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? 
 A: 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".
A: The eigenvalues don't have to be unique, but you have to take into account their multiplicities, or what is nearly the same thing, the dimension of the entire eigenspace corresponding to each eigenvalue, to make sure you obtain/account for every direction in which the given function is decreasing.  Recall that a Hessian matrix at a critical point is symmetric; thus the peculiarities associated with generalized eigenvectors etc. do not occur in this situation; the dimension of an eigenspace is the multiplicity of its eigenvalue.   In your problem, the index is $3$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
