Consider two real matrices:

  • the $n\times n$ matrix $A$

  • the $n\times m$ matrix $B$ of rank $m$, with $m<n$.

Let, for $a\in\mathbb{R}$, $$S_a=A-aI_n,$$ and denote by $P_a$ the orthogonal projection onto the orthogonal complement of $\operatorname{col}(S_a B)$, that is, $P_a=I_n-S_aB(S_aB)^\dagger$, where $(\cdot)^\dagger$ denotes the Moore-Penrose pseudoinverse.

I'm interested in whether the following is correct.

Conjecture: For an arbitrary real eigenvalue $\lambda$ of $A$, $$P_\lambda S_\lambda =0$$ implies $$\lim_{a\to \lambda}P_a S_a =0.$$

My attempts :

Note that $\operatorname{rank}(S_a B)=k$ for any $a$ different from an eigenvalue of $A$. The conjecture is correct when $\operatorname{rank}(S_\lambda B)=k$, because in that case $$\lim_{a\to \lambda}P_a=P_\lambda.$$ (since the Moore-Penrose pseudoinverse of a matrix is continuous at some point if the rank of the matrix does not change at that point). From my numerical examples so far it seems to me that the conjecture may be correct even when $\operatorname{rank}(S_\lambda B)=k$, but I cannot see why.

I'll also state some facts that might be relevant, or at least might help to clarify my question. I can see that

  1. $\operatorname{rank}(S_\lambda B)=k$ iff $\operatorname{null}(S_\lambda)\cap \operatorname{col}(B)=\{0\}$, that is, iff $\operatorname{col}(B)$ does not contain any eigenvectors of $A$ associated to $\lambda$;

  2. $P_\lambda S_\lambda =0 \text{ iff }\operatorname{col}(S_\lambda)=\operatorname{col}(S_\lambda B),$ because $\operatorname{col}(S_\lambda B)$ is trivially a subset of $\operatorname{col}(S_\lambda)$ and $P_\lambda S_\lambda =0$ iff $\operatorname{col}(S_\lambda)\subseteq\operatorname{col}(S_\lambda B)$;

  3. $P_a S_a \neq0$ for any $a$ different from an eigenvalue $\lambda$ of $A$.

I think that your remark 2 is the key here. It implies the existence of a matrix R such that $S_\lambda (I - BR) = 0$, where $I$ is the $n \times n$ identity matrix.

[To construct $R$, let $e_1, \ldots, e_n$ be a basis for $\mathbb R^n$, and pick $w_i$ such that $S_\lambda e_i = S_\lambda B w_i$. Then define $R$ by $Re_i = w_i$.]

Since $a \mapsto S_a$ is continuous, this means that $\lim_{a \rightarrow \lambda} S_a (I - BR) = 0$. Since $P_a$ always has norm bounded by 1, this means that

$$ \lim_{a\rightarrow \lambda} P_a S_a (I - BR) = 0\text.$$

On the other hand, as a property of the Moore-Penrose pseudoinverse, we have $S_aB(S_aB)^\dagger S_a B = S_a B$, which means that $P_a S_a B = 0$. This means that

$$ P_a S_a B R = 0$$

for all $a$.

Adding the two displayed equations together gives $\lim_{a \rightarrow \lambda} P_a S_a = 0$.


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