Question about the Moore-Penrose pseudoinverse

Question

Suppose I have a matrix $A$ that is not invertible, but such that $A + \epsilon I$ is invertible for all $\epsilon$. I'm wondering whether we can say something like $$\underset{\epsilon \to 0}{\lim} (A + \epsilon I)^{-1} = A^{\dagger}$$ where $A^{\dagger}$ is the Moore-Penrose pseudoinverse of $A$. In essence, can we plug $\epsilon = 0$ into the limit, so long as we swap out the inverse for a pseudoinverse? It seems intuitive to me, but I don't see a way to prove or disprove it. Any insight would be appreciated.

Answer 1

No. Consider the diagonal matrix $A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ its pseudo inverse is also $ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.

However $(A + \epsilon I)^{-1} = \begin{pmatrix} {1\over 1+\epsilon} & 0 \\ 0 & {1\over \epsilon} \end{pmatrix}$ and the limit is singular.