Maximum/Minimum of a quadratic form on the unit sphere

Question

Prove that, the maximum/minimum of the quadratic form $$ f(x, y, z)=A x^{2}+B y^{2}+C z^{2}+2 D y z+2 E z x+2 F x y $$ on the unit sphere surface $$ x^{2}+y^{2}+z^{2}=1 $$ is exactly the corresponding max/min eigenvalue of the matrix $$ \Phi=\left(\begin{array}{lll} A & F & E \\ F & B & D \\ E & D & C \end{array}\right) $$

I don't even know where to start. Can anyone help?

Answer 1

The key fact is that every real symmetric matrix is orthogonally diagonalisable, i.e. we can express $\Phi$ as $Q\Lambda Q^T$, where $Q$ is an orthogonal matrix and $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$ is a diagonal matrix whose diagonal entries the eigenvalues of $\Phi$. The proof of this fact can be found in most introductory texts on linear algebra.

Since $Q$ is orthogonal, $(x,y,z)\mapsto(u,v,w):=(x,y,z)Q$ is a bijective mapping on the unit sphere. Hence $$ \max_{x^2+y^2+z^2=1}f(x,y,z) =\max_{x^2+y^2+z^2=1}\pmatrix{x&y&z}\Phi\pmatrix{x\\ y\\ z} =\max_{u^2+v^2+w^2=1}\pmatrix{u&v&w}\Lambda\pmatrix{u\\ v\\ w} $$ and the rest should be straightforward.