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

Question

How would you proceed if you were asked in an interview to show that B is a nonsingular matrix (in an elegant way)?

$$B= \begin{pmatrix} 1& 1.25& −0.50& 0.15\\ 0.15& 2& 1.25& −1.50\\ −0.45& 0.25& 3& 1.25\\ 0.25& −0.15& 0.25& 4\\ \end{pmatrix}$$

In my opinion, taking the time to compute the determinant of this $4\times4$ matrix during the interview would not be appreciated by the interviewer.

Answer 1

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

Answer 2

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.

Answer 3

Another possibility:

Use Gaussian elimination on $B$ to yield an upper-triangular matrix. Nonzeros on the diagonal of the resulting upper-triangular matrix means that $B$ is nonsingular.

(OP asked for an elegant way to do this; however, "elegant" was not defined. IMHO, Gaussian elimination is way more elegant than computing the determinant.)

Answer 4

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.