@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
@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.
@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'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)
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.
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?
@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.
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 $2$
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 $1$)
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 $2$
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
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
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
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)
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?
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 :(
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