Let $A,B \in \mathbb{R}^{n \times n}$ and prove that $AB$ and $BA$ have the same eigenvalues. Also prove that if $AB$ and $BA$ are symmetric then $\|AB\|_F \leq \|BA\|_F$.
For the first part, we look at the characteristic polynomial: $P_{AB}=\det(AB-\lambda I)=\det(ABAA^{-1}-\lambda IAA^{-1})= \det(A(BA-\lambda I)A^{-1})=\det(A)\det(BA-\lambda I)\frac{1}{\det(A)} =\det(BA-\lambda I)$.
This holds under the assumption that $A$ is invertible. What about the second part? I know that the Frobenius norm for an arbitrary matrix $A$ is $\sqrt{\sum_{1\leq i\leq n} \lambda_i(A^TA)}$ and in my case $$\|AB\|_F=\sqrt{\sum_{1\leq i\leq n} \lambda_i((AB)^TAB)}=\sqrt{\sum_{1\leq i\leq n} \lambda_i(ABAB)}=\sqrt{\sum_{1\leq i\leq n} \lambda_i((AB)^2)}=\sqrt{\sum_{1\leq i\leq n} \lambda_i((BA)^2)} = \sqrt{\sum_{1\leq i\leq n} \lambda_i(BABA)}=\sqrt{\sum_{1\leq i\leq n} \lambda_i((BA)^T BA)}=\|BA\|_F.$$ Is this ok? I get equality, but maybe I made some mistake.