# If A, B, C, D are non-invertible $n \times n$ matrices, is it true that their $2n \times 2n$ block matrix is non-invertible?

Is it true that

$\left( \begin{array}{ccc} A & B \\ C & D \\\end{array} \right)$

is non-invertible?

Assume that the matrix is over a field.

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Try the $2\times 2$ matrices that make up $$\begin{pmatrix}1&0&0&0\\ 1&0&1&0\\ 0&1&0&0\\ 0&1&0&1\end{pmatrix}.$$ (This example works over all fields, no matter what characteristic).
No. I think the simplest counterexample is the permutation matrix $$P=\left[\begin{array}{cc|cc} 1&0&0&0\\ 0&0&1&0\\ \hline 0&1&0&0\\ 0&0&0&1 \end{array}\right].$$ Edit: In general, for every $n\ge2$, the $2n\times2n$ permutation matrix containing ones at positions $(k,2k-1)$ and $(n+k,2k)$ for $k=1,2,\ldots,n$ will serve as a counterexample.
This is not true. Assume $A$ has half $1$ on the upper diagonal and $B$ have half 1 on the lower diagonal, $C$ has half $1$ on the lower diagonal and $D$ has half $1$ on the upper diagonal. Then $A,B,C,D$ are all singular, but after some shifting the matrix become $I$.