Method of Lagrange mulitplier

Question

I have a function $F(x,y,z)=x^2+xy+y^2+yz+z^2$. I have to find its extreme value using Lagrange multiplier with the given constraint $x^2+y^2+z^2=1$. We can write $F(x,y,z)={\bf{x}}^TB {\bf x}$ where

$$ B=\begin{bmatrix} 1&1&0 \\ 0&1&1 \\ 0&0&1 \end{bmatrix}. $$

The eigenvalues of this matrix are all $1$. I know the maximum and minimum of the function occur on the corresponding eigenvectors. How do I determine which eigenvector gives max or min? How do I determine if there is any saddle point on the unit sphere or not?

Thank you

Answer 1

It doesn't quite work like that. You should write $F=x^T Bx$ with a symmetric matrix $$ B = \left[ \begin{matrix} 1 & \frac12 & 0 \\ \frac12 & 1 & \frac12\\ 0 & \frac12 & 1 \end{matrix} \right]$$ with evals $1\pm \sqrt{2}/{2}$ and 1. It is non-trivial to see that you have a saddle point for $\lambda_2=1$. But that you have a max for $\lambda_1=1+\sqrt{2}/{2}$ and min for $\lambda_1=1-\sqrt{2}/{2}$ follows directly from standard theory for symmetric matrices.

Answer 2

Since the question says Lagrange Multiplies, and you seem to be unclear on what that means...

maximize $F(x,y,z)=x^2+xy+y^2+yz+z^2$ constrained by: $x^2+y^2+z^2=1$

$F(x,y,z, \lambda)=x^2+xy+y^2+yz+z^2 -\lambda(x^2+y^2+z^2-1)$

$\nabla F(x,y,z, \lambda) = \mathbf 0$

$2(1-\lambda) x + y +0 z= 0\\ x + 2(1-\lambda)y+ z = 0\\ 0x+ y + 2(1-\lambda)z = 0\\ x^2 +y^2 + z^2 = 0$

And then solve this system of equations.

But it jumps out that that does look a lot like the symmetric matrix $B$

$\frac 12 \begin{bmatrix} 1-\lambda &\frac 12 & 0 \\\frac 12 & 1-\lambda& \frac 12 \\ 0 & \frac 12&1-\lambda\end{bmatrix}\mathbf x = \mathbf 0$

And that is only possible if the matrix is singular. Which means that $\lambda$ equals an eigenvalue, and $\mathbf x$ is the corresponding eigenvector.

And we have come around to the same place where you wanted to start.