Lagrange multipliers and calculus

Question

I'm trying to solve this problema but I don't know what else to do:

Let A be a nonzero symmetric 3x3 matrix. Thus, its entries satisfy $a(ij) = a(ji)$. Consider the function $f(x) = (1/2)(Ax)\cdot x$. What are the maxima and minima for $f$ on $B = {(a, b, c) | a^2 + b^2 +c^2 \leq 1}$? I have learned to find the gradient of both of these and use lagrange multipliers.

I find out that $\nabla f=A \cdot x$, and I define $g(x)=\| x \|^2$ so $\nabla g= 2 x $ where x is a vector in $\mathbb{R^3}$.

Answer 1

Note the restriction is the whole unit sphere (and not just the surface) $$B=\{(x,y,z)|x^2+y^2+z^2\leq1\}$$

This means that first you have to search all critical points inside the sphere, and then use Lagrange multipliers to find critical points on the surface of the sphere. Remember that, when you have all critical points, you have to find wich of them give max/min values of $f$, using the second derivatie criteria (the Hessian matrix).

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For the first part, take all critical points that satisfy $\nabla f(\vec{x})=\vec{0}$ and $\|\vec{x}\|<1$, and then find which of them give max/min values.

$$\begin{array}{rcl} \nabla f(\vec{x})&=&\vec{0}\\ A\vec{x}&=&\vec{0}\\ \end{array}$$

This is equivalent to finding the null space (or Kernel) of $A$. From this point, I can't go further without knowing $A$.

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For the second part, use Lagrange multipliers on $$f(\vec{x})=\frac{1}{2}(A\vec{x})^T\vec{x}$$ $$g(\vec{x})=\|\vec{x}\|^2-1=0$$

Then $$\begin{array}{rcl} \nabla f(\vec{x})&=&\lambda\nabla g(\vec{x})\\ A\vec{x}&=&2\lambda\vec{x}\\ \end{array}$$ with $$\|\vec{x}\|=1$$

Using the substitution $\mu=2\lambda$, this expression is equivalent to finding the eigenvectors of $A$ (all the eigenvectors have unitary magnitude): $$A\vec{x}=\mu\vec{x}$$

Note that $A$ is real symmetric, so all its eigenvalues are real. But I can't go further without knowing $A$ (more than have a lot of symbolic expressions).

Answer 2

$$ f_{x_i} = \frac{1}{2} ( Ae_i\cdot x + Ax\cdot e_i ) =Ae_i\cdot x \Rightarrow \nabla f = Ax $$

Step 1 : What is critical point ? $$ \ast\ x\in {\rm Ker}\ (A) \ or\ x=0$$

Let $X=\{\lambda_i\}$ be a set of eigenvalues of $A$.

At $\ast$, $f(x)=0$.

Step 2 : As you define $g$, we consider the equation $$g(x)=1,\ \nabla f = c \nabla g $$

So $(A-2cI)x=0$. That is if $2c\in X$, $$ f(x)= \frac{1}{2} \lambda_i $$

So we have $$ {\rm min}\ X\cup \{0\}\leq 2f(x) \leq {\rm max}\ X\cup \{ 0\} $$