# Find the saddle point of $F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2+(-x_1+x_2)y_3$

Finding the saddle points of $$F(x_1,x_2,x_3,y_1,y_2,y_3)=(x_1-2x_2+x_3)y_1+(2x_1-2x_3)y_2$$+$$(-x_1+x_2)y_3$$ subject to the constraints $$x_1+x_2+x_3=1, y_1+y_2+y_3=1$$. Show that the saddle point is $$x=(\frac{1/3}{1/3},\frac{1/3}{1/3},\frac{1/3}{1/3}),y=(\frac{2}{7},\frac{1}{7},\frac{4}{7})$$.
I know how to find the saddle point of a function with two variable by using $$\Delta=(f_{12})^2-f_{11}f_{22}$$, then when $$\Delta>0$$ the point is a saddle point.
But for this question, by using the constraints, we can reduce the variables from 6 to 4, but still can not use the $$\Delta$$ formula, and this question requires not to use lagrange multiplier. Thanks.

At a saddle point, all of the partial derivatives are 0. This leads to a system of equations: \begin{align} x_1 -2x_2 +x_3=0,\tag{1}\\ 2x_1 -2x_3=0,\tag{2}\\ x_2-x_1=0,\tag{3}\\ y_1+2y_2-y_3=0, \tag{4}\\ -2y_1 +y_3=0,\tag{5}\\ y_1-2y_2=0. \tag{6} \end{align} Adding 1 to both sides of equation (4) leads to $$$$2 y_1 = 1-3 y_2. \tag{7}$$$$

Solving for $$y_1$$ from equation (6) and substituting in (7) leads to $$y_2= \frac{1}{7}$$ and consequently $$y_1=2y_2=\frac{2}{7}$$ and $$y_3=2y_1=\frac{4}{7}$$.

Similarly, we can subtract 1 from both sides of equation (1) and use the constraint for x, which leads to $$x_2=\frac{1}{3}$$ and consequently from (3), $$x_1=x_2=\frac{1}{3}$$. Finally, from (3), we have $$x_3=x_1=\frac{1}{3}$$.

EDIT: Computing the Hessian matrix yields: $$$$H = \begin{pmatrix} 0 & 0 & 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & -2 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 0\\ 1 & -2 & 1 & 0 & 0 & 0\\ 2 & 0 & 0 & 0 & 0 & 0\\ -1 & 1 & 0 & 0 & 0 & 0 \end{pmatrix}.$$$$ Diagonalizing gives us the characteristic polynomial $$$$\lambda^6-12\lambda^4 +27\lambda^2 -4=0.$$$$

This polynomial has a $$\lambda \rightarrow -\lambda$$ symmetry. So if we can find one non-zero real solution, we automatically show that the Hessian is indefinite (There are both positive and negative solutions). Therefore the critical point above is indeed a saddle point.

EDIT 2:

In case you find it difficult to actually find the roots of the polynomial, substitute $$\rho=\lambda^2$$, this gives $$$$\rho^3-12\rho^2 +27\rho -4=0.$$$$

Now, using the fact that complex roots come in pairs and that we can see that zero is not a solution, we can deduce that there is at least one real solution of the cubic $$\rho_0$$ and therefore two opposing sign solutions $$\lambda_{+/-} = \pm \sqrt{|\rho_0|}$$.

• Thanks, but this only shows that the given point is a critical point, not proves it's a saddle point? – Cathy Jun 3 at 20:31
• That's true. You can try the Hessian test after you have shown that it is a critical point? Seems like you should get an indefinte matrix - I haven't tried. – Stratiev Jun 3 at 20:44
• I haven't learnt Hessian test yet, so there may exist some way to prove it's a saddle point without using Hessian test? – Cathy Jun 3 at 20:54
• The expression that you defined as $\Delta$ is the determinant of the Hessian for a function of two variables. It's just that in this case you'd get a 6x6 matrix defined by the second order partial derivatives. It will be a sparse matrix with a few integer entries. – Stratiev Jun 3 at 21:02