Show $AA^+$ is symmetric 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!

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. 
