$A$ square matrix,nonsingular $\implies $ all submatrixes of $A$ are also nonsingular?

Question

If a square matrix $A$ is nonsingular, then every submatrix of $A$ is also nonsingular.

I am trying to come up with a counter example. But most involve really difficult examples, so I am starting to think this is actually true and probably something to do with a non zero determinant through each sub matrix.

Also, does taking a zero entry in say an identity count as a non singular matrix? (1 x 1 matrix)

Answer 1

Consider the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$. The submatrix $\begin{bmatrix} 0 \end{bmatrix}$ isn't invertible. If you now want to consider just the principal submatrices, then consider $\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$.

Edit: The following is slightly related to your question.

It is well known that for any hermitian matrix, the characterization below holds:

The matrix is positive-definite $\iff$ All its eigenvalues are (strictly) positive $\iff$ All its principal minors (i.e., the determinants of its principal submatrices) are (strictly) positive.

Since any matrix with strictly positive eigenvalues is nonsingular, the above characterization guaranteers that if it is hermitian, then its submatrices will also be nonsingular.