Let A be a non-square matrix, say m*n for m not equal to n, then C=A*A^t and D =A^t*A are both square matrics, but of different sizes. Show that C and D both have the same non-zero eignvalues.
let $\displaystyle \lambda \neq 0$ be an eigenvalue of $\displaystyle C.$ then $\displaystyle AA^Tv=Cv=\lambda v,$ for some non-zero vector $\displaystyle v.$ note that $\displaystyle w=A^Tv \neq 0,$ because otherwise we'd have $\displaystyle 0=Aw=\lambda v,$ which is impossible.
now $\displaystyle Dw=A^TAw=\lambda A^Tv=\lambda w,$ which means that $\displaystyle \lambda$ is an eigenvalue of $\displaystyle D.$ so every non-zero eigenvalue of $\displaystyle C$ is an eigenvalue of $\displaystyle D.$
proving that every non-zero eigenvalue of $\displaystyle D$ is an eigenvalue of $\displaystyle C$ is identical and is left for you.