Multivariable Second derivative test with Hessian matrix

Question

In my book, I am seeing a contradiction in saddle type definition for multivariable second derivative test with Hessian matrix:

"In case the determinants of the diagonal submatrices are all non-zero, but the hessian matrix is not positive- or negative definite, the critical point is of saddle type."

On the next page:

"If Hessian matrix is neither positive- nor negative-definite, but its determinant is nonzero, it is of saddle type."

The former definition requires all determinants of diagonal submatrices to be non-zero, the second definition only requires the determinant of Hessian matrix to be nonzero.

For example f(x, y) = x^2 + y^2 + z^2 + 2xyz has hessian matrix at (-1, 1, 1):

2 2 2

2 2 -2

2 -2 2

Which has determinant -16.

Determinant of the below submatrix is 0:

2 2

2 2

By the former definition it is not a saddle point because one of the submatrices is zero. By the latter definition, the point is a saddle point because the Hessian matrix is nonzero, -16 (from the example). What is going on here?

Answer 1

Actually I misinterpreted the words, but to be fair the way it was written is quite misleading. The former statement doesn't explain what happens if any of the determinants of the submatrices are 0, it only covers the case where all of the submatrix determinants are non-zero. The latter definition is the better one and the one I should use.