[Math] Proving a determinant = 0

determinantslinear algebramatricesra.rings-and-algebrasrecreational-mathematics

The two most elementary ways to prove an N x N matrix's determinant = 0 are:

A) Find a row or column that equals the 0 vector.

B) Find a linear combination of rows or columns that equals the 0 vector.

A can be generalized to

C) Find a j x k submatrix, with j + k > N, all of whose entries are 0.

My minor question is: Is C a named theorem that one can easily reference?

My major question is: Are there are other canonical ways of proving a determinant = 0?

The context is that I'm trying to solve the generalized form of what was, as stated, a very easy Putnam Exam problem, and I last took a linear algebra class in 1974.

In response to comments below, let me say:
– Thanks!
– This isn't about computational
efficiency.
– Frobenius-Koenig looks very helpful.

Best Answer

The Lindstrom-Gessel-Viennot Lemma is sometimes useful if you can find a nice enough graph corresponding to your matrix. I guess I wouldn't call this canonical though.

Edit: Well, I don't totally understand the downvote. To illustrate with a nice example (which I think is in the original Gessel-Viennot paper), if one takes $A$ to be the $n\times n$ matrix with $[A]_{i,j}={i+j-2\choose i-1}$, then the lemma provides a very elegant way to prove that the determinant of this matrix is 1, and if we then subtract $1$ from the $(n,n)$-entry, then the determinant is 0.