Let $W \in \mathbb R^{n\times d}$ and $\mathbb M = \{\Theta \in \mathbb R^{n\times d} | \text{rank}(\Theta)=1, \|\Theta\|_F = 1\}$ where $\|\cdot\|_F$ is the frobenius norm. Show that the operator norm has the following variational representation:
$$\|W\|_2 = \sup_{\Theta \in \mathbb M} \langle W, \Theta \rangle$$
My guess is that we can write the SVD of $W$ then apply properties of trace and the matrices of left and right singular vectors to show that the supremum is equivalent to pulling out the max singular value.
Some hints: If $\operatorname{rk} \Theta =1$ then it can be written as $\Theta = uv^T$. Show that $\|\Theta\|_F = \|u\|_2 \|v\|_2$. Note that $\langle W, uv^T \rangle_F = u^T Wv$.