I wonder how to diagonalize a matrix that is non-normal, and does not have distinct eigenvalues.
Let $\lambda_i$ be the eigenvalue, and $v_i$ be the eigenvector with that eigenvalue.
I think the process would go like this:
- Determine if $\dim(\mathrm{span}(v_i)) = $ multiplicity of $\lambda_i$. If no, then it is not diagonolizable. If yes, go to 2
- Is the eigenvectors linearly independent? If yes, we can diagonalize. If no, … I don't know.
Best Answer
Non-normal matrices may or may not be diagonalizable.
Given $n\in \mathbb N$ and $A\in \mathcal M_{n\times n}(\mathbb C)$, it holds that $A$ is diagonalizable (over) $\mathbb C$ if, and only if, there exists a basis $\{v_1, \ldots ,v_n\}$ of $\mathbb C^{n\times 1}$ such that $v_1, \ldots ,v_n$ are all eigenvectors of $A$.
If $A$ is diagonalizable, then to find a diagonalizing matrix $P$, you just have to find the vectors $v_1, \ldots ,v_n$ above and define $P$ by letting its columns be the eigenvectors found. This implies that $P^{-1}AP$ is a diagonal matrix.
This works whether $A$'s eigenvalues have all multiplicty $1$ or not.