Show $AA^+$ is symmetric

Question

Can somebody show me how $AA^+$ is symmetric if $A^+$ is the pseudoinverse of $A$?

All I can muster is:

$(AA^+)^T => (A^+)^TA^T$

I know: $(A^+)^T = (A^T)^+$ but that doesn't really seem like it gets us anywhere.

Thanks!

Answer 1

The comment from user759562 is correct, it is Hermitian by definition. But in the spirit of the question, lets do the computation with the definition provided here. That is, when $A$ has linearly independent columns, $A^+$ can be expressed as $A^+ = (A^*A)^{-1}A^*.$ Note that for an invertible matrix $B$, we have $(B^*)^{-1} = (B^{-1})^*.$ Also, if you are working in the reals, just replace $B^*$ with $B^T$.

We wish to show that $(AA^+)^* = AA^+.$

We have \begin{equation} \begin{split} (AA^+)^* &= (A^+)^*A^* \\ &= [(A^*A)^{-1}A^*]^*A^* \\ &= A[(A^*A)^{-1}]^*A^* \\ &= A[(A^*A)^*]^{-1}A^* \\ &= A(A^*A)^{-1}A^* \\ &= AA^+. \end{split} \end{equation}

This can be shown for the 'other' definition (ie. when A has linearly independent rows) using a similar sequence of steps.