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.

  • $\begingroup$ Ah okay, I thought it was positive definite, since the main diagonal was strictly positive. But I'd need to check the leading principle minors, I just realized. Silly! Thanks for the correction - duly noted $\endgroup$ – stromae Feb 10 '15 at 17:13
  • $\begingroup$ Yes, that's the standard Sylvester's criterion. Though if you see that the matrix is diagonally dominant then there is no need to check that too. $\endgroup$ – Samrat Mukhopadhyay Feb 10 '15 at 17:45
  • $\begingroup$ Sorry - this is not clear to me. Are you saying I don't need to check the leading principal minors if the matrix is strictly positive or strictly negative? $\endgroup$ – stromae Feb 11 '15 at 0:38
  • $\begingroup$ 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 $\endgroup$ – Samrat Mukhopadhyay Feb 11 '15 at 3:56

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