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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 $B$?

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    $\begingroup$ Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute. $\endgroup$ – Jochen Glueck Oct 12 '19 at 18:51
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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 $B$.

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    $\begingroup$ And just in case others have the same next thought that I did: the question’s conjecture can’t be fixed by adding a condition that $A,B \neq 0$, since taking the direct sum of this answer’s counterexample with an identity matrix gives a counterexample with $A,B \neq 0$. $\endgroup$ – Peter LeFanu Lumsdaine Oct 13 '19 at 13:11

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