TITLE: Consistency of matrix norm — $\| A x \|_2 \leq \| A \|_{F} \| x \|_2$
QUESTION [5 upvotes]: I'm trying to show that
$$\| A x \|_2 \leq \| A \|_{F} \| x \|_2$$
where $A$ is an $n \times n$ matrix, $x \in \mathbb R^n$, $\| \cdot \|_2$ is the Euclidean norm, and $\| \cdot \|_F$ is the Frobenius norm.
I actually wrote down what $Ax$ is, it is
$$Ax =\begin{pmatrix} \sum_{i=1}^n a_{1i}x_i \\\sum_{i=1}^n a_{2i}x_i \\ \vdots \\ \sum_{i=1}^n a_{ni}x_i \end{pmatrix}$$
And so:
$$\|Ax\|_2 = \sqrt{\sum_{j=1}^n \left|\sum_{i=1}^n a_{ji}x_i \right|^2}$$
I need to show that that is smaller than
$$\sqrt{\sum_{j=1}^n \sum_{i=1}^n |a_{ij}|^2\sum_{k=1}^n |x_k|^2}$$
But this seems very complicated. I'd love a hint in the right direction.

REPLY [11 votes]: Since the matrix norm $\|\cdot\|_2$ is an operator norm, we have $\|Ax\|_2\leq\|A\|_2\|x\|_2$. The rest follows from the fact that $\|A\|_2\leq\|A\|_F$. This is true simply because 
$$
\|A\|_2^2=\rho(A^TA)=\lambda_{\max}(A^TA)\leq\sum_{i=1}^n\lambda_i(A^TA)\leq\mathrm{trace}(A^TA)=\|A\|_F^2.
$$