linear-algebra

[undone garnet]

well, then $rank(A^k) \le rank(A) \forall k \ge 1$

[stoic python]

henry:

[dusky epoch]

yup!

[undone garnet]

wow

this's new for me 😄

[brittle juniper]

( ^_^)

[undone garnet]

thank you

[brittle juniper]

you're welcome ( ^_^)

[indigo cradle]

why cant i invert this matrix by each 2x2 block

even though an inverse matrix

and i have seen this method done for other 4x4 matrices

[dusky epoch]

blockwise inversion works only in the block-diagonal case

[indigo cradle]

such as this

[dusky epoch]

this isn't

[dusky epoch]

yes like that

that's a block-diagonal matrix

[indigo cradle]

so its special?

[dusky epoch]

"it"?

[indigo cradle]

yea

[dusky epoch]

what's "it"

[indigo cradle]

the block-diagonal matrix

[dusky epoch]

wdym by "special"

[indigo cradle]

it has this characteristic

[dusky epoch]

wdym by "characteristic"

[undone garnet]

$A,B \in M_n, A is invertible and satisfies A^2 + B = BA and B^{2016} = 0$
prove that $A^{2015} + B^{2015} is invertible$

[stoic python]

henry:

[undone garnet]

this's how I do

is it right?

$A$ is invertible

[stoic python]

henry:

[dusky epoch]

oof bad tex

hang on a minute though

ok wait alright this seems ok

[undone garnet]

Did I do right?

[dusky epoch]

your argument looks ok to me

[undone garnet]

thank you

[indigo cradle]

may i know where does 4a-3b come from?

[dusky epoch]

$\begin{bmatrix} 4 & -1 & -1 & -1 \ -1 & 4 & -1 & -1 \ -1 & -1 & 4 & -1 \ -1 & -1 & -1 & 4 \end{bmatrix} \cdot \begin{bmatrix} a & b & b & b \ b & a & b & b \ b & b & a & b \ b & b & b & a \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix} $

[stoic python]

Ann:

[dusky epoch]

@indigo cradle

[indigo cradle]

o

omg thx

haha i tot elimination

[dusky epoch]

?

[indigo cradle]

i started using gaussian elimination

then it got tedious so i tot smt was up

[dusky epoch]

oh

sorry, i didn't recognize it when you spelled thought as tot

[indigo cradle]

oops

i tot its like internet slang

mayb only my place

[dusky epoch]

it is very likely more local than you think

[indigo cradle]

welp my country is literally a dot on the map

so i guess it cant get any more local

[dusky epoch]

if you don't mind sharing... what country

[indigo cradle]

singapore

[dusky epoch]

right

[urban spruce]

How to present such complex numbers in polar form? Ive tried doing it casually, looking at it just like it would be a+bi

[broken hawk]

there’s a general conversion formula: $$r = \sqrt{x^2 + y^2}$$ and $$\theta = \arctan\left(\frac{y}{x}\right)$$ (with the arctan you have to pay attention to which solution you take)

[stoic python]

Sascha Baer:

[broken hawk]

though for the second, you’re almost in polar form already

for (e) I don’t immediately see a clever trick. but for (f) consider that $\cos \alpha + i \sin \alpha = e^{i \alpha}$ and figure out what the minus does here

[stoic python]

Sascha Baer:

[broken hawk]

I also just realized you posted this in #linear-algebra
this is not linear algebra. go to #real-complex-analysis if you have further questions

[clever relic]

$\begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & k+18 \end{bmatrix}$

[stoic python]

GyroW:

[clever relic]

when does this matrix have only the zero solution? when k = -18 or -17?

I think -17 but I'm not sure

[brittle juniper]

what do you mean by a matrix having solutions?

[clever relic]

uh

they're my solution of using gaus jordan on 3 vectors that need to be linearly indepentend

It's not the right word

[vague belfry]

Do you mean when the rank of the matrix is 3 and when 2?

[clever relic]

I guess?

[vague belfry]

The three vectors will be linearly independent when k + 18 != 0

since then you will have three pivots

[clever relic]

ah yes

[vague belfry]

And linearly dependent when k = -18

Ah, now I get what you're asking

it's not only for k = -17

[brittle juniper]

Maybe you wanted to know when the matrix was inversible?

[clever relic]

That's the same thing really

but yeah thanks

[clever relic]

k so no I'm stuck again, I need vectors (1, -2, -3), (-3, 5, 7) and (-4, 5, k) to be linearly independent, when I do it via Gauss-Jordan I get k != 18 but when I calculate the determinant it says k != 6

It's defo 6

as I'm dumb and can't do basic counting

[vague belfry]

I think you made a mistake when you reduced it with the Gauss-Jordan method

Somewhere

[native lodge]

Was thinking that too

Would have to confirm using a calculator though lol

[clever relic]

yeah I did

[gritty ore]

Let T= [p;q;s] be an arbitrary triangle in R2. Without loss of generality, one can assume that s is the origin and that q lies on the x axis, say q= (q1;0). This will greatly simplify the calculations. Find the equation of the perpendicular bisectors of the three sides of this triangle.Show that these three lines intersects each other at the same point. Find an explicit formula for this point and let us call itQ

[wintry steppe]

I'll find the vertical one for you

x = q1/2;

you know that the line is vertical because it is perpendicular to a horizontal line, and you know that it has to bisect the horizontal side, so there's a point at (q1/2, 0)

now see if you can find the rest

find slopes of the other two perpendicular bisectors; the points are given by the midpoints of the two points that make up a side

[spiral kelp]

How do I organize this word problem?

[cedar solar]

The matrix A should have the nutrients in the rows and fruits in the columns, then solve Ax=b

The way to think about it is, x is the quantity of fruits, right? So when you do Ax what happens?

Thats how you check if your matrix is filled out properly

[spiral kelp]

I don’t understand this^

[slow scroll]

You're going to want to find some matrix A such that when you plug in a vector (oranges, apples) you get out (price, sugar) aka (400, 30000). Once you find that matrix, you want to solve the equation A(o, a) = (400, 30000) using ur preferred method.

For finding the matrix, think of it this way: what is the matrix A such that
A(1, 0) = (.15, 9) and A(0,1) = (.2, 19)?

[clever relic]

Supposes 2 non-zero vectors v, w in inproductspace V, suppose v+3w and 2v-3w are orthogonal and v-4w and v+2w are orthogonal, find the angle betweeen v and w

So I put those pairs of vectors in inproducts and got
2<v, v>+ 3<v,w> - 9<w,w> = 0
<v,v> - 2<v,w> -8<w,w> = 0
and I suppose I'm looking for <v, w> / (||v|| * ||w||)

<@&286206848099549185>

[cunning hedge]

could try subtracting one equation from the other

[clever relic]

yeah then I get <v,v> + 5<v, w> + <w, w> = 0 but that doesn't give me an extra equation since it'll just result in a 0 row if I throw it in a matrix

I don't know if there is a single solution or if I need to write in in terms of the other

[stoic python]

you:

[cunning hedge]

yeah nah 2 equations 3 variables

[clever relic]

unless the info that v and w are non-zero vectors

[rancid iron]

this is a dumb question but what would the matrix [u1, u2, u3, u4, u5] look like

would the rows of the matrix be made up from the vectors or the columns

[clever relic]

yes

[rancid iron]

uhhh

that doesnt answer my question XD

[clever relic]

yeah I read it wrong, the columns

so like imagine squeezing those colums together

[rancid iron]

ok how do i prove span(u1, u2) = span(u1, u2, u5)

im not sure

[clever relic]

check if u5 is linearly dependent on u1 u2

is there any a, b so that u5 = au1 + bu2

this is easily checked by putting u1, u2 and u5 in a matrix and row reducing and if you come up with a zero row, they're linearly dependent

[rancid iron]

oh i see

[clever relic]

u5 = -u1 + u2

[rancid iron]

how do i explain that though

since u5 is a linear combination of u1 & u2

uhh

[clever relic]

yes

[rancid iron]

idk what to say next

[clever relic]

u1 and u2 form a base from which u5 can be constructed

so it's not needed in the span

idk if that's good though, I have my linear alg exam tomorrow

[rancid iron]

wym by base?

a basis?

[clever relic]

basis

yeah I'm not native english

[quartz flax]

If I want to learn LA for machine learning purposes what book should I use

[violet dragon]

Googles ur best friend

[quartz flax]

There's both Strang and Axler but idk what is most practical in my case

[violet dragon]

A lotta good research papers online, reading em is fun

[slow scroll]

I'd say Axler is very theoretical

and not too applicable to comp sci

[vague belfry]

I think the new spring 19 course on MIT is more oriented towards being less theoretical, not sure tho

I mean not less theoretical in a way

[violet dragon]

U guys do machine learning stuff too?

[slow scroll]

nope xd

[violet dragon]

Lol k

I write a research paper on the maths behind machine learning for my 12th grade project

Was fun

I found online guides github and videos most helpful tho tbh

Coursera has amazing videos on machine learning

Not too theoretical, incomplete maths

But if u just wanna get the gist of how stuff works, coursera is good

[quartz flax]

Yeah but I want to learn p much everything lol

I'm looking for something comprehensive and still applicable

Hmm maybe Introduction to Linear Algebra by Strang?

[slow scroll]

strang's book is definitely more applied than axler. There is an overview of it in #books-old
Also, there are online lectures to go with it.

[quartz flax]

Ok that then lell

[gray lichen]

Hello

[slow scroll]

hi

[gray lichen]

My friend and I were trying to simulate some physics on a game engine and we figured we can not work without Linear Algebra. Initially we decided to make our own basic LA library to fit our needs, however, we settled on making a general purpose simple LA library which others can also use. While I've just a beginner understanding of LA and most of my knowledge comes from the internet, my friend has taken a basic LA course but combined, we still lack the rigor to decide what functions or aspects of LA should we implement.

Is there any resource for such a scenario or can someone guide us on their own. Thanks a lot in advance.

tldr; I need help with what features do people expect from a basic Linear Algebra library

[sonic osprey]

Don't reinvent the wheel? That's a thing cs people say right

[gray lichen]

Two issues. First of all, we sort of want this to be a learning experience for ourselves both in mathematics and programming concepts. Secondly, the game engine I'm using is not comprehensive enough and there's not a variety of libraries to select from.

[sonic osprey]

What game engine is it?

[gray lichen]

oof

ROBLOX

[sonic osprey]

.

is that a joke

or

[slow scroll]

im trying to think what kind of linear algebra you really need to do physics. All i've ever had to use were vector products

[gray lichen]

No, its genuinely ROBLOX.

[sonic osprey]

But honestly, you could look at some well-established linear algebra packages, like numpy for python for example, and see what methods they have there

Oh it's Lua

[gray lichen]

Yeah.

im trying to think what kind of linear algebra you really need to do physics. All i've ever had to use were vector products

Its not physics related anymore, we just want to make a simple library.

[sonic osprey]

Matlab has a ton of linear algebra stuff too

[gray lichen]

But honestly, you could look at some well-established linear algebra packages, like numpy for python for example, and see what methods they have there

Nice suggestion definitely going to check it out.

Someone else suggested Matlab too lol

[slow scroll]

the coding train has some videos where he make a linear algebra library in js. Maybe you could reference those

for his machine learning thing

[gray lichen]

Aha, I forgot about them. I've watched a lot of his other tutorials. very good tutor

Thanks guys.

[slow scroll]

np

[plain fjord]

At least d) is correct

Anyway, I have my own problem, if A is a positive definite square matrix, with possible complex numbers, then show that diagonal values of A are always positive.

I know that A is hermite matrix, thus A^* = A, and it's eigenvalues are all positive. Also A = UDU^*, where D has the eigenvalues

Also square root matrix for A exists and is unique.

Which has eigenvalues, which are the same from A except they are square roots from A's eigenvalues

Also I know the diagonal values of A are real valued numbers

[sonic osprey]

What did you use as the definition of a positive definite matrix?

[plain fjord]

Hmm, matrix A is hermitic and has eigenvalues which are all positive.

[sonic osprey]

Oh, there's another definition that I think is more popular that would be of use

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

[plain fjord]

Hmm, you mean x^*Ax > 0? That is also in the material I read, but not sure if that is the definition...

[sonic osprey]

They're equivalent definitions

But yeah that's what I was talking about, that's a hint

[plain fjord]

Okay, good... I did write that down, among a lot of other things, wasn't sure which would be helpful.

[indigo cradle]

how i would do this if i cant jus invert the blocks inside it

[vague belfry]

You couldn't either way since they don't lie on the diagonal?

I mean, there's block matrices outside of those on the diagonal

[indigo cradle]

yea cos that only works for block diagonal matrices

so i have no idea what to do for these

[sonic osprey]

Think about how block diagonal matrices multiply

[lean maple]

does the order of liner combinations matter aka is AB=BA

[brittle juniper]

that doesn't look like a linear combination 🤔

[lean maple]

this is the question i took it out of

my at least my prof said you can write it like that

[brittle juniper]

it's not about linear combinations, it's about transformation composition

[lean maple]

oh sorry

am i wrong in assuming it work like Function composition

[brittle juniper]

linear transformations are to be thought as vector space to vector space functions

[quaint heart]

Hmm

I think there actually is a right and wrong answer to this question

[brittle juniper]

consider having three vector spaces $E$, $F$ and $G$\
if $A$ is a transformation $F\to G$ and $B$ is a transformation $E\to F$,
then you should be wondering if $BA$ does even make sense

[stoic python]

Tuong:

[quaint heart]

Probably they mean linear functions A-> A

[lean maple]

so if they are in the same vector space it would be true

[brittle juniper]

why would you assume that?

[lean maple]

im just wondering

from the question you cant know

[quaint heart]

Cause otherwise the question would be stupid

If you assume that they are A-> A the question is mildly interesting

[vocal breach]

(very new to linear algebra; no formal lectures) I know how to solve it using system of equations, but I was wondering if there is a better way to solve it using linear algebra?

[dusky epoch]

other than solving the resulting system of eqs using linear-algebraic methods like gaussian elimination, no

[vocal breach]

Something like this?

But would you first solve it with a zero vector on the rightside?

(I also want to understand the connection between matrixes and this problem...)

I mean is there a geometric understanding that gives you this equation?

[dusky epoch]

nnno not quite, the matrix is wrong

at least it doesn't look right to me

at first glance

[vocal breach]

It looks like I got the right answer... But that might be luck

But does this look like the right way to compute x and y?

[dusky epoch]

idk i'd first determine the equations of those lines, in the form ax + by = c

then go off there

make a system of eqs, turn it into a matrix eq, and then do the thing

[wintry steppe]

How would I multiply out something like this?

[sonic osprey]

Do you understand how to multiply a matrix and a vector?

[wintry steppe]

No xD I know how to do vector and vector but I don't think I've been introduced to matrix-vector multiplication @sonic osprey

[sonic osprey]

You can't multiply a vector and a vector what do you mean?

Also, vectors are just special cases of matrices so it's really about matrix and matrix multiplication

[wintry steppe]

I'm studying through brilliant and afaik they haven't really explained

Or maybe I just forgot

Idk ;(

I'll have to see

Let's look at Khan academy

[vocal breach]

@sonic osprey You multiply the matrix by the vector so that you get vector and vector

[sonic osprey]

But that's addition of vectors

Not multiplication

So not what we were talking about

[vocal breach]

(@sonic osprey, I don't know much about this subject, but my "guess" is that it is like this)

If you could crosscheck the answer you could figure out if it is correct or not...

[plain fjord]

This one is tough:

Let A = [aᵢⱼ] be square positively definite Hermitian matrix with complex values. Show that aᵢᵢaⱼⱼ > |aᵢⱼ|² when i ≠ j. I have already shown that entries in the diagonal of A are all positive (meaning aᵢᵢ for all i) , which is part of that exercise, and might be useful to know in this one too. Given tip is, think of vector x = eⱼ - (aᵢⱼ/aᵢᵢ)*eᵢ.

In the earlier exercise I showed the positivity of diagonal entries using eᵢ, in which e means standard basis, and that (Aeᵢ | eᵢ) is a diagonal entry, which was greater than zero, since for positively definite Hermitian matrix, A > 0 <=> x^* Ax > 0 <=> (Ax | x) > 0 , so that is probably something I need to do here too... hmm.

Hmm, formatting got little messed up

[tacit sigil]

Why c-1?

[sonic osprey]

Because arithmetic

They subtracted v_3 from both sides

@plain fjord

[tacit sigil]

ah ok thanks

[sonic osprey]

Use the vector in the hint and use

The xAx > 0 thing

[plain fjord]

The tip vector would become something like (0,0, ... aⱼⱼ, 0, ... -aᵢⱼ, 0,0, ..., 0) if i use the same logic

Assuming the -aij/aii * aii works like thast, so the aii gets cancelled out

Since earlier, (Aei | ei) became just one diagonal entry, the aii, which had to be greater than zero.

So I wonder if i replace the ej and ei with ajj and aii here too

If that were to happen, basically I would have inner product ajj - aij, which doesnt help much i suppose...

[tacit sigil]

This might be a dumb question but how do we know where the 1s are?

[sonic osprey]

The drawing is kinda bad

The last box of 3's, should be two separate boxes

One with $\begin{pmatrix} 3 & 1 \ 0 & 3 \end{pmatrix}$

[stoic python]

Zopherus:

[sonic osprey]

and then one with just the last 3

[plain fjord]

Yeah if the block size is one, there is no 1 above the entry

[tacit sigil]

wait how do we know the block size? I thought 3's block size is three

[plain fjord]

Yeah in that picture above, it shouldn't be 3 like that, if it doesn't have 1 above the last entry. There is either two blocks, or the picture is drawn incorrectly

[tacit sigil]

ah ok thanks

[plain fjord]

The last 3 in the matrix is it's own block in that case

[sonic osprey]

Yes it's that

[earnest jacinth]

If you have some matrix A with characteristic polynomial (x-2)^2(x-1) for example with a basis (b1,b2) for the eigenspace of eigenvalue 2 and b3 for the eigenspace of eigenvalue 1, does the order in which you put those basis vectors into M matter when you're finding the JNF of A? like can M be (b1,b2,b3) or (b3,b1,b2). I think you do need to keep b1 and b2 together because they are from the same eigenspace.

[wintry steppe]

Yes you need to put them in the order you put the eigenvalues

And the basis of the eigenspace together

B1 b3 b2 wouldn't be correct

[earnest jacinth]

ye I get that but (b1,b2,b3) and (b3,b1,b2) would both work right?

because then only the order of the blocks would change in the JNF matrix

I think

they would both je JNF matrices of A right?

but the M matrix for M^-1JM=A would have some columns interchanged

[dusky epoch]

JNF is defined up to permutations of blocks

[wintry steppe]

Yup

But its convention to sort from the smallest block to the biggest

[dusky epoch]

isn't the convention to sort by eigenvalue first and only then by block size

[wintry steppe]

? You get the eigen values then the algebraic multiplicities, you find the eigenspaces and their dimensions are the geometric multiplicities
Using that you sort from smallest to biggest iirc

[dusky epoch]

yeah but like idk don't you usually arrange the blocks so that the ones with the same eigenvalues are together

[earnest jacinth]

well you have to right

if you have an eigenvalue with multiplicity n you have to have an n by n block

you cant split it up

because then you'd be missing out on 1's

[wintry steppe]

Yea and we like to put the smaller blocks first

Again this is convention and you can do whatever you want

[dusky epoch]

i mean like

$\begin{bmatrix} 3 \ & 2 & 1 \ & & 2 \ & & & 3 & 1 \ & & & & 3 \end{bmatrix}$ is a valid JNF, i guess

[stoic python]

Ann:

[red pike]

so if i have a linear system and I manipulate it so that one of the rows is "0 0 0 | 5" for example, that should prove that it is inconsitent and has no solution right?

[vague belfry]

Yes

[dusky epoch]

yes

[earnest jacinth]

If you have some bilinear form $\phi (x,y)=y^TAx$, how would you diagonalize A? I know you have a relation $B=P^TAP$ but I dont know how to find P

A is given btw

[stoic python]

Casper:

[wintry steppe]

?

you need more info on A

[earnest jacinth]

and we're working in R^3

[wintry steppe]

as you see, the matrix is symmetric

hence always diagonalizable

you find the eigenvalues

then you find the bases of the eigen spaces

those are then your P

and the diagonal matrix has the eigenvalues in order

[earnest jacinth]

but thats for $P^{-1}AP$ right? not when you have $P^TAP$

[stoic python]

Casper:

[wintry steppe]

yes

but since A is symmetric

there always exists an orthonormal basis for the eigenspaces

you take the vectors in your P

and you apply gram schmidt orthonormalization algorithm

you end up with another matrix, that is equivalent but additionally P^T=P^-1

this holds for hermitian matrices in general

you diagonalize as normal, and apply gramschmidt to the basis vectors of the eigenspaces

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

[earnest jacinth]

A has the ugliest eigenvalues

[wintry steppe]

has?

[earnest jacinth]

even though the final answer should be a nice answer

[wintry steppe]

non integer ya

[earnest jacinth]

but this should be the basis for which A is diagonal

so idk how you get those from non integer eigenvalues

[wintry steppe]

were you asked to diagonlize?

[earnest jacinth]

yes

find a basis for R^3 such that A is a diagonal matrix

[wintry steppe]

good luck with that lol

[earnest jacinth]

it was an exam question last year

[wintry steppe]

did you find the eigenvalues?

care to share?

[earnest jacinth]

I asked wolfram cuz they are non integer

the char poly is x^3-6x^2+5x-1

[wintry steppe]

seems odd to put that in an exam

i hate it when they focus on calculation instead of actually understanding the algorithm

[earnest jacinth]

there is plenty of that too

[wintry steppe]

if i had that in my exam ill just write how to do it and not do it

did it in my ODE exam

[earnest jacinth]

speaking of ODEs, I didnt get this one

I need to find a vectorspace of all solutions

[wintry steppe]

dont jusrt just use the characteristic equation

since its a homogenous linear equation

$x^3-8x^2+21x-18=(x-3)^2(x-2)$

[stoic python]

grade a failure:

[earnest jacinth]

I know you had to make $\vec{f}=\begin{pmatrix}
f\f' \f''\end{pmatrix}$ and then have $\vec{f}'=A\vec{f}$ with $A=\begin{pmatrix}
0&1&0\
0&0&1\
18&-2&8\
\end{pmatrix}$ and the solutions would be the first entry of $\vec{f}$ after you solve $\vec{f}=ce^{Ax}$

[stoic python]

Casper:

[earnest jacinth]

ye u can get A into JNF

[wintry steppe]

i dont remember much but you can just the characteristic equation

[earnest jacinth]

This is the general solution

Does the extra t have anything to do with the multiplicity?

since eigenvalue 3 has multiplicity 2

[wintry steppe]

literally what i wrote

yes

with algebraic multiplicity deterimines the order of the coefficient

[earnest jacinth]

u had the t in front of a e^2x not e^3x

[wintry steppe]

typo

[stoic python]

grade a failure:

[native lodge]

That factor of t is made clear when using the matrix exponential to solve the system

But most people are happy with knowing they need to place a t there lol

[earnest jacinth]

ye I tried to solve it but I got something completely different

[wintry steppe]

the matrix exponential is not easy to do by hand

you have to turn it into a system of first order

then exponentiate and possibly reduce the matrix

[earnest jacinth]

ye and you can't split A up into aI+bN directly

[wintry steppe]

which is why people result to the characteristic equation

[earnest jacinth]

so you need to get the JNF first

and split that up

to deal with the powers

[wintry steppe]

yeah you need a nilpotent term

[earnest jacinth]

@wintry steppe I now found a way to diagonalize a symmetric bilinear form without the use of eigenvalues

so I can now diagonalize it nicely without using those non integer eigenvalues

[wintry steppe]

how?

[earnest jacinth]

by first finding a vector for which $\phi (e_1,e_1)\neq 0$

[stoic python]

Casper:

[wintry steppe]

what is phi and what are the e's

oh

[earnest jacinth]

phi is the form

[wintry steppe]

the bilinear form defined by your A

[earnest jacinth]

e_1 is just some vector

[wintry steppe]

yeah

[earnest jacinth]

and then find a basis for the space $e_1^\perp$

[stoic python]

Casper:

[earnest jacinth]

lets say that basis is $\text{span}{e_2,e_3}$, then construct a matrix with $(e_1,e_2,e_3)$ as column vectors

[stoic python]

Casper:

[wintry steppe]

bro

this is literally

gram schmidt

which i mentioned

[earnest jacinth]

call it $P$ or something, then evaluate $P^TAP$, it might be diagonal already if not keep going with sub matrices

[stoic python]

Casper:

[earnest jacinth]

gram schmidt is a method to find an orthonormal basis

[wintry steppe]

so you take a random vector

[earnest jacinth]

not to diagonalize a bilinear form

its used

but thats not the general purpose

[wintry steppe]

suppose it is an eigen vector, and use gramschmidt to form an orthonormal basis

[earnest jacinth]

that would work but in this case the eigenvalues were retarded

and this method works regardless of the eigenvectors

[wintry steppe]

when youre finding a basis for the orthogonal space for e, that is gram schmidt

the problem you have is

when you choose e

your first vector

it needs to be an eigenvector

[earnest jacinth]

no it doesnt

any vector $v$ for which $\phi (v,v)\neq 0$ works

[stoic python]

Casper:

[wintry steppe]

then how will it be in P, for P^TAP to give u D

[earnest jacinth]

U dont need D, u directly construct P

[wintry steppe]

ok

but youre finding P for the diagonal form

so D has to be diagonal

for that to be the case

P needs to be the basis change matrix to the basis in which A is diagonal, aka D

[earnest jacinth]

@wintry steppe

[wintry steppe]

the last sentence is the thing

D is diagonal so we can stop else we continue

so your way doesnt always result in a diagonal D you might have to continue

how do you continue

[earnest jacinth]

yes but it always does

you might get something thats not diagonal there

and you repeat the entire process on the 2 by 2 matrix

[wintry steppe]

alright

[earnest jacinth]

and these will always be zero after the first time

so for a 3 by 3 you only need to repeat it twice at max

[wintry steppe]

so finding the eigenvalues from the ONB instead of the otherway around

when you redo that for the 2x2

what happens to your P?

[earnest jacinth]

the first row and colomn stay the same

but that 2 by 2 becomes diagonal

when you have $A=\begin{pmatrix} 0&1&2\ 1&0&3\ 2&3&0\end{pmatrix}$ for example, you get $P=\begin{pmatrix} 1&0&0\ 0&-2&-4\ 0&-4&-20 \end{pmatrix}$

[stoic python]

Casper:

[earnest jacinth]

so you need to repeat it once more

[wintry steppe]

reminds me of QR decomp

possibly the same idea

[earnest jacinth]

you can prove this method using induction

[wintry steppe]

nice

ill keep that in mind

[earnest jacinth]

but I do physics so i dont many proofs

[cobalt tree]

how would i find the x intercept of y= -1/2x + 4

[sonic osprey]

Use #prealg-and-algebra

[cobalt tree]

ok thanks

[wanton zenith]

could someone help me with QR iteration? i need to learn single and double shift

[slender shale]

anyone here

test in 23 mins and im starting to panic lol

why does the same type of determinant for each matrix yield different types of solutions

are the rules flipped for homogeneous/non-homogeneous systems

nevermind i got it

they are

[buoyant pike]

well glad you figured that out before your test

[stone drum]

Please do not post the same question in multiple channels, per the rules in #❓how-to-get-help

[mossy furnace]

Oops sorry I wasn’t getting a reply so I thought maybe the chat room was not being visited, will remove

[indigo cradle]

can i have help making sense of this

why does it conclude that both sides are diagonal

when it just said each side is triangular

[stone drum]

If an upper triangular matrix is equal to a lower triangular matrix, then both matrices must be diagonal.

Because the only elements they can have in common are those on the diagonal.

[indigo cradle]

OOOO

das rite

thanks

[vague belfry]

So I was solving this problem and I don't understand how to get the second and the fourth column

Why they need to be the way they are

I got this system of equations x1=x2+x4, x2 = x2, x3 = -x2+x4, x4 = x4

and I don't know how to progress further to get the same solution

[vague belfry]

Ok, now I multiplied the columns of the matrix first by the first independent vector

and then the second one and got the same result as them

is that the correct way to do it?

[wintry steppe]

I think one easy way to do it is to recognize that you are given $\text{Ker}(A)$, and therefore $\text{Im}(A^\top)^\perp$

[stoic python]

Sadly Lachrymose:

[wintry steppe]

Well, actually, since they ask for a particular matrix, perhaps that's not the best way.

You can have it so that $A[1\ 0\ 1\ 0]^\top = [1\ 2\ 1]^\top$ and then use the other two columns to ensure that the two other vectors are in the kernel.

[stoic python]

Sadly Lachrymose:

[wintry steppe]

What is F_w again? yikes

[gritty ore]

@wintry steppe

[wintry steppe]

so pretty much you're trying to find a matrix $B$ such that $A = I - 2BB^T$

[stoic python]

Sadly Lachrymose:

[wintry steppe]

where B has one column: w

[gritty ore]

so B is column vector?

[wintry steppe]

eh lemme make things simpler

We know that $A = I - 2ww^\top$

[gritty ore]

what about vector x?

[wintry steppe]

what about it

you haven't really defined x

oh sorry, I messed up

[gritty ore]

there's a typo in the notes i posted btw, rm is a typo and the top of the summation should be k not n

[wintry steppe]

hmm lemme thonk about this one

I'm familiar with the theory of orthogonal projections, but haven't really done anything with reflections per se

F_U(x) = x - (2 Proj_Uperp x) = x - (2 (x - Proj_U x)) = x - (2x - 2Proj_U x) = -x + 2Proj_U(x)

therefore, for a unit length w, we have F_w(x) = -x + 2ww^T x = (2ww^T - I)x

[gritty ore]

sorry what happened to 2 in front of Proj_U(x)

nvm lol

[wintry steppe]

forgot about it

of course, the little conundrum of sorts here that I haven't figured out is why A+I isn't rank-1

this is actually a bit strange, because one would expect that the reflection applied to w would result in w

therefore, it would have an eigenvalue of 1, but this is actually not the case, unless a and b are constrained such that sqrt(a^2 + b^2 ) = 1

[gritty ore]

yea w will be one of the eigenvectors right?

[wintry steppe]

yes

[gritty ore]

and then line perpendicualr to w should be other one

[wintry steppe]

yes

but for general a and b, the matrix does not necessarily have an eigenvalue of 1

[gritty ore]

yes no constraints for a and b were specified in problem

but to find w is the equation you presented correct?

[wintry steppe]

so, assuming that sqrt(a^2 + b^2) = 1

we know that A = 2ww^T - I and therefore I+A = 2ww^T

for unit length w

we can observe that the range of (ww^T) = span of w

and therefore the range of (I+A) = span of w, so just take one of the columns of I + A and you're done

[gritty ore]

huh that simple?

[wintry steppe]

not really simple tbh

I would ask a TA/prof about this, since it doesn't work for arbitrary a and b

for arbitrary a and b, I don't think the matrix corresponds to a reflection

[fringe cave]

So, you know what $F_w$ is yes?

[stoic python]

Darkrifts:

[fringe cave]

have you tried writing it out more matrix-like?

for a particular w that you know of course

[wintry steppe]

Yeah, you get $F_w = 2ww^\top - I$ for a unit vector w

[stoic python]

Sadly Lachrymose:

[wintry steppe]

The problem is that there doesn't necessarily exist a $w$ such that $A = F_w$, because that requires $A+I$ to be rank-1

[stoic python]

Sadly Lachrymose:

[wintry steppe]

which only happens if -1 is an eigenvalue, which requires that sqrt(a^2 + b^2) = 1

of course, I'm being lazy here when I say to consider only unit vectors, but we're not losing any generality here, since reflections are defined only by the subspaces

[spiral kelp]

Is there a way to check my work when calculating an angle between two functions w(x) and p(x)

Like a calculator online or something

Because I got theta = 8.328 and I'm doubting it. Not sure if this is an out of the ordinary result when using arccos

[sonic osprey]

calculating angles between functions seems weird, what's your dot product?

[gritty ore]

yea i will ask prof about it thanks @wintry steppe

[fringe cave]

inb4 his product is $\langle f, g \rangle = \int f\bar{g}\dd{x}$

[stoic python]

Darkrifts:

[spiral kelp]

I wish I knew how to use the bot here so I could type out everything

That's the formula you use for the inner product, yes @fringe cave

Although I'm not sure what the g bar is supposed to mean

[fringe cave]

lmao there we go

the bar is the conjugate my dude

which, if you're doing real numbers, is the same thing

[spiral kelp]

Weird, my professor never mentioned or used g bar

[fringe cave]

I mean, you're probably in a real case

[spiral kelp]

He did mention he'd back off with complex numbers since we're going too fast and it'd get complicated

[fringe cave]

There ya go

[spiral kelp]

Anyway, using that formula, I calculated <w,p> and got sqrt(3)/6

w(x) = x^2 - sqrt(6)x+sqrt(6)/2

p(x) = sqrt(3)(2x-1)

[spiral kelp]

When finding an orthogonal basis, do you write the result as V = span{v1,v2,...,vm}?

Sorry, just making sure for notation.

[half ice]

You need write "span" if you know it's a basis

Unless that actually spans a basis?

[dusky epoch]

"spans a basis"

[spiral kelp]

Yeah, I got confused with that one lol

v1 through vm span the subspace V

[half ice]

True, if the basis is spanned by something, then that would itself be the basis

Nvm me

[spiral kelp]

How can I tell if a set of vectors is an orthogonal basis already?

I got v1 = x1 for the first vector (obviously) but then I got v2 = x2 as well because the dot product <x2,v1> was equal to 0.

V = span {v1, v2,...,vm} (orthogonal set) and span{x1, x2,..., xm} is the original set

So when I used the Gram-Schmidt process on the set span{x1,x2} I got v1=x1 and I got v2=x2

[dusky epoch]

How can I tell if a set of vectors is an orthogonal basis already?

by checking if it satisfies the definition of an orthogonal basis

[spiral kelp]

lmao I should have gotten the hint when the dot product was 0

I totally overlooked that.

[meager tinsel]

Hello everyone
Asked this on SM, but they don't wanna talk to me
https://mathoverflow.net/questions/334148/an-explicit-formula-for-characteristic-polynomial-of-matrix-tensor-product
Maybe somebody here knows?
<@&286206848099549185>

[wintry steppe]

Can I post here a linera algerbra worded question that I cant figure out?

<@&286206848099549185>

[dusky epoch]

yes, but you shouldn't ping helpers until 15 minutes after your q

[wintry steppe]

Miriam has a collection of 10 cent and 20 cent stamps, she has 5 more 20 cent stamps than 10 cent stamps, the total of all the stamps is 4 dollars and 60 cents. How many 10 cent stamps are there?

Can somone please provide a worked solution for me

[dusky epoch]

we aren't going to solve this for you

[wintry steppe]

why?

this is a math discord

@dusky epoch

[dusky epoch]

math discord ≠ do-math-for-you discord

we're happy to help if you are stuck somewhere

[wintry steppe]

Im asking for help because I dont get it

[dusky epoch]

but you need to put in some effort on your part

[wintry steppe]

I am trying to make the linera equation for it

I can work it out from there

[dusky epoch]

what have you done so far

[wintry steppe]

tried to make the linera equation

and that failed

[dusky epoch]

do you have any work you could show?

[wintry steppe]

I tried,2x+5=4.60

that ended with a negative x

[dusky epoch]

okay, the first question is

what is x meant to be

[wintry steppe]

so the amount of 10 cent coins is x

[dusky epoch]

ok

[wintry steppe]

the amount of 20 cent coins is x + 5

[dusky epoch]

ok

so where did you get 2x + 5 = 4.60 from?

[wintry steppe]

the word equation

Miriam has a collection of 10 cent and 20 cent stamps, she has 5 more 20 cent stamps than 10 cent stamps, the total of all the stamps is 4 dollars and 60 cents. How many 10 cent stamps are there?

[dusky epoch]

can you tell me what the expression 2x+5 represents?

[wintry steppe]

the amount of 10 cent coins and the amount of 20 cent coins

1x + 1x+5

[dusky epoch]

yes, so 2x+5 is the total AMOUNT of stamps.

while 4.60 is the total VALUE of the stamps.

[undone knot]

shouldn't this be in #prealg-and-algebra ?

[dusky epoch]

yes

it should

but whatever

[wintry steppe]

these are linear algebra equations thoigh

though*

[undone knot]

no.?

[dusky epoch]

"linear algebra" is a bit of a bad name

[wintry steppe]

yes.

[dusky epoch]

it does not mean "algebra with linear equations"

[wintry steppe]

linear equations using algebra

[undone knot]

Linear algebra is matrices and vectors

[dusky epoch]

it's a bad name ok

[wintry steppe]

ok well i didnt know that

[undone knot]

The terminology is bad, I agree

[dusky epoch]

ok but back on topic

2x+5 is the total COUNT of your stamps

4.60 is their total VALUE

as such it makes no sense to equate the two

[wintry steppe]

can you please tell me what the equation would look like?

[dusky epoch]

well

[wintry steppe]

because I am so confused

[dusky epoch]

you want to use the fact that 4.60 is the total VALUE of the stamps

[wintry steppe]

yes

[dusky epoch]

you said that you have x stamps each worth 10 cents

[wintry steppe]

yes

[dusky epoch]

what is the total VALUE of your 10 cent stamps?

[wintry steppe]

Thats what I would work out from figuring how many 10 cent stamps there were

[dusky epoch]

no

you should give an expression

in terms of x

[wintry steppe]

ok

[undone knot]

Can I post the valid equation in between ||'s?

[wintry steppe]

how?

yes

[dusky epoch]

no

[undone knot]

But you will cheat and look at it tho

so no

[wintry steppe]

?

what

this isnt about me cheating

[undone knot]

Okay

So you have x 10 cent stamp

The value of x 10 cent stamp is ???

[wintry steppe]

I dont have that value

[dusky epoch]

ok let's approach it this way

[wintry steppe]

x10 * 10

x*10

10x

[undone knot]

yes

[dusky epoch]

not quite

0.10x

[undone knot]

Ann
just

[wintry steppe]

huh

[dusky epoch]

unless you want to do everything in cents

[undone knot]

Yea

cents

[wintry steppe]

oh yes

ok

[dusky epoch]

in which case you'll need to write the value of your whole thing as 460 cents

ok

[wintry steppe]

ok we will use 0.10

[dusky epoch]

so 10x is the value of your 10 cent stamps

ok FINE

[wintry steppe]

yes

[undone knot]

And you have x+5 20 cent stamps, right?

[dusky epoch]

^

[wintry steppe]

yes

[undone knot]

So what is the value of the 20 cent stamps in total?

[wintry steppe]

20x+100

[undone knot]

0.20x+1

CENTS!

[wintry steppe]

oh yea I forgot

[undone knot]

Okay

and these two equal 460, right?

[wintry steppe]

yes

[undone knot]

Boom

you have your equation

[wintry steppe]

waitttttt

[undone knot]

now write it

[wintry steppe]

so,

[dusky epoch]

wow

y'all keep going back and forth between counting in dollars and counting in cents

[wintry steppe]

0.1x + 0.20x+1 = 4.60

[undone knot]

Yes!

[wintry steppe]

okkkk

[undone knot]

Now solve this and boom

[wintry steppe]

so thats the linear equation to work out the qunatity of 10 cent stamps

[undone knot]

yea

[wintry steppe]

ok wait lemme solv

[undone knot]

x is the amount of 10cstamp

[wintry steppe]

12 ten cent stamps

[undone knot]

,calc 3.6/0.3

[stoic python]

Result:

17.142857142857

[wintry steppe]

and 17 20 cent stamps

no

[undone knot]

wait

[wintry steppe]

you use inverse operations to get rid of the 1

[undone knot]

,calc 3.6/0.3

[stoic python]

Result:

12

[wintry steppe]

then your left with, 0.3x/0.3 = 3.60/0.3

yea

[undone knot]

I got this
I just made a typo

[wintry steppe]

ok lit

thanks for your help, and @dusky epoch

[undone knot]

Your welcome
And dont forget

#prealg-and-algebra

[wintry steppe]

ok

[meager tinsel]

anybody gonna help me?
I probably should've posted it in some other channel

[vague belfry]

Let A be a mxn matrix and P a matrix of orthogonal projection onto column space of A. If X is from R^n to which fundamental subspace does ((I - P)^2)*x belong.

I have a problem with this problem xD

So I have computed that I-P squared is the same as I-P

or P-I

now what bugs me is that P has size mxm

but then I can not compute Px

I figured out that the answer is the left nullspace

since X can be represented as a unique sum of y and z where y is in the left nullspace and z is in the column space, and then we just subtract the projection of x onto the column space which is z

so we are left with y

tho I still don't understand the dimensions

[oak briar]

I just found on wikipedia that there's a $O(n^{2.373})$ complexity algorithm for the calculation of a determinant. https://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations#Matrix_algebra

How can matrix diagonalization have a cost of $O(n^3)$ ?

[stoic python]

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

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

[undone knot]

Oof

[sonic osprey]

To do matrix diagonalization, at least the way it's normally taught

You have to calculate a determinant to calculate the characteristic polynomial

But then there are a ton more things you have to do, like solve this polynomial, and then find the eigenvectors associated with these eigenvalues

[oak briar]

Yes but calculating the determinant is the step with highest scaling complexity, solving the polynomial is linear.

[sonic osprey]

Where do you have that matrix diagonalization has a cost of n^3

[stone drum]

https://en.m.wikipedia.org/wiki/Jacobi_eigenvalue_algorithm

Often the best numerical algorithms look nothing like how you'd do it by hand.

[proper crescent]

@mint anvil let's talk here actually there's already a convo

Okay so

[stoic python]

Dami:

Dami:

Dami:

Dami:

Dami:

Dami:

Dami:

Dami:

Dami:

[sonic osprey]

wait this is cool

[hushed igloo]

what is the "origin"

[slow scroll]

(0,0) and this is #prealg-and-algebra ffr

[hushed igloo]

alrighty

[sand hedge]

"We can write any real matrix A=UXV’, where U and V are orthogonal matrices and X is diagonal using the singular value decomposition. The way one can view this decomposition is that any matrix is the composition of a rotation U, a stretch and embedding/projection X and another rotation V’.

Taking the transpose we get A’=VX’U’. The transpose of the rotation matrices “reverses the direction” of the rotation, so we get that the transpose of a matrix A reverses the order of the rotations and changes the direction of them." saw this on the maths subreddit and although it's not a question still nifty tbh

[chilly dragon]

i'm confused on some stuff

we can solve linear equations using gaussian elimination

or using LU decomposition by forward substitution or back substitution

they way i see it that all of these yield the same result

the question is: is it a matter of choice? or are there any advantages for using one over another ?

[dusky epoch]

sometimes certain methods may perform better than others in terms of computational speed if extra information is known about the matrix's structure

[chilly dragon]

thanks

[half ice]

@chilly dragon
These substitutions are very fast. Once a matrix has been LU decomposed, it can solve any linear system quickly.

So, if you have multiple b in Ax = b to solve for, then an LU decomposition can get all of them, rather than Gaussian for each one

[pallid rampart]

When it says $a_0, a_1, a_2, a_3,...,a_n$ are linearly independent over the $\bbQ$, does it mean that no $a$ is the sum of rational multiples of other $a$?

[stoic python]

Whoever:

[quaint heart]

Yes

Equivalently $\sum_{i=0}^n r_i a_i = 0$ iff all the $r_i$ are 0

[stoic python]

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

[pallid rampart]

Yeah I can see that

[quaint heart]

Do you know what a field is?

This definition of linear Independence works for any field

[viscid apex]

a question just to see if i understood:
A linear transformation T from Fn to Fn transforms base B1 to base B2. Therefore the representing matrix $$[T]^{B1}_{B2} = I$$

[stoic python]

kakamaika:

[viscid apex]

so for example if im asked for a matrix representation of a linear transformation from V to V I can just choose a base, choose the base it transforms to as the second base and then give I as the matrix

yes?

[sonic osprey]

No

Because the vectors that the matrix spits out are meant to be in B1

So you need to write out the basis vectors of B2 in terms of the B1 basis

And use those vectors to write your matrix

[wintry steppe]

I asked this some time ago already

But, what is a good introductory book to linear algebra?

I know about Axler and Hoffman, Kunze

But they look like a second year of linalg and seem more advanced than just an introductory book

[proper crescent]

They're both fine to start with, I personally don't like Axler but I feel HK is basically the correct book to learn linear algebra with

There's also Linear Algebra Done Wrong

Which is a direct response to Axler and better

[wintry steppe]

Ok thank you

[lone cove]

done wrong is awful

[quaint heart]

Just read the relevant parts of aluffi (or of Lang)

[plain fjord]

Done wrong? Isn't it Linear Algebra Done Right? 😃

I've heard of that book, haven't really read it

[slow scroll]

i learned from that book

[pliant berry]

Done wrong is a response to Done right, afaik

[plain fjord]

How about Linear Algebra and it's Applications, by David C. Lay, Steven R. Lay, Judi McDonald?

[gritty ore]

the matrix from A is very ugly and calculating the determinant to find eigenvalues for part b is even worse, is there another way to solve this

[dusky epoch]

yes

there are two very obvious eigenspaces

[sage mantle]

hey if the scalar product of two vectors is 0 , it means they are orthogonal but what if the result is something else than 0. should i simply write that they are not orthogonals or does that result imply something else?

[native lodge]

Writing that they are not orthogonal is fine

You can find the cosine of the angle formed by the two vectors if you wish

[sage mantle]

i see , thank you !

[wintry steppe]

Hello, I need to find c1, c2 and c3 for

with these vectors

i'm on c1 = 3, c2 = 9, c3 = 3 but that gives me <0, 51, -39> so just one off, and i'm stuck now

[wintry steppe]

@native lodge i see ur being busy with linear algebra, so a cheeky ping shrug

[native lodge]

(c1 * 0) + (c2 * 0) + (c3 * 0) = 0
hmmmmmm

[wintry steppe]

i'm going into the fourties now

c1 = 43, c2 = 49, c3 = -2 gives me

<0, 51, -34>

always so close yet so far

[native lodge]

oh why am I being dumb lol

let's assemble a matrix and get the null space solution to this

$\begin{bmatrix} 0&0&0\-3&4&8\6&-6&-1 \end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} 0\51\-40 \end{bmatrix}$

I messed that up

or did I?

[wintry steppe]

was checking it out but looked good to me

[native lodge]

let me check again

$\begin{bmatrix} 0&0&0\-3&4&8\6&-6&-1 \end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} 0\51\-40 \end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[native lodge]

we're good, yea lol

[wintry steppe]

yeah

[native lodge]

well we have to find a particular solution and a null solution

just take c1 to be free variable and then you can get c2 and c3

[wintry steppe]

that's how far i've gotten yeah, previously there was 1 given so one of the 3 was static

[native lodge]

actually I would exchange so the zero row is on bottom then c3 is free variable

so then choose c3 = whatever you want

[wintry steppe]

so that basicly makes it

[native lodge]

they just need one combo of c1,c2, and c3?

if so then just say c3=0

[wintry steppe]

$\begin{bmatrix} 6&-6&-1-3&4&8\ 0&0&0\end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} -40\51\ 0\end{bmatrix}$

[native lodge]

makes your life easy

you need \\

discord renders \\ as just \

[wintry steppe]

ah

$\begin{bmatrix} 6&-6&-1-3&4&8\ 0&0&0\end{bmatrix}\ \begin{bmatrix} c_1 \ \c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} -40\51\ 0\end{bmatrix}$

$\begin{bmatrix} 6&-6&-1-3&4&8\ 0&0&0\end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} -40\51\ 0\end{bmatrix}$

idk what im doing wrong sigh

[native lodge]

$\begin{bmatrix} 6&-6&-1\-3&4&8\ 0&0&0\end{bmatrix} \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}=\begin{bmatrix} -40\51\ 0\end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[wintry steppe]

ty

[native lodge]

finding the rref will get you

1 0 0
0 1 0
0 0 0

actually there may be stuff above the 1's

[wintry steppe]

the ref? english isn't my first language so i only know my own country's terms etc

[native lodge]

reduced row echelon form

[wintry steppe]

oh no

i hate that

[native lodge]

to find the general solution, you will need to augment the matrix and take it to rref

[wintry steppe]

gauss-jordan fun

[native lodge]

that will give you are particular solution

then for the null, you use the rref without the augmented part and solve rref(A)x=0

[native lodge]

$\begin{bmatrix} 6&-6&-1&|&-40\-3&4&8&|&51\ 0&0&0&|&0\end{bmatrix}$

[stoic python]

⚡Amphy⚡:

[native lodge]

my best attempt at an augmented matrix lol

but you take this matrix to reduced row echelon form

and for the particular solution, just copy the vector that appears on the right of the line I drew

[native lodge]

so x_{p} = [73/3 31 0]^T
(I just wrote it as a row vector and transposed that to make it a column)

now for x_null you have to solve:

[1 0 22/3] [c1] [0]
[0 1 15/2 ] [c2] = [0]
[0 0 0 ] [ 1 ] [0]

I took c3 as the free variable and assigned it a value of 1

[wintry steppe]

solve it using what?

[native lodge]

in the end you get x_null = [(-22/3) (-15/2) 1]^T

[wintry steppe]

as in (-22)/3

or

-(22/3)

[native lodge]

$-\frac{22}{3}$

[stoic python]

⚡Amphy⚡:

[wintry steppe]

so that's c1?

[native lodge]

the general solution is x = x_{p} + (any multiple of x_null)

kind of weird because we have to introduce one last variable, call it s

then the general solution is x_{p}+ (s * x_null)

[wintry steppe]

and s is?

[native lodge]

any number

any multiple of x_null will still cause Ax to be 0

so then c1 = first component of x_{p} + (s * first component of x_null)

and then c2 is so forth

if they just want one combo out of all possible combos then just take s=0

makes it pretty easy then

[wintry steppe]

they want a full digit combo

like: 1

[native lodge]

then say s=0

[wintry steppe]

kk

[native lodge]

that means c3=0

then you can solve for c1 and c2

but as you saw, there are infinitely many sets of c1,c2, and c3 that will work

[wintry steppe]

x_{p} is the column right?

[native lodge]

well I was giving very fast run down on null space and solving a system that is not full rank

I don't exactly know where you are in linear algebra at the moment

maybe vc I can try to do a better job?

[wintry steppe]

erm not that experienced, and i can't talk myself but can listen

[native lodge]

you don't have to talk, I'll just try to explain it again

[wintry steppe]

its 2:25 in the morning

sure

came for a quick answer honestly but explanation is always nice

i can

yeah

yeah, that's why i came for help rofl

yeah, sadly i need whole numbers

*3 ?

219/3

93

73, 93

i don't think they want that

it's a scuffed program

it's the site

it gives everyone random shit

and grades us

yeah, before i had a row that had a stuck number

and stuff that made no sense

nah that's fine

always better to expand your knowledge

even though it goes out of my head instantly

thanks for your time though

[native lodge]

sure sure

[wintry steppe]

i'll be sure to file a complaint

[native lodge]

lol they just choose whatever at this point

[wintry steppe]

Is it okay if i hit you up later? Have a test coming in soon and you explained quite clearly, should do youtube vids

[native lodge]

just ping in math server later then

[wintry steppe]

Will do

Cheers mate

[sage mantle]

hey there , could someone please explain to me in details the linear independence/relation with basis and spanning i'm getting confused

[fringe cave]

which part of it?

[sage mantle]

i know that a linear independent set of vectors is a set in which no vector is a multiple of another vector (using a scalar) or the result of a linear combination of two other vectors in the set

but then i see this definition online of a basis : A basis of a vector space is any linearly independent subset of it that spans the whole vector space. In other words, each vector in the vector space can be written exactly in one way as a linear combination of the basis vectors.

[fringe cave]

${x_n}$ is linearly independent if $a_1 x_1 + a_2 x_2 \dots a_n x_n = 0$ is satisfied only by the condition where, for all i, $a_i = 0$

[stoic python]

Darkrifts:

[fringe cave]

A linearly independent set can be a basis if the span covers the entire vector space

Not every linearly independent set is a basis of your space, but they are a basis of a space

[feral mountain]

yatori in your first message it should be: a linear combination of two or more other vectors in the set

[sage mantle]

i see thanks @fringe cave

@feral mountain i'm used to working in R3 but yeah mb

[fringe cave]

a set of only one vector is clearly linearly independent (so long as it's not the 0 vector)

[sage mantle]

and a set of two vectors is only linearly independent if they are not multiples of each other right?

[feral mountain]

yeah

[vague belfry]

if AtransposeA has the same eigenvalues as BtransposeB, how does that imply that A and B have the same eigenvalues

I was thinking that because the null space of A and AtransposeA is the same

and the same for B

[sonic osprey]

How are the eigenvalues of transpose A related to the eigenvalues of A

[vague belfry]

they're the same

hm

lemme think this through

[sonic osprey]

Ok

[vague belfry]

I got nothing xD

does that mean that if one eigenvalue of AtransposeA is x, then one eigenvalue of A is x/2

ATA

I think not because they're not the same vectors

I can't connect the eigenalues of ATA and A

[lone cove]

its not x/2, they multiply not add

and theres an easy counter example in 1x1 matrices

[quaint heart]

Lmao 1x1 matrices

[vague belfry]

😐

but how can I prove that

they multiply

[winter reef]

det A = detA^T

[lone cove]

dont knock 1x1 matrices, they all equal their transpose

[quaint heart]

@vague belfry think about what taking the transpose does to det(A-\lambda I)

Like dog said

[lone cove]

I think hes confused since the eigenvectors arent necessarily the same as in A

[quaint heart]

[vague belfry]

oh they are?

oh ok

[lone cove]

typo

[vague belfry]

yeah, that's the part that's confusing me xD

[quaint heart]

Yeah

Because of dogs statement

[vague belfry]

and then it becomes det(A^T - \lambda I)

[quaint heart]

Yes

[vague belfry]

Now I don't understand how does it relate to A^T*A

if they have the same eigenvalues

[patent cloak]

If you take a matrix A and diagonalize it, the diagonal-matrix entries are the eigenvalues of A.

Eigenvalues do not change under transposition, I believe.

So you square the entries when you do A.A

And thus merely square the eigenvalues

However, sign becomes ambiguous

[slow scroll]

diagonalizing is not really helpful here. Even though the eigenvalues do not change under transposition, in general, the eigenvectors do

[vague belfry]

Then you can not kind of nullify the middle parts

You would have SDS^-1 * BDB^-1

where B is consisting of eigenvectors of A transponed

I did not know what letter to choose xD