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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)

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    $\begingroup$ This proposition has simple conuterexample $$A=\begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$ Since $\det A=-1$, so $A$ is nonsingular, but its submatrix $\begin{pmatrix} 1&1\\1&1\end{pmatrix}$ has determinant 0. $\endgroup$
    – Hanul Jeon
    Mar 1 '13 at 7:17
  • $\begingroup$ @tetori Why not post it as an answer? $\endgroup$
    – Git Gud
    Mar 1 '13 at 7:17
  • $\begingroup$ The following is a true statement,if $A$ is a non-singular matrix then there exists two permutation matrix $P$ and $Q$ such that $PAQ$ has non-singular principal sub-matrices. $\endgroup$ Aug 26 '19 at 9:36
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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.

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  • $\begingroup$ Okay so my example worked? $\endgroup$
    – Hawk
    Mar 1 '13 at 7:23
  • $\begingroup$ @sizz Sorry, I hadn't read your question until the end. The matrix $[0]$ is singular. I don't really understand what you mean with the "example working". $\endgroup$
    – Git Gud
    Mar 1 '13 at 7:31
  • $\begingroup$ is there actually a theorem that guarantees the possibility of non singular submatrices? $\endgroup$
    – Hawk
    Mar 1 '13 at 7:33
  • $\begingroup$ @sizz Under very strict assumptions, maybe. Otherwise I don't think so. Necessary conditions are easy to find, for instance the matrix musn't have any null entry. Sufficient conditions on the other I can't see any which aren't too restricting. $\endgroup$
    – Git Gud
    Mar 1 '13 at 7:37
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    $\begingroup$ It looks like this answer misquotes Sylvester's criterion (about leading, principal minors). Note that the identity matrix is positive definite, but has minors that are entirely zero. $\endgroup$
    – acr
    Apr 4 '17 at 20:58

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