Invertibility of a block matrix I need to prove that the following matrix is invertible
$$\left( {\begin{array}{*{20}{c}}
{{B_{n \times n}}}&{{I_{n \times m}}}\\
{{I_{m \times n}}}&{{0_{m \times m}}}
\end{array}} \right),$$
where $B_{n \times n}$ is an invertible $n\times n$-matrix, $I_{n \times m}$, $I_{m \times n}$ are identity matrices and $m < n$, for example
$${I_{3 \times 2}} = \left( {\begin{array}{*{20}{c}}
1&0\\
0&1\\
0&0
\end{array}} \right),\qquad{I_{2 \times 3}} = \left( {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0
\end{array}} \right).$$
I will be grateful if someone help me to prove it.
 A: The claim is false; take $m=1$, $n=2$ and $B=\tbinom{0\ 1}{1\ 0}$ to get the matrix
$$\begin{pmatrix}
0&1&1\\
1&0&0\\
1&0&0
\end{pmatrix},$$
which is clearly not invertible. In the same way, for any value of $m$ and $n$ with $m<n$, the matrix $B$ with $1$'s on the antidiagonal and $0$'s elsewhere yields a singular matrix.
On the other hand, if $m=n$ then it is easily verified that
$$
\begin{pmatrix}
B&I_n\\I_n&O
\end{pmatrix}
\begin{pmatrix}
O&I_n\\I_n&-B
\end{pmatrix}
=I_{2n},$$
and so the matrix is indeed invertible. Here $I_k$ denotes the $k\times k$ identity matrix.
A: In general, if $M$ can be divided in four blocks $A$, $B$, $C$, and $D$ as follows
$$ M = \begin{bmatrix}
A&B\\C &D\end{bmatrix}$$
where $A$ is invertible, then
$$\det(M) = \det(A)\det(D-CA^{-1} B).$$
We can use this to say that, since $B_{n\times n}$ is invertible, the matrix
$$\begin{bmatrix}
B_{n\times n} & I_{n\times m} \\ I_{m\times n} & 0_{m \times m} \end{bmatrix}$$ is invertible whenever the matrix
$$I_{m\times n} B_{n\times n}^{-1} I_{n\times m}$$
is invertible, in other words, if the $m$th principal minor of the matrix $B_{n\times n}^{-1}$ is nonzero.
