11:00 PM
@Alessandro "it turns out that if a linear transformation has "enough" eigenvectors than it can be rewritten in another basis as a diagonal matrix" dam thats cool

now suppose you need to iterate the linear map a lot of times, matrix exponentation is computationally expensive, but it's MUCH easier to do that with a diagonal matrix

yeah i was waiting for you to link the diagonal matrix theory with the eigenvalues

@isaac9A that's why we're interested in eigenvectors :) which brings us back to the question of how we find them

thanks sef, you rock

@isaac9A programs don't find eigenvectors and eigenvalues in order to apply a single matrix to a single vector one time, so your insinuation that they do that or that that is the point of eigenstuff is misleading.

11:01 PM
gotcha

@0celo7 I didn't know it was the only way so I said the easiest just to be sure

@Alessandro Is there a proof somewhere for: it turns out that if a linear transformation has "enough" eigenvectors than it can be rewritten in another basis as a diagonal matrix
I would love to try and understand that one

@Alessandro It's the only way I know to do it systematically...but I don't know any linear algebra
I guess the hard way is to just stare at the matrix until God tells you what the eigenvectors are

@0celo7 hardest way is trial and error XP

I'm getting to it @isaac, back too our example, we now know that there are solutions to the equation $(T-2I)v=0$ (as well as $(T-3I)v=0$ but let's focus on the first one)

11:03 PM
@isaac9A Almost all vectors are not eigenvectors, that will take a long time.

Google's PageRank algorithm for instance, to simplify matters, models the entire internet by an nxn matrix, where n= the number of webpages it has indexed, and then calculates what would happen if that matrix were to be applied over and over again infiintely. but it doesn't actually multiply the matrix out over and over again, at that wouldn't be computationally even possible - instead, the theory of eigenvalues allows them to calculate it.

@Alessandro oh shit ur gonna explain it <3333 marry me please
ok im following you so far

there are some methods in numerical analysis to approximate eigenvectors and eigenvalues, or so I've heard

@Alessandro Probably

but $(T-2I)v=0$ is just a linear system and we know how to solve them!

11:04 PM
@arctictern thats awesome

@isaac9A your next question should be what happens when the characteristic equation has no solutions

@0celo7 and that's when we start assuming an algebraically closed field :P

sure I would love someone to explain that too, I love learning about most things math theory but first I would like to understand how we can rewrite it in another basis
using a diagonal matrix
@arctictern how does google find the corresponding eigenvectors?
for such a huge matrix

@isaac9A the key fact is that since the number of hyperlinks on any given webpage is small (compared to the total number of webpages on the internet), the matrix will be "sparse," i.e. it will be mostly comprised of 0s. numerical analysis has special methods for calculating eigenthings of sparse matrices, but I am not familiar with their techniques.

ok thanks

11:09 PM
now, since the vectors with $2$ as an eigenvalue are the solution to this linear system we know they must be a subspace of the vector space (actually we need to add the zero vector since we decided that eigenvectors must be nonzero), the set of the vectors with $2$ as an eigenvalue, plus the zero vector, is called the eigenspace associated with $2$

ok still follwing you

now, we need to introduce a couple of new terms, the algebraic multiplicity of an eigenvalue is it's multiplicity as a root of the characteristic polynomial (in this case the algebraic multiplicity of $2$ is $1$)

we are adding the zero vector so we have a vector space?

yes
(and we don't want the zero vector to be an eigenvector, because it satisfies $T0=\lambda0$ for every $\lambda$)

Not really following "algebraic multiplicity of an eigenvalue is it's multiplicity as a root of the characteristic polynomial"
there were 2 roots

11:12 PM
each with multiplicity $1$

?
so we divide 2 (the eigenvalue) by 2 (the number of roots)

from college algebra, we know roots of polynomials have "multiplicity"
if $z$ is a root of $p(x)$, we say $z$ has multiplicity $e$ if $(x-z)^e$ appears in the complete factorization of $p(x)$

suppose you have a $3\times3$ matrix with characteristic polynomial $(2-\lambda)^2(4-\lambda)$ (by the way it can be shown that the degree of the characteristic polynomial is the dimension of the matrix), then $2$ is a root "repeated" twice, ot with multiplicity $2$

so like x2-2x+1 the only eigenvalue would be 1
but it has multiplicity 2 right?

right

11:14 PM
right

ok still following you now
sorry my english blows

The library's copy of Husemoller smells terrible
It's a smelly book, holy jeez

ok, now the geometric multiplicity of an eigenvalue is the dimension of the associated eigenspace

so since the eigenspace is a subspace of R^2, the geometric multiplicity is 2?

no
R^2 is not the eigenspace
the geo mult. is the dimension of the eigenspace, not the dimension of R^2

11:17 PM
I thought the eigenspace was a subspace of R^2

it is a subspace of R^2
but again, the geo. mult. is the dim of the eigenspace, not the dim of R^2

isn't the dim of the eigenspace 2?

yes the eigenspace is a subspace of R^2, but that doesn't mean the eigenspace is R^2

since it is a subspace of r^2?

the eigenspace is the space of the solution to the linear system $\begin{pmatrix} 0 & 0 \\ 4 & 1 \\ \end{pmatrix}v=0$

11:18 PM
@isaac9A no. do you know what dimension means?
for instance, the x-axis is a subspace of R^2, and that subspace has dimension 1, not dim 2

oh ok

(the matrix minus $2I$), and by Kronecker-Rouché-Capelli's theorem that space of solutions has dimension $1$

right a 2x2 is not necessarily in R^2 if the vectors are not linearly independent

exactly

where did you get that matrix @Alessandro?
where did you get the linear system (0, 0)

11:20 PM
it's the example I used before $\begin{pmatrix} 2 & 0 \\ 4 & 3\\ \end{pmatrix}$ after subtracting $2I$

(4, 1)
ok following now
so the dimension would be zero? or 1?

now the geometric multiplicity is always at least $1$ since if the system has some nonzero solution it must have a whole subspace of solutions and we know it has some solutions

right the only way it could be zero is if you just have a matrix of zeroes correct?
it could be zero = dimension could be zero

no, that can happen, remember we are interested in the dimension of the space of vectors solving the system, if the matrix is the zero matrix every vector is a solution

lol
so if the matrix was all zeroes, it would mean every vector in the eigenspace is an eigenvector for the corresponding eigenvalue

11:25 PM
yes
it can be shown, but I'm not going to that now because I don't have enough time, that the geometric multiplicity of an eigenvalue is always smaller or equal to the algebraic one
\begin{pmatrix}
2 & 0 \\
3 & 2 \\
\end{pmatrix}

ok cool im with you now so far

this matrix has only one eigenvalue, namely $2$, with algebraic multiplicity $2$ but geometric multiplicity $1$ (calculate it and you'll see, assuming I didn't make any stupid mistake)

ok im going to work it out right now
im getting the algebraic multiplicity
bout wouldn't the geometric also be 2
calculating the eigenvalue to be 2

\begin{pmatrix}
0 & 0 \\
3 & 0 \\
\end{pmatrix}

oh the 3 is on the bottom
shit

11:32 PM
that's the matrix associated to the system, right?

yeah
i fucked up and zeroed the 3 when subtracting
im with you now

ok
do you know about change of basis matrices and how to rewrite a linear transformation in a different basis?

not really
when you say different basis, can it be in a different dimension?

uh, that's a problem, but I don't have enough time to explain everything... the main point is that if all of the geometric multiplicities are equal to the corresponding algebraic multiplicities then you can take a basis for each eigenspace and their union will be a basis for the whole space and the matrices turns out to be diagonal in this base, but I need some facts about similar matrices and change of basis to justify this
a basis is a set of linearly independent vectors spanning the whole space
like $\{(0,1),(1,0)\}$ is a basis of $\mathbb{R}^2$

so is a change of basis spanning a different vector space? or is it just changing the basis vectors to different basis vectors that span the same vector space?

11:38 PM
the second option
we're choosing a different basis for the same vector space

ok so like {(0, 2), (3, 0)} is a different set of vectors that span the same vector space

yep

ok

and we can build a linear transformation that sends the coordinate of a vector in a base into its coordinate in another base
and the matrix associated with that transformation is the so called change of basis matrix

ok that makes sense

11:42 PM
now I'd need to explain a few facts about change of basis, but's it's 1:40AM here and I really don't have time, I'm sorry to leave you with an incomplete story :(

haha thats ok
thanks a lot man

anyway those things are contained in any linear algebra books, the keyword you're looking for are "change of basis" and "diagonalization"

I kind of understand this block and why it would make sense, but I don't have the knowlege/vocab to completely explain my intuition.
uh, that's a problem, but I don't have enough time to explain everything... the main point is that if all of the geometric multiplicities are equal to the corresponding algebraic multiplicities then you can take a basis for each eigenspace and their union will be a basis for the whole space and the matrices turns out to be diagonal in this base, but I need some facts about similar matrices and change of basis to justify this

yes, I wrote this part hurriedly, it deserves a deeper explanation
anyway you're welcome :)
I'm going now, I hope to find you in chat again!

have a good night :)
hope to see you again too