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.]
Best Answer
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 $M$.