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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.

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  • $\begingroup$ If we define $x_\epsilon$ to be the minimizer for the function $(1/2) \| Ax - b \|_2^2 + \frac{\epsilon}{2} \| x \|_2^2$, then I think we have $\lim_{\epsilon \to 0^+} x_\epsilon = A^\dagger b$. (Is that correct?) This is intuitive because $A^\dagger b$ is the least norm solution to $Ax - \hat b$, where $\hat b$ is the projection of $b$ onto the column space of $A$. $\endgroup$ – littleO May 11 '20 at 6:45
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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.

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