What will $A^+A$ and $A^gA$ actually or exactly get if $A$ is not invertible?

Question

I know if $A$ is invertible then $A^{-1}$ is the inverse of $A$, and $AA^{-1}=A^{-1}A=I$.

I just learnt the concept of Generalized inverses and Moore–Penrose pseudoinverse. For a matrix $A$ that is not invertible, it has a unique pseudoinverse $A^+$ and many generalized inverses $A^g$.

What will the following four actually or exactly get if $A$ is not invertible?

• $AA^+=?$
• $A^+A=?$
• $AA^g=?$
• $A^gA=?$

Which one can I get an indentity matrix as result like invertible ones?

Answer 1

Generalized inverse can be defined in many ways so IMHO there is no a precise answer to you question for this case. But pseudo inverse is defined precisely. So we can decompose $A=UDV^T$ where $U,V$ ar left/rhgit orthonormalized singular vectors and $D$ is positive diagonal matrix of singular values with dimension equal to the rank of $A$. Thus $$ A^+=VD^{-1}U^T, ~A^+A=VV^T,~ AA^+=UU^T $$

Answer 2

It's worth pointing out that if $A^+$ is the pseudoinverse of $A$, then $A^+A$ is projection onto $\operatorname{Row}(A)$ and $AA^+$ is projection onto $\operatorname{Col}(A)$.

Of course, if $A$ has full column rank, then $A^+A=P_{\operatorname{Row}(A)}=I$. Likewise, if $A$ has full row rank, then $AA^+=P_{\operatorname{Col}(A)}=I$.