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

Question

Is it true (and under which conditions) that the products of an non-square matrix $A$ and its transpose and vice versa (so the product of the transpose and $A$) share the same eigenvalues (multiplicities omitted)?

Answer 1

They have equal all eighevalues different from zero because if $v$ is an eighevector of $A^tA$ of eighenvalue $\lambda\neq 0$ then

$A^tAv=\lambda v$

and if you apply the matrix A you get

$AA^t(Av)=\lambda Av$ because $Av\neq 0$

(If $Av=0$ then $A^t(Av)=\lambda v=0$ and it is not possible because $v$ is an eighenvector of $A^tA$ and so it is different from zero. )

Answer 2

They share all nonzero eigenvalues. Basically, you "pad" the shorter list with zeroes.

One way to see it is to first note that $\ker A^TA=\ker A$. Now, if $A^TAx=\lambda x$, with $\lambda\ne0$, then $$ AA^T(Ax)=\lambda Ax. $$ And $Ax\ne0$ since $A^TAx\ne0$ (which follows from $\lambda\ne0$). This shows that every eigenvalue of $A^TA$ is an eigenvalue of $AA^T$. By exchanging the roles you get that eigenvalues are the same.

Alternatively, if $A=UDV$ is the singular value decomposition, then $$ A^TA=V^TD^TDV,\ \ AA^T=UDD^TU^T. $$ Since $D$ is "diagonal" (though rectangular), it is easy to see that all nonzero entries of $D^TD$ and $DD^T$ are the same.

Answer 3

More generally, for any $n \times m$ matrix $A$ and $m \times n$ matrix $B$, $AB$ and $BA$ have the same nonzero eigenvalues.