I want to prove that $\|uv^T\|_2 = \|uv^T\|_F = \|u\|_2\|v\|_2$ where $u$ and $v$ are $u\in\mathbb{R}^m$, $v\in\mathbb{R}^n$. The RHS equality can be easily verified by applying the definition of $\|\cdot\|_F$. The problem is that I'm not sure how to prove LHS equality by proving $\|uv^T\|_2= \|u\|_2\|v\|_2$.
The upperbound is clearly obtainable as follows:
$\|uv^T\|_2^2 \le \max_{\|x\|_2=1}\|uv^Tx\|_2^2 = \sum\limits_{i=1}^m \sum\limits_{j=1}^n(u_iv_jx_j)^2 \leq \max_{\|x\|_2=1}\sum\limits_{i=1}^mu_i^2 \sum\limits_{j=1}^nv_j^2x_j^2$.
Since $\|x\|_2 = 1$, we have $\max_{\|x\|_2=1}\sum\limits_{i=1}^mu_i^2 \sum\limits_{j=1}^nv_j^2x_j^2 \le \sum\limits_{i=1}^mu_i^2 \sum\limits_{j=1}^nv_j^2 = \|u\|_2^2\|v\|_2^2.$
It only remains to show that we can construct such $x\in\mathbb{R}^n$, where $\|x\|_2 = 1$ for this to be an equality, but it doesn't strick me as easy.
The definition of the operator resp. matrix norm yields: $$\|uv^T\|_2 = \sup_{x \not=0} \frac{\|uv^Tx\|_2}{\|x\|_2} = \sup_{\|x\|_2=1} \|uv^Tx\|_2 \underbrace{\geq}_{x = \frac{v}{\|v\|_2}} \|uv^T\frac{v}{\|v\|_2}\|_2 = \|v\|_2\|u\|_2$$
And for the other bound I think using again the definition above and Cauchy-Schwarz would be much shorter, but your way seems to be fine as well ;)