# Verifying stationary points - check my answer please - Has a hessian

Can someone check this for me;

For $f(x_1,x_2,x_3) = x_1^2 + x_2^2 + x_3^2-x_1x_2+x_2x_3-x_1x_3-x_1+x_2$

the stationary point occurs at $\nabla f(x)^T =\left[ \begin {array}{c} 0\\ 0\\ 0\\ \end{array} \right]=$ $\left[ \begin{array}{c} 2x_1-x_2-x_3-1 \\ 2x_1 -x_1+x_3+1 \\ 2x_3+x_2-x_1\end{array} \right].$

After some quick algebra, which doesn't concern me too much, the SP occurs at $x_1= \frac{1}{3} = -x_2$ and $x_3 =0$

Part which concerns me:

For the hessian I have the main diagonal as $(2,2,2)$

Some of the off diagonals are negative.

But is this sufficient proof to say the Hessian is Positive definite, and hence the Stationary point is a strict global minimizer?

The Hessian is $$H=\begin{bmatrix} 2 & -1 & -1\\ -1 & 2 & 1\\ -1 & 1 & 2 \end{bmatrix}$$You can check by Gershgorin's circle theorem to see that the eigenvalues are $\ge 0$, and since the matrix is symmetric, it is Positive semi-definite. In fact $[1\quad 1\quad 1]$ is in its kernel. So the stationary point may not be a strict global optimizer.
• Yes, if the matrix is diagonally dominant, with positive diagonals then $|a_{ii}|\ge \sum_{j=1}^n |a_{ij}|\ \forall j$. And if the matrix is symmetric, the eigenvalues are always $\ge 0$, hence the matrix is positive semi-definite – Samrat Mukhopadhyay Feb 11 '15 at 3:56