I'm sorry if what I am asking is unclear, but I was hoping that somebody might be able to help me understand diagonalization a bit more by explaining it through the lens of a change of basis.
I understand the material for change of basis fairly well. The way I think of it is as a way translating vectors into an alternative perspective.
Now for diagonalization I see that the form $B = S^{-1}AS$ has returned and so I believe it must be analogous to change of basis in some way. For the change of basis, $S$ consisted of the basis vectors of $B$ and changed them into the basis vectors of $E$. Now, $S$ consists of the eigenvectors of the matrix. Therefore those vectors must form some sort of basis. However, I cannot wrap my head around what it means that $B$ (from $B = S^{-1}AS$) consists of the eigenvalues. I think it must stem from me having a shaky understanding of all things eigen to begin with.
Best Answer
The eigenvalue equation is given by
$$Av_i=\lambda_i v_i \quad \text{ for all } i=1,\ldots,n.$$
If you combine all equations for all eigenvalues you can rewrite the previous equations in a single matrix equation as
$$A[v_1,\ldots,v_n]=[v_1,\ldots,v_n]\text{diag}[\lambda_1,\ldots,\lambda_n].$$
For simplicity assume that we have $n$ distinct eigenvalue and eigenvector pairs. If we introduce $V=[v_1,\ldots,v_n]$ and $\Lambda = \text{diag}[\lambda_1,\ldots,\lambda_n]$ solve for $\Lambda$ to obtain
$$\Lambda = V^{-1}AV.$$
Hence, by multiplying in this fashion with the eigenvector matrices we obtain a diagonal matrix with the eigenvalues on the diagonal.
You can think of this as a change of basis for the system. We first used an arbitrary coordinate system. After transforming the system we are using a system that has a coordinate axis aligned with the eigenvectors.