Prove that Frobenius matrix norm is compatible with the vector norm

Question

Show that, the Frobenius matrix norm $||.||_F$ is compatible or consistent with a vector norm $||.||_2$ , that is, $||Ax||_2 \leq ||A||_F ||x||_2, \forall x \in \mathbb{R}^n$. Where $||A||_F = \sqrt{ \sum_{i,j=1}^N |a_{ij}|^2} $

Answer 1

$(Ax)_i=\sum_{j=1}^na_{i,j}x_j$, hence $$\|Ax\|_2^2=\sum_{i=1}^n(\sum_{j=1}^na_{i,j}x_j)^2$$ But $(\sum_{j=1}^na_{i,j}x_j)^2\leq (\sum_{j=1}^n{a_{i,j}^2})(\sum_{j=1}^nx_j^2)=(\sum_{j=1}^n{a_{i,j}^2})\|x\|_2^2$ We get $$\|Ax\|_2^2\leq \sum_{i=1}^n\sum_{j=1}^na_{i,j}^2\|x\|_2^2=\|A\|_F^2\|x\|_2^2$$ i.e $$\|Ax\|_2\leq \|A\|_F\|x\|_2 $$