How does diagonalizing a matrix help with eigenvalue calculation? So in linear algebra class we learned that a matrix $A$ is diagonalizable if it can be written in the form:
$$A=PDP^{-1}$$
The useful part was that $A^k$ can be easily computed with $PD^kP^{-1}$. The diagonal entries of $D$ were simply the eigenvalues of $A$, and the corresponding eigenvectors were the columns of $P$. If you made $P$ orthogonal, $P^{-1}=P^{T}$ which is also easy.
This all makes sense, but the part I don't get is how this somehow makes finding eigenvectors easy. Our professor hinted that it involved iterative methods i.e. $\vec{x}\mapsto A\vec{x}\mapsto A^2\vec{x}...$ which is made easy with diagonalization. But this seems unintuitive because we got the diagonalization of $A$ by using the eigenvectors/eigenvalues, so what's the point of this method then? Or am I misinterpreting the use of diagonal matrices?
 A: I think what your professor was referring to is the power iteration algorithm for eigendecomposition.
The basic idea is to apply $ A $ repeatedly to a random vector $ \vec{v} $. If you visualise the action of $ A $ on the vector space, every application of $ A $ stretches $ \vec{v} $ more in the direction of the eigenvector with the largest eigenvalue than in the direction of the other eigenvectors. With a sufficiently large $ k $, you should be able to approximate that eigenvector as much as required. There's a visualisation of this here. Variants of this algorithm can be used to find more than just the dominant eigenvector, and are generally faster than full diagonalisation if you only want a few dominant eigenvectors.
While you don't need to perform diagonalisation in power iteration, you do need the idea of diagonalisation to show that it works. Here, we are effectively computing $ A^k\vec{v} = PD^kP^{-1}\vec{v} $. If you think of $ P $ as a coordinate transformation, the matrix $ D^k $ essentially serves to "pick out" the largest eigenvector. To be precise, as $ k $ increases, the ratio between the largest entry in $ D^k $ and any other entry increases exponentially, causing the dominant eigenvector to quickly dominate.
In other words, diagonalisation, in your case, should be thought of as a concept (specifically the concept of transforming to the eigenbasis of the matrix) rather than an algorithm.
A: It's easy to find eigenvalues of a diagonal matrix(i.e. if original matrix was diagonal). Here I think you have to find the diagonal matrix through change of coordinates and not by the equation of diagonizable matrix.
