How can I find the maximum and minimum value of the following quadratic form
$$Q(x) = x_1^2+3x_2^2+10x_1x_3+25x_3^2$$
subject to the equality constraint $\|x\|_2 = 3$? The norm is the Euclidian one.
Normally if I was given the constraint $\|x\|_2 = 1$ I would find the matrix A:
$$
A = \begin{bmatrix}
1 & 0 & 5\\
0 & 3 & 0\\
5 & 0 & 25
\end{bmatrix}
$$
Where I would calculate the eigenvalues of $A$ , which are $\lambda_1 = 26, \lambda_2 = 3, \lambda_3 = 0.$ Then the maximum value would be 26, and and the minimum value would be 0. But how am I supposed to do this with a constraint such as $\|x\|_2 = 3$ ?
Any guidance will be much appreciated.
Best Answer
This problem is homogeneous then calling $x_2 = \lambda x_1, x_3 = \mu x_1$ we arrive at
$$ \min(\max) \left(x_1^2\left(1+2\lambda^2+10\mu+25\mu^2\right)\right), \ \ \ \text{s. t.}\ \ \ x_1^2(1+\lambda^2+\mu^2)=k^2 $$
so we can follow without the constraint as
$$ \min(\max)_{\lambda,\mu} \frac{\left(1+2\lambda^2+10\mu+25\mu^2\right)k^2}{1+\lambda^2+\mu^2} $$
and deriving we find the conditions
$$ \cases{ 2 \lambda (10 \mu + 23 \mu^2-1) k^2=0\\ 2 ( 5 \mu^2-5 - 5 \lambda^2 - 24 \mu - 23 \lambda^2 \mu) k^2=0 } $$
giving the real solutions $(\lambda=0,\mu = -\frac 15)$ and $(\lambda=0,\mu = 5)$ with respective minimum and maximum $(0, 26k^2)$