0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.