linear-algebra

[cursive narwhal]

So making the rows into columns etc

[pale shell]

nottt really i get what you mean

[gray dust]

everything gets lost in what i said

[pale shell]

I am just saying

In terms of columb

[gray dust]

when you do this for each entry, the effect is that the matrix's rows become its columns and its cols become its rows

[pale shell]

oh so you can just swap either wta then

[gray dust]

i think you have the gist, it's just you have a tendency to misconstrue some stuff i say and run really far with it

[pale shell]

o

no i mean just

you are saying swap position

but i am saying swap row columb

[gray dust]

i never said swap position

[pale shell]

Am i allowed to run /nick

[gray dust]

the swap of position you refer to is the swapping of each entry's row position with its col position

[pale shell]

yes

But

Can

You also swap the cols with roes

[ocean sequoia]

if you swap the cols with the rows

then you are swapping the rows with the cols

but you switch the position

more specifically

[pale shell]

oh

[ocean sequoia]

you are misunderstanding how transposing works i think

[pale shell]

bruh what

How

[gray dust]

if you have to preface your question with "but" even though it doesn't make any counter to what i said, something's gone over your head

[pale shell]

ooof

[gray dust]

perhaps work through an example to get a better feel

[pale shell]

yeah i should do that

[ocean sequoia]

you move positions

not columns and rows

[pale shell]

ok but bro

You move positions by moving columns and row

[ocean sequoia]

no you move the entries

when you transpose, the entry in the (i,j) spot is sent to the (j,i) spot
@gray dust

[pale shell]

Yes

[gray dust]

both of you stop for a sec, earl just post a matrix you want to transpose

[pale shell]

But you are doing that

By

Okay lemme think of a good one

Hmm

[cursive narwhal]

$\begin{pmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \end{pmatrix}$ Here try transposing this one

[stoic python]

Abhijeet Vats:

[pale shell]

Ok

Uhh

Is just becomes

Flipp

So is three ones down and same on the other

Okay so that whole row

Becomes a columb

So is

Veryicle

So then its not even a matrix

Its a vectis

[gray dust]

it's a matrix

[pale shell]

No

Not the oen he posted

Then it became a vectie

[gray dust]

it's a matrix

[pale shell]

how what

how is a

one single vector

A matrix

[gray dust]

it's a matrix

[pale shell]

Can u

Explain why

[gray dust]

it's a matrix with one column

there's no why

[pale shell]

But what does that even do

In terms of gemeotriczly

[gray dust]

matrices sit around and do nothing

[pale shell]

no they map vector to other vector

matrices=functions

[gray dust]

i guess you missed the defn of matrix

[pale shell]

........

how

what

[gray dust]

1st sentence of wiki page

[pale shell]

k then

but wikipedia

[sonic osprey]

It maps one dimensional vectors to n dimensional vectors

[pale shell]

Isnt a very good surce

[sonic osprey]

It's a fine matrix

[pale shell]

Zopherus solved

Thanks

I am understznd nowo

I have transcended

[gray dust]

i still think you should read the 1st sentence

[pale shell]

But

Can you refer a different source please

[gray dust]

no, i'll be stubborn too

[pale shell]

waht

ok

but for advixe

Dont trust wikipedis

[cursive narwhal]

Do you really not spellcheck yourself?

[pale shell]

I mean

[ocean sequoia]

Dude are you trolling?

[pale shell]

...

No

[sonic osprey]

Wikipedia is a fine source for math

[pale shell]

I spell check a lot actually

And

Im not a troll

Im just french ok

[cursive narwhal]

Ohhhh

[ocean sequoia]

ok sorry

[cursive narwhal]

I mean, i know you're not a troll

It's just that you should really spellcheck yourself so that you're saying shit that makes sense.

[pale shell]

Ok sorry about

[gray dust]

even with spellcheck, some of it doesn't make sense

[pale shell]

What

I am also trying to learn american math so its hard

And other

Because

[cursive narwhal]

? American math?

[pale shell]

Not lot of rescource in french

[cursive narwhal]

I didn't know something like that existed

[pale shell]

In other countries they have resources for math

[ocean sequoia]

I think he means the information is in english

[pale shell]

Like american textbooks

[ocean sequoia]

which makes it hard to understand

[cursive narwhal]

Why not just use french materials to learn?

Use bourbaki's textbooks, they're really good.

[ocean sequoia]

earlten wikipedia is a good source of math also symbolab can be helpful if you want to see steps

[gray dust]

wiki yes, symbo no

[ocean sequoia]

really you dotn like symbo

oh

nvm

[pale shell]

I am use khan academis

[ocean sequoia]

why is symbo bad

[cursive narwhal]

Just use bourbaki

It's very good

[pale shell]

ok but

Is pdf

Or what

[cursive narwhal]

Yeap, just search it up on libgen

Best resource possible for math

[pale shell]

Thank

[cursive narwhal]

I'm only joking lol

[pale shell]

Oh

[cursive narwhal]

Use Klaus Janich's Linear Algebra, i really like that book and I'm sure you'll enjoy it

[pale shell]

Is ferench

[cursive narwhal]

Nope, german but it has an english translation

[pale shell]

o

but i am heard

Gilbert strank

[cursive narwhal]

Yes, gilbert stank

[ocean sequoia]

I liked Strang

[pale shell]

Sounds like good

[ocean sequoia]

I have a package for Tony Stank

[pale shell]

Its like a very good oke

[ocean sequoia]

whats oke?

[cursive narwhal]

I don't like gilbert's strang book but it might work for you

[pale shell]

Sounds like a very good one

[cursive narwhal]

Didn't really work for me

[pale shell]

Well noe

Bo

No

[cursive narwhal]

@ocean sequoia HAHA i remember that scene

[ocean sequoia]

😄

[pale shell]

I am looking a lectur

[ocean sequoia]

OH

3Blue1Brown

is great too

[pale shell]

who is

[ocean sequoia]

youtube

good for visuals

[pale shell]

hmm

[ocean sequoia]

like great

[cursive narwhal]

There's a series of linear algebra lectures by Dr Aviv Censor from Technion. I found his lectures rather enlightening on many issues.

You can watch those. They're theoretically-oriented, though. So, you don't really get matrices and such thrown your way every lecture. He builds the subject up from vector spaces.

[pale shell]

Hmm

So not like a computeration

[cursive narwhal]

Well, there is computational material in there but it's limited to very basic stuff.

[pale shell]

o

sounds line a very good one

[cursive narwhal]

It's meant to be rigorous. That might be what you need

[pale shell]

Well

Like a how rigerous

One

[cursive narwhal]

Because, as of right now, you seem to be struggling with formulating mathematical statements or questions, hence why jintarou has had to make many clarifications about the questions you ask. That course is likely to teach you how to ask questions properly or formulate statements properly.

Janich's book will help you do that as well and I suggest you use both of them in conjunction with each other.

[subtle walrus]

and multiplying λ by x gives the SAME value that you would get by transforming x by A anyways

that is exactly what Ax = λx means

your explanation is correct

you are right

multiplication from the left by a matrix A is applying a linear transformation

multiplication by a scalar \lambda is however just scaling

first case is matrix vector multiplication, second is scalar vector multiplication we just use same notation for it

wait

its only scaling of this particular vector

or actually the subspace spanned by this vector

like, the standard example for this happening is an rotation in 3 dimensions around some axis

the axis of rotation will be an eigenvector with eigenvalue 1

do you have a globe at home?

basically spinning a globe

so yeah, as long as the axis of rotation goes through 0

it has to go through 0 just because of the fact that all linear transformations have to fix 0

(thats also the reason why you have to disallow the zero vector from qualifying as an eigenvector, because otherwise all scalars would be eigenvalues)

[gloomy arrow]

Can you use eigenvalues and eigenvectors as another way to solve a system of equations?

[subtle walrus]

in what sense?

[gloomy arrow]

Like can I use it as an alternative in any way to rref

Idk if that is even a thing

But with the emphasis on eigenvalues I thought it would be another way to solve a system

I know you can use it in diff eq but I mean like a regular system of linear equations

[static bison]

Suppose A ∈ Mn×n, λ is an eigenvalue of A, and A2 = I. Prove that λ = ±1.

I've done some work and established some proofs but I don't know how to finish it

a^2 +bc = 1
bc + d^2 = 1
ab + bd = 0
ac+cd = 0
b = c
a = -d
ad - bc = -1

lambda^2 - (a+d)*lambda + (ad-bc) = lambda^2 - 1

hold up

i think i got it

[gray dust]

a decent proof for this doesn't require playing with so many junk constants

[low plank]

Hello, can someone solve this ?

Not sure if I'm stupid

Or its actually hard

Kinda new to the linear algebra scene so ye

Ye I know how to find the inverse

[cursive narwhal]

Kinda new to the linear algebra scene so ye
You're not stupid, it gets easier over time

What's the trouble you're having?

[low plank]

Should be 1/A^-1

If I'm not wrong

[cursive narwhal]

[low plank]

Guess who lost hope in linear algebra

🙃😂👍

[wintry steppe]

to find the inverse, you need to first set up Matrix | A ones(3)|

[cursive narwhal]

What is 1/A^{-1} lul

If you have an n x n matrix A, then its inverse, if it does exist, is denoted by $A^{-1}$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

If you multiply A with its inverse, you must get the identity matrix back

So, do you know how to calculate the matrix inverse?

[low plank]

Ig not

I better get back to my books

[cursive narwhal]

Yeap, do some simple examples

Learn the theory before doing the problems

[low plank]

Yeah I'm trying

I'm really bad with maths in general

[cursive narwhal]

Don't think people emphasize this enough but learning it is necessary. If you have specific problems with the theory, you can come to us.

You're not as bad as you think

[low plank]

I study computer science for the programming part of it, but still gotta pass through some maths

[cursive narwhal]

Anyways, if you're approaching linear algebra by starting off with matrices, I would actually recommend looking at The Theory of Linear Spaces by Georgi Shilov. It begins with a wonderful treatment of determinants that makes sense and will teach you how to do the theory and compute as well.

[low plank]

I'll do my best, thanks again for your time

[static bison]

@wintry steppe i couldn't comprehend what you're saying but I figured it out

a = -d
b = c
d = sqrt(1-c^2)
if c = 2,
-sqrt(3)i 2
2 . sqrt(3)i

[uneven cypress]

By any chance would u be willing to help me?

[limber sierra]

see #prealg-and-algebra

also dont ping individuals

[uneven cypress]

ok thanks

[river jasper]

Can someone explain to me why a matrix needs to be full rank for it to be invertible?

[limber sierra]

there are a ton of ways to prove this

have you been exposed to elementary matrices and how they relate to row operations?

alternatively, have you been exposed to the determinant?

[river jasper]

are you referring to things like gaussian elimination?

[limber sierra]

yes

[river jasper]

yeah but I still don't quite fully understand determinant

[limber sierra]

alright i'll go with the first one

so, you can represent gaussian elimination as right-multiplication by "elementary matrices"

each operation in gaussian elimination corresponds to multiplication by an elementary matrix

if we have a square matrix that has a nonzero rank, that means we can eventually row reduce it to an RREF matrix that is not the identity matrix

i.e. that has a zero row

....

ok there we go

this means we can write $A = E_{1}E_{2}E_{3} \cdots E_{n}B$ where $B$ is our original matrix and $A$ is its RREF form

that's better

[stoic python]

Namington:

[limber sierra]

ugh i really cant type today

wtf

anyway

E_1, E_2, ... E_n here are elemtnary matrices

now suppose B is invertible

that means there's a $B^{-1}$ such that $BB^{-1} = I$

[stoic python]

Namington:

[limber sierra]

where I is the identity matrix

in that case we should be able to write

$AB^{-1} = E_1E_2\cdots E_n BB^{-1} = E_1E_2 \cdots E_n$

[stoic python]

Namington:

[limber sierra]

so $AB^{-1}$ must be a product of elementary matrices

[stoic python]

Namington:

[limber sierra]

that is to say, we should be able to row-reduce $AB^{-1}$ to get $I$

[stoic python]

Namington:

[limber sierra]

but because we assumed that $B$ has unfull rank

[stoic python]

Namington:

[limber sierra]

$A$ must have a zero row

[stoic python]

Namington:

[limber sierra]

so the product $AB^{-1}$ must have a zero row

[stoic python]

Namington:

[limber sierra]

which means it can't be reduced to the identity matrix

contradiction

so, $B$ is not invertible

[stoic python]

Namington:

[limber sierra]

and thus our theorem is proven

[river jasper]

yeah that makes sense

thank you!

Wait in the beginning didn't we assume that B was invertible, how are we allowed to assume that later on in the proof that B has an unfull rank? wouldn't the assumption contradict before we fully prove the theorem?

[limber sierra]

we assume B is invertible so we can contradict it later

[river jasper]

oh nvm proof by contradiction

yeah

[limber sierra]

sorry, i shouldve been more explicit

was half-recalling the proof from my head

[river jasper]

nah mb I should have read more carefully

[hollow finch]

Hm I'd like to prove that $A$ has the same rank as $A^TA$ but I can't think of quite where to start. How exactly do you show rank?

[stoic python]

nix:

[gray mason]

Show that they have the same null space.

First start with the fact that if $x \in \mathcal{N}(A)$, then $x \in \mathcal{N}A^\top A$. To show that if $x \in \mathcal{N}A^\top A$, then we have that $A^\top A x = 0\implies x^\top A^\top A x = 0 \implies |Ax|_2^2 = 0$. We know that this implies that $Ax = 0$. Therefore $x \in \mathcal{N}(A)$.

[stoic python]

Stephen James:

[spiral sonnet]

how would I find two vectors that are orthogonal to a given vector? and those two vectors have to be linearly independent

[wintry steppe]

a vector is orthogonal to another vector if the dot product between your new vector and your given vector = 0

and a vector is linearly independent from another vector if it can't be defined as a linear combination of the other vectors

[slow scroll]

@spiral sonnet you can complete to a basis and do gram-schmidt

or well, complete to however many linearly independent vectors you need

[wintry steppe]

How would you do this problem?

[hidden sable]

I do not understand why the solution is not unique

[wintry steppe]

i think it has something to do with your matrix being rank-deficient.

since it's rank deficient, it's got a column that depends on another column

[hidden sable]

What about a 3x3 matrix in which it has row 1 being (1,0,0), row 2 being (0,1,0), and row 3 being (0,0,0)?

The rank is 2, and I don't think that it is linearly dependent on each other

[limber sierra]

every set containing the zero vector is trivially linearly dependent

[hidden sable]

Hmm ok

[limber sierra]

since you can change the coefficient of the zero vector arbitrarily and it doesnt change the linear combination

i..e you can always make it a nonzero coefficient

anyway, the approach to this sort of question depends on what you've already covered in class

perhaps the most elementary explanation is:

if a matrix is rank-deficient, then its corresponding system, when row reduced, will have free variables

we know it has at least one solution, so it can't be inconsistent

and you can adjust the free variables

of course, you need to justify why it has free variables

and what that actually means

[hidden sable]

So if b = [1,1,0], x can be [1,1,x] for any x?

[limber sierra]

i'm... not sure what you mean

Ax = b is a matrix equation representing a system of linear equations

A is the matrix representing the "left hand side" of the equations, x is the vector consisting of variables in the system, and b is the vector representing the "right hand side" of the equations

when we talk about "solutions" to the system, we're talking about values of x

[hidden sable]

So, is it the fact that x_3 can be anything, since it is a free variable, that the is not unique?

[limber sierra]

yes... you have less equations than variables

[hidden sable]

Ok

[limber sierra]

the key term here is "underdetermined"

[hidden sable]

So if the solution for x is multiplied by a scalar, is the original solution still "unique"?

[limber sierra]

no, that's a different solution vector.

[unless the scalar is 1]

[spiral sonnet]

So I'm given a vector [2, 6, 5]. I need to find two vectors that are orthogonal to it and lineraly independent. A vector is orthogonal to another vector if their dot product is 0. So I have 2x + 6y + 5z = 0. I solve for x and get x = -3y -5/2(z). I plug in an y and z, but apparently my vectors still aren't correct. I checked the dot product of the vector I got and the given vector and I get 0. Not sure what I'm doing wrong

[hidden sable]

So is a non-unique solution defined by when a element of it changes the other values continue to stay the same?

[limber sierra]

no... a nonunique solution is just a different set of values for the variables

for example, consider the system:
x + y + z = 0
x + 2y = 4

this has multiple (indeed infinitely many) solutions, such as:
x = 0, y = 2, z = -2
x = 2, y = 1, z = -3
x = 4, y = 0, z = -4

and so on

meanwhile, the system:
x + y = 0
x + 2y = 4

would only have one solution, namely:
x = -4, y = 4

[wintry steppe]

@spiral sonnet one of your vectors can be [6,-2,0] and the other one can be [8, -3, -2]

[hidden sable]

So if there is only one solution to a system, then it is unique? While if there is multiple, it is not unique?

[limber sierra]

yes, that's what "unique" means

"unique" means "only one"

[hidden sable]

Ok, I thought that unique meant that there are multiple solutions to the system, where as the values of the variables differ

[wintry steppe]

i think the solution is not unique (if a solution exists) because you have two columns (vectors) that are dependent on one another

[hidden sable]

So if a column is the zero vector, it is automatically dependent because you can multiply any vector by 0 to get it?

Is that correct?

[half ice]

Any set that contains the zero vector is linearly dependent, yes

[hidden sable]

Ok

One more question, isn't it row vectors that are depend on each other?

[wintry steppe]

row vectors are just the transpose of column vectors and vice versa, so both can be dependent

[half ice]

A set of vectors are linearly dependent if you can make any one out of a linear combination of the others

[river jasper]

Is the set of eigenvectors for a nxn matrix necessarily independent?

[wintry steppe]

yah

[river jasper]

wait why? I would assume no because if the matrix has two non distinct eigenvalue then it would have two of the same eigenvector which could result in linear dependence if one of the element is 0

[wintry steppe]

ah, then in that case i think you're right; it would be no

i hadn't thought of that. my bad

[river jasper]

yeah but I'm still not sure

[limber sierra]

that is indeed not the case

you should be able to find a counterexample

as mentioned, it holds if all the eigenvalues are distinct

but nondistinct eigenvalues may have dependent eigenvectors

[river jasper]

alright cool thanks

[spiral sonnet]

So I'm given vectors w1, w2 and v. The set {w1, w2} is an orthogonal basis of a subspace W = span(w1, w2) of R^3. How can I find a vector u that is orthogonal to W and v - u is in W?

[slow scroll]

the projection of v onto W lies in W, and v - that projection is orthogonal to W.

[spiral sonnet]

Okay yeah I just figured it was the projection haha

[placid oracle]

Can somebody explain the Zassenhaus algorithm to me
i found an example of its use on the wikipedia page https://en.wikipedia.org/wiki/Zassenhaus_algorithm
Zassenhaus algorithm
In mathematics, the Zassenhaus algorithm
is a method to calculate a basis for the intersection and sum of two subspaces of a vector space.
It is named after Hans Zassenhaus, but no publication of this algorithm by him is known. It is used in computer algebra systems.
but i dont follow the steps
how did they just decide on the 4x8 matrix there, and what is going on with the 0 vectors on the bottom right entries

[limber sierra]

they describe the algorithm in the "algorithm" section

it calls for creating this matrix

where the notation is defined here

[placid oracle]

my bad, thanks 🙂

[limber sierra]

this colour coding might help

[placid oracle]

got it, i was just confused by the zeroes as i didnt see where they were coming from.

thats a super useful algorithm

[limber sierra]

yeah they're just fixed

[wintry steppe]

Anyone got an idea of how to do this?

I'm not confident with basis so don't know what to do

[wintry steppe]

I think the last one is e

[wintry steppe]

plug in the standard basis vectors of R^3 and u'll see 1 is d, 2 is c, 3 is b and 4 is e

[gray mason]

@wintry steppe Do you still need help with that question

[wintry steppe]

yea

[gray mason]

If $S$ has full rank, it is clear that $S = I$. There are two cases that we have to check. If it has a rank of $0$, then it is clear that $S = 0$. If $S$ has a rank of 1 and ${v_1, v_2}$ is the basis for the domain, we have that either $S(v_1) = \alpha v_1$ and $S(v_2) = 0$ or the other case. If it is the first case, we are done because we can just choose ${\alpha v_1, v_2}$ as our basis, but if its the other case, then we can just choose the basis ${\alpha v_2, v_1}$.

[stoic python]

Stephen James:

[wintry steppe]

I understand most of it, but why is it clear that S = I if S is full rank?

how do you show that

Can I do this?

S(a) = S^2(a) so S(Sa-a) = 0, but since ker(S) = {0}, it must be Sa = a, and so S = I

[gray mason]

You have that S^2=S

That also works

[plucky hull]

Excuse me?

Can someone help me with slope intercept form

it's been pretty easy for me but there's like 5 questions that I Just don't get

[slow scroll]

@plucky hull this is the wrong channel for that. ask #prealg-and-algebra

[plucky hull]

Oh

but

linera

linear

lines

Am I just stupid or is that what that means

[vague cedar]

different thing

[slow scroll]

Nah, people make this mistake all of the time. linear algebra is the study of vector spaces and the linear maps between them (think matrices). So yeah

[plucky hull]

Oh ok

I have no idea what that is so

[wintry steppe]

actually @gray mason I don't know why S(v1) = av1

and s(v2) = 0

[wintry steppe]

Can someone help me with this, I am struggling so much

With this problem

I broke it into 3 cases, but I couldn't think of a way to create a basis if rank(S) = 1

I tried to use S(Sa-a) = 0 for all a in R^2, but can't find anything interesting

[wintry steppe]

oh i figured it out

[wintry steppe]

any idea to do b) better

i just used matrix multiplication

wondering if theres better way

[lime granite]

Hello, this is the first time I post here, pleased to meet you all, I wanted to know whether a question about linear algebra should be asked here or in one of the question rooms

[quasi vale]

[lime granite]

Thank you, I'll write my question and then post some references, wait!

If I have an operator over a complex vector space, with finite dimension, I know, because of the result in the image, that I can always find a basis with respect to which its matrix is upper-triangular. I also know that this isn't always the case for operators over real vector spaces. So my question is: if I have an operator A over a real vector space V with dimension at least 3, that has only one real eigenvalue, whose corresponding eigenspace has dimension 1, can I find a basis that makes A triangular?
Or in another way, can I find other 2 vectors that are linearly indipendent and whose span is A-invariant, in the situation I stated above?

Images from "Linear algebra done right" by Sheldon Axler

My doubts come from this operator/matrix that has, correct me if I'm wrong, one real eigenvalue 1 and its eigenspace is span{(1,0,0)}

[wintry steppe]

no you're correct

[lime granite]

From what I understood, if I had an eigenvalue with eigenspace of dim 2 or two eigenvalues with eigenspaces dim 1 respectively, I could just complete a basis of R^3 with a vector that is linearly indipendent to the other two and have a basis for an upper-triangular matrix

But in this case, if I randomly complete a basis, with first vector (1,0,0), I don't necessarily get an upper-triangular matrix, just by taking the standard basis of R^3 I get D again, which, I forgot to say, is the matrix of an operator associated with the standard basis

That's where I'm stuck: should I immediately give up on finding a basis as soon as I see that the real eigenvalues are "not enough"?

[river jasper]

can someone help me with finding the eigenvalues of this matrix? v1={0 0 1} v2={1 0 0} and v3={0 1 0}. Note: these are column vectors. I got the characteristic polynomial of (-lambda)^3 + 1 but I couldn't simplify it.

[bitter glade]

@river jasper use the sum of cubes

[eternal fern]

how do I show this?

[subtle walrus]

just choose a basis

[gray dust]

you can start by expressing an arbitrary $x\in\bR^n$ wrt whatever ordered basis, like $\brc{v_1,\dots,v_n}$, for $\bR^n$. there exist scalars $x_1,\dots,x_n$ where
$$x=x_1v_1+\dots x_nv_n$$
using the fact that $A$ is linear, show that $A(x)$ can be rewritten as a matrix-vector product

[stoic python]

RokettoJanpu:

[vital swallow]

That's where I'm stuck: should I immediately give up on finding a basis as soon as I see that the real eigenvalues are "not enough"?
@lime granite Axler only means that it won't always work if the vector space is real. it might work for some matrices anyway

[unique pewter]

Hey guys I know I should show some thinking I did of my own, but I honestly don't know how to approach the question. Can someone give me a hint, on what I have to do ?

Thanks in advance!

[lime granite]

@lime granite Axler only means that it won't always work if the vector space is real. it might work for some matrices anyway
@vital swallow I understand, but then at what point do I know that a particular matrix cannot be put into upper-triangular form, using only real eigenvalues in the process?

[vital swallow]

if it has only real eigenvalues, you can put the matrix into upper triangular form

the eigenvalues will end up lying on the diagonal if a matrix can be put into upper triangular form by a similarity transformation, so the eigenvalues would have to be real

[quaint sun]

Can anyone prove this theorem? The book just gives it but doesn’t give any proof that it is true

[placid oracle]

If I have a subspace U = <(1,1,0,0),[1,0,1,1]> and W = <[0,0,1,1],[0,1,1,0]. I

I've claculatred the basis of U+W to be <[1,0,0,0], [0,1,0,0],[0,0,1,0]> and the basis of their intersection to be [0,-1,-1,-1]

i did this using this algorithm: https://en.wikipedia.org/wiki/Zassenhaus_algorithm

which then implies than the dimension of U+W is three, and U intersect W is 1

but in my answer key, it says that their dimensions are 4 and 0 since the basis of their intersection is the zero vector

not sure what im missing or what i did wrong, i can post a pic of the work i did if needed

[wintry steppe]

posting it would help

[quaint sun]

I posted mine, just waiting for a response

[placid oracle]

@wintry steppe

[vital swallow]

The vector (0,1,-1,-1) is not in your subspace W

[wintry steppe]

wdym the dimensions are 4 and 0 @placid oracle

[vital swallow]

I mean [0,1,-1,-1] is not in the span of [0,0,1,1] and [0,1,1,0]

[placid oracle]

the dimension of U+W is 4 and th edimension of U intersect W is 0 since its basis is the zero vector. Thats whats in my answer key

am i doing that algorithm wrong then? i dont see what im missing in the calculations

[vital swallow]

I have never used that algorithm, so I'm not going to try to follow the calculations. But I checked that what you gave as a basis for U \cap W is not in fact contained in W (though it is contained in U).

[placid oracle]

@vital swallow how would you set this up then

is "basis of U+W to be <[1,0,0,0], [0,1,0,0],[0,0,1,0]>" right at least

or did i get the whole thing wrong

[vital swallow]

I really don't know. sorry!

[placid oracle]

@wintry steppe ?

[wintry steppe]

um...

it looks like you followed the steps correctly

[placid oracle]

do you know how to set this problem up in a differeent way

<@&286206848099549185>

yes

and from there their dimensions, but thats pretty simple

@bitter saddle isnt that only if dim of their intersection is 0 => the basis of the their intersection is the zero vector

which i know to be true, but im not sure how to actually show that

@bitter saddle do youy know how to find the basis of their intersection, i think thats all i need an then i know how to figure the rest of the problem out

thats what i was doing actually

the second thing you said

whats the basis for U+W ?

[wintry steppe]

https://math.stackexchange.com/questions/25371/how-to-find-basis-for-intersection-of-two-vector-spaces-in-mathbbrn does this help?

[placid oracle]

im looking at that thread rn actually

ok i think i know what to do

[wintry steppe]

you have to make linear combinations of your given vectors equal 0

[placid oracle]

@wintry steppe yeah, for U int W, it clearly => that all constants in the equation for here have to be 0

which means the basis is just {0}

=> dimension is 0

whats confusing me is I've put both vectors in matrix form and using row reduction i found that every column has a pivot => the basis is the set of vectors [1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1] this is for U+W

@wintry steppe does that part make sense?

[wintry steppe]

maybe you didn't row reduce correctly

[placid oracle]

@wintry steppe i put it in a calculator, thats the row reduction

which would be the basis right?

whats wrong about that

[wintry steppe]

that seems correct

[topaz linden]

How do I find the basis for which the matrix $\begin{matrix} 2& 0 & 0\ 0& 2 & 0\ -1& 1 & 2 \end{matrix}$ is in Jordan form?

[stoic python]

Have a Banana Bitch:

[steady fiber]

find the eigenvalues, then the eigenvectors

if there aren't enough eigenvectors

find the generalized eigenvectors

and put all the eigenvectors and generalized eigenvectors as column vectors in a matrix

[topaz linden]

The eigenvalue is 2 w/ multiplicity 3. The eigenvectors are (1,1,0) and (0,0,1)

Since the only eigenvalue is 2, the 3rd generalized eigenvector is (0,1,0)

is this the required basis?

[lime granite]

the eigenvalues will end up lying on the diagonal if a matrix can be put into upper triangular form by a similarity transformation, so the eigenvalues would have to be real
@vital swallow if the matrix has only one real eigenvalue with eigenspace of dim 1, and the vector space is R^3 (dim 3) I don't know how to find a basis to transform it into upper-triangular and whether such a basis exists in the first place, if it was always possible the theorem I posted above would be valid not only for complex vector spaces but for real ones too

[topaz linden]

@steady fiber is the list of the generalized eigenvectors the required basis

[vital swallow]

if it only has one real eigenvalue, it will not be upper triangular

[lime granite]

I think we should also add the fact that the eigenspace of that single eigenvalue is of dimension 1, if we are talking of real vector spaces of dim > 2

With dim 1 it's not even necessary to do anything obviously

With dim 2 you complete a basis with another linearly indipendent vector and you're done

With dim=n, n>2, the eigenspace of that single vector has to be at least of dim=n-1 to easily complete a basis to make the matrix upper-triangular or maybe for the basis to exist at all

Or that's what I'm getting from all of this

@vital swallow thanks for the answers!

[topaz linden]

@vital swallow Think you can help me out with the jordan form problem?

[lime granite]

Still, the best way to check is to find all the eigenvalues and as soon as you find a complex one you're done, that is, you can't do it

[vital swallow]

yep TitorP

For the Jordan form, you first find eigenvectors, then pseudoeigenvectors if necessary

[topaz linden]

yup

I have those

Eigenvectors: (1,1,0) and (0,0,1); Generalized eigenvectors: (0,1,0)

what do I do next?

[vital swallow]

just throw them all together into a matrix as the columns

[topaz linden]

Alright, thanks

[keen reef]

help would be greatly appreciated

[ocean sequoia]

I for the life of me cant figure this out... does anyone have a hint i could use

[pallid rampart]

If it's not a subspace it must violate other axioms for a subspace

[ocean sequoia]

It has to contain the zero vector for it to closed under multiplication I believe so I cant go there

So i have to have it violate closed under addition

[pallid rampart]

Yes

[ocean sequoia]

{(x,y) E R^2: x + y = 1} that wont work cause its not closed under scalar multiplication

[pallid rampart]

Yeah

it violates all three axioms

xD

[ocean sequoia]

XD

So im looking for a set that f(x) + f(y) =/= f(x + y)

[pallid rampart]

Ok think about this visually

What does it mean for a subset in R^2 to be closed under multiplication

What does it mean visually

[ocean sequoia]

You can scale up and down the line

shrink/strech it

[pallid rampart]

Yes

So if a point p is in the subset, then the line through the origin and p is also in the subset

But what if you have two different points

[ocean sequoia]

So something like

0 = a +b, 0 = a -b?

[pallid rampart]

Yes

[ocean sequoia]

actually that exactly works right

lol

it has the zero vector

[pallid rampart]

It's closed under scalar multiplication

[ocean sequoia]

yep thanks!

[pallid rampart]

I was thinking of something more simple

Like $\brc{(x,0):x\in\bR}\cup\brc{(0,y):y\in\bR}$

[stoic python]

Whoever:

[pallid rampart]

ig they're the same

rotated 45 degrees

[ocean sequoia]

what does ig mean?

[pallid rampart]

i guess

[ocean sequoia]

ah

true

thanks for the help i appreciate it

[pallid rampart]

sure

[keen reef]

@pallid rampart do you have any idea on what to do with the problem i sent earlier?

so k=+-i?

[pallid rampart]

For part b, the solution of a differential equation in the form of ${\bf x}'=A{\bf x}$ is ${\bf x}(t)=e^{At}x_0$ where ${\bf x}_0={\bf x}(0)$

[stoic python]

Whoever:

[pallid rampart]

where $e^{At}=\sum_{n=0}^\infty\frac{A^nt^n}{n!}=I+At+\frac{A^2t^2}{2!}+\frac{A^3t^3}{3!}+\cdots$

[stoic python]

Whoever:

[pallid rampart]

and use the fact that A^2=-I

To reduce the power series into the power series of familiar functions

Then it follows that $\lim_{t\to\infty}\frac{t!^t}{\text{Whoever}(t)}=0$

[stoic python]

Whoever:

[pallid rampart]

Yes

Keep in mind this is #linear-algebra

@keen reef

Since you pinged me personally

[keen reef]

yup just analyzing ty

how id you from the summation to the limit?

[pallid rampart]

Try to simplify the summation

$I+At+\frac{A^2t^2}{2!}+\frac{A^3t^3}{3!}+\cdots=\br{I+A^2\frac{t^2}{2!}+A^4\frac{t^4}{4!}+A^6\frac{t^6}{6!}+\cdots}+\br{At+A^3\frac{t^3}{3!}+A^5\frac{t^5}{5!}+\cdots}=\br{I-I\frac{t^2}{2!}+I\frac{t^4}{4!}-\cdots}+\br{At-A\frac{t^3}{3!}+A\frac{t^5}{5!}+\cdots}$

[stoic python]

Whoever:

[pallid rampart]

@keen reef

Then obviously factor out A and I

and you get sin(t) and cos(t)

Then idk

Probably write A as [a,b;c,d]

and compute the limit explicitly

It's probably 0 I feel like

[keen reef]

Any ideas on the second one? Thank you guys

[keen reef]

ples, sorry for ping @pallid rampart

[pallid rampart]

I don't know how to solve it

[keen reef]

o damn ok, thank u for help on first one and so sorry for ping

[pallid rampart]

I just learned about it since you brought it up, but I don't know how to solve the second one

[keen reef]

ah ok, thank you!

[hollow goblet]

Question 1

[limber sierra]

"Use calculus"

hmmm where should i post this question

#linear-algebra seems right!

[unique pewter]

I am reposting my problem

Can someone help me?

[dusky epoch]

the $(i,j)$ element of $A^k$ counts the number of paths of length exactly $k$ from vertex $i$ to vertex $j$.

[stoic python]

Ann:

[dusky epoch]

if you need me to, i can write out an inductive proof of this

@unique pewter

[unique pewter]

So then why do I have to sum every matrix and not just keep the A^k?

[wintry steppe]

ive a really short question: Why is (f+g)(x) = f(x)+g(x)?

[unique pewter]

If it's not too much effort that would be great! @dusky epoch

[dusky epoch]

@wintry steppe because that's how we define addition of functions

[wintry steppe]

ic, thanks

[dusky epoch]

@unique pewter alright so for the base case k=1

a 1-path is simply a single edge

and A is the adjacency matrix of your graph, so its (i,j) entry is 1 if there's an edge from i to j and 0 if there isn't

which can be looked at as counting the number of edges from i to j

makes sense?

[unique pewter]

Yeah so far

[dusky epoch]

alright

now suppose we already know that $A^k$ counts the $k$-paths (in the sense that $(A^k)_{ij}$ is the number of $k$-paths from $i$ to $j$) and we wanna show that $A^{k+1}$ counts the $(k+1)$-paths in the same sense

[stoic python]

Ann:

[dusky epoch]

so how can we construct a (k+1)-path from i to j?

we can pick some arbitrary vertex which i'll call $m$, walk from $i$ to $m$ along a $k$-path (which gives $(A^k){im}$ options), and then go along the edge from $m$ to $j$ if it's present (which gives $A{mj}$ options)

[stoic python]

Ann:

[dusky epoch]

and we simply sum that for all m from 1 to...

...well however vertices there are in the graph, let's say it's N (so that A is an N by N matrix)

and so we get $(A^k){i1} A{1j} + (A^k){i2} A{2j} + \cdots + (A^k){iN} A{Nj}$

[stoic python]

Ann:

[dusky epoch]

which is the definition of the (i,j) entry of A^k * A

like straight up it comes down to the definition of matrix multiplication

and so our sum is $(A^{k+1})_{ij}$ as desired

[stoic python]

Ann:

[unique pewter]

Gimme a sec to process the info

Probably stupid question but aren't we looking for the sum of all matrixes until A^k and not A^(k+1)?

ahh wait no.

$(A^k){i1} A{1j} + (A^k){i2} A{2j} + \cdots + (A^k){iN} A{Nj}$

[stoic python]

Baroque:

[unique pewter]

this is basically it

[dusky epoch]

_

anywayg

i'm specifically talking about the fact that A^k counts the paths of length exactly k.

if you want the paths of length at most k, you'll wanna add up A + A^2 + A^3 + ... + A^k

[unique pewter]

I don't get why we have to multiply the two A's together at the previous equation. I am sorry but can you dumb it down a bit further?

Like I get the summing part, but not how you formed the equation

[tight glade]

IDK if this falls under this or some other channel, but in order to do a cross product, your vectors have to be in R^3, don't they?

[wintry steppe]

yea

cross product is like the determinant except the first row is variables i,j,k

[limber sierra]

the cross product is only defined on R^3

there's a certain sense where it can be viewed as a specific case of the composition of the hodge dual with the exterior product mapping V * V -> Lambda^2 V

but this generally isnt a helpful generalization.

[wintry steppe]

imo the specific creation of cross product seems to be because of physics than math

[limber sierra]

absolutely

mathematically the cross product is a random thing from a fairly mundane vector space

but it's certainly quite useful in physics (and 3D computer engines and whatnot)

[unique pewter]

Thanks @dusky epoch I get it now. Your explanation helped a lot!!!

[spiral sonnet]

So I'm given a vector V = [2, 6, 5]. I need to find two vectors W1, W2 such that both W1 and W2 are orthogonal to V and w1 and w2 are linearly independent. I got [6, -2, 0] and [8, -3, -2], but apparently my answer is incorrect. Anyone know what I did wrong? I setup 2x + 6y + 5z = 0. Then solved for x and I plugged in numbers to get my two vectors

[subtle walrus]

is it possible to write every element of $\mathrm{SL}_n\mathbb{R}$ as the product of 2 elements of $\mathrm{SL}_n\mathbb{R}$ having eigenvalue 1

[stoic python]

Lochverstärker:

[split heart]

Hi, I've a question. In the topic of linear transformations, the dimension of a matrix is the numbers of elements it has?

[hasty estuary]

no, a matrix has 2 dimensions

[split heart]

Always?

[hasty estuary]

yes

[split heart]

Right, and in the case I've a vector like this:
((x, y)). The dimension now depends on the number of elements?

[hasty estuary]

a vector lives in a vector space

for example that 2d vector could be in R^2

a 2 dimensional space

[split heart]

So, yes?

number of elements = dimension

[hasty estuary]

yeah

[split heart]

Thank you 🙂

[rose flower]

How do I even start on part B

[dusky epoch]

here's a hint: suppose v is an eigenvector of A with eigenvalue λ. show that v is also an eigenvector of A^-1. what must its eigenvalue be?

[rose flower]

1/eigenvalue

I have D^-1

[dusky epoch]

1/λ

anyway

this will be enough for you to determine the EVs of A^-1

[rose flower]

I have those, but I meant the Basis part

That is where I am utterly stumped

I can't even find eigenvectors

[wintry steppe]

If I have a Matrix $T=V V^T$ with $V \in \mathbb{R}^{n\times m}$, is it possible to get a symmetric factorization $ LL^T = T$ with $L\in \mathbb{R}^{n\times n}$ in some way that leverages the existing factorization instead of just applying cholesky on $T$?

[stoic python]

c3aa:

[wintry steppe]

(with T being symmetric positive definite)

[spiral sonnet]

So I'm given a vector V = [2, 6, 5]. I need to find two vectors W1, W2 such that both W1 and W2 are orthogonal to V and w1 and w2 are linearly independent. I got [6, -2, 0] and [8, -3, -2], but apparently my answer is incorrect. Anyone know what I did wrong? I setup 2x + 6y + 5z = 0. Then solved for x and I plugged in numbers to get my two vectors

[half ice]

@spiral sonnet
If two vectors are orthogonal then their dot product will be zero

With that in mind, it's quick to check that [8,-3,-2] doesn't work

But something like [0, 5, -6] would

[spiral sonnet]

Oh lol. I'm dumb

Thanks!

[ocean sequoia]

Prove that the interesction of two subspaces U1,U2 is also a subspace? I know how this works i think im just having some trouble expressing it mathematically and not in english... So because the points in the intersection belong to both U1 and U2 and those are subspaces those points must also exhibit the axioms of a subspace

[limber sierra]

just check each of the requirements

in order for a subset of a vector space to be a subspace, it must satisfy the following:

• It contains 0
• It is closed under vector addition
• It is closed under scalar multiplication

this is both necessary and sufficient

so go one-by-one

[Hint: Use the fact that, since U_1 and U_2 are subspaces, they are also closed under addition and multiplication]

[ocean sequoia]

Hm. it has to contain the zero vector because both U_2 and U_1 must contain the zero vector.

Which means the interesction must contain it as well.

[cursive narwhal]

Go on

[ocean sequoia]

I think im stuck on scalar addition f(x) + f(y) = 0 = f(x + y) where x, and y are in the intersection of U1 and U2?

that feels... wrong tho

[cursive narwhal]

You have to show that if $v,u \in U_1 \cap U_2$, then $v+u \in U_1 \cap U_2$.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

No. Why did you introduce a random f in there? We are not talking about linear transformations/functions.

[ocean sequoia]

oh yea

shoot

because im bad at this trying to get better

[cursive narwhal]

The condition I stated above is what namington meant by 'closed under vector addition'

Of course you are. That's why you're putting in effort into understanding this. So, go ahead and prove that it is closed under addition.

[ocean sequoia]

Because U1 is a subspace then v + w must be an element of U1 and the same goes for U2 which thus means it must be in the interesction

I can probably make that more formal

[cursive narwhal]

How did you deduce that v+w is an element of U_1?

Make it formal.

[ocean sequoia]

Because its given that U_1 is a subspace

So it has to be closed under addition

[cursive narwhal]

Sure, but what makes you say that w is in U_1?

[ocean sequoia]

If v,w is an element of U1 intersection U2 and U1, U2 is a subspace they are closed under addition thus v+w is still in the U1 and U2 thus it is still in the intersection of U1 and U2

[cursive narwhal]

No.

That argument is messy at best.

[ocean sequoia]

Shoot

[cursive narwhal]

Let $v,w \in U_1 \cap U_2$. Then what can you say about $v$ and $w$? Forget about $v+w$ for just a second.

[stoic python]

Abhijeet Vats:

[ocean sequoia]

They belong to both U1 and U2

[cursive narwhal]

Yeap. So, $v \in U_1$ and $w \in U_1$. So, $v+w \in U_1$ and the same can be said for $U_2$. Hence, $v+w \in U_1 \cap U_2$.

[stoic python]

Abhijeet Vats:

[ocean sequoia]

Ah thats what i was trying to say is that enough?

[cursive narwhal]

That second sentence where I wrote 'the same can be said for U_2' is basically another way of saying 'write the same thing down for U_2'.

[ocean sequoia]

and we know that v + w is in U1 because U1 is closed under addition right? Or no?

[cursive narwhal]

Yes.

You can include that as well. Generally speaking, there comes a point where you stop writing down details that explicitly. Since you're just starting out, you should just write all of them out.

[ocean sequoia]

Ok i will I appreicate the help thanks!

[cursive narwhal]

In the same way, you can prove that the intersection is closed under scalar multiplication. That proves that it is a subspace.

[ocean sequoia]

v E U1 so cv E U1 and v E U2 so cv E U2 thus v E U1 intersection U2 and cv E U1 intersection U2

I need to learn how to use the TeXit

[cursive narwhal]

No. You have to show that $c \in \mathbb{F}$ and $v \in U_1 \cap U_2$ implies that $cv \in U_1 \cap U_2$.

So, start with that and work your way through each step.

[stoic python]

Abhijeet Vats:

[ocean sequoia]

I think im a bit confused. Since, U1 and U2 are subspaces that means that if v is an element of U1 and c is an element of F than cv is an element of U1 . Also if v is an element of U2 then cv is an element of U2. Doesnt that imply that cv is in the intersection?

[cursive narwhal]

Indeed, that's correct.

[ocean sequoia]

I need to be more explicit I cant "math language" what i mean

[cursive narwhal]

'math language'?

[ocean sequoia]

symbols im mostly just being tongue and check

bad joke

sorry lol

[cursive narwhal]

Symbols are just as much a part of math as anything else that you might use in your proof. You could write the entire proof in symbols or you could use English. It really doesn't matter so long as you're presenting your work properly.

Lul

In any case, let $c \in \mathbb{F}$ and $v \in U_1 \cap U_2$. So, $v \in U_1$ and $v \in U_2$. Since $U_1$ and U_2$ are subspaces, it follows that $cv \in U_1$ and $cv \in U_2$. Hence, $cv \in U_1 \cap U_2$.

[stoic python]

Abhijeet Vats:
Compile Error! Click the reaction for details. (You may edit your message)

[cursive narwhal]

That should be the way you present the argument. Since you're just starting out with this, you could make it even more detailed by writing out, explicitly, why cv has to belong to U1 and U2.

[ocean sequoia]

because U1 and U2 are closed under scalar multiplication. Im writing this down thank you.

[cursive narwhal]

You're welcome.

It should be noted that you don't actually have to check if 0 belongs to the subspace. If it is closed under scalar multiplication and vector addition and if you've proved that 0v = 0 for any vector v in the vector space, then any subset that is closed under scalar multiplication and vector addition must contain the 0 vector. It's just a condition placed in there for.........actually, i'm not quite sure why but oh well

[ocean sequoia]

huh i didnt know that

but that makes sense...

[limber sierra]

@cursive narwhal its just to make sure its nonempty

you could technically check for the existence of any vector

but 0 is generally most convenient

[cursive narwhal]

I do remember a text I was reading where it stated that a subspace has to be non-empty anyways.

[cold topaz]

how do you practice linear algebra proofs? What are the fundamentals?

[cursive narwhal]

There are problem books on linear algebra available online, you can check them out. For example, Linear Algebra Problems Book by Ikramov has some nice problems. Each chapter focuses on problems related to a particular topic and the author summarizes what you need to know before attempting the problems.

It's not meant as an introduction to linear algebra, it's meant to supplement a linear algebra text that lacks problems.

@cold topaz

^I should also say that a lot of them are proof problems. There are relatively few computational problems. A lot of them go through the basic proofs you should've encountered in the main body of a linear algebra text. A lot of them go outside of those proofs so you'll have plenty to contend with.

[cold topaz]

is det(λI-A)=0 applicable to ALL matrices to find their eigenvalue?

[wintry steppe]

you can only find eigenvalues on square matrices

so no

[cold topaz]

it doesnt work on this one. right? because i get this for det: λ(λ^2 - 11λ + 39) + 3.

[wintry steppe]

for matrix A ?

[cold topaz]

yes

[wintry steppe]

it should work for that

[cold topaz]

i have tried to get its determinant different way and i end up with this all the time: λ(λ^2 - 11λ + 39) + 3.

[wintry steppe]

your eigenvalues should be 3 (with a multiplicity of 2) and 5

the determinant is correct btw

[spiral sonnet]

If I want to find the projection of a vector b onto a column space a. Would I just do summation of the projection of b onto each column?

Nevermind! I think I got it!

nevermind I do not have it!

[wintry steppe]

https://math.stackexchange.com/questions/1774595/what-is-the-difference-between-the-projection-onto-the-column-space-and-projecti

does this help

[spiral sonnet]

MEVERMIND

I got it. I keep leaving out the - signs.......

[spare crystal]

Why is ker(A) the set of all vectors orthogonal to the rows of A? Shouldn't it be the set of all vectors orthogonal to the columns of A?

[wintry steppe]

Yes it should be set of all vectors orthogonal to the columns of A

[sonic osprey]

uh wait no

it is the rows

[spare crystal]

why's that?

the way im intuitively thinking of it the kernel is orthogonal to the image of A

[sonic osprey]

uh

the kernel and image live in different spaces

the kernel lives in the domain

the image lives in the codomain

[cursive narwhal]

@spare crystal This is just looking at the definition of orthogonality and matrix multiplication. Let A by an m by n matrix. So, Ker(A) is the set of n-dimensional vectors x such that Ax = 0, where 0 is the zero vector in m-space. So, take the i-th row of A and multiply that with this vector x that supposedly belongs in Ker(A). You can see that it just produces the i-th 0 in that zero vector. Now, you can just view this as the standard inner product on R^n so the inner product of the i-th row of A with x just gives 0. In other words (A_i,x) = 0 and so, they are orthogonal.

Let me know if I'm misinterpreting anything you've said.

[thorn narwhal]

Hi, I have a quick question that I want cleared up that I believe is pretty basic. If anyone could help a bit that would be great. I'm kind of confused about how matrices can be interpreted as linear transformations

[cursive narwhal]

What are you specifically confused about with regards to that?

[thorn narwhal]

I know how a vector can be interpreted as a linear transformation, but I have no idea how a matrix can

I have an example right here hold on

So here you can alter a through g to transform the icon on this graph

I know that c does horizontal transformations and f does vertical transformations but I have no idea how to interpret the rest

or why c and f do that

[cursive narwhal]

If you multiply the rows of the matrix by the vector [x,y,1], then you have the following equations relating x and y to x' and y':

ax+by+c = x'
dx+ey+f = y'

Geometrically, c does affect the value of x' after the transformation occurs. f does affect the value of y'. The other constants a,b,d,e also affect x' and y' in exactly the way described by the equations above.

[thorn narwhal]

oh so if I interpret the rows as equations, the variables are affecting x' and y' which do the transformations?

[cursive narwhal]

You do not interpret the rows as equations. The matrix is, indeed, transforming x and y into x' and y'.

Ultimately, this is going back to a more fundamental question about why matrices are multiplied to vectors in that specific way. Do you agree that that's the root cause of your confusion right now? I just want to make sure I'm understanding where you're confused.

[thorn narwhal]

oh so the x and y in the first vector are the coordinates of the icon before the transformation, and the x' and y' are the coordinates after?

and the matrix determines how it transforms

I think I was looking at everything from a different perspective than I should have

[cursive narwhal]

Yeap, the matrix controls that. So, depending on the entries in the matrix, the vector [x,y,1] transforms in a very particular way that is a consequence of how matrix multiplication with vectors is defined.

In any case, if you're concerned about why matrices can be 'interpreted' as linear maps, which was your original question, there is a very nice proof that shows that there is a bijection between the set of m by n matrices and set of linear maps between F^n and F^m. In other words, every linear map has a matrix associated with it and a matrix is always associated to some linear map. If you're interested, you can find a proof for this result. The proof of this that I have seen shows you exactly why matrix multiplication is defined in the way that it is and why you can 'interpret' a matrix as a linear map.

Anyways, if you didn't understand any of that, it's fine. It was just extra information directed towards the first question that you asked.

[thorn narwhal]

Thank you very much

[cursive narwhal]

You're welcome.

[low plank]

hey

i want someone to cofirm if i solved this question right ?

Here's a better pic

[wintry steppe]

For a) my bottom row is (3, 2, 1)

[low plank]

Hmmm

[cold topaz]

from where did the I appear here?
https://www.slader.com/textbook/9781118879160-elementary-linear-algebra-applications-version-11th-edition/301/exercises/34/

[low plank]

how about b ?

all good ?

[wintry steppe]

For b) my first row is (-5,-3,-2)

@low plank

[spare crystal]

@cursive narwhal ah, thank you, that makes sense!

[low plank]

Hmmm

So my steps are correct but the calculations are wrong

Well anywho, thanks man. I'll recheck it

[wintry steppe]

ah np

[cursive narwhal]

You're welcome.

[coral sage]

oh my god what is the name of hte set of linear functionals that map everything to 0

[wintry steppe]

Kernel?

[sonic osprey]

Uh, there's only one linear functional that maps everything to 0

[sand maple]

I'll be taking Linear Algebra next year, is there a way to prepare before it?

[cursive narwhal]

Are you taking a proof course on LA or a computational course? Do you have a list of things you'll be covering in that course?

[pallid rampart]

If $M=\begin{bmatrix}A&B\0&D\end{bmatrix}$ where $A,B,D$ are square matrices, how do I prove that $\det M=(\det A)(\det D)$

[stoic python]

Whoever:

[pallid rampart]

I can prove it using Leibniz's formula for determinant, the one with permutations

[sand maple]

It's a proof course @cursive narwhal

[pallid rampart]

But how do I prove that without using permutations?

[onyx tundra]

how are you defining it?

I would've expected a definition involving permutations

[pallid rampart]

Expansion by minors

[limber sierra]

my hunch would be inducting on the size of A, B, D

which seems like it really fucking sucks

probably a better way

[slow scroll]

maybe you could use the way determinants behave under row operations. There exists some invertible transformation E_A that reduces A to something triangular and some other transformation E_D that reduces D to something triangular, and we know how E_A and E_D affect the overall determinant. maybe thats not rigorous enough for you idk.

[limber sierra]

that's nicer but AFAIK those properties dont have nice proofs either from just the minors definition

besides inductive ones

i mean, in general, most things involving the minors definition will require induction to prove

so maybe thats unavoidable

[pallid rampart]

I have this here

$\begin{bmatrix}A&B\0&D\end{bmatrix}=\begin{bmatrix}I&0\0&D\end{bmatrix}\begin{bmatrix}A&B\0&I\end{bmatrix}$

[stoic python]

Whoever:

[pallid rampart]

I is identity

[agile gyro]

fIx a kxk matrix A and prove by induction that det([A B][O I]) = det(A)

similarly for [I O][O D]

[onyx tundra]

the induction for this is much simpler because most minors are going to give 0 det

[lavish jewel]

you can use a schur complement and invert blockwise

[viscid kernel]

What is a pre hilbert space ?

[limber sierra]

a space with an inner product

it's often used in contrast to "hilbert space" which is a complete space with an inner product

you might know the term "inner product space", which is equivalent but is often used in a different "context"

in the sense that calling it a "prehilbert space" emphasizes that it can be completed to become a hilbert space

[viscid kernel]

💀

I didnt understand anything

[sonic osprey]

If you didn't understand this, you should go learn these things before trying to think about pre-hilbert spaces

[viscid kernel]

I dont know what an inner product is

I do know what a dot product is but google says that the inner product is a generalisation of the dot product. What do they mean with “ generalization”

[dusky epoch]

an inner product is a function < , > that takes two vectors as inputs and produces a scalar, satisfying the following for all x, y, z in the vector space and all real scalars a, b:

1. <ax + by, z> = a<x,z> + b<y,z> (linearity in the first argument)
2. <x,y> = <y,x> (symmetry)
3. <x,x> >= 0 for all x, and <x,x> = 0 if and only if x = 0 (positive definiteness)

the dot product on R^n satisfies these axioms

it is a trivial matter to show that an inner product must be linear in its second argument despite that not being an axiom

this is the definition of a real inner product. the complex case is a bit different

[viscid kernel]

Thanks, I only want to stick to the reeel case for now.

[dusky epoch]

for example: $\ang{f, g} := \int_{-1}^1 f(x) g(x) \dd{x}$ defines an inner product on the space of real-valued continuous functions $[-1,1]$

[stoic python]

Ann:

[dusky epoch]

i invite you to show this yourself

[spiral sonnet]

So I'm given an eigenvector v which is an eigenvector of Matrix A. How can I find the corresponding eigenvalue?

Nevermind!

[autumn kraken]

Does anyone know how they want me to show Symmetry and Transitivity? I wrote down the definitions but I don't know how I should explain or prove it.

[subtle walrus]

the equivalence relation is defined in a way that all scalar multiples are equivalent, so $(a, a') \sim (b, b') \iff (b, b') = \lambda\cdot (a, a')$ for some $\lambda$

[stoic python]

Lochverstärker:

[autumn kraken]

wouldn't that only be true if lambda is 1 ?

[subtle walrus]

what

why?

[autumn kraken]

no nevermind, sorry I mixed it up with reflexivity

[subtle walrus]

$(\frac{1}{2}, \frac{1}{4}) \sim (2, 1)$

[stoic python]

Lochverstärker:

[autumn kraken]

They want us to show Reflexivity, Symmetry and Transitivity seperately I think. This is like one of the earliest exercises

[subtle walrus]

yes, its not hard

and there is not much more i can say without giving you the solution

[autumn kraken]

I see, maybe that's the problem. I am overcomplicating it

((a, a') ~ (b, b') => (b, b') ~ (a, a'))
For symmetry I wrote this down because it's how it's defined, but I don't understand how or if I have to formally prove this.

Am I on the wrong track?

[subtle walrus]

you did not show anything

[autumn kraken]

yeah

[cursive narwhal]

"don't understand how or if i have to formally prove this" The question is asking you to prove it. There is a definition given for this relation so start by using that.

[subtle walrus]

you start with $(a, a') \sim (b, b') \iff (b, b') = \lambda \cdot (a, a')$, and have to show that that also $(b, b') \sim (a, a')$, i.e. that there exists $\lambda_2$, such that $(a, a') = \lambda_2 \cdot (b, b')$

[stoic python]

Lochverstärker:

[subtle walrus]

(this is just unraveling the definitions)

(also, this is for proof of symmetry)

[autumn kraken]

I don't know how to do this in a general way quite often. For proving things for sets I say let x be any element of M and then go with that, but here I can't just prove it for 1 example right, it has to be in general

[subtle walrus]

yes, hence why you let (a, a') and (b, b') be arbitrary elements of Q^2

i mean, this is just the same as the example you gave 🤔

you let (a, a') and (b, b') be any elements of Q^2, then further assume (a, a') ~ (b, b'), and (try to) show (b, b') ~ (a, a')

[autumn kraken]

yeah

would it be sufficient to say that any element of Q\{0} can be represented as the quotient of two elements of Q\{0}?

[cursive narwhal]

In the problem above, you've been given that (a,a') ~ (b,b') so there exists an x such that x(a,a') = (b,b'). Now, you're asked to show that (b,b') ~ (a,a') so there needs to exist an x' such that x'(b,b') = (a,a'). Relate x' and x.

You are guaranteed the existence of x, now show the existence of x' by relating x' to x. Since x exists, x', then, has to exist.

[subtle walrus]

would it be sufficient to say that any element of Q\{0} can be represented as the quotient of two elements of Q\{0}?
@autumn kraken i would just comment "nothing shown" and give you 0 marks

you have to find a constant such that your tuples are related in the way that i (or abhijeet) wrote

[autumn kraken]

that was kind of my thought process too but I don't know how to prove that x' exists. I will try it now though, thanks

[subtle walrus]

you can explicitly write down the x'

in terms of x

which you know exists

[cursive narwhal]

Look at what you started out with and what you have at the end. Compare them and see if there's an obvious way in which x' related to x.

Over here, you sort of have to work backwards like in an epsilon delta proof. You need to look at what you get at the very end so that you can make a clever choice of the stuff you need in the middle.

[autumn kraken]

I am not familiar with the tuple notation, so I write them down as lambda * a = b and lambda * a' = b' but that's the same thing right?

[cursive narwhal]

Tuple notation is very similar to what you do with, say, a 2D or 3D coordinate system. But yes, it does come down to the same thing.

In fact, the original question defined the relation as what you have just written, not in terms of tuples. That's something that loch introduced into the problem because it's easier to read in that way.

[autumn kraken]

okay I see

I didn't find a way yet to express x' as x, but I have x = b/a and x = b'/a' for the first one and x' = a/b and x' = a'/b' for the second one. Doesn't that prove the existance of x' already?

or can you not do this

because they're connected with and

[cursive narwhal]

I mean, look very closely at how x and x' are related there. Can you really not see how they're connected? [x = b/a and x' = a/b]

[autumn kraken]

1 = x * x' ?

so x' = 1/x

[cursive narwhal]

Yea precisely. Since x is a nonzero rational, it has a multiplicative inverse so that proves the existence of our desired x'

[autumn kraken]

damn that was complicated

lol

thanks for the help, I will try transitivity now ...

[cursive narwhal]

Yeap. Work straight from the definitions.

If you need any help, just come here.

[autumn kraken]

thanks I appreciate it

Is my reflexivity proof correct? I said (a; a') ~ (a; a') is true because for lambda = 1 it becomes 1*a = a and 1*a' = a'

and the definition asks for at least one lambda

[wintry steppe]

How do I find the determinant of a 2x3 matrix again

[cursive narwhal]

You don't. Determinants are only defined for square matrices.

[wintry steppe]

Oh wtf

[cursive narwhal]

Madmike, that works.

[wintry steppe]

How do I find the wronskian when I only have two solutions for y then

[cursive narwhal]

[wintry steppe]

😭

[autumn kraken]

btw. does lambda have any special meaning or do mathematicians just like weird symbols ? 😛

[cursive narwhal]

It has a very special meaning

It's called the suckondeeznuts constant of an equivalence relation

[autumn kraken]

[subtle walrus]

how is it a weird symbol

[cursive narwhal]

It actually looks like some person showing off their leg to someone else

like those fashion models and such

[subtle walrus]

$\lambda$ $\Lambda$

[stoic python]

Lochverstärker:

[subtle walrus]

[autumn kraken]

I like the lambda symbol

but it's just not something I have seen before university

[cursive narwhal]

it's okay, the suckondeeznuts constant is prevalent in math

[subtle walrus]

it just so happens that the roman alphabet doesnt have enough letters

and at some point you stop making a distinction between the alphabets

there are some "conventions", like lambda is often used for eigenvalues, but your example shows that conventions are made to be broken

[autumn kraken]

interesting

[lime granite]

A quick (maybe) question, if I have an nxn real matrix A that is similar to a diagonal matrix B, can I say that A is symmetrical?

Wait I figured it out, sorry

[proven garden]

Hi is anyone here familiar with spectral decomposition?

[pale shell]

In a vector space

Does every vector have to have an inverse such that

v+(-v)=0

Or do inverses not have to exist

[subtle walrus]

additive inverses do exist

[half ice]

4th down, a vector space has to have additive inverses for every element

[pale shell]

So a vector space

Of all positive vectors

Can’t really exist then?

[half ice]

What's a positive vector? Haha

Much easier to digest is the three axioms of a subspace. If you know V is a vector space, and W is a subset of V, then W is also a vector space if

• Closed under addition
• Closed under scalar multiplication
• Has zero vector

[pale shell]

Oh so you could just say

Is not closed under scalar multiplication

Also when you prove a vector space

Do you have to prove all ten axioms?

[cursive narwhal]

@pale shell Something like $\bR$ is, indeed, a vector space over $\bR$. The set of real numbers does have an ordering associated with it and you can, therefore, talk about positive vectors. In general, however, there is no such notion of ordering for vector spaces in general.

What you're confused about is the fact that we're using + and -, the same symbols that we use for addition and subtraction on the reals.

[stoic python]

Abhijeet Vats:

[pale shell]

o

[cursive narwhal]

There's no such thing as 'proving a vector space'

[pale shell]

wot

[cursive narwhal]

You prove that a set is a vector space over some field and in order to do that, you need to verify that that set has elements that satisfy the vector space axioms.

[pale shell]

Yeah...

Proving a set is a vector space..

Is what I mean

[half ice]

99.999% of the time, in order to prove a set is a vector space, you'll actually use the three axiom version

[pale shell]

Mk

[half ice]

And take it as a subspace of a larger space

Or you'll use other theorems you've learned in class

[pale shell]

Also what do you think

Would be the best way to prove like

Basic transpose properties

[cursive narwhal]

Use the definition of transposition.

[pale shell]

Okay so kinda draw out a matrix and show that

[subtle walrus]

how did you define transpose?