5
$\begingroup$

Let $M$ be a (real) non-diagonalizable square matrix, and $S=\frac{1}{2}(M+M^T)$ the symmetric part of $M$. Is the matrix $M+S$ always diagonalizable? This is easily verified to be true if $M$ is a $2\times 2$ matrix. Can someone provide a proof or a counterexample for $n\times n$ matrices with $n>2$?

Note: Let $M=S+A$, with $A=\frac{1}{2}(M-M^T)$ the antisymmetric part of $M$. Let $Q$ be an orthogonal matrix that diagonalizes $S$, and put $Q^TSQ=\Lambda$ and $Q^TAQ=B$. The question can then be restated as: given an antisymmetric square matrix $B$ and a diagonal matrix $\Lambda$ with $B+\Lambda$ non-diagonalizable, is $B+2\Lambda$ diagonalizable?

Thanks!

[This question arises naturally in research work I am doing; a positive answer to the question would lead to considerable progress.]

$\endgroup$
8
  • 1
    $\begingroup$ To be clear, when you say "diagonalizable", do you mean over $\mathbb{R}$ or over $\mathbb{C}$? $\endgroup$ Dec 7, 2019 at 23:01
  • 1
    $\begingroup$ Assuming that you mean "diagonalizable" with potentially complex eigenvalues: my hunch is that we can guarantee that $B + \alpha \Lambda$ will be diagonalizable for "most" values of $\alpha \geq 0$, but not necessarily that this will hold for $\alpha = 1$. $\endgroup$ Dec 7, 2019 at 23:06
  • 1
    $\begingroup$ @Lode I suspect that we can deduce that $\{t \in \Bbb R | A + t\Lambda\}$ is finite, and that one can prove this as a consequence of the fact that diagonalizability is generic (i.e. the diagonalizable matrices form a dense and open set in the Zariski topology over $\Bbb C^{n \times n}$). I'm not fluent in the tools of algebraic geometry though, so I'm not confident about this. $\endgroup$ Dec 8, 2019 at 10:45
  • 1
    $\begingroup$ @Lode I meant the set of $A$ such that $A + t\Lambda$ fails to be diagonalizable. $\endgroup$ Dec 8, 2019 at 10:59
  • 1
    $\begingroup$ @Lode Ah, the discriminant is a great way to go here. So with that being said, it suffices to select an $\alpha > 0$ that is sufficiently large. To find such an $\alpha$ systematically, you might want to look at the eigenvalue inequalities for the sum of Hermitian and skew-Hermitian matrices given in Bhatia's Matrix Analysis. $\endgroup$ Dec 8, 2019 at 12:53

1 Answer 1

6
$\begingroup$

This is false in general. Fix a non-diagonalizable matrix $M$ with symmetric part $S$ and antisymmetric part $A$. Now consider the block matrices $$S'=\begin{pmatrix} S & 0 \\ 0 & S/2\end{pmatrix}$$ and $$A'=\begin{pmatrix} A & 0 \\ 0 & A\end{pmatrix}.$$ Then $M'=S'+A'$ is not diagonalizable because its top left block is $M$, but $M'+S'$ is not diagonalizable either since its bottom right block is $M$.

$\endgroup$
3
  • $\begingroup$ Thanks a lot, this indeed provides a nice counterexample! At the same it raises a new question: your construction gives counterexamples for even $n$; could one construct a counterexample for odd $n$ as well? $\endgroup$ Dec 8, 2019 at 10:43
  • 1
    $\begingroup$ @Lode We can simply add a final row and column of zeros to both matrices $\endgroup$ Dec 8, 2019 at 11:00
  • $\begingroup$ @Omnomnomnom: oh...yes, of course. Meanwhile I figured out that Eric's construction can be straightforwardly generalized: let $M_1=S_1+A_1$ and $M_2=S_2+A_2$ be two square non-diagonalizable matrices of dimension $n_1$ and $n_2$ respectively, written in their symmetric/antisymmetric decomposition. The diagonal block matrices $S'=\textrm{diag}(S_1,S_2/t)$ and $A'=\textrm{diag}(A_1,A_2)$, with $t\in\mathbb{R}_{\neq 0}$, are of dimension $n=n_1+n_2$, and we have that $S'+A'$ and $S'+tA'$ are both non-diagonalizable. $\endgroup$ Dec 8, 2019 at 12:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .