# Limit of the pseudo inverse of a matrix as this matrix goes to zero

This might be a silly question, compared to the level of the questions here, but is it okay to ask the following:

For a square positive semi definite matrix $\pmb{A} \in \mathbb{C}^{N \times N }$, where $\pmb{A} \xrightarrow[N \to \infty]{} \pmb{0}$ What is the following limit: $$\pmb{A}^{+} \xrightarrow[N \to \infty]{} \ \ ??$$ where $\pmb{A}^{+}$ is the Moore-Penrose pseudo inverse.

If we were to use the definition of Moore-Penrose pseudo inverse, then we have to compute the following: $$\pmb{A}^{+} = (\pmb{A}\pmb{A}^{H} + \epsilon \pmb{I})^{-1}\pmb{A} \xrightarrow[N \to \infty \ \ \ \epsilon \to 0]{} \ \ ??$$ which i guess does not exist

• Is the size of the matrix getting bigger, or did you accidentally use $N$ twice? – Omnomnomnom Aug 10 '16 at 13:45
• Thank you for pointing that out. I meant to say $\pmb{A} \in \mathbb{C}^{M \times M}$ and the rest is correct. $M$ is fixed. But $N$ is a parameter in $\pmb{A}$ that lets it go to zero as $N$ grows large. – Ahmad Bazzi Aug 10 '16 at 15:25
• The Moore-Penrose pseudo inverse agrees with the inverse for invertible matrices. So the limit doesn't exit in $\mathbf R^{M\times M}$ in general. – user251257 Aug 10 '16 at 18:07