linear-algebra

[robust pond]

like in this specific matrix

since theyre all linearly independent of any ONE other columnn

[wintry steppe]

he means that if you peak any two random vectors and put them in the matrix then they are linearly independent iff det isnt 0

[robust pond]

then you know that the whole thing cant be

[wintry steppe]

but by doing that youre not guaranteed the right answer, unless you do it twice

[robust pond]

but that just seems more involved

yea

also some might be and some might not

you just get lucky with this one

[gritty frigate]

But, why not?

[wintry steppe]

cause its not fast, requires calculating determinants lol

[robust pond]

$\begin{bmatrix} 1 & 2 & 1 & 1 \ 2 & 4 & 3 & 4 \end{bmatrix}$

[stoic python]

jan Niku:

[gritty frigate]

Oh yep but it is 2x2

[robust pond]

here youd have to pick the right two

you could get different answers

[gritty frigate]

It is kinda fast

[robust pond]

is checking a determinant faster than checking rank

[wintry steppe]

is it faster than saying 3 vectors cant be independent in r2?

[gritty frigate]

No of course not hahaha

[wintry steppe]

I hate determinants tbh

[robust pond]

lol

[gritty frigate]

But I guess he had to provide another answer hahaha

[robust pond]

determinants are dumb

what are they even good for

[wintry steppe]

like integrals and shiet only

[gritty frigate]

Determinants are amazing

[robust pond]

lets spend a whole week learning

how to take arbitrary determinants

[gritty frigate]

I love them and I dont know what the hell they are

[wintry steppe]

laplace

[robust pond]

laplace is

[brisk fractal]

determinants are dumb
@robust pond CRINGE

[gritty frigate]

L.A vs Calculus

Who wins ?

[wintry steppe]

I remember this one problem in my lin alg exam when I was only one of the few who got it correct, it was this very tricky laplace usage

[robust pond]

idk they didnt seem really useful to me @brisk fractal

[wintry steppe]

I like laplace since then

[robust pond]

other than showing up in the formula of an inverse

which is just like

factoring out a common factor anyways

[wintry steppe]

LA is better if its pure, calculus is better if you mean analysis

[gritty frigate]

Well, have you seen vectors in R3 right ?

[brisk fractal]

they sort of lay the groundwork for spectral theory

which is uh sort of important

[robust pond]

i like laplace since you get write the fancy L

and use table

[wintry steppe]

WHAT?

what L and what table

[gritty frigate]

hahahaha

[robust pond]

$\mathcal L$

[stoic python]

jan Niku:

[robust pond]

thats not it 🤔

[wintry steppe]

BRUH do you mean laplace transform?

[robust pond]

yea

[wintry steppe]

I meant laplace way to calculate determinants

any determinant you wish with that

,w laplace expansion

[stoic python]

Results provided by WolframAlpha

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

Requested by Godel#2664

[gritty frigate]

That definition is amazing

[wintry steppe]

looks scary but very ez

[gritty frigate]

I love how that is written

[wintry steppe]

ive never learned laplace transforms and I wish I could write this L @robust pond

[gritty frigate]

That L is pretty sexy

[wintry steppe]

so are you

[robust pond]

its a sexy L

i cant remember

$\mathscr L$

[stoic python]

jan Niku:

[robust pond]

is this it?

@wintry steppe

it looks like this

[gray dust]

$F(s)=\L\brc f(s)$

[stoic python]

RokettoJanpu:

[robust pond]

wrong L 🤔

[gray dust]

$F=\mathscr L\brc f$

[stoic python]

RokettoJanpu:

[robust pond]

wrong L

but an L nonetheless

wait godel what the hell is a laplace determinant

i got sidetracked

[gray dust]

wrong laplace

laplace expansion for finding det involves computing dets of smaller matrices (minors of the original matrix) as you iterate over a chosen row/col

[wispy copper]

Basic stuff but right handedness depends on order?

For example if i,j,k are righthanded

j,I,k is left handed?

cyclic permutations conserve handedness?

<@&286206848099549185>

[quartz compass]

yeah that's right

@wispy copper

[wispy copper]

@quartz compass just to clarify then, when texts say “the right handed set a,b,c” they mean the ORDERED set ...? Thanks :)

[quartz compass]

yeah

@wispy copper

[cunning arch]

can someone walk me through how they simplified in step 2?

i keep getting \begin{pmatrix}1+m^2&0\ 0&-m^2-1\end{pmatrix} when I multiply, I'm not sure how they got there

[stoic python]

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

[limber sierra]

i'm not sure what you're doing - it seems you're using rows for both matrices?

should be using rows of the first, columns of the second

for example, for the top-right entry (the green image)

it's in the first row, second column

so we expand on the first row of the first matrix (1 m), and the second column of the second (m 1)

1 * m + m * 1 = m + m = 2m

hence the top-right entry (green) is 2m

a similar process holds for all 4 positions

the bottom-right for example

it's in the 2nd row and 2nd column of the product (bottom) matrix

so we expand on the 2nd row of the first matrix, and 2nd column of the second

m * m + (-1) * 1 = m^2 - 1

[cunning arch]

wait give me a moment to read through it haha, thank you!!!

OH

I DON'T KNOW WHAT I WAS DOING EITHER

thank you so much

[limber sierra]

in the future, visualizing like this can be helpful

also, in case you're wondering about the 1/(1+m^2):

scalar-matrix multiplication commutes

[cunning arch]

oh nah i got that part, thank you so much

[limber sierra]

so we can just swap around the first matrix with the 1/(1+m^2)

and be fine

(but of course, we can't do this with two matrices)

[grizzled folio]

Hey, I am writing a paper on linear algebra at a high school level, can anyone here read through a few pages (less than 5 minutes would be 👌 ) and give me some feedback?? Thanks!

[native rampart]

Sure

[hollow finch]

@grizzled folio I'd like to see it too

[native rampart]

Just post it here

[grizzled folio]

Ok, I'd rather DM both of you

[foggy lodge]

Hmm, not sure how to approach this one

Any pointers?

[native rampart]

Suppose there is a x,such that x is not an eigenvector of T, Then Tx and x are linearly independent(Making V a T cyclic subspace)

If not,Then all vectors in the space will be eigen vectors of T,implying T=cI

[foggy lodge]

Not sure if I follow

[native rampart]

Just take 2 cases:There is a vector,which is not eigen and There are no such vectors

[foggy lodge]

if its not an eigenvector, then Tx, x are lin indep, but how does it show that T is a cyclic subspace?

*V

oh wait

[solid bough]

hi, idk if this is suppose to be trival,

Suppose S is a subspace for R(nxn) has a basis {b1 , b2 ,... br},
Denote B = [b1 b2 ... br], rank(B) = r

For any W in R(rxr) , denote BW = U = [u1 u2 ... ur]
If W is non-singular, then {u1, u2,..., ur} also form a basis for subspace S.

Can someone enlightened me?

[grand mesa]

Can someone solve this?

[modern palm]

@grand mesa

[hollow finch]

Pretty sure that's #prealg-and-algebra

@solid bough In case you still need help with your question: this is a case where it helps to think of matrix multiplication in terms of "column perspective"

$$\begin{bmatrix}\mid&\mid\c_1&c_2\\mid &\mid\end{bmatrix}\begin{bmatrix}x_1\x_2\end{bmatrix}=
x_1\begin{bmatrix}\mid\c_1\\mid \end{bmatrix}+x_2\begin{bmatrix}\mid\c_2\\mid\end{bmatrix}$$

[stoic python]

nix:

[hollow finch]

this happens in all matrix multiplication for each individual column. so the columns of a matrix product AB are a linear combination of the columns of A (similarly the rows of AB are a linear combination of the rows of B which is "row perspective")

so the columns of BW are a linear combination of the columns of B, which means that the columns of BW are in the span of your basis B

if W is nonsingular then its a linear combination which leaves the columns linearly independent

since the span is the same and they are linearly independent, its just as valid a basis as the columns of B

[narrow moth]

does anyone know what the qu means by working over Z3 and Z5? do we simply have to convert each entry to Z3 or Z5 at each stage?

[narrow moth]

<@&286206848099549185>

[soft burrow]

basically yes

it implies that e.g. 3 or 6 wouldn't be invertible when working in Z_3

since they're zero

[narrow moth]

so if say i multiply the second row with 2 in z3 will i get ( 2 1 1 1 0 ) ?

[soft burrow]

yeah

[narrow moth]

so at each stage i just change numbers to z3 or z5 right?

[soft burrow]

yes

[narrow moth]

ok ty!

[grizzled folio]

Hey, as I said before, I am writing a paper on linear algebra at a high school level, can anyone here read through a few pages (less than 5 minutes would be 👌 ) and give me some feedback?? Thanks!

[wintry steppe]

Sure post it

[grizzled folio]

Can anyone else here look through this report?? 👆

[soft burrow]

I might check it out in my free time

[rugged basalt]

Could somebody help me out with this?

I understand how I would solve it if it was just: y = au + Bv

but how do I solve it given the added y variable

[trail mulch]

Have you tried appending a column full of 1's to match with the gamma coefficient?

[glass meteor]

i tried working through this but im not sure where i went wrong

[ivory moon]

@glass meteor www.math.odu.edu/~bogacki/cgi-bin/lat.cgi

[solid bough]

@solid bough In case you still need help with your question: this is a case where it helps to think of matrix multiplication in terms of "column perspective"
@hollow finch Thank you for the explanation, i think it becomes clearly when i view it as a linear combination. Just to clarify, if column vectors in B form a basis subspace S , then BW is still a basis for subspace S. Would this be viewed as a change of basis?

[hollow finch]

Typically we use change of basis for a basis of a vector space, but in the sense that we are changing a basis for a subspace i think that's appropriate. Not 100% sure but I think as long as W is invertible then coordinate vectors should stay unique and consistent

Even then I don't think I've considered coordinate vectors for a lower dimensional subspace. I feel like it should work though

[acoustic path]

Hey there. Is it true that the row operations are just multiplying by matrices/vectors without writing the entire thing?

[native rampart]

I don't know how you multiply with vectors

But, yea you could see matrix multiplication as doing row operations

Look up elementary matrices

[acoustic path]

Ye i meant matricez

Thx

[oblique rune]

What does adjoint of a matrix really mean

I only know to calculate it

[pallid rampart]

You can define the adjoint operator of a general linear operator over a (potentially infinite dimensional) complex vector space

So let $V$ be a complex inner space, $T$ be a linear operator on $V$, let $T^*$ be the adjoint of $T$. Then for each $v\in V$, $T^*v$ is the unique vector such that $\brk{Tv,v}=\brk{v,T^*v}$

[stoic python]

Whoever:

[pallid rampart]

So basically the adjoint operator satisfy $\brk{Tv,v}=\brk{v,T^*v}$ for all $v\in V$

[stoic python]

Whoever:

[pallid rampart]

If V happens to be finite dimensional, then you can find an orthonormal basis (it HAS to be an orthonormal basis) and you realize that the matrix of T* is the conjugate transpose of the matrix of T

@oblique rune

Hopefully that helps

[oblique rune]

What is an inner space?

[native rampart]

Transpose of the cofactor matrix is also called adjoint

[oblique rune]

And what does ⟨Tv, v⟩ mean

Yes that's how I was taught

And I was told that it's used for finding inverses

And that was it

[native rampart]

[oblique rune]

Ikr

[pallid rampart]

Um

Well

Here’s the definition of inner product, and an inner product space is a vector space with an inner product

But I suggest you to read axler chapter 6,7

[oblique rune]

Oh I have a PDF of that book

[pallid rampart]

They talk about inner product and adjoint operator

[oblique rune]

Ok

[pallid rampart]

And much more

This is 3rd edition btw

[oblique rune]

Ok, I'll check it out and ask doubts if I have any, bye

[pallid rampart]

Oh trueee

@oblique rune yeah I think you meant a different adjoint matrix

I’m sorry

https://youtu.be/JyISylNXGzE

[rose umbra]

does this means infinite solutions?

[honest token]

Yes @rose umbra

there is 1 free variable

which you can choose to be any value

so infinite solutions

[rose umbra]

@honest token is it always the case with free variable?

[honest token]

yes

every time you have a free variable, it can take an infinite number of values

[rose umbra]

and when I dont have free variable, is it always single answer?

unless i have contradiction(?)

[honest token]

If the system is consistent then you have a unique solution

If you have a contradiction you have 0 solutions

[rose umbra]

unique solution = one solution?

[honest token]

yes

[rose umbra]

thanks a lot !

[honest token]

welcome!

[native rampart]

It is enough to define an inner product on a finite-dimensional real space by specifying an orthogonal basis for the space, correct?
Yes

I mean,same for complex spaces too

[honest imp]

im having a hard time answering t/f questions like these, can someone guide me through the thought process of figuring this out? I've gone as far as identifying the set as an Ax = 0 equation to further help, but not sure if thats something i should be doing. that would make A 6x4

and it can have 4 pivots at most

but how does any of that help me

[wintry steppe]

Do you know what a basis is.

[honest imp]

ya

just learned about it

[wintry steppe]

I get very confused about lattice in linear algebra, can someone help me understand what is it?

[half ice]

@honest imp
Consider
(1,0,0,0,0,0)
(0,1,0,0,0,0)
But 4 of those

You seem to be eager to think of a matrix, but linear independence has nothing to do with matricies right away

[honest imp]

@half ice ty for the response!, this was one of the conclusions i came to when thinking about this problem. 4 of the vectors can be e1-e4 which are basis and are lin. independent

[half ice]

Glad you got it!

Wait no e1-e4 is not a basis for R⁶

But they are linearly independent yes

[honest imp]

no i meant basis vectors

but would be for R4

sorry, that mightve confused you LOL

[half ice]

Details aside, that works as an answer to this question

[honest imp]

i have one more question if you dont mind helping while youre here :o

[half ice]

Go for it!

[honest imp]

for these types of questions, idk if its because i have a hard time visualizing it, but what would i be looking for when doing this problem?

[dusky epoch]

think about the concepts of dimension and span

[half ice]

Do you know what dimension is?

[honest imp]

yes, im relatively new to the concept though

[half ice]

I'll bring this down to a lower dimension to really discuss the concept. The span of two vectors can be, at most, a plane

You can never get any 3D space by using the span of two vectors. That make sense?

[honest imp]

ya

[half ice]

R³ has dimension 3. That means,

• You need 3 vectors in order to span it
• More than three vectors "won't have enough room" and will be forced to be linearly dependent

[honest imp]

that makes sense too

[half ice]

R⁶ is the same game. Has dimension 6, you need at least 6 vectors in order to span it

So the question you posted above is false, 4 can't do it

[honest imp]

that seems a lot simpler than what i was thinking in my head maybe its because i was over thinking it

if i had 4 vectors, they would be able to span r3 then but one of those vectors will become linearly dependent of the other?

[trail mulch]

yes, exactly

[honest imp]

https://tenor.com/view/cat-nervous-anxious-cute-gif-15464674

ahhhh ty!

[oblique rune]

@oblique rune yeah I think you meant a different adjoint matrix
@pallid rampart

Ohh yes I didn't specify that

Why is the inner product over a vector field of polynomials written like this?

I'm guessing it has something to do with how integral can be thought of as infinite summations of p(x)q(x)* but I don't know.

[pallid rampart]

there is no the inner product

[oblique rune]

oH yes

[pallid rampart]

that is only one inner product

[oblique rune]

My bad

[pallid rampart]

But you can just check the axioms

If this function (a function that assigns two vector to a scalar) satisfy all the axioms of an inner product then it is an inner product

[oblique rune]

Why is inner product written like this in this case?

[pallid rampart]

written like what

the inner product isn't even written here

[oblique rune]

That integral

[pallid rampart]

well that's the definition

that's the definition of one example of an inner product

[oblique rune]

Hmm ok, so since it returns a scalar, it's an inner product. But why is the integral evaluated from 0 to 1? Why can't it be something like 0 to 2

[pallid rampart]

well it's just an example

but you are right

the integral from 0 to 2 also defines an inner product

in fact the integral from a to b for any a≤b defines an inner product

[oblique rune]

But wouldn't the value change

[pallid rampart]

right

so that's why this is only one example of an inner product

there are many many inner products

[oblique rune]

So different inner products can give different values?

I see

[pallid rampart]

i mean yeah of course

by definition the same inner product means they produce the same value on every pair of vectors

therefore different inner product must give different values

that is what it means for two functions to be different

because afterall, inner products are just functions

[oblique rune]

Hmm I see

[uncut barn]

is $mat^{-1}$ the inverse matrice ?

[stoic python]

Cewein:

[wintry steppe]

Yes

Inverse matrix of mat

[pallid rampart]

No it's the inverse of t multiplied by ma on the left

[grizzled folio]

nevermind I figured it out

@wintry steppe I am currently writing a paper on linear algebra, would you be willing to give me some feedback??

[royal ore]

Hello! I have to show <u+v, w> = <u,w> + <v,w> for any u,v,w ∈ ℝ²

[half ice]

Let u = (u1,u2) et cetera

Compute <u+v, w>
And compute <u,w> + <v,w>

Each computation should come out to the same thing, proving the claim

@royal ore

[wintry sphinx]

but isn't <> an inner product? isn't that true by defn?

[royal ore]

yeah just needed to prove it but figured it out

now how would i show that <v,v> = 0 if and only if v = 0

[half ice]

Nah it's likely the dot product here

Same idea, evaluate <v,v> for v = (v1,v2) @royal ore

[royal ore]

what would <v,v> be evaluated as to show it's 0

[half ice]

What do you get for <v,v>? Lol

[royal ore]

wouldn't it be v1v1+v2v2

[half ice]

Exactly, or v1² + v2²

Because of the way squares work, this is always 0 or positive

So that means this is only ever 0 if v1² = v2² = 0

Which is just v1 = v2 = 0

Ergo <v,v> = 0 iff v = 0

[royal ore]

thank you

now I have a question that involves matrix multiplication

i got the product of B * Z

but now how does 3rd row of it get written as a linear combination of rows of Z

there's 4 for the 3rd row but 3 rows in Z

[half ice]

These are steps that you take in the multiplication

For example, in order to get that bottom left 3, you did
1(0) + 2(3) + 3(-1)

Note that the numbers I put in parentheses are the first digit of each of the rows of Z

And indeed:
1(0,1,-1,2) + 2(3,0,2,-1) + 3(-1,1,1,4)
Gives the bottom row of BZ

That's just doing multiple of your calculations at one time

[royal ore]

so just do 3(0) + 3(1) + 3(-1) + 3(2) ...

and then replace 3 with 4 for next row

or no

because it only does 3,4 and 6

[royal ore]

anyone know how to finish?

[wintry steppe]

$\begin{bmatrix} 0 & 1 & -1 & 2 \ 3 & 0 & 2 & -1 \ -1 & 1 & 1 & 4 \ \end{bmatrix}$

[stoic python]

DrifAssault:

[wintry steppe]

@royal ore good?

[royal ore]

yeah that’s Z, but now how do I do a linear combination of 3 4 6 12 with it

[wintry steppe]

i am not sure tbh

you can try ping helper role

[royal ore]

<@&286206848099549185>

[wintry steppe]

I've been trying to reduce this matrix, and find the conditions to make it consistent:

$$\begin{pmatrix}1&-3&a\ :1&-2&b\ :1&1&c\ :1&-4&d\ :1&5&e\end{pmatrix}$$

[stoic python]

ryаn:

[wintry steppe]

Symbolab and others use some wacky technique to remove several columns

Anyone know what the heck they just did?

Oh nvm..

You can multiply by any constant, so I guess that's fair game.

Though, this feels a lot like multiplying with zero? How does this help find conditions for consistency?

[honest token]

@wintry steppe the equation is consistent iff all the rows containing 0s (except the last term) have a last term that is 0

so the rightmost column needs to have 0s in its 3rd and 5th rows to be consistent

[wintry steppe]

Right, though I'm slightly confused with what they did with multiplying R3 with some insane constant?

I've never seen this trick before, is there a name for it?

[honest token]

Gauss-Jordan elimination

[native rampart]

Row operations

[honest token]

or row reduction

[wintry steppe]

Well sure, it's row reduction, but I've only seen it with normal constants.

[native rampart]

why should it be any different?

[wintry steppe]

I guess not, but I'd personally not sure how I'd come up with such an expression.

Is it multiplying by a conjugate or something?

Oh nvm, yeah I see they're multiplying and subtracting with some conjugate.

[honest token]

yes

[wintry steppe]

Right, I think I have a better grasp, thanks!

[honest token]

welcome

[tacit sonnet]

In a given vector space there can be n number of basis vectors right?

[marble lance]

Huh?

[tacit sonnet]

I just meant to ask how many basis vectors are possible in a given vector space?

[native rampart]

how many elements can a set have?

[marble lance]

R^n can always have n basis vectors. So you can have any number of basis vectors.

[magic acorn]

i went through all the questions but then i realized that in order to continue on to b-e i would have to complete a) at least which i dont understand at all

i dont get how you would use the quadratic equation

to find a and b

[subtle walrus]

you can even turn any set into the basis of a vector space

[magic acorn]

meaning i would have to use a cartesian plane for x, y and z?

or would that be unrelated

[subtle walrus]

my remark was to the previous discussion

[magic acorn]

ah mb mb

[magic acorn]

alright so i completed a-c now but now idk how you would graph positive values on that certain function

[surreal thistle]

hey does anyone here know anything about the rhombic tilings of spheres

theres two infinite families that im trying to find

im writing a paper about it

any resources or anything helps

[dusky kayak]

In a right-angled triangle, one catheter is three times as long as the other. The hypotenuse is
4.0 m longer than the longest catheter. Calculate the circumference of the triangle. Round your answer to tens
meter.

Can i get som healp hear

sory for bad english

[patent python]

can someone help me out with this

[dusky kayak]

prety ezzy math i know, but a friend recomended me to send here

[patent python]

i dont really understand how we get that part

@dusky kayak try doing pythagorean theorem and you get an equation from that with negative and positive length of each triangle leg. From there just add everything up. Just my idea though

equation looks like this i believe
16 + 12x + 9x^2 = x^2 + 9x^2

[novel sedge]

I'm not sure my question belonged to multivar-calc-and-diffeq, so i posted there... might be linear algebra too though

[severe cedar]

"Let H = {x \in R, x_1 + x_2 + x_3 + x_4 = 0} be a hyperplane and T: R^4 -> R^4 be a reflection on H. Find all eigenvalues and eigenvectors of H. Is T diagonizable?"

How tf do I solve this?

[patent python]

oh boi..

i had fun with those :'')

[severe cedar]

I've never encountered a problem like this before and it just showed up on my exam...

[patent python]

eigenvalues and eigenvectors are foudn on matrices

eigenvector would be a matrix with n rows 1 column

and eigenvalue is 1 value in that row

[severe cedar]

So failure is already a fact but I'm curious how to solve it

[patent python]

honestly dont remember how you find eigen~val/vectors. If i were you i would have just swapped the values around in the matrix seeing as its a reflection the values would be the same just in different positions

so X axis would have the value of another axis etc lol

[honest token]

@severe cedar you can solve the problem without knowing the matrix for T, i think

the eigenvectors are the vectors unchanged in direction and the eigenvalues are the factor they change magnitude by

so for the eigenvalue of 1: you can get 3 eigen vectors that span the plane

and the eigenvalue of -1: you can use the normal to the plane

because T has 4 eigenvectors, it is diagonalisable I think

[severe cedar]

Thanks. Then I might have gotten something right on that question at least

[honest token]

I'm not completely sure but that would get some marks

[severe cedar]

I "guessed" that I would get eigenvalue 1 of multiplicity 3 and -1 of multi 1 on the assumption it was a Householder reflection

[honest token]

that's fair

that seems like a really good question to throw off students though

[olive atlas]

say i have the eigenvalues and the eigenvectors for a nxn matrix is there a way to get the original matrix?

[dark valley]

whats the matrix rank?

[olive atlas]

im not too sure

i feel like it has something to do with diagonalization

[hard coral]

the amount of linearly independent rows/cols

[severe cedar]

say i have the eigenvalues and the eigenvectors for a nxn matrix is there a way to get the original matrix?
@olive atlas

Given some n x n matrix A you can write it as A = BDB^-1 where D is a diagonal matrix with the eigenvalues diag(λ_1,λ_2...) and B is a matrix with the eigenvectors as columns, and B^-1 it's inverse. Be careful to match the eigenvectors and eigenvalues in the matrices. That is, if you have λ_1 in the first column of D, you should have the eigenvector corresponding to λ_1 in the first column of B etc.

This obviously requires the eigenvectors to be linearly independent, otherwise you won't be able to define the inverse of B.

[olive atlas]

yeah thats what i was thinking

thanks

[olive atlas]

also how do i determine the order for the eigenvalues i get from the roots of the characetestic polynomial like if i have a 3x3 matrix and i get -3,-2,-1 as the values how do i know which is value1, value2 and value3?

[olive atlas]

nvm order doesnt matter as long as the order of the eigenvectors matches the order of the corresponding eigenvalues

[robust pond]

what a weird username

@gilded solstice do you have a formula for inverse

you can just do this directly

[stoic python]

jan Niku:

jan Niku:

[robust pond]

a lot can be inferred from this; its just algebra though, albeit a lot of it

:p

[wintry sphinx]

you could also look at like A^2 = I and check the spectrum

[robust pond]

😮

good idea

this seems weirder

[wary moss]

https://cdn.discordapp.com/attachments/763251860156317696/770322290092539904/Capture.png

this was on my linalg hw and I was able to prove that a second order homogenuous equation has at least two solutions by cases but generalizing to nth order... i have no clue

is there any hint to what direction i could go in?

since it seems that if i just try to prove it the same way i did before it would be impossible to account for every time a repeated root comes up

[robust pond]

might want to try #multivariable-calculus , although ig it works in both

[wary moss]

yeah idk where to put but i was hoping maybe the explanation would be more linalg-y if i put it in here

😔

[topaz geode]

I'm doing Linear programming and I'm trying to understand this task. I have been looking all over my textbooks, but I can't find the definitions for these letters... Anyone that can help me out here?

[torpid horizon]

can someone help me

with a math queston please

im willing to donate

im having troble with tis hw problemcan someone please help me

if someone can please help me i dont mind making a donation

[wary sinew]

Can you explain your problem

What’s R_4

[torpid horizon]

its

[wary sinew]

..?

@torpid horizon

[torpid horizon]

im double checking

[wintry sphinx]

he's [checking]_2

[primal rapids]

maybe you should think before you post

[torn hornet]

whats R_n

[torpid horizon]

means sunset

[torn hornet]

subset of what, and isomorphisms of what?

[torpid horizon]

yes

[wary sinew]

Pixie join the voice call

[torpid horizon]

how?

[wary sinew]

Scroll down, it’s “mathematics” in the server

[torpid horizon]

where it says math voice?

[wary sinew]

Yeah

[wintry sphinx]

$\mathbb R^3$?

[stoic python]

Saccharine:

[wintry sphinx]

is that what you mean

[torn hornet]

$R_2$

[stoic python]

JohnDS:

[wintry sphinx]

$\mathbb R$

[stoic python]

Saccharine:

[slow scroll]

I vaguely remember a my linear algebra class using $\bR_n$ to mean rows

[stoic python]

kxrider:

[slow scroll]

pretty dumb ik lol

[torn hornet]

,rotate

[half ice]

@wary moss
It's not really an interesting question oop. Basically, given a linearly independent f and g, there's nothing preventing af + bg from being a solution

It is much more interesting to note that af + bg + ch could never be a solution

[wary moss]

right sure but what about for nth order diffeq?

[torn hornet]

$\mathbb{Z}$

[stoic python]

JohnDS:

[half ice]

A diffeq of order n has n linearly independent solutions. Perhaps they don't want a proof of this.

[wary moss]

i hope not

but i feel like they do 😔

my classmates say that's what it is asking though

[half ice]

From the sounds of it, they didn't do the first part well enough to generalize oop

[wary moss]

The instructor or my classmates?

and thank you for helping

[jolly dome]

Can somebody plz help me with this?

[austere cedar]

Write it as an augmented matrix and perform row operations until you have the column with entries 1 0 0.

[jolly dome]

idk how

thats the problem

[dense girder]

how can I expand on 1/x^(a+1) ?

brain is not working

[half ice]

So if you do a matrix multiplication but only fill out the bottom row of BZ, you'll do
1(0,1,-1,2) + 2(3,0,2,1) + 3(-1,1,1,4)
In order to find it

Giving the third row of BZ as a linear combination of the rows of Z

[royal ore]

ohhhh

i was doing it really different

thanks

i thought the 3rd row of B*Z had to be used

[royal ore]

not sure how to connect this or put into formal writing

[wintry sphinx]

there are a lot of ways to interpret this

you have shown that any vector of the form u = [blah, blah, z], where z is nonzero does not have a vector v such that Av = u, so therefore A is not surjective. Trying to find the inverse matrix will fail when you attempt to invert the basis vector [0, 0, 1], by the previous part. If A is not surjective, then the inverse cannot be defined for the elements which A does not hit.

[wintry steppe]

can anyone help me with a least squares parabola question?

[wintry steppe]

how do i do this

[wintry steppe]

i don't understand where i'm going wrong

[limber sierra]

well, L doesn't have ones on the diagonal

it has a 2 right there

what technique (if any) have you been taught to find LU decompositions? gaussian elimination?

[arctic hazel]

i have a question about a linear algebra problem that i'm doing, but it's more about writing up the latex than the actual problem, if that's ok

i'm solving an augmented matrix, and i have this block of code to express it:

$$\left[\begin{array}{cc|c}{4 & 1 & 0 \ -3 & -1 & 0}\end{array}\right]\to\left[\begin{array}{cc|c}{4 & 1 & 0 \ 1 & 0 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \ 4 & 1 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \ 0 & 1 & 0}\end{array}\right]$$

[stoic python]

Snodlop:

$$\left[\begin{array}{cc|c}{4 & 1 & 0 \\ -3 & -1 & 0}\end{array}\right]\to\left[\begin{array}{cc|c}{4 & 1 & 0 \\ 1 & 0 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \\ 4 & 1 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \\ 0 & 1 & 0}\end{array}\right]$$
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.54 $$\left[\begin{array}{cc|c}{4 &
1 & 0 \ -3 & -1 & 0}\end{array}\right]...
I've put in what seems to be necessary to fix
the current column of the current alignment.
Try to go on, since this might almost work.

[arctic hazel]

texit gives the same error that my other latex editor gave, but i don't know how to fix it

here's the text again so you can see what it looks like

$$\left[\begin{array}{cc|c}{4 & 1 & 0 \ -3 & -1 & 0}\end{array}\right]\to\left[\begin{array}{cc|c}{4 & 1 & 0 \ 1 & 0 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \ 4 & 1 & 0}\end{array}\right]\to \left[\begin{array}{cc|c}{1 & 0 & 0 \ 0 & 1 & 0}\end{array}\right]$$

[stoic python]

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

[arctic hazel]

if this is the wrong place to ask a question like this, i'm sorry but i don't know where else to ask

[limber sierra]

uh

the contents of your arrays dont need to be wrapped in {}s

its an environment like any other

the \begin{array}{cc|c} and \end{array} suffice as a wrapping

$$
\left[
    \begin{array}{cc|c}4 & 1 & 0 \\-3 &-1 & 0\end{array}
\right]\to \left[
    \begin{array}{cc|c}4 & 1 & 0 \\ 1 & 0 & 0\end{array}
\right]\to \left[
    \begin{array}{cc|c}1 & 0 & 0 \\ 4 & 1 & 0\end{array}
\right]\to \left[
    \begin{array}{cc|c}1 & 0 & 0 \\ 0 & 1 & 0\end{array}
\right]
$$```

[stoic python]

Namington:

[limber sierra]

this should compile fine

@arctic hazel

[arctic hazel]

oh that's absolutely perfect

thank you so much

this one error was driving me crazy and i had no clue what i did wrong

[limber sierra]

to clarify what was going on

normally you can insert {}s randomly in latex and itll kind of ignore them

the problem is that

because youre separating the entries of your array using &s and/or \\s

you're putting a { in one entry, and a } in the other entry

this confuses latex

since its expecting the { to be completed in the same entry (LaTeX is not very good at looking ahead)

so it tries to insert the "missing" } at the end of the entry with the {, and gives an error

the solution, as shown above, is just to remove the useless {}s that are confusing latex.

[arctic hazel]

ok that makes sense

i didn't realize that array was its own environment, i thought i still had to wrap the insides in braces

[limber sierra]

everything within \begin \ends is an environment

[arctic hazel]

good to know, thanks

i'm still very new to latex

[severe cedar]

Given a linear operator on C^3, T(z_1,z_2,z_3) = (2z_2 + iz_3, iz_2, z_1), how would you going about finding the adjoint of this operator? Would you produce a matrix using the standard basis and find the conjugate transpose? Or is it possible to find the adjoint using <T(z),y> = <z,T^*(y)>

[native rampart]

The standard basis might not be orthonormal. Preferred method would be to convert to a orthonomal basis and take conjugate transpose

Sure,You can find the adjoint with that definition

(The adjoint being conjugate transpose of matrix of T in an orthonormal basis is based on that)

@severe cedar

[severe cedar]

With which definition, the last one?

[native rampart]

Yes

[severe cedar]

Thanks. I hope I got it right then.

What confused me too is that the standard inner product of C uses the conjugate of the second factor. So I'm not sure if the adjoint "removes" the conjugate in <z,T^*(y)>.

[native rampart]

Do you know how you get the representation matrix is conjugate transpose ?

[severe cedar]

sorry what

Do I know how to find the conjugate transpose of a representation matrix? Yes

and yeah the conjugate transpose matrix uses the conjugate so i guess that does it

[native rampart]

No,I am asking whether you know why taking conjugate transpose of a matrix representation of T gives you the matrix representation of T*?

[severe cedar]

hmm i'm not totally sure

[native rampart]

Let ($e_1,e_2...,e_n$) be an orthonormal basis .Do you know if $Te_1=c_1e_1+c_2e_2...c_ne_n \implies \langle Te_1 \mid ej \rangle=c_j$

[stoic python]

DrunkenDrake:

[reef ridge]

to solve a system of linear equations by EROs do i need to have the matrix completely reduced i.e.
[1 0 0] [1 a b ]
[0 1 0] or can it be [0 1 c ]
[0 0 1] [0 0 1 ]

in my lectures it was always completely reduced but looking at some other revision materials it shows this

does having it completely reduced just make it easier?

[native rampart]

Yes

[reef ridge]

okay so are either able to solve a system of linear equations

Man I wanna understand why gaussian elimination actually works but i gotta catch up on a lot of stuff and i've no time

[knotty blaze]

@reef ridge for gaussian elimination you have to put it in "row echelon form" which looks for a system of 3 eq-ns looks like this

[reef ridge]

I understand what it is, i'm more wondering why elementary row operations and stuff to get there actually solve linear equations

if that makes sense

[knotty blaze]

okay the way I imagine it

is that each row is a variable

you're able to solve linear eq-ns because

you effectively get a solution for 3rd variable in the bottom row

z=constant

and this immediately allows you to solve the rest of the matrix as the rest is:
y+z=constant
but we already know z=const

[reef ridge]

right

[knotty blaze]

so this

is the same as

1x-2y+1=4
0x+1y+6z=-1
0x+0y+1z=2

[reef ridge]

yeah

[knotty blaze]

is that what you were wondering

[reef ridge]

not exactly but i'll come back to it later cause i need to study some other stuff

like i have no idea what this means

[native rampart]

What is a net?

[reef ridge]

i think it's supposed to say set

[knotty blaze]

https://math.stackexchange.com/questions/871145/proving-a-subset-is-a-subspace-of-a-vector-space

[blissful coral]

Can someone help me with this expresison

the result is ln(2)

[native rampart]

Wrong group

#calculus

[tacit girder]

anyone can help

https://gyazo.com/4a9552d4828389f08d91006dacfc9f59

[honest token]

@tacit girder the rank is 1 meaning the columns must be linearly dependent

so the second column has to be a multiple of the first one

[tacit girder]

so a would =2

[honest token]

yes

[tacit girder]

thanks

[honest token]

welcome

[stuck stratus]

'Let A be an orthogonal matrix. Show that A^2 is an orthogonal matrix, too.' How do I do this?

[steady fiber]

we know $AA^T = I$,
let's find $A^2(A^2)T$
$$
AA(AA)^T = AAA^TA^T = A(I)A^T = AA^T = I
$$

[stoic python]

PorosInMyAshe:

[steady fiber]

so A^2 has to be orthogonal too

[stuck stratus]

Got it thanks :)

Suppose that for a symmetric matrix A it holds that A = QRQ−1 with Q an orthogonal matrix and R an upper triangular matrix. Prove that R must be a diagonal matrix then

Is this correct: A = QDQ^-1 = QDQ^T. Then A^T = (QDQ^T)^T = (Q^T)^T D^T Q^T = Q D^T Q^-1. Then both D and D^T have to be upper diagonal matrixes so D is diagonal

?

[gray dust]

why do you think

B's col count must match A^T's row count

[thorny hemlock]

im stupid

ye

whats a unit matrix?

it can have zero as entries aswell right

[gray dust]

in most cases it's synonymous with identity matrix

[thorny hemlock]

[0 1] is a unit matrix

?

[gray dust]

an identity matrix is a square matrix with 1s on the diagonal & 0s off it

[thorny hemlock]

oh

ok

that makes sense

[cunning arch]

if i'm given a transformation reflection matrix, how can i "prove" that it is rigid?

[stoic python]

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

[cunning arch]

close enough. ^ i got this transformation matrix, which is a reflection acorss the y=tantheta x line, and ik that reflections are all rigid, but idk my prof is asking me to prove it?

[native rampart]

|T(a)-T(b)|=|a-b|

Show that for all a,b

[cunning arch]

like just find the length of transformation matrix?

sorry i'm a bit lost, kinda don't remember the notatino

[native rampart]

Yes,distance

I guess that would just be $\sqrt{x^2+y^2}$ for (x,y) in your case

[stoic python]

DrunkenDrake:

[cunning arch]

ooooh ok, thank you!

[tacit flume]

i got this problem
The chance certain seeds grow into decent plants
depends on the time between buying the seeds and planting.
of it. If the seeds are planted after a week, there is a 92% chance of the seeds being planted.
that the seeds hatch. After four weeks, the chance is only 79% and after an
year this is still 6.7%
and i need to calculate this
The chance that a seed will hatch immediately after it has been planted is about 97%. Calculate this probability to one decimal accurate.

[thorny hemlock]

how to find adjoint of A?

[gray dust]

@thorny hemlock https://www.mathsisfun.com/algebra/matrix-inverse-minors-cofactors-adjugate.html

[warped garden]

hey all, I understand that matrix A must be a symmetric matrix for it to be orthogonally diagonalizable. But will it still be possible if matrix A is a diagonalizable matrix? Since during computation, the bases will be transformed to orthonormal bases. Or am I having some misconception in my thought process?

[golden dagger]

im having trouble proving the following are equivalent

can anyone hint as to what I have to do

[quasi vale]

multiply and divide by conjugate, that is $ (\sqrt{x+1} + \sqrt{x})$

[stoic python]

Sup?:

[quasi vale]

starting with f(x)

[golden dagger]

ah got it, cheers

[cunning arch]

i'm just... so confused??? can someone walk me through one of them?

like i know what it means when something is one-to-one and/or onto, but like

[hollow finch]

@cunning arch can you describe what it means to be one-to-one/onto?

[cunning arch]

basically one-to-one is if each vector in like space D has a unique image/gets mapped to different places in R. onto is if each vector already has like at least 1 image in D

i think

or it's onto if the columns span R, and on-to-one if the columns of transofrmation matrix A is linearly dependent

[wintry sphinx]

so one way of doing this, if you're more familiar with the idea of using the columns of a matrix

is writing down the transformation matrix for each of them

[wintry steppe]

I get confuse to solve this

Problem say : 'Prove this equation'

[jagged marsh]

so for an $n\cross n$ matrix $\mathbb{A}=[a_{ij}]$, where $a_{ij}=i+j$, how do you find det($\mathbb{A})?

[stoic python]

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

[native rampart]

The same way you do for any other matrix

Write the rows and try row reducing

[jagged marsh]

hhh it doesnt work

or im too dumb

[rich dove]

If it's a factorization you need, think of 5 as $\sqrt{5}^2$

[stoic python]

A~Z:

[native rampart]

I think you get 0

Because if you subtract 2nd row from 1st row,you get 1,1,1...

Similarly row3- row 1 gives you 2,2,2...

@jagged marsh

[thorny hemlock]

i know how to find the echelon form, just not in this case? what do i do?

[dusky epoch]

construct the augmented matrix

rref it

[thorny hemlock]

of which matrix

theres 3

@dusky epoch

[dusky epoch]

you have your system written down in matrix form

Ax = b

the augmented matrix is literally just A but with b stuck to it on the right

[thorny hemlock]

what does that even mean

augmented? stuck?

[honest token]

A|b

like the top row will be 0 1 1 0 1

[thorny hemlock]

i hope its legible

but

you mean like that?

[honest token]

you can't let the x vector equal that

but the matrix is right

[thorny hemlock]

ok

so now, i can find do the echelon algorithm as normal right

[honest token]

yes

[thorny hemlock]

ok after finding the echelon form

how to solve?

[honest token]

what's your matrix

[thorny hemlock]

one second

[honest token]

yes

[thorny hemlock]

idont think its right, cuz ive got elements on the left most collumn

i cant eliminate those

[dusky epoch]

this is not in row-echelon form

row switching is a thing, just sayin'

[thorny hemlock]

um how is it done then

it cant be done?

[dusky epoch]

yes it can

[thorny hemlock]

ive got elements below 0 first collumn

[dusky epoch]

you just need to stop being one track minded for a moment and like

switch rows 1 and 2

and then do rref again

[thorny hemlock]

wait waht

[dusky epoch]

what

[thorny hemlock]

switch rows 1 and 2 before or after

[dusky epoch]

you can continue from here

[thorny hemlock]

i can continue from where i am?

[dusky epoch]

yes you can continue from where you are right now

[honest token]

the 0 0 0 0 -3 row looks kinda sus

[dusky epoch]

as long as you didn't fuck anything up in getting here

[thorny hemlock]

lemme check it with matlab

nvm

matlab only does rref

ehm

shall i try switching rows 1 and 3

in the original matrix

and redo ing it

[dusky epoch]

if you want

[thorny hemlock]

with switching rows

do we want the largest element row in the top row

that doesnt make sense but hopefully you understood lmao

[dusky epoch]

it doesn't mater

so long as the top left element isn't zero and you can actually do the elimination

[thorny hemlock]

@dusky epoch does this look good?

[dusky epoch]

idk

the hand(mouse?)writing is certainly horrid

but i dont feel like checking this

[thorny hemlock]

um

ok

assuming that im correct

how would i work out x1 x2 x3 x4 now

@dusky epoch @honest token

[honest token]

you can't

[dusky epoch]

recall how to convert augmented matrices back into equations

[thorny hemlock]

is that necessary ?

the rank of the augmented and coefficiant matrix are different so that means no solutions right

[honest token]

its no solutions because the bottom row has all 0s except the last column

which is non 0

[thorny hemlock]

why does that matter

isnt it an issue of rank

(idk?)

[honest token]

@thorny hemlock the bottom row says that 0x1 + 0x2 + 0x3 + 0x4 = 5

[thorny hemlock]

ok for this one

thats the row echelon form of the augmented

what next

@honest token any idea?

this one seems possible

[hidden ember]

factor out 2 from the second row?

[honest token]

That's definitely possible

[thorny hemlock]

why factor tho

is there a need?

[honest token]

divide the second row by 2

because it makes it easier

[thorny hemlock]

so i can just change the matrix ?

[honest token]

no, you can divide the entire row by 6

[honest token]

so that you get 1 as the pivot

[thorny hemlock]

ok

looks good?

[honest token]

yes

[thorny hemlock]

what next

[honest token]

now you can convert the equations back to x y z form

and let z be a free variable

[hidden ember]

or you could have just subtracted the second row from the first row

[honest token]

fair point

[thorny hemlock]

um how do i convert the equations back?

[honest token]

the bottom one is 0x + y + 4/3 z = 2/3

[hidden ember]

you write x and y in terms of z

your rows represent linear equations

[thorny hemlock]

so $x + 3y + 4z = 1$ and $y + 4/3 z = 2/3$

[hidden ember]

you will get a parametric form such that for any Z that is a real number there is a solution that satisfies the system

[stoic python]

Yes:

[honest token]

yes

z is a free variable

so solve for x and y in terms of z

[thorny hemlock]

why is z free

[hidden ember]

if a variable corresponds to the pivot column it's called basic variable

but here

z does not correspond to a pivot column

[thorny hemlock]

we are talking horizontal correspondence ?

[honest token]

there is no pivot in the 3rd column from the left

in row echlon form

[thorny hemlock]

ok i see

im not exactly sure on how to solve

im getting x = -1

[hidden ember]

you dont solve anything after that

you get a parametric form

[thorny hemlock]

and 3y + 4z = 0

[honest token]

yes

[thorny hemlock]

what does this parametric form mean

[hidden ember]

it means writing ur basic variables in terms of ur free variables

[hidden ember]

such that for ur free variable, a real number, there is a solution that satisfies the system

[thorny hemlock]

hm ok

so what did they do with y and z there?

[hidden ember]

they wrote y in terms of z

but z = u

so y in terms of u

now if u take any real value say 1, you will get a solution that will satisfy ur linear system

[thorny hemlock]

ahhhhh

thats makes sense

thanks alot @hidden ember

[hidden ember]

((:

[tacit girder]

anyone can help
https://gyazo.com/6d6f09d98acbf55370ba2023b46c32d1

[hidden ember]

hm

[honest token]

$\begin{bmatrix} 1 & 1 \ 0 & 0 \ 1 & 1 \end{bmatrix}$

[stoic python]

Frosty_:

[honest token]

works

[tacit girder]

how did u get tehre

[honest token]

you know its 3 by 2

[tacit girder]

yeah

[honest token]

then assume its like $\begin{bmatrix} a & b \ c & d \ e & f \end{bmatrix}$

[stoic python]

Frosty_:

[honest token]

and multiply x with it

compare coefficients of alpha and constant terms

[hidden ember]

yeah sounds good

[tacit girder]

thanks

appreciate it

[honest token]

welcome

[tranquil hazel]

Can someone explain this, and why it would be true or false?

[honest token]

@tranquil hazel think of A as $\begin{bmatrix} a & b \ c & d \ e & f \end{bmatrix}$ and u as $\begin{bmatrix} a \ c \ e \end{bmatrix}$

[stoic python]

Frosty_:

[honest token]

is that vector in R^3

[dark gale]

does it have 3 entries?

[tranquil hazel]

yes

[honest token]

then...

[tranquil hazel]

it's in R3 got

oh that was pretty straightforward

thankss

[honest token]

welcome

[wintry steppe]

which step is wrong?

where is the mistake

i think its 3 cus thats not how the rule works

[pallid rampart]

Yes

You are correct

[shy totem]

very quick question, why is it multiplying everything by 6 here?

this was shit i had years ago and fully forgot

[honest token]

common multiple of 2 and 3

[shy totem]

ohhh aite

does common multiple only apply to fractions?

think it does

[honest token]

To get rid of the denominator, yes

[mild igloo]

I can solve for t2 and t3 but how would I get t4?
tbh I dont understand the picture or how the sectioning is working here

[lucid cedar]

I bodied my LA test today thanks to this discord. Thanks to all the Helpers

[dim venture]

consider not @ ing all of them like that then...

[lucid cedar]

I changed it but i think it still pings them

oh well

they get to see my gratitude

[south sedge]

gotta find the angle between the vectors u and v.
u + 3v is orthogonal to 2u - v. And u +7v is orthogonal to 2u + v.

which gives

looking at a solution

how did it get to the red marked one?

[wintry sphinx]

solve system of equations

for the unknowns |u|^2, u dot v, and |v|^2

[acoustic zodiac]

can someone explain to me what a generator space is, i didn't really get it

[limber sierra]

as in a generating set of a vector space?

im not familiar with the precise term "generator space"

[acoustic zodiac]

generator set yeah sorry

[limber sierra]

a generating set is just a set of vectors; the idea behind a generating set is that we can make every vector in a certain space out of vectors from that set

let me give an analogy

if i asked you "what words can you spell with only the letters a, c, e, and f?", the set {a, c, e, f} would be the generators of some "space" of words

[acoustic zodiac]

hmm alright makes sense

[limber sierra]

and some words in this space would be "ace", "face", "cafe", "efface"

of course we're dealing with vectors

not letters

the way we construct vectors out of other vectors is by taking linear combinations

[nocturne jewel]

Is there an easy way to determine if a block matrix is invertible?

[limber sierra]

essentially, adding vectors, and multiples of vectors, to each other

a proper intuition for this can only really be developed by practicing it but

as an example

lets suppose our generating set is $\left{\begin{pmatrix}1\0\1\end{pmatrix}, \begin{pmatrix}-1\2\2\end{pmatrix}\right}$

[stoic python]

Namington:

[acoustic zodiac]

hmm we use a different notation but ok

[limber sierra]

then the vector $\begin{pmatrix}4\-2\1\end{pmatrix}$ would be in the space generated by this set

[stoic python]

Namington:

[limber sierra]

since $\begin{pmatrix}4\-2\1\end{pmatrix} = 3\begin{pmatrix}1\0\1\end{pmatrix} - \begin{pmatrix}-1\2\2\end{pmatrix}$

[stoic python]

Namington:

[acoustic zodiac]

hm yeah alright

lemme translate a theorem real quick though, it's about a "generated subspace", this is what i don't really get

heard about it?

[limber sierra]

the generated subspace is a subspace generated by a given (generating) set

so the subspace generated by my example of $\left{\begin{pmatrix}1\0\1\end{pmatrix}, \begin{pmatrix}-1\2\2\end{pmatrix}\right}$ would be ALL vectors of the form:

[
a\begin{pmatrix}1\0\1\end{pmatrix} + b\begin{pmatrix}-1\2\2\end{pmatrix} = \begin{pmatrix}a - b\2b\a + 2b\end{pmatrix}
]

[stoic python]

Namington:

[limber sierra]

for any a, b

as an example, $\begin{pmatrix}1\0\0\end{pmatrix}$ would NOT be in this subspace, since it would need to solve:

[ \begin{pmatrix}1\0\0\end{pmatrix} = \begin{pmatrix}a-b\2b\a+2b\end{pmatrix}]

but $2b = 0$ implies $b = 0$, but then we must have $a = 1$ and $a = 0$ simultaneously, which is impossible

[stoic python]

Namington:

[limber sierra]

(assuming you're dealing with real-number-valued or complex-number-valued vectors)

[acoustic zodiac]

huhh i'm getting trouble typing with "" "" on latex

these things _

they disappear

[limber sierra]

underscores?

[acoustic zodiac]

yeah

[nocturne jewel]

Latex still sees them there

it just makes text go weird

[acoustic zodiac]

yeah but in this case is the subspace generated by $\sum_{i=1}^{n}{\alpha^i\cdot v_{i}$

[stoic python]

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

[acoustic zodiac]

which is all linear combinations

[limber sierra]

indeed

so the generating set of that is the set of vectors v_i

wait

what's the full statement of the theorem

[acoustic zodiac]

Let $W\subset E$. The set of all linear combinations ($\sum_{i=1}^{n}{\alpha^i\cdot v_{i}$) of the vectors of $W$ is $<W>$, the generated subspace, and it's the smallest subspace that contains W.

oops

there's probably something wrong with what i copied now that i'm reading it out loud

[limber sierra]

okay, it seems this is a definition-combined-with-a-theorem

[stoic python]

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

[limber sierra]

it defines "generated subspace" as a set of all linear combinations of elements of W

which is a definition (and the most common one)

then it states the result that this is the SMALLEST subspace containing W

which is a theorem

[acoustic zodiac]

yes, that's the part i don't really get, what does "smallest" even refer to?

[limber sierra]

if a subspace of E contains every element of W, it must contain <W>

i.e. <W> must be a subset of that subspace

[acoustic zodiac]

yes

[limber sierra]

as an example, if $W = \left{\begin{pmatrix}1\0\0\end{pmatrix}\right}$, then the set $\langle W \rangle$ is the set of all linear combinations of this vector, i.e. vectors of the form

[
\begin{pmatrix}a\0\0\end{pmatrix}
]

for some $a$.

now if we define a new subspace $S \subset E$, if $S$ contains $W$, then $\langle W \rangle \subset S$

[stoic python]

Namington:

[limber sierra]

in particular, every vector of the form $\begin{pmatrix}a\0\0\end{pmatrix}$ (i.e. every vector in $\langle W \rangle$) must also be in $S$.

[stoic python]

Namington:

[limber sierra]

this is what we mean by "smallest"

<W> being the "smallest" subspace containing W means that every subspace containing W also contains <W>

[acoustic zodiac]

oh, well that makes sense

[nocturne jewel]

ad-bc != 0 means a matrix is invertible right?

[limber sierra]

if we took another subspace, for example the subspace $S$ generated by $\left{\begin{pmatrix}2\1\0\end{pmatrix}, \begin{pmatrix}1\1\0\end{pmatrix}\right}$, then since:

[
\begin{pmatrix}2\1\0\end{pmatrix} - \begin{pmatrix}1\1\0\end{pmatrix} = \begin{pmatrix}1\0\0\end{pmatrix}
]

the subspace contains $W$, which means that $\langle W \rangle \subset S$.

[stoic python]

Namington:

[limber sierra]

this ends up being a very important fact for many proofs

basically any proof of a big result involving "dimension" relies on it in some way

(though often, it'll be through citing another theorem whose proof used this fact.)

@nocturne jewel for a 2x2 matrix, yes.

[acoustic zodiac]

ohhh now with your examples my prof's proof makes more sense, thanks

[nocturne jewel]

What about a block matrix that's 2x2?

[acoustic zodiac]

he was doing it for any W and any <W>, so it got a bit confusing

[limber sierra]

oh you mean like

a matrix composed of 4 blocks arranged in a 2x2 pattern?

the statement isnt as nice im afraid

[nocturne jewel]

Like I did the ad-bc thing and got AD

[limber sierra]

yeah you cant just apply that without proof

[nocturne jewel]

proof of the ad-bc != 0 => invertible?

[limber sierra]

since the ad - bc works for 2x2 matrices whose entries are, you know, numbers

that block matrix will not be 2x2

(unless each block happens to be 1x1)

[nocturne jewel]

yeah it's m+n x m+n

But yeah i have no idea how to go about showing it's invertible lol

[limber sierra]

well, it suffices to find its inverse, and then show (upon multiplication) that their product is the identity matrix

of size (m+n) x (m+n)

are you familiar with block matrix multiplication?

[nocturne jewel]

Yeah (did it once but I can look at the textbook and notes)

[limber sierra]

of note is this

assuming A_1, D_1, A_2, D_2 are square

(which they are in this case)

[nocturne jewel]

why do a and d have to be square?

[limber sierra]

otherwise those products dont all make sense!

[nocturne jewel]

right the dimension thing

ok ima give it a shot then

ty

[limber sierra]

anyway, given C_1 = 0, see if you can find a set of values for A_2, B_2, C_2, D_2 that makes the bottom matrix there an identity matrix

(so A_1A_2 + B_1C_2 and C_1B_2+D_1D_2 will have to be identity matrices, while A_1B_2+B_1D_2 and C_1A_2+D_1C_2 will have to be zero matrices)

(substituting C_1 = 0 will greatly simplify these computations; it essentially becomes a system of equations.)

[nocturne jewel]

Ok I got entries I expected, then one I didnt

Top right is -A^-1BD^-1

Nvm it worked out

[slow scroll]

"the general solution to this system is: the set of all (1,0,-1,0) + t(-1,1,0,0) + s(-2,0,-2,1) such that t, s are in R"

by R i mean $\bR$.

[stoic python]

kxrider:

[slow scroll]

@main nacelle

[wheat prairie]

curiousity question: Can one say let S be the set of all vectors of all dimensions? If so how would it be denoted?

[native rampart]

Just S

[wheat prairie]

so if they were let's say vectors of real numbers can't i call it R ^ n where n is a variable?

[native rampart]

Wdym vectors of real numbers?