If an n by n square matrix 'W' has at least one r by r sub-matrix that is singular and (n-1) > r > n/2 what 'conditions' are necessary so that W also singular? Or is there no way to tell if given a matrix 'W' has at least one singular sub-matix if the matrix 'W' itself is singular without calculating it's determinant?
2026-04-15 12:50:22.1776257422
A faster way to tell if a matrix is not non-singular.
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$$\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ is non singular but has plenty of singular submatrices like $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.