# 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?

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.