The Lagrangian is:

$L(x,\lambda) = x_1x_2-2x_1-\lambda (x_1^2-x_2^2)$

Taking the derivatives and setting it equal to zero gives:

$x_2-2\lambda x_1-2=0$

$x_1+2\lambda x_2=0$


The points satisfying the conditions above are found to be: $(1,1)$ with $\lambda = -\frac{1}{2}$ and $(-1,1)$ with $\lambda = \frac{1}{2}$

This is the point where I get lost, how can I found out which one of them is a maximizer or minimizer? I usually used the Hessian to find the eigenvalues to see if it is greater or less than zero.

But the Hessian for the first point becomes

$\begin{bmatrix} -2\lambda & 1 \\ 1&2\lambda \end{bmatrix}$= $\begin{bmatrix} 1 & 1 \\ 1&-1 \end{bmatrix}$ which has eigenvalues

$\lambda_1 =1.41$ and $\lambda_1 = - 1.41$ but this is not correct. I get the same eigenvalues for the other point. I really need someone's help !

  • $\begingroup$ It is neither. If the Hessian has both positive and negative eigenvalues then $x$ is a saddle point. $\endgroup$ – Winther Aug 4 '15 at 21:50
  • $\begingroup$ Thanks for you response but I'm first of all not sure if the Hessian is correct. Second, I saw once that a Hessian was [-4/3 0;0 2/3] and it still turned out to be a minimizer when working with the tangent space. $\endgroup$ – Steven Aug 4 '15 at 21:54
  • $\begingroup$ The Hessian is correct... And never mind my previous (deleted) comment, I did a silly mistake in my calculation:) $\endgroup$ – Winther Aug 4 '15 at 22:14
  • $\begingroup$ So if the Hessian has positive and negative eigenvalues it is indefinite and thus (1,1) and (-1,1) are saddle points and thus there are two Saddle point solutions to this problem, correct? $\endgroup$ – Steven Aug 4 '15 at 22:23
  • $\begingroup$ Yes that is correct. $\endgroup$ – Winther Aug 4 '15 at 22:29

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