Diagonalizability of Non-Hermitian Matrices

Question

We know that if $A \in \mathcal{M}_{n\times n}(\mathbb{C})$ is Hermitian, then it must be diagonalizable.

However, the converse of the above is not true. There are diagonalizable matrices that are not Hermitian. But when exactly is a non-Hermitian matrix diagonalizable? There are a couple of straightforward observations using the spectral theorem for matrices.

Given that $A$ is diagonalizable, $A$ is not Herimian if: $$A \text{ has a complex eigenvalue.}$$ $$A \text{ is unitary but not Hermitian.}$$ $$A \text{ is normal but not Hermitian.}$$

Clearly, these are not enough to characterize ALL non-Heritian matrices that are diagonalizable.

So how do we classify ALL diagonalizable matrices that are not Hermitian?

An example of what I mean by "classify" is:

$$\{A \in \mathcal{M}_{n\times n}(\mathbb{C})\ | \ A \text{ is normal} \} = \{ UDU^* \ | \ U \text{ is unitary}, D \text{ is diagonal}\}$$ That is, the set of normal matrices can be exactly classified by the set $\{ UDU^* \ | \ U \text{ is unitary}, D \text{ is diagonal}\}$.

Thanks in advance.

Answer 1

Note that Hermitianness is quite special to $\mathbb{C}$ or $\mathbb{R}$ but the question of diagonalizability makes sense over any field. Here is a precise characterization over any algebraically closed field.

Definition-Theorem: Over an algebraically closed field $K$, a matrix $X \in M_n(K)$ is diagonalizable iff it satisfies any of the following equivalent conditions:

1. $X$ can be written in the form $PDP^{-1}$ where $P \in GL_n(K)$ is invertible and $D$ is diagonal.
2. $X$ admits a basis of eigenvectors.
3. The algebraic and geometric multiplicity of each eigenvalue $\lambda$ agree. Explicitly, if $\lambda$ is a root of the characteristic polynomial $\chi_X(t)$ of multiplicity $m_{\lambda}$, then $m_{\lambda} = \dim \ker(X - \lambda)$ for all eigenvalues $\lambda$.
4. The minimal polynomial $m_X(t)$ is squarefree.
5. $p(X) = 0$ for some squarefree polynomial $p(t) \in K[t]$.

This is a good exercise. The equivalence of the first three conditions is more or less chasing through the definitions but the equivalence with 4 and 5 require a bit of work. It all follows from the Jordan normal form theorem but that actually isn't necessary and it can be proven directly (see e.g. here). If $K$ is not algebraically closed then 3 and 4 need an additional hypothesis that the characteristic polynomial splits over $K$, and 5 needs an additional hypothesis that $p$ splits over $K$.

This set of equivalences is not amazingly useful in practice (although 5 has some nice applications) because these conditions are not so easy to verify. An easier-to-verify sufficient condition is that the characteristic polynomial $\chi_X(t)$ has distinct roots (in $K$). This condition is in turn equivalent to the condition that the discriminant $\Delta$ of the characteristic polynomial is nonzero (and, if $K$ is not algebraically closed, that $\chi_X$ splits over $K$). For these matrices it's easy to see that we can find a basis of eigenvectors, since the eigenvectors associated to distinct eigenvalues are linearly independent; equivalently, the algebraic multiplicities are all $1$, so the geometric multiplicities are also all $1$.

When $K$ is algebraically closed this accounts for "almost all" matrices (e.g. over $\mathbb{C}$, the set of matrices not satisfying this condition has measure $0$), so in other words "almost all" matrices over an algebraically closed field are diagonalizable, and this "almost all" can be made precise over an arbitrary field using the Zariski topology.

The fiddly business here has to do with matrices which are diagonalizable but have repeated eigenvalues, which are in some sense "barely diagonalizable," and hard to distinguish from matrices which are not diagonalizable; see Why is there not a test for diagonalizability of a matrix for more details.