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?


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

$${\bf S} = \left[\begin{array}{rrr}1&0&0\\0&0&-2\\0&-3&-1\end{array}\right],{\bf PD} = \left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-3&0\end{array}\right]$$ 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?


1 Answer 1


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.

We consider only the real case. The proof for the complex case is similar (and is even simpler).

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 $\mathbb R$.

  • $\begingroup$ 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? $\endgroup$ Apr 30, 2017 at 20:16
  • $\begingroup$ @mathreadler Yes, every complex diagonalisable real square matrix can be written as $S(PD)S^{-1}$ over $\mathbb R$. $\endgroup$
    – user1551
    Apr 30, 2017 at 20:19
  • $\begingroup$ Wow and can be directly constructed within the generalized eigenspaces as you mention too. $\endgroup$ Apr 30, 2017 at 21:13

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