Prove that $A$ is nonsingular

Question

Problem: Let $M$ be a $n \times n$ nonsingular matrix, and $$M = \begin{bmatrix} A \quad B \\ C \quad D \end{bmatrix} \in \mathbb{K}^{n \times n}$$ with $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, $A \in \mathbb{K}^{k \times k}$, $D \in \mathbb{K}^{q \times q}$, $k<n$. Prove that $A$ is nonsingular.

My attempt: Since $M$ be a nonsingular matrix so every leading principal submatrices of $M$ nonsingular and $A$ be the leading pricipal submatrix of order $k$ of $M$. Q.E.D

Is that true? Thank all!

Answer 1

Well, the identity matrix $I=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right)$ is nonsingular but it contains 1x1 submatrices (each of which with entry $0$) which are singular.

Answer 2

The claim in the problem is false: $M=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ is a counterexample.

Answer 3

Let $$M =\begin{pmatrix} 0& 0 & 0 &1 \\ 0& 0 & 1 &2 \\ 0& 1 &2 &1 \\ 1 &2 & 3 &2 \end{pmatrix}$$ It's clear that $\det(M) = 1 \neq 0$, so $M$ is invertible. But $\begin{pmatrix} 0& 0\\ 0 & 0 \end{pmatrix}$ is not invertible.