Prove ${\rm tr}\ (AA^T)={\rm tr}\ (A^TA)$ for any Matrix $A$ Prove ${\rm tr}\ (AA^T)={\rm tr}\ (A^TA)$ for any Matrix $A$
I know that each are always well defined and I have proved that, but I am struggling to write up a solid proof to equate them. I know they're equal.
I tried to show that the main diagonal elements were the same but if I say that $A$ is $n\times m$ then 
$$(AA^T)_{ii} = (a_{11}^2+\dots +a_{1m}^2) + \dots + (a_{n1}^2+\dots +a_{nm}^2)$$
and
$$(A^TA)_{ii} = (a_{11}^2+\dots +a_{n1}^2) + \dots + (a_{1m}^2+\dots +a_{nm}^2)$$

All approaches to the question made so far are pretty good. I just want to add a small observation that will make the proof solid.
trace($AA^T$) or trace($A^TA$) is sum of squares of all elements of the matrix $A$ (which is the same as the sum of squares of all elements of the matrix $A^T$).
