In pseudoinverse of a matrix, we have a special case when the columns are linearly independent.
It is mentioned in that article and in other articles that
It follows that $A^+$ is then a left inverse of $A$: $A^+A=I_{n}$.
But I think it should be right inverse because:
\begin{align}
(A^*A)^{-1} A^*A = I_{n} &\implies (A^*A)^{-1} A^*AA^+ = I_{n}A^+ \\
&\implies (A^*A)^{-1} A^*I_{m} = A^+ \\
&\implies (A^*A)^{-1} A^* = A^+.
\end{align}
From this we see that $AA^+= I_{m}$, then $A^+$ here is right inverse, so why is it mentioned that $A^+$ is left inverse?
Best Answer
If an $m\times n$ matrix $A$ has linearly independent columns, then $m\ge n$ and the matrix has a left inverse (which is also a right inverse if $m=n$, of course).
If $m>n$, then $A$ cannot have a right inverse.
In this case the matrix $A^*A$ is invertible, because its rank is the same as the rank of $A$, so we can consider $(A^*A)^{-1}A^*$ and this is a left inverse of $A$, because $$ \bigl((A^*A)^{-1}A^*\bigr)A=(A^*A)^{-1}(A^*A)=I $$ Set $A^+=(A^*A)^{-1}A^*$ and prove it satisfies the requirements for being the pseudoinverse in this special case.
Indeed, $AA^+A=AI=A$; also $A^+AA^+=IA^+=A^+$. Next $(A^+A)^*=I^*=I=A^+A$. The last condition to verify is $(AA^+)^*=AA^+$; now $$ (AA^+)^*=(A(A^*A)^{-1}A^*)^*=A((A^*A)^{-1})^*A^* =A((A^*A)^*)^{-1}A^*=A(A^*A)^{-1}A^*=AA^+ $$