How can you prove that: $$\|A\|_2 \le \|A\|_F$$ I cannot use: $$\|A\|_2^2 = \lambda_{max}(A^TA)$$ It makes sense that the 2-norm would be less than or equal to the Frobenius norm but I don't know how to prove it. I do know:
$$\|A\|_2 = \max_{\|x\|_2 = 1} {\|Ax\|_2}$$
and I know I can define the Frobenius norm to be:
$$\|A\|_F^2 = \sum_{j=1}^n {\|Ae_j\|_2^2}$$ but I don't see how this could help. I don't know how else to compare the two norms though.