Let $A$ be an $n \times n$ matrix. We want to calculate
$$\|A\|^2 = \max_{v \in \mathbb{R}^n, \, |v|^2 =1} |Av|^2$$
a) Show that the attempted problem has a solution.
b) If $f(v) = |Av|^2$ and $h(v) = |v|^2$, what are the gradients of $f$ and $h$?
c) The square root of the value of the optimisation $\|A\|$ is called the operator norm. What would be the operator norm if $A$ is symmetric?
I don't actually know where to start as I am not sure exactly what I am calculating. Any start off hints would be greatly appreciated.
a) The set $\mathbb{S}^n = :\{v \ | \ |v|^2 = 1\}$ is compact and matrix multiplication is continuous so the map $$f = h\circ T_A : \mathbb{S}^n \mapsto \mathbb{R}^n$$ $$v \mapsto |Av|^2$$ where $h(v) = |v|^2$ must attain a maximum.
b) This is just chain rule $$\nabla h = \nabla (v\cdot v) = 2 (\nabla v) v = 2 Iv = 2v $$ $$\nabla f = 2\nabla (Av)^T(Av) = 2A^TA v $$ see here
c) If $A$ is symmetric you will have a full set of orthogonal eigenvectors, so consider what happens to those in regards to their corresponding eigenvalues. You can show that $A$ and $A^2$ will have the same eigenvalues.