A theorem in matrix theory states that if $A\in \mathbb{F}^{m\times n}$ , $B\in \mathbb{F}^{n\times m}$, where $m,n\in \mathbb{N^*}$ and $m>n$, then $\det{(AB)}=0$.
Here comes my thought: Consider $\mathcal{A}$ as a linear transformation from a $n$-dimensional linear space $V$ to an $m$-dimensional space $W$, while $\mathcal{B}$ as a linear transformation from $W$ to $V$, symmetrically. If the theorem mentioned above is correct, can we make following statement?(which can be considered as a geometrical explanation of the theorem )
Suppose $\mathcal{A}:V\rightarrow W$, $\mathcal{B}:W\rightarrow V$, $\dim V=n, \dim W=m$, $m>n$. Then the operator $\mathcal{A} \circ \mathcal{B}\in \mathcal{L}(W)$ must have at least one eigenvalue equals $0$.
Can anybody give me a brief prove of the statement geometrically?(Do not use much matrix theory) Or convince me that the statement was wrong.
Because $m > n$, $B$ must have a non-trivial kernel (AKA nullspace). This can be seen, for instance, as a consequence of the rank-nullity theorem. Any vector in the nullspace of $B$ is an eigenvector of $AB$ associated with the eigenvalue $0$.