On approximate simultaneous diagonalization

Question

It is well known that two $n\times n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable.

My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let $\{A_k\}$, $\{B_k\}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis $\{v_i^k\}$ of eigenvectors of $A_k$ and a basis $\{w_i^k\}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that $\{c_1,\dots, c_n\}$ form a simultaneous basis of eigenvectors for $A$ and $B$?

Answer 1

The answer is no in general.

For a $2\times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k \in \mathbb{N}$ and $$ A_k = \frac{1}{k} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$ for each $k \in \mathbb{N}$. Then each sequence $(v_k) \subseteq \mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.

However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.