Quadratic Form - find a minimal scalar $m \in \Bbb R$ such that $q(x,y,z) \le m(x^2+y^2+z^2)$

Question

Let $q (x,y,z)$ be a quadratic form, $$q(x,y,z)=2zx+4yz-2xy $$ $$V=\Bbb R^3$$

Find a minimal scalar $m \in \Bbb R$ such that $$q(x,y,z) \le m(x^2+y^2+z^2)$$ for all $x,y,z \in \Bbb R$.

Actually I don't where to start. How do I find such a scalar?

Answer 1

to start, find the Gram matrix $G$ for the quadratic form, which is $1/2$ the Hessian matrix of second partial derivatives of $q.$ The result is that, when we have a column vector $$ v = \left( \begin{array}{c} x \\ y \\ z \end{array} \right) $$ with row vector $v^T,$ then we get $$ q(v) = v^T G v $$

extreme behavior comparison with the identity form, occurs at eigenvectors of $G.$

So, maybe you know partial derivatives or maybe you don't, find symmetric matrix $G$ such that $ q(v) = v^T G v $

If you have not had partials, write $$ G = \left( \begin{array}{ccc} a & f & e \\ f & b & d \\ e & d & c \end{array} \right) $$ then write out $v^T G v $ in symbols, and adjust $a,b,c,d,e,f$ to make it all fit your $q(x,y,z)=2zx+4yz-2xy $

EDIT: I once taught linear algebra out of Elementary Linear Algebra with Applications by Anton and Rorres (1987). Theorem 1 on page 345 is proved on pages 348-349. I will give the version for a three by three real symmetric matrix $G:$ Let $$ \lambda_1 \leq \lambda_2 \leq \lambda_3 $$ be the eigenvalues of $G.$ Then $$ \color{blue}{\lambda_1 (x^2 + y^2 + z^2) \leq \; \; v^T G v \; \; \leq \lambda_3 (x^2 + y^2 + z^2)} $$ Equality, on either side, can occur only at the relevant eigenvector. Their proof does not use calculus (Lagrange multipliers), instead it is essentially that real quadratic forms (real symmetric matrices) can be orthogonally diagonalized.

For our exercise, $\lambda_1 < 0$ while $\lambda_3 > 0.$

Hmmm. I think I taught out of that book; I have a copy. It was a long time ago, though, not sure.