Show that for any $ A\in R^{m\times n}$ the matrices $A^TA$ and $AA^T$ possess the same nonzero eigenvalues

Question

Show that for any $ A\in R^{m\times n}$ the matrices $A^TA$ and $AA^T$ possess the same nonzero eigenvalues.

To be honest, I do not have any clue. I know $A^TA$ and $AA^T$ are symmetric and we can apply our eigenvalue equation like $A^TAx = \lambda x$ and $AA^T=\lambda x$.

Please provide an accurate explanation and all steps, thanks.

Answer 1

Suppose $\lambda\ne 0$ is an eigenvalue of $A^TA$. Then, $$A^TAx = \lambda x$$ for some $x\ne 0$. Left-multiplying by $A$, we get $$AA^T(Ax) = \lambda (Ax)$$ So, $\lambda$ is an eigenvalue of $AA^T$ also. Note that $Ax\ne 0$, for if $Ax = 0$ then $\lambda x = 0 \implies \lambda = 0$ which is a contradiction.

Similarly, suppose $\mu \ne 0$ is an eigenvalue of $AA^T$. Then, $$AA^T x = \mu x$$ for some $x\ne 0$. Left-multiplying by $A^T$, we get $$A^TA(A^Tx) = \mu (A^Tx)$$ So, $\mu$ is an eigenvalue of $A^TA$ also. Note that $A^Tx\ne 0$, for if $A^Tx = 0$ then $\mu x = 0 \implies \mu= 0$ which is a contradiction.