply it to (5,-3)

[nocturne jewel]

You can

[minor bluff]

To find the image

[nocturne jewel]

left multiply the matrix representation to the vector in question gives the same as just applying T

$T[v]=(T)_\beta^\alpha v$

[stoic python]

Mosh

[nocturne jewel]

for bases alpha and beta

[minor bluff]

Hmm I've not actually seen that notation yet

The most I've done is T(x)=Ax

Is this the same thing

[nocturne jewel]

if A is the matrix representation of T, yes

[minor bluff]

Ah right okay

So I realised what I did

I got confused because I found the matrix A for e1 to y1 and it was different for the one for e2 to y2

But its a linear combination of the two isn't it

Which is the same as a linear combination of e1 and e2 due to linearity

Yeah I see now, thanks for your help

[nocturne jewel]

$A=\begin{bmatrix}2&-1\5&6\end{bmatrix}$ in this instance

[stoic python]

Mosh

[nocturne jewel]

not sure how you got a matrix that represents T with only 1 of the vectors

[minor bluff]

Basically I got the same as you x_1(2,5) +x_2(-1,6)

Which is the same as what you got

[wintry steppe]

is it always favorable to represent a systems of equations in augmented matrix form?

[nocturne jewel]

what do you mean favorable?

[wintry steppe]

in my textbook it says i can solve a system of equation using EEO(Elementary Equations Operations) but it seems as tho i cant find much information about this online, however it also says i can perform ERO(Elementary row operations) if i convert the system of equations into matrix form

[nocturne jewel]

They should be synonymous I'd think

[wintry steppe]

ero

but if they synonymous why do we need to convert the linear system into matrix to do gaussian elimination?

[nocturne jewel]

you dont need to

[wintry steppe]

oh its just easier?

[nocturne jewel]

augmented matrix just removes the fluff of writing out the variables every line

You can keep it as a system, but each line is going to have every variable

[wintry steppe]

so i could just do the matrix form instead of equation form typically

[nocturne jewel]

yes

[wintry steppe]

so forward elimination can also be performed on a system and not just matrices

[nocturne jewel]

anything you can do to an augmented matrix you can do to a system.. as they are equivalent representations

[wintry steppe]

gauss elimination consists of forward elimination and then doing back substitution right?

[nocturne jewel]

Gauss Jordan, yes

[wintry steppe]

isnt gauss jordan forward elimination then backward elimination

or am i mixing them up

[nocturne jewel]

that's what you just said. .

[wintry steppe]

my book has different definitions of backward elimination and back substituin

[hollow garnet]

Gauss part is the Row Echelon part whereas the Jordan part is the Reduced Row Echelon part

[wintry steppe]

is there a difference between forward elimination and forward reduction?

[native rampart]

How does your book define forward reduction and forward elimination?

[main ivy]

Hey everyone, first time linear algebra learner here. I'm trying to cram as much as I can conceptually before starting a grad program that I have no idea how I got accepted to. I'm learning about linear algebra through a udemy course, but sometimes I feel like they leave some important rules or concepts out. My question is in regards to finding the determinant of this 5x5 matrix. https://imgur.com/a/fqMoNK3

From my udemy course, if two rows in a matrix are the same then the determinant is 0. If i need to bring my matrix into row echelon form and in this example I do R4 - R5 -> R4 and R5 - R4 -> R5 then these last two rows are just the negation of each other. Then if I multiply either row by -1 the two rows will be the same and I would have a determinant of 0. I know this is not the case, but WHY is it not the case? What rule prevents me from doing this for any matrix determinant?

[native rampart]

When you do R_4-R_5->R_4 your R_4 changes

So R_5-(new R_4) != R_5-R_4 in most cases

You can't do two elementary row operations simultaneously

You have to do one at a time

[main ivy]

An amazingly important detail my course has left out - row operations happen one at a time.

[native rampart]

Row reducing a matrix is the same as solving a system of linear equations in a sense

[main ivy]

Another thing I've wondered, must I start a row operation with the row I intend to supplant? So I can do R_3 - R_2 -> R_3, but can I do R_2 - R_3 -> R_3?

[native rampart]

You can

Think of linear equations

If you have
x+y=1 R1
2x+y=2 R2
Then
x+y=1 R1
(x+y)-(2x+y)=-1 R1-R2
Is an equivalent system

A matrix is just this but in a more abstract form

A system of linear equations

By equivalent system I mean the new system and original system have the same solutions

If a matrix A is row equivalent to matrix B,then
Ax=0 and Bx=0 will have the same set of solutions if a solution exists

[main ivy]

Got it, this makes sense

[native rampart]

That's not one elementary row operation tho. It's a combination of -R_3->R_3 and R_3->R_3+R_2

An elementary row operation is always of 3 forms:

1. swap rows
   2)R_i+cR_j->R_i
   3)R_i->cR_i

If it's not in these three, it's not elementary

[main ivy]

the second point can be addition or subtraction (addition of a negative) no?

Or is that then two operations?

[native rampart]

c can be positive or negative

So,included in that

Same for (iii)

[main ivy]

Then I'm lost on how R3 - R2 -> R2 is not one operation. I am subtracting a row from another row and storing within a row occurring in the operation.

[native rampart]

This has to be 0,not for any general c mb

[native rampart]

You have to fix R_2 and add a multiple of R_3 for that to count as a row operation

Remember it's R_i ->R_i + c R_j

Here i=2,j=3

[main ivy]

Meaning then I have to perform some operation on R_2 and store within R_2 for it to remain as one single row operation.

[native rampart]

Do you include R_3-R_2->R_2 when you say "operation on R2"?

[main ivy]

no, because r_3 is the row in which an operation is being performed on, but the result is being stored into r_2, which I've been wondering if it is acceptable as a row operation

[native rampart]

Well,There are many operations you could do on R_2,like you could do R_2+aR_1+bR_3->R_2 if there are 3 rows

Not elementary

[main ivy]

right, so that's the key term here. r_3 - r_2 -> r_2 is not an elementary row operation?

[native rampart]

Yes,it is not

[native rampart]

Check if the row operation is in this format. If not,It is not elementary. There are no other ways in which you get a elementary row operation

Mostly because we define only those operations as "elementary" row operations

[main ivy]

When bringing a matrix into row-echelon as a simple method to find the determinant, what effect do elementary operations have on the determinant?
Suppose in my operations I perform 2R_1 - 3R_2 -> R_1. Do I then need to multiply my end result determinant by 1/2 and 1/3 to cancel the effect of having multiplied my rows by the inverse constants?

[native rampart]

That's 2 elementary operations but w/e(2R_1->R_1 and R_1-3R_2->R_1)

Swapping of rows changes the sign of determinant

R_i+cR_j->R_i doesn't affect the determinant

cR_i->R_i scales the determinant by c

That's det(new matrix)=c det(old matrix) for cR_i->R_i

[main ivy]

got it

[modern relic]

Hey guys, can someone help me on this problem?

https://i.imgur.com/mcANeUl.png

[native rampart]

Take a wild guess

[modern relic]

thanks

[wintry steppe]

what does ran mean

[dusky epoch]

do you have some context

[wintry steppe]

for vectors like we have ker

kernel

[dusky epoch]

it probably means range but i want to make sure

[wintry steppe]

Linear transformation

is the range also called image?

[dusky epoch]

yes...

[wintry steppe]

not good with the english terms

[dusky epoch]

idk why you're reluctant to show me exactly where you saw the ran notation

[dusky epoch]

okay yeah it's range

[verbal wedge]

for a positive definite matrix, is it true that all the entries down the main diagonal must be positive?
aka the trace of a positive definite matrix is always > 0

[dusky epoch]

those are not equivalent

the trace, being the sum of the eigenvalues, is of course positive

but that alone isnt enough to conclude that each diagonal entry is positive

however it is true that the diagonal entries are positive

because $a_{ii} = \ang{Ae_i, e_i}$

[stoic python]

Ann

[verbal wedge]

ah yeah my second statement was incorrect

[stoic python]

!superficialsicko

!superficialsicko

[dusky epoch]

are you confused as to why there's a conjugation there?

or is it something else

[stoic python]

!superficialsicko

!superficialsicko

[dawn fractal]

x and y being column vectors

[dusky epoch]

$x^*y = \sum_{j=1}^n \overline{x_j} y_j$ actually. and under your definition, $\ang{x,y} = y^*x$, not $x^*y$.

[stoic python]

Ann

[dusky epoch]

conventions differ as to which argument in the inner product gets conjugated. i think physicists generally conjugate the first and mathematicians the second? but don't quote me on that.

[dawn fractal]

basically they are complex conjugates of each other

that's where i was confused

so idk how they are equal

[dusky epoch]

are you sure the theorem was $x^*y = \ang{x,y}$?

[stoic python]

Ann

[dawn fractal]

yes

[dusky epoch]

and are you sure it's the y's that got conjugated in the definition?

[dawn fractal]

number 5

third pic talks about how Euclidean inner product is the f in Example 36 but with all a_i = 1

[dusky epoch]

third line in remark 37 should have F = R...

this book's full of misprints it looks like

[dawn fractal]

it's actually my prof who typeset it

lmao

[dusky epoch]

then talk to your prof

[dawn fractal]

so it should be a typo

[dusky epoch]

thm 45 item 5 should have y*x

[stoic python]

!superficialsicko

[dawn fractal]

right?

[dusky epoch]

yeah, sure, that works

i'd rather it be y^* x tho

[dawn fractal]

the pdf did have a few other typos so im not surprised at this point

ty

[stable kindle]

let infinity = x

then x = (1)x = (2 - 1)x = 2x + -x = x + -x = 0

[dusky epoch]

careful, you're implicitly assuming this addition they've defined is associative

which it is not

[stable kindle]

ok i don't see how i'm assuming that lol

[dusky epoch]

well ok no

[stable kindle]

all i see is distributivity

[dusky epoch]

i guess you're assuming the distributive law

which doesnt hold either

because of the infinities

[stable kindle]

right

[dusky epoch]

but i think explicitly breaking assoc is easier

try the two different ways of parenthesizing $\infty + \infty + (-\infty)$

[stoic python]

Ann

[stable kindle]

so if we assume distributivity, contradiction because x = 0 = -x, but by assumption x and -x aren't in R, so does that do it

oh ok

yeah, there's no associativity

[dusky epoch]

$(\infty + \infty) + (-\infty) = \infty + (-\infty) = 0 \ \infty + (\infty + (-\infty)) = \infty + 0 = \infty$

[stoic python]

Ann

[dusky epoch]

these are not equal, therefore addition isnt associative, therefore no hope to be a vector space

[stable kindle]

yes

alright, onwards

ty

[stoic python]

!superficialsicko

!superficialsicko

[dusky epoch]

normal means A commutes with A* right

[stoic python]

!superficialsicko

[dawn fractal]

indeed

also that too

i tried using associativity

[dusky epoch]

hermitian => real eigenvalues does not require normality iirc

[dawn fractal]

indeed

[dusky epoch]

so really we need the other way around

real eigenvalues => hermitian

[dawn fractal]

ive already proven that in a previous item

[dusky epoch]

hm

[dawn fractal]

i was wondering whether to use inner products or not

if not, how i can make use of these three facts

using associativity perhaps, but..

[native rampart]

If T is normal and c is an eigenvalue,then c* is an eigenvalue of T*

[dawn fractal]

we dont have that theorem sadly

but if i can prove that then good

[dusky epoch]

isnt c* an eigenvalue of T* anyway tho

regardless of normality

[dawn fractal]

sooo

[native rampart]

How would that finish the proof?

[dawn fractal]

not implying that

so how

[native rampart]

Do you know spectral theorem?

[dawn fractal]

1. $AA^* = A^*A$\
2. $Ax = \lambda x$ for some nonzero (eigen-)vector $x\in\C^{n\times 1}$\
3. $\lambda = \overline{\lambda}$

yep

[stoic python]

!superficialsicko

[dusky epoch]

there are ways of doing lists in TeX

specifically you want the enumerate environment

here's how to do it:

[dawn fractal]

ik that lmao

[dusky epoch]

oh

[native rampart]

Then write your matrix of T in an orthonormal Eigenbasis

[dawn fractal]

its faster this way tho

[native rampart]

Take conjugate transpose to get T*

[dawn fractal]

A is unitary diagonalizable iff A is normal

[native rampart]

If c=c* for all eigenvalues you have that the two matrices are equal

[dawn fractal]

is that the spectral theorem?

[native rampart]

Yes

[dawn fractal]

can i make use of $\C^{n\times n}$ standard basis?

[stoic python]

!superficialsicko

[native rampart]

Not in general

Unitarily diagonalisable means your Eigenbasis is orthonormal

[dawn fractal]

i dont think we have that theorem

[stoic python]

!superficialsicko

[dawn fractal]

would i need to use gram-schmidt for that?

im lost lmao

[native rampart]

You know U*AU is the matrix of operator in some basis?

The basis consists of columns of U

[dawn fractal]

what exactly is a "matrix of operator"

so i might not know

[native rampart]

You know representation of operators as matrices requires you to choose a basis?

And change of basis formula helps you to find the representation in a different basis

[dawn fractal]

ah u mean representations of linear operators, yes

[native rampart]

Yea that

[dawn fractal]

yes

[native rampart]

You ok with this?

Now this being unitary means the basis is orthogonal

And the basis is Eigenbasis because the matrix is diagonal in that representation

[dawn fractal]

why is that

also are u talking about this

[native rampart]

Yes,That

[dawn fractal]

where U = B2PB1

so how did the basis consist of U's cols

[native rampart]

Yes

Try constructing B2PB1

[dawn fractal]

each col of U corresponds to an element from B1 written with a coordinate matrix relative to B2

but those are the same bases

?

[native rampart]

Also,This should be U^-1 I think

If that's confusing,Take the columns of U

Since U is invertible they form a basis

Let this basis be B_2

Now try seeing what B_2PB_1 will be where B_1 is your standard basis

I think you will end up with U or U^-1 not sure which one

[dawn fractal]

oh, are you actually talking about since A is (unitary) diagonalizable, there's a basis for C^{nx1} containing A's eigenvectors?

[native rampart]

Yes

[dawn fractal]

oh u should've led with that lmao

i was confused as to how u obtained an eigenbasis from U

or U*

[dawn fractal]

ok, i.e. a basis for the colspace of U

[dawn fractal]

so yes

[native rampart]

You know U is unitary implies the basis formed from col space is orthogonal?

[dawn fractal]

was also confused since u didnt specify what the basis is a basis for

[native rampart]

mb

[dawn fractal]

nope

i dont see that directly

[native rampart]

https://math.stackexchange.com/questions/1688950/why-do-the-columns-of-a-unitary-matrix-form-an-orthonormal-basis

[dawn fractal]

ok, but im afraid id have to prove that too along with A being Hermitian

just to refresh, heres what i need to prove

maybe post the entire problem for context, but this channel is currently occupied

[stoic python]

!superficialsicko

[native rampart]

Your thing follows easily if you show your version of spectral theorem is equivalent to "every normal operator has an orthogonal Eigenbasis"

[dawn fractal]

operators having bases? that's new for me

i.e. linear transformations (on the same vector spaces) having bases

[native rampart]

Ok,Just ignore that I am being dumb

[dawn fractal]

idk if this proof is supposed to be long anymore lmao

[native rampart]

idts

[native rampart]

as in there is a basis such that the representation matrix in it is diagonal

Just read up on the various forms of spectral theorem

This follows directly from one particular form

[dawn fractal]

ok, ill continue this in 14 hours

need time to digest all this

[native rampart]

Just read this when you come back

@dawn fractal

(ping so that you can find this easily)

[dawn fractal]

will do

[wintry steppe]

what does this sign mean?

context?

probably "not equals"

post the full image

nvm you got your answer in #math-discussion

[quartz compass]

it's $\rotatebox{-90}{$\pi$}$

[stoic python]

Meρρa

[wintry steppe]

ye

[quaint flint]

Is calc workshop best resource for learning linear algebra

[dusky epoch]

>calc workshop
>linear algebra

[quaint flint]

https://calcworkshop.com/linear-equations/

Check the link

[dusky epoch]

okay so their name is just misleading, got it

[quaint flint]

Any thoughts

I’ll have an midterm this Friday afternoon and want to get a very good score

Chapters 1 and 2 in lay and McDonald

[quartz compass]

I think you should ask a specific question about a linear algebra problem you've attempted to work on but can't figure out

[quaint flint]

I don’t have any specific question

My question to begin with was a broad open ended one

In that what’s a good online resource to get a comprehensive understanding of linear algebra for my midterm

@wintry steppe

[vast iron]

I liked Gilbert's lectures, you can probably find (at least elementary stuff) at MIT open courseware. It's better to work through a textbook though imo.

[short coral]

Suppose for matrix A which is of dimension n*n, it's given that COL SPACE = NULL SPACE

What can you say about A^2 ?

What can you say about A^2 * X , where X is a non zero vector?

[wintry steppe]

i can say a lot about A^2, and also a bit about A^2 * X where X is a non zero vector

wbu?

do you know what COL SPACE and NULL SPACE mean?

[short coral]

Col space = Set of Linearly independent column vectors
Null space = Set of Linearly dependent column vectors

[lavish jewel]

that's not right

[wintry steppe]

column space = span of the matrix's columns
null space = the set of x with Ax = 0, A being your matrix

[dusky epoch]

a space is not the same as its basis

[lavish jewel]

gang appearance of tteppann

[wintry steppe]

considerable

[coarse marsh]

is a zero matrix Symmetric?

[wintry steppe]

Do you know what a symmetric matrix is?

[coarse marsh]

it's transpose is equal to the matrix itself

right?

[wintry steppe]

A matrix is symmetric iff it is equal to its transpose yes

[coarse marsh]

so this should be true?

cause i just got an 0 for saying it is

[wintry steppe]

The transpose of a nxn zero matrix is itself

so those are symmetric

if it was a nxm zero matrix then its not symmetric

[coarse marsh]

it's a 2x2 zero matrix

[wintry steppe]

talk to your teacher then

[coarse marsh]

sure will

thanks

[wintry steppe]

Is there any weird trick for this? ik this is true cuz T-9I injective else 9 would be eigenvalue, therefore also surjective cuz operator and finite dim. Is there another way or are the specific numbers just red herrings?

[verbal pivot]

pretty sure the numbers are just red herrings (This is a question from Axler's book right?)

[wintry steppe]

yea Axler

[wintry steppe]

lol that's a troll question

[minor bluff]

is orthogonally diagonalizing a matrix the same as diagonalizing it as normal?

because P^-1 = P transpose right

[minor bluff]

<@&286206848099549185>

[weak needle]

An orthogonal diagonalisation of a matrix will also be a normal diagonalisation, but not the other way around

[minor bluff]

ah i seeeee

makes sense, thanks

[misty storm]

how would I go about finding uA and uB?

feels wrong to ping helpers for what I think is an easy question

[verbal pivot]

uhh, what does uA even means?

[misty storm]

well, u is a vector, and A is a base for the linear space V

I'm not entirely sure what to make of it

a similar practice exercise showcases this

[verbal pivot]

ah, so by $u_A$ you mean u in terms of the basis A?

[stoic python]

b2unit

[misty storm]

so coordinates in respect to a basis, maybe?

[misty storm]

I suppose, yes

the test is in a week and I'm trying to make sense of all of this stuff

[verbal pivot]

well supposing it is (you should revise if this is indeed correct), then the straightforward way of solving for the coordinates, is to solve the system of equations created by the equality $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = u$ (where the lambdas are the coordinates and the v's are the basis vectors), if you express the vectors in terms of the canonical vectors (in this case, those are ${1,x,x^2}$), then you will obtain a system of three equations and three unknowns (the three lambdas)

[stoic python]

b2unit

[verbal pivot]

but sometimes, finding the coordinates is easier than that

[misty storm]

thanks

[misty storm]

1. What should I be googling for?
   2.The coordinates which lambda represent would be x^2,x^1 and x^0? basically?

so, in the case of this exercise: 3 -6 +5?

[verbal pivot]

uh rather than googling it would be better to compare it with your class notes or the textbook that you are using, but if you want to google anyways,a good search term to begin with would be "coordinate and change of basis" I think

[verbal pivot]

the lambdas would be the coordinates you want to find, if you are saying 3 -6 +5 because those are the coefficients of u then that reasoning would be incorrect, 3 -6 +5 are the coordinates of u with respect to the canonical basis {1,x,x²}

[misty storm]

right

and I want to find u with respect to A rather than it's canonical basis, is that right?

[verbal pivot]

well, assuming that u_A means that, then yes

that's the system

(I wrote it out on paper since I forgot how to do the vectors in latex lol)

[misty storm]

omg, thanks a lot

I really appreciate it

[bitter cape]

I'm having a little trouble figuring out bijection. Can someone help me understand two examples? (That'd be [T(x, y) = (0, 0) and T(x, y) = (2x, y)]

[Please. Also, not saying both are, but two examples to understand the reasoning, cause I've been unable to so far]

[lilac stratus]

I don't understand what you're asking ? You're asking if both of these maps are bijections ?

[bitter cape]

Uhum

[lilac stratus]

ok so a map f: X -> Y is bijective if:

• f is injective, meaning that two distinct elements can't have the same image, i.e f(x) = f(y) implies x = y
• f is surjective, meaning that for any y € Y, there exists x € X such that f(x) = y, so basically "every element is reached"
  Here, I assume T goes from R² to R² ?

[bitter cape]

It does, yes

I understand it on a theoretical level, but been having trouble matching this to the actual questions/examples

[lilac stratus]

Ok so first let's look at T: R² -> R², (x, y) |-> (0,0).
Does every element of R² seems reached ?

[bitter cape]

No 😂

[lilac stratus]

Right, so that means it is not surjective.
So it can't be bijective, right ?

(not it isn't injective either, do you see why ?)

[bitter cape]

Uhum. This one is a little easier since it's so 'final' with the (0, 0)

[lilac stratus]

ok then let's look at the second map T: R² -> R², (x, y) |-> (2x, y)

[gray adder]

Hi, Can anyone guide me with polyphase filters used for decimation and interpolation?

[lilac stratus]

First is it surjective ? If I take (a, b) € R², is there some (x, y) € R² such that T(x, y) = (2x, y) = (a, b) ? @bitter cape

(hint: ||this reduce to solving two distinct one variable equations||)

[thorny plume]

can someone tell me how they converted that equation into that vector form?

[dusky epoch]

it's a bit of a sleight of hand, but it appears they are implying that b should satisfy 3b_1 + b_2 = 0

which can be considered as a system in its own right, with 2 unknowns and 1 equation

[wise wyvern]

what does the equals sign with dot above mean?

isomorphic to ?

[dusky epoch]

i think it means the same as :=

i.e. 'defined as'

[wise wyvern]

aha

[dusky epoch]

also is this translated from another language

because "non-negatively defined" just sounds super wrong to me

[wise wyvern]

yea most likely.

On wikipedia it says non-negative means > 0

positive would be >= 1

[lavish jewel]

backwards

[wise wyvern]

i mean all elements in the matrix would be > 0

[dusky epoch]

that's so bad

[dawn fractal]

nonnegative means >=0

[dusky epoch]

no, this isnt what positive-definite means

you can have a matrix contain some negative entries and yet remain positive-definite

[wise wyvern]

oh sorry yes, >= 0

[dusky epoch]

positive means >0, nonnegative means ≥0

[wise wyvern]

correct

this class isnt fun. The professor translated all the materials and homework so it is very confusing

the class is like programming and neural networks, but the final exam practice is only linear algebra

[dusky epoch]

translated from what language, if you don't mind sharing

[wise wyvern]

polish

[dusky epoch]

...okay that was closer than i thought itd be

i thought it might've been translated from russian

[lavish jewel]

i mean, ML deals very much with linalg

[wise wyvern]

it does, but these problems weren't part of the course

[lavish jewel]

aha

[wise wyvern]

my only complaint is that the exam is really obtuse and not very related to the course.

[stoic python]

!superficialsicko

[dawn fractal]

i ended up doing this but encountered the above problem

[stoic python]

!superficialsicko

[spark marlin]

I had a question, I worked through a proof sketch but ended up with $c_1$ in place of $c_0$. Did my lecturer make a typo or am I wrong here. I used induction on k for $P_k(z)$

[stoic python]

Panda_Bear59

[dusky epoch]

never

$v^*v = \nrm{v}^2$

[stoic python]

Ann

[dawn fractal]

oh right, from definition of norm

but <v,v> is only 0 when v is 0

so that is not 0

ty

[dusky epoch]

yeah but you said v ≠ 0

yeah

[dusky epoch]

you should have c_n not c_1

[spark marlin]

ah i was thinking that too, but was trying to make c_0 work. So it is a typo?

[dusky epoch]

it is a typo yes

[thorny plume]

is orthogonally diagonalising this matrix the same as diagonalizing it as normal, since its symmetrical?

oh wait, nvm its not symmetrical is it

[dusky epoch]

it's not symmetric, unless you insist 6 = 0

[thorny plume]

now that its symmetrical, diagonalizing it as normal is the same as orthogonally diagonalizing right

[wintry steppe]

you have to adjust the eigenvectors to have norm 1

and if you have repeated eigenvalues, you have to choose orthogonal eigenvectors

otherwise theyre already orthogonal

[wintry steppe]

A^k means AxAx...A k times

What do you think they are asking?

They first ask you to calculate a few small powers of those matrixes

so makes sense they ask you for a general expression for their powers after

[lavish jewel]

that's the whole point, no?

v*v is not zero, but the eigenvalue minus its complex conjugate is

[wintry steppe]

for example if D is the following matrix
[0 1]
[0 0]
then D⁰ is identity, D¹ is D and D^k for k>=2 is the 0 matrix

[dawn fractal]

how do i prove that if a square matrix A is diagonalizable, then A has a square root?

[wintry steppe]

if you had a diagonal matrix, can you find a square root of it?

[dawn fractal]

i do know that $I_nI_n = I_n$

[stoic python]

!superficialsicko

[dawn fractal]

but diagonal matrices are simply identity matrices multiplied by scalars

[wintry steppe]

nah

that is false

[dawn fractal]

oh right

so how

[wintry steppe]

Well, the whole point of diagonalizing

is that it's easier to work with diagonal matrices

do you know how to calculate the square of a diagonal matrix?

[dusky epoch]

a real square root?

[dawn fractal]

simply squaring the diagonal entries

[dusky epoch]

you will need its eigenvalues to be ≥0 if you want a real square root

otherwise it's not guaranteed

[dawn fractal]

this is the theorem

that i need to prove

[native rampart]

Take the matrix in the Eigenbasis,now consider the diagonal matrix whose diagonal entries are square root of the original one

i.e.,$b_{ii}=\sqrt{a_{ii}}$

[stoic python]

Buncho Dragons

[dawn fractal]

not sure which eigenbasis u are referring to here

[native rampart]

Diagonalise A

Say D=PAP^-1(D is diagonal)

Now consider C a diagonal matrix such that c_{ii}=\sqrt{d_{ii}} for all i

[dusky epoch]

ah, so you're working in C

[native rampart]

P^-1 B P will be your square root

by sqrt{d_{ii}},I mean an element c such that c^2=d_{ii}

[dawn fractal]

ofc.

[dawn fractal]

im not bright enough to easily get this though

[native rampart]

C^2=D

Because of how we defined it

(P^-1 C P)(P^-1 C P)=(P^-1 C^2 P)=(P^-1 D P)=(P^-1 P A P^-1 P)=A

[lavish jewel]

this looks ok, but you made the notation cursed

they already used B for something else in the problem statement

just be careful with that

[native rampart]

One sec,I will change it

[dawn fractal]

no need, i get it

ty

[dawn fractal]

ok i have tried using the fact that cols of AB are l.c.'s of cols of A and others but i am lost

im thinking it has something to do with which columns correspond to columns with leading 1s in the RREF

but idk how to determine that

WLOG suppose rank A <= rank B. So min(rank A, rank B) = rank A

i need to prove that rank AB <= rank A then

[native rampart]

A(Bx) is in range of A

That's range (AB) is a subspace of range(A)

[dawn fractal]

i don't follow.

are u referring to the operator it represents

[native rampart]

range (A),where A is a matrix is the set of vectors y such that y=Ax for some x

range(A) also happens to be a vector space

[lavish jewel]

when doing matrix-vector products, you can represent the result as a linear combination of the columns of A, where the weights of the columns are the elements of a vector x

so Ax is a linear combination of column vectors in A. these column vectors are the spanning set of range(A)

[dawn fractal]

wdym by 'weight'

[dusky epoch]

coefficients

[wintry steppe]

is there such a thing as elementary equations operations?
Like we have elementary row operations?

[limber sierra]

the elementary row operations represent the common ways you manipulate equations in a linear system

so theyre kind of the same thing

but no, no one uses the term

elementary equations operations

[wintry steppe]

my book does in a small section but i cant find any info on it online

also is forward elimination and forward reduction the same?

[limber sierra]

well its just the general principle of

you can do something to both sides of an equation

[stable kindle]

_ _

feeling unchallenged, might prove later idk

[coarse sandal]

Im not sure how to approach this problem

What does it mean that A and B are matrices whose reduced row echelon forms are both Identity?

does it mean that A and B are both identity matrices?

or just when we RREF do they become Identity matrices

[weak needle]

This.

[rustic moon]

Oh I just did this one, this is a good one to do

[coarse sandal]

How would you go about proving this?

Would you show that A and B are invertible because they are can be RREF into Identity matrix?

[rustic moon]

Are you familiar with what RREF means for matrices? Basically if I have the augmented matrix [A | b] this represents the x such that Ax = b

[nocturne jewel]

they are invertible by that property

[rustic moon]

^^

[coarse sandal]

oh

[rustic moon]

To find the inverse you rref until you get the identity on one side and the inverse in the other

[coarse sandal]

oh

[A|I] -> [I | A^-1]

that property

[nocturne jewel]

yes, fundamental theorem of invertible matrices

[coarse sandal]

because both A and B are invertible, then AB is also invertible

thus (AB)x = 0 only has a trivial solution

[weak needle]

Yes you got it

[coarse sandal]

thus exactly one solution

[rustic moon]

👏

[coarse sandal]

thank you

[nocturne jewel]

well it says $E_{32}E_{21}b$

[stoic python]

Mosh

[nocturne jewel]

multiply them together

what?

what you just said makes no sense

E_32 is a 3x3...

i have no clue how you've gotten a matrix is a scalar

[stable kindle]

lemma (the previous exercise): let A and B be subs of V; then A union B is a sub of V iff B is a sub of A wlog, as:
the if is trivial
conversely assume A union B is a sub of V and that there exists a in A but not in B, and b in B but not in A.
then a+b must be in A union B
but if a+b is in A, then (a+b)-a is in A, ie. b is in A. similarly if a+b is in B, then (a+b)-b is in B, ie. a is in B. contradiction.
therefore there must be either no a in A not in B, or no b in B not in A. qed

_ _
now to prove the actual statement: let A, B and C be subs of V; prove that A union B union C is a sub of V iff two of A, B, C are subs of the last.

case 1: one of the subs is a sub of another. wlog let C be the sub of B. then A union B union C is just A union B.
case 1a: then if B is a sub of A, then B and C are the subs of A and the union is a subspace, in accordance with the statement.
case 1b: if A is a sub of B, then A and C are subs of A and the union is a subspace, in accordance with the statement.
case 1c: if A is neither a sub of B nor B a sub of A, then by the lemma the union is not a subspace, in accordance with the statement.

wlog, if two are the subs of one, then the union is a subspace, in accordance with the statement. however, if one is the sub of another but the last is neither a sub of the second nor the second a sub of the last, then the union is not a subspace, in accordance with the statement.

then the only case left is:

_ _
case 2: none are the sub of any other; then neither A union B nor A union C nor B union C are subs of V. we must show there is a+b+c not in A union B union C to fulfil the prophecy.

lemma: there is a in A not in B or C, and b in B not in A or C, and c in C not in A or B
this is because if every a in A is in B or C, then A is a subset of B union C.
if A is a sub of neither B nor C, then there is g in B but not C and h in C but not B.
then g+h is in neither B nor C, so A is not in B union C, contradiction.
therefore A is a sub of either B or C, but this is also a contradiction as this is case 2.

wlog, by the second lemma we can have non-empty sets D, E and F:
let D be all a in A not in B or C; let E be all b in B not in A or C; let F be all c in C not in A or B.

case 2a: if e+f is not in A, given that e+f is never in B union C, d+e+f is not in A union B union C. then the union is not a subspace, in accordance with the statement.
case 2b: if e+f is in A, e+2f cannot be in A. given that e+2f is never in B union C, d+e+2f is not in A union B union C. then the union is not a subspace, in accordance with the statement.

this proves the statement.

i really hope this is right lol

this was sorta fun but i am almost certain there was a better, shorter way without so much case juggling

[wintry steppe]

christ

[sudden narwhal]

[heavy crown]

if I know that A is matrix of order m x n , and rank(A) = n
how do I show that C(A) = R^m ?

[lavish jewel]

what is C(A)?

[heavy crown]

column space of A

[lavish jewel]

then the column space cannot be R^m, unless m = n

[heavy crown]

true, because rank(A)=n then A is invertible, and because it's invertible it means it's square matrix

[lavish jewel]

that's not true in general

you can have rank n, but m > n

and then the matrix is not invertible

[heavy crown]

hmm

I missed one detail in the question like

The Given was: A is of order m x n, and for every Ax=b there's singular solution.
Prove that A is square matrix and that C(A) = R^m

[lavish jewel]

and what are you told about A?

[heavy crown]

that's it hmm

so I tried answering this like this:

specifically for b={0} vector there's also singular solution.
That means rank(A)=n

right?

[lavish jewel]

you still haven't told me what they tell you about A

so i can't say anything

the question alone is not enough

[heavy crown]

that's all they said it's just one liner 😦
I'll try translating it word by word sorry one moment

Let A be a real matrix of the order m x n, and assume that for every b ∈ R^m, the linear system Ax = b has a singular solution.
Show that C(A) = R^m and that A is square matrix.

[tribal moss]

Nice name username

[heavy crown]

thanks I tried my best

[lavish jewel]

ah, they gave you the extra info that every Ax = b has a singular solution

that was missing before

[heavy crown]

sorry, my bad.

yes

this is how I started: Specifically for b={0}^m there's also singular solution, which means rank(A)=n
and then stuck

[lavish jewel]

every b having a single solution means not only that the rank is n, but also that every b in R^m is in the span of A

that the system is consistent, i.e. appending the column b to A does not increase the rank