How one can prove that the Moore-Penrose pseudoinverse can be calculated by using the limit concept as $$H^+=lim_{\delta \to 0^+} (\delta I+H^*H)^{-1}H^*$$, where $H^*$ denotes the Hermitian matrix.

I know this limit exists even if $A^*A$ is not invertible. Moreover, since $(H^*HH^*+\delta H^*)=H^*(HH^*+\delta I)=(H^*H+\delta I) H^*$, and

$(H^*HH^*+\delta H^*)$ and $H^*(HH^*+\delta I)H^*$ have inverses when $\delta > 0$

hence $$H^+=lim_{\delta \to 0^+} (\delta I+H^*H)^{-1}H^*$$ $$=lim_{\delta \to 0^+} H^*(\delta I+HH^*)^{-1}.$$

But, I couldn't go further to prove the limit definition is also Moore-Penrose pseudoinverse.i.e. $lim_{\delta \to 0^+} (\delta I+H^*H)^{-1}H^*=H^+=(H^*H)^{-1}H^*.$

I was wondering if anyone could help me?


1 Answer 1


You show that the limit gives $H^+$ by showing it satisfies the properties that define $H^+$; e.g. here's a list on Wikipedia.

The specific formula $H^+ = (H^* H)^{-1} H^*$ only applies in special cases.


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