# Why does calculating Moore-Penrose pseudoinverse require to add $\alpha I$?

Moore-Penrose pseudoinverse are limits: $A^+ = \lim_{\alpha \searrow 0} (A^T A + \alpha I)^{-1} A^T$.

I have noticed that: $A^+ A = \lim_{\alpha \searrow 0}(A^T A + \alpha I)^{-1} A^T A = I$, even $A$ is not invertible. But I wonder why does calculating the pseudoinverse require to add $\alpha I$ even $\alpha$ converged zero.

• because $A^TA$ may not be invertible? - Also note that for some $A$, $\lim_{ \alpha\to 0}\bigl((A^TA+\alpha I)^{-1}A^TA\bigr)\ne I$. Just try $A=0$. – Hagen von Eitzen Apr 1 '17 at 10:42
• An analogy with the scalar case: $\lim_{\alpha\to0}(a^2+\alpha)^{-1}a$ is defined even when $a=0$. – Yves Daoust Apr 1 '17 at 10:49
• $A^TA$ is not sure be full rank. Any 0 eigen/singular value of $A$ will be 0 for $A^TA$ – mathreadler Apr 1 '17 at 11:09

## 1 Answer

$A^TA$ is not sure be full rank. Any $0$ eigen/singular value of $A$ will be $0$ for $A^TA$, but any non-zero eigenvalue of $A$ will be $>0$ for $A^TA$ and $+\alpha I$ moves all eigenvalues by $+\alpha$ which is positively so any previous $0$ will be $>0$ and any previous non-zero will be even bigger so sure to not become a $0$.