# Pseudoinverse of a diagonal matrix

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?

• What exactly does "have $k$ diagonal elements" mean? Aug 19 '20 at 11:06

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 $$A$$.