$\mathrm{rank}(I_n-BA)=\mathrm{rank}(I_n-AB)$

Question

Is it always true for $A,B \in \mathcal{M_{n}}(\mathbb{C})$ that $$\mathrm{rank}(I_n-BA)=\mathrm{rank}(I_n-AB)\ ?$$

I'm looking for a matrix theoretic proof, not involving dimension theory.

Thanks in advance

Answer 1

Using block multiplication, one has by direct computation $$\begin{pmatrix}I & -A \\0 & I\end{pmatrix} \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix} = \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix} \begin{pmatrix}I & -A \\0 & I\end{pmatrix}$$

So $$ \begin{pmatrix}AB & 0 \\B & 0\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}0 & 0 \\B & BA\end{pmatrix}$$ are similar, so $$ \begin{pmatrix}AB-I_n & 0 \\B & -I_n\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}-I_n & 0 \\B & BA-I_n\end{pmatrix}$$

are similar, so they share the same rank, and this implies that $$\mathrm{rank}(AB-I_n)=\mathrm{rank}(BA-I_n)$$