TITLE: Diagonalization for sums of Hermitian matrices
QUESTION [6 upvotes]: I found an interesting question about diagonalizable matrices,
Let $A,B\in \mathcal{M}_n(\mathbb{C})$ Hermitian, such that $AB\neq BA$.
Do there exist complex numbers $u\neq v$, such that $A+uB$ and $A+vB$ are not diagonalizable ?
It seems to be a consequence of  Motzkin-Taussky theorem , which is difficult to prove, perhaps one can prove it directly without Motzkin-Taussky theorem  ?

REPLY [2 votes]: The following is a sketch of the proof.
A pair of complex matrices $(A,B)$ has property L if there exist orderings of the eigenvalues $(\lambda_i)$ and $(\mu_i)$ of $A$ and $B$ s.t., for all complex $(x,y)$, $spectrum(xA+yB)=(x\lambda_i+y\mu_i)$. According to the proof of Theorem 3 in "Pairs of matrices with property L. II " (Motzkin and Taussky): if $\lambda A+\mu B$ is a pencil in which all matrices, except possibly one, are diagonalizable, then $(A,B)$ has property L.
According to Theorem 2 in "Pairs of matrices with property L." (Motzkin and Taussky): if $A,B$ are hermitian and have property L, then $AB=BA$.
Here $AB\not= BA$ ; thus $(A,B)$ has not property L and the associated pencil contains at least $2$ non-diagonalizable matrices.