The characteristic equation is det(A − λI) = 0. Why does setting a determinant to zero find exactly the eigenvalues, rather than being an arbitrary computational trick?

C. Because (A − λI) must send a nonzero vector to zero, so be singular.

The rearrangement is the whole argument: Av = λv becomes (A − λI)v = 0 with v nonzero, so A − λI has a nonzero kernel and is singular, and a square matrix is singular exactly when its determinant vanishes. Note det(A − λI) is a polynomial that is not zero for most λ; the eigenvalues are its roots, which is why there are at most n of them.