Eigenvalues and index If I have a set of eigenvalues for a Hessian matrix of a non-degenerate critical point, and I have to determine its index, must I consider all negative eigenvalues, or must the negative eigenvalues be unique? I mean, $$(\lambda + 1)(\lambda + 2)^2 = 0 $$ is the determinant of the diagonalized Hessian of a function with three degrees of freedom. The values are $\lambda =-1, \lambda =-2, \lambda =-2$. Is the index $3$ or $2$? Alternatively, is the critical point a local maximum or a saddle point? 

I think the answer to your question lies in this paragraph quoted from Wikipedia.
http://en.wikipedia.org/wiki/Critical_point_(mathematics)#Several_variables 
"For a function of n variables, the number of negative eigenvalues of the Hessian matrix at a critical point is called the index of the critical point. A non-degenerate critical point is a local maximum if and only the index is n, or, equivalently, if the Hessian matrix is negative definite; it is a local minimum if the index is zero, or, equivalently, if the Hessian matrix is positive definite. For the other values of the index, a non-degenerate critical point is a saddle point, that is a point which is a maximum in some directions and a minimum in others".
