Determinant of block matrices with non-square blocks Let $A$ be $m \times n$ matrix, and $B$ be $n \times m$ matrix. Then

*

*Show that $$\det\begin{bmatrix}I_{n} & B\\ A & I_{m} \end{bmatrix}=\det\begin{bmatrix}I_{m} & A\\ B & I_{n}\end{bmatrix}$$


*Show that $$\det(I_{m}-AB)=\det(I_{n}-BA)$$


*Is it true that $I_{m}-AB$ and $I_{n}-BA$ have the same rank?

Someone please show me a way to start proving. I tried to use formulas for calculating block matrix with square matrices, but it didn't help me on this one. I really don't know where to start.
 A: Hint: for the first one, note that
$$
\begin{bmatrix}
0 & I_m\\
I_n & 0
\end{bmatrix}
\begin{bmatrix}
I_n & B\\
A & I_m
\end{bmatrix}
\begin{bmatrix}
0 & I_n\\
I_m & 0
\end{bmatrix} =
\begin{bmatrix}
I_m & A\\
B & I_n
\end{bmatrix}
$$
A: Since $\begin{bmatrix}I_{n} & 0\\ -A & I_{m}\end{bmatrix}
\begin{bmatrix}I_{n} & B\\
A & I_{m}\end{bmatrix} = \begin{bmatrix}I_{n} & B\\ 0 & I_{m}-AB \end{bmatrix}$, and $\det \begin{bmatrix}I_{n} & 0\\ -A & I_{m}\end{bmatrix} = 1$, we have
$\det\begin{bmatrix}I_{n} & B\\
A & I_{m}\end{bmatrix} =  \det \begin{bmatrix}I_{n} & 0\\ 0 & I_{m}-AB \end{bmatrix} = \det (I_{m}-AB) $.
Similarly, $\begin{bmatrix}I_{n} & -B\\ 0 & I_{m}\end{bmatrix}
\begin{bmatrix}I_{n} & B\\
A & I_{m}\end{bmatrix} = \begin{bmatrix}I_{n}-BA & 0\\ A & I_{m} \end{bmatrix}$, and we have 
$\det\begin{bmatrix}I_{n} & B\\
A & I_{m}\end{bmatrix} =  \det \begin{bmatrix}I_{n}-BA & 0\\ 0 & I_{m} \end{bmatrix} = \det (I_{n}-BA ) $.
Hence we have 2.
Swapping the roles of $A,B$ above shows that 
$\det\begin{bmatrix}I_{m} & A\\
B & I_{n}\end{bmatrix} =  \det (I_{n}-BA) $, hence we have 1.
Finally, 3. Let $r_m = \operatorname{rk}(I_{m}-AB)$, $r_n = \operatorname{rk}(I_{n}-BA)$. The identities on Lines 1 and 3 above show that $n+r_m = m+r_n$, hence if $m\le n$, then $r_n = r_m+n-m$. 
It follows that $r_n=r_m $ iff $n=m$.
Note: The above argument shows more. Suppose $n\ge m$, and $\lambda \neq 0$. Then
\begin{eqnarray}
\det (\lambda I -BA) &=& \det (\lambda (I -\frac{1}{\lambda}BA)) \\
&=&  \lambda^n \det ( I -\frac{1}{\lambda}AB) \\
&=& \lambda^n \det ( \frac{1}{\lambda} ( \lambda I -AB)) \\
&=& \lambda^{n-m} \det ( \lambda I -AB)
\end{eqnarray}
This shows that the characteristic polynomials of $AB$ and $BA$ are the same except for $n-m$ roots at $0$.
