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Let matrix $A \in \Bbb R^{n \times n}$ have $k$ diagonal elements, where $k < n$, and rest of the elements are zero. I am trying to find the pseudoinverse of $A + \lambda I$ when $\lambda$ approaches zero.

Then $\frac{1}{a_i + \lambda}$ would be the diagonal elements for $i$ going from 1 to $k$ of the pseudo inverse and $\frac{1}{\lambda}$ would be the rest of the diagonal elements. If I put $\lambda$ equal to zero then the pseudo inverse would a matrix with elements of $A$ matrix inverted, but there would be elements going to infinity. But that does not sound right. What is wrong in this logic?

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  • $\begingroup$ What exactly does "have $k$ diagonal elements" mean? $\endgroup$ Aug 19 '20 at 11:06
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The problem is that the pseudo inverse is not a continuous function on the space of matrices as exactly you've shown. Consider the 1d matrix $(x)$ for $x\in\mathbb R$. Then the pseudo-inverse map is $$ (x)\mapsto\begin{cases}1/x&\text{ if }x\neq 0,\\0&\text{ otherwise.} \end{cases} $$ This is not a continuous at zero, and so we would not expect it to preserve a limit of an element to zero. The same happens with your example when we restrict to the kernel of $A$.

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