Show that $\dim V = \dim W$ $V$ and $W$ are subspaces of $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. $m, n$ be positive integers. Let $A$ be a $(n,m)$ matrix and $B$ be a $(m,n)$ matrix assume that $\lambda \neq 0$
$V= \{x\in \mathbb{C}^m |$ for a positive integer $k$ , $(BA-\lambda I_m)^k x=0$ holds}
$W= \{y\in \mathbb{C}^n |$ for a positive integer $l$ , $(AB-\lambda I_n)^l y=0$ holds}
Show that $\dim V= \dim W$
I have proved $BA$ and $AB$ have the same eigenvalue $\lambda$ but then how do we involve these statements, any hints would be appreciated.
 A: I'm editing the post after johnny10 pointing out my mistake in the previous post.
If we denote for fix $k$:
$V_k= \{x\in \mathbb{C}^m |  (BA-\lambda I_m)^k x=0 \}$
and
$W_k= \{x\in \mathbb{C}^n |  (AB-\lambda I_n)^k x=0 \}$
then $\dim V_k =\dim W_k$.
For the proof, I just extend the answer that johnny10 gave above (but the comment doesn't allow me to type too many things so I make a post here):
The following is the same as that given by Christiaan Hattingh in the post with adapted size:
From $\begin{bmatrix} I_n & -A \\ 0 & I_m\end{bmatrix}\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}\begin{bmatrix} I_n & A \\ 0 & I_m\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$, we see that $\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$ are similar so they have same Jordan form.
In particular, $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda  I_{m+n}\right )^k$ for all $k$.
On the other hand, we have $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (AB-\lambda I_n)^k & 0 \\ D & (-1)^k \lambda^k I_m\end{bmatrix} \right ) = \text{rank}(AB - \lambda I_n)^k +m$
and
$\text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (-1)^k \lambda^k I_m & 0 \\ D' & (-1)^k (BA-\lambda I_m)^k\end{bmatrix} \right ) = \text{rank}(BA - \lambda I_m)^k +n$
Therefore, $n-\text{rank}(AB - \lambda I_n)^k=m-\text{rank}(BA - \lambda I_m)^k$ for all $k$ and it implies the statement above.
Hence, $\dim V =\dim W $ since in general $V$ and $W$ are just $V_k$ and $W_k$ for $k$ large (we can take $k$ to be the algebraic multiplicities).
A: As mathmath has already posted a handmade solution, I feel free to write a more sophisticated proof.
Recall that if $T$ is an endomorphism and $\sigma$ is an eigenvalue, its geometric multiplicity is the dimension of $\ker (T-\sigma I)$, while its algebraic multiplicity is the multiplicity of $\sigma$ a a root of the characteristic polynomial of $T$, $p_T(x)$ and you know that the first one is less than or equal to the second one.
When studying the Jordan normal form, one needs to consider the generalised eigenspaces $\ker (T-\sigma)^k$ and they form a tower that stops at some point $$\ker (T-\sigma I)^1 \subsetneq \ker (T-\sigma I)^2 \subsetneq \ldots \subsetneq \ker(T-\sigma I)^N = \ker(T-\sigma I)^{N+1}=\ldots$$And it turns out that the algebraic dimension of $\sigma$ is the dimension of this top space $\ker (T-\sigma I)^N$.
Now, turning back to your problem, and using your notation, we have $$V= \ker (BA-\lambda I) \oplus \ker(BA-\lambda I)^2 \oplus \ldots \qquad W= \ker (AB-\lambda I) \oplus \ker(AB-\lambda I)^2 \oplus \ldots$$Therefore, $\dim V$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $BA$ and $\dim W$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $AB$.
It is well known that if $A$ and $B$ are square matrices of the same dimension, then the characteristic polynomials of $AB$ and $BA$ are the same (I'll let you think about this or look it by yourself). What is not so well known is that a similar result holds when $A$ and $B$ are not square, as in your case. If $m<n$, for example, add zeroes at the matrices $A$ and $B$ to get two square $n \times n$ matrices $A'$, $B'$, and use that $p_{A'B'}(x) = p_{B'A'}(x)$ to obtain $p_{AB}(t) = t^{n-m}p_{BA}(t)$ and from there and all what I have commented, the result follows. Note the importance that $\lambda \neq 0$
