= Solution
Fix one common <eigenvector> $v\ne0$, and write $AA^\top v=\lambda v$ and $BB^\top v=\mu v$. By (d),
$$
AA^\top BB^\top=A(A^\top B)B^\top=0,
$$
so $\lambda\mu v=0$ and \b[at least one of the two <eigenvalues> is zero].
To exclude both being zero, use the <kernel of a Gram product> identity:
$$
v^\top AA^\top v=\|A^\top v\|^2,\qquad
v^\top BB^\top v=\|B^\top v\|^2.
$$
If both products annihilate $v$, then $A^\top v=B^\top v=0$. By (b), $v\in\operatorname{im}L$, and by (e), this is $\operatorname{im}B$. On the other hand, $B^\top v=0$ makes $v$ <orthogonal> to every <vector> in $\operatorname{im}B$. In particular $v$ is <orthogonal> to itself, so $\|v\|^2=0$ and $v=0$, a contradiction.
Therefore \b[each specified common <eigenvector> belongs to exactly one of $\ker(AA^\top)$ and $\ker(BB^\top)$]. Both <eigenvalues> are nonnegative by the displayed squared-<norm> identities, so the other <eigenvalue> is strictly positive. This is the spectral mechanism behind <orthogonal null-space completion of a square matrix>.
Back to article page
