Reference Request; Existence of Limit

Question

Background/thoughts

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Since $GL_d(\mathbb{R})$ is a topological group then the maps $A\mapsto A^{-1}$ and $(B,A)\mapsto A+B$ are continuous. Therefore, in the case where $X \in GL_d(\mathbb{R})$ then $GL_d(\mathbb{R})$, it follows that $$ \lim_{k \downarrow 0} (X+k I_d)^{-1} =X^{-1} . $$

Question:

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In general however, for $k>0$ and $X$ any $d\times d$ matrix $X+kI_d\in GL_d(\mathbb{R})$ but what is the limit equal to (since $GL_d(\mathbb{R})$ is open as a subset of $\mathbb{R}^{d^2}$ therefore it need not contain all its limits...) $$ \lim_{k \downarrow 0} (X+k I_d)^{-1} =? . $$ Does it equal to the Moore-Penrose pseudo-inverse?

Answer 1

The limit may not exists in all cases.

Without loss of generality we first check whether this limit exists for our scalar field $\Bbb C$. Suppose $X$ has $0$ as its eigen value. Then it has a Jordan-Block $B_{n\times n}$ corresponding to $0$. But note that, $\lim_{k \to 0} (B+k I_n)^{-1}$ does not exists : $\text{det}(B+k I_n)=k^n$, so finding inverse using classical adjoint techniques we have the result, actually $(1,1)$ entry will be $\frac{1}{k}$.

So the above limit exists if $X$ has no zero eigen value, hence invertible.

Answer 2

In general the limit does not exist. For example let $X=0$. Then

$$(X+kI_d)^{-1}=\frac{1}{k}I_d.$$