Method of Lagrange mulitplier 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
 A: 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.
A: 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. 
