# Which matrices are permutation diagonalizable?

Are there some results for what must hold for a matrix of complex (or real) entries $\bf M$ for it to be similar to a permuted diagonal matrix $\bf D$: $${\bf S(PD)S}^{-1} = \bf M$$

where $\bf P$ is a permutation matrix, with exactly one 1 each row and column and rest 0s.

It is easy to by hand construct matrices which behave like that, but the opposite problems :

1. when are they sure to exist, and
2. how to calculate them?

Example:

$${\bf A} = \left[\begin{array}{rrr}1&0&0\\0&1&-2\\0&2&-1\end{array}\right]$$

$${\bf S} = \left[\begin{array}{rrr}1&0&0\\0&0&-2\\0&-3&-1\end{array}\right],{\bf PD} = \left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-3&0\end{array}\right]$$ where we manually can figure out $\bf P$ and $\bf D$, gives $${\bf SPDS}^{-1} = {\bf A}$$

And we can also check that $\bf A$ is not diagonalizable (over $\mathbb R$). So it is also a counterexample that the set of real diagonalizable matrices would be equally large as the set of real permutation-diagonalizable matrices.

Own thoughts

Since the trivial permutation ${\bf P=I}$ gives us the set of diagonalizable matrices, reasonably the set of permutation diagonalizable matrices must be at least as big. Can we show how big?

We may write $M=S(PD)S^{-1}$ if and only if all nontrivial Jordan blocks in the complex Jordan form of $M$ are nilpotent. This is true no matter the ground field is real or complex.
Suppose $M=S(PD)S^{-1}$. Since some positive $k$-th power of $PD$ is a diagonal matrix (e.g. when $k$ is the lowest common multiple of all the lengths of all disjoint cycles in the permutation represented by $P$), some positive power of $M$ is diagonalisable over $\mathbb R$. It follows that all nontrivial Jordan blocks (if any) in the complex Jordan form of $M$ are nilpotent.
Conversely, suppose all nontrivial Jordan blocks in the complex Jordan form of $M$ are nilpotent. Then the real Jordan form of $M$ is the direct sum of a real diagonal matrix, some nilpotent Jordan blocks and some real $2\times2$ submatrices of the form $C=\pmatrix{a&-b\\ b&a}$ (each representing a conjugate pair of eigenvalues of the form $a\pm ib$). Since one can shift the rows of a nilpotent Jordan block cyclically to form a diagonal matrix and one can interchange two rows of $C$ to get a real symmetric matrix that is diagonalisable over $\mathbb R$, it follows that $M$ can be written in the form of $S(PD)S^{-1}$ over $\mathbb R$.
• Very nice observation about the $k$th power being diagonal. So for example all complex diagonalizable (where Jordan blocks are 1x1) will be sure to have one? Apr 30, 2017 at 20:16
• @mathreadler Yes, every complex diagonalisable real square matrix can be written as $S(PD)S^{-1}$ over $\mathbb R$. Apr 30, 2017 at 20:19