Zero eigenvalues of $AA^T$ and $A^TA$

Question

Given that for any matrix $A \in \mathbb R^{m\times n}$, if $v$ is an eigenvector of $A^TA$ with eigenvalue $\lambda$ not equal to zero, then $Av$ is an eigenvector of $AA^T$ with the same eigenvalue.

Show that if $\lambda = 0$ then the statement above is false

How can I prove this?

Thanks and regards.

Answer 1

If $A^TAv = 0$ then also $AA^TAv = 0$ and $Av$ would be an eigenvector of eigenvalue $0$ unless $Av$ is the zero vector. So this is what you need to prove, that if $A^TAv = 0$ then $Av$ is the zero vector.

You have that $x = Av$ is in the column space of $A$. But $A^TAv = A^Tx = 0$ which means that $x$ is also in the left nullspace of $A$. By the fundamental theorem of linear algebra the only vector satisfying both is the zero vector.

Edit: There's a more straight forward proof. $A^TAv = 0 \Rightarrow v^TA^TAv = 0 \Rightarrow (Av)^T(Av) = 0 \Rightarrow ||Av||^2 = 0 \Rightarrow Av = 0$

Answer 2

If $\lambda=0$ the statement may be true or false, depending on the matrix $A$.

The easiest example is for $A=0$ a null matrix: the only eigenvalue of both $A^T\!A$ and $AA^T$ is $0$ and every nonzero vector is an eigenvector.

For a case when the statement is false, consider $A=[1\ 0]$. Then $$ AA^T=[1],\qquad A^TA=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ so $0$ is not even an eigenvalue of $AA^T$, but it is of $A^T\!A$.

If you consider the full statement, that is,

if $0$ is an eigenvalue of both $A^T\!A$ and $AA^T$ and $v$ is an eigenvector of $A^T\!A$ relative to $0$, then $Av$ is an eigenvector of $AA^T$ relative to $0$,

then this statement is indeed false, because the assumption is that $A^T\!Av=0$, so $v^T\!A^T\!Av=0$ as well, which implies $Av=0$.