Show that for every $A\in M_{n}(\mathbb{R})$ with rank(A)=1 that $||A||_{op}=||A||_{HS}$. What happens in the general case?

Question

Show that for every $A\in M_{n}(\mathbb{R})$ with rank(A)=1 that $||A||_{op}=||A||_{HS}$. What happens in the general case?

Just to clarify the notations: $||A||_{op}= \sup\left\{\frac{\|Av\|}{\|v\|} : v\in V \mbox{ with > }v\ne 0\right\}.$ $\|A\|^2_{HS}={\rm Tr} (A^{T}A) := \sum_{i \in I} \|Ae_i\|^2$

So using this inequality that I proved before: $\|AB\|_{HS} \leq \|A\|_{\mathrm{op}} \|B\|_{HS} $ If I Substitute B for I I get the weak inequality, but I can't see how rank=1 helps me here..

Any help?