Show that B is a nonsingular matrix (not that obvious).

$B^T$ is non-singular because it is a strict diagonally dominant matrix. So $B$ is non-singular as well.


Another approach:

Multiply by a constant to make it an integer matrix: in this case, consider the matrix $20B$, which is an integer matrix. Show that its determinant is non-zero modulo some small prime, eg: 2 or 3.


Something that will always work with low complexity and not dependent on which matrix:

  1. Take a random vector
  2. Loop over columns / rows:
    • Remove projection with current row/col

If after three (or $n-1$) removed projections is not parallel to last vector ( up to precision ) then there is a non-zero sized null-space.

I'm very sure that the chance of accidentally hitting a subspace in the value space is abysmally small.

Tags:

Linear Algebra

Matrices

Matrix Rank

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