a) Suppose $AA^T x=tx$ where $||x||=1$. Then $t=(AA^Tx,x)=||A^Tx||^2>0$ since $tx\not=0$. b) Suppose that $AA^T$ has eigenvalues $0<t_1\le \cdots\le t_n$ and that $A^TA$ has eigenvalues $0<s_1\le \cdots\le s_n$. Then $\{t_1^p,\cdots,t_n^p\}=\{s_1^q,\cdots s_n^q\}$ hence $t_k^p=s_k^q$ for all $k$. It is well-known that $AA^T,A^TA$ have the same eigenvalues (with mpl) hence $t_k=s_k$ for all $k$. Write $t=t_1=s_1$ then $t^p=t^q \implies t^{p-q}=1\implies t=1$. Hence $AA^T$ has eigenvalues $1,\cdots,1$ and it is symmetric. Hence $AA^T=I$.