If $A$ is a real $n \times n$ matrix and $x \in \mathbb{R}^n$ then is it true that $|Ax| \le |A||x|$?
Here $|x|$ is the usual euclidean norm for vectors and $|A| = \sqrt{\Sigma_{i,j} a_{ij}^2}$
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$\begingroup$Yes it's true. Indeed, let $y=Ax$ then $y_i=\sum_{j=1}^n a_{i,j}x_j$ and so
$$\vert Ax\vert^2=\vert y\vert^2=\sum_{i=1}^ny_i^2=\sum_{i=1}^n\left(\sum_{j=1}^n a_{i,j}x_j\right)^2\le \sum_{i=1}^n \left(\sum_{j=1}^n a^2_{i,j}\sum_{j=1}^n x^2_j\right)=\vert A\vert^2\vert x\vert^2$$ where the inequality in the middle is the Cauchy-Schwarz inequality
$\endgroup$ 1 $\begingroup$For matrix norms, you have the so-called induced norm, which is defined by
$$\|A\|=\min_{x \in \mathbb R^n, x \neq 0} \frac{|Ax|}{|x|}.$$
Here, $| \cdot |$ can denote any norm on domain/codomain (and you could even use different norms on domain and codomain).
For these, you have by defintion $|Ax| \leq \|A\| |x|$.
In your particular case, where the vector norm is the euclidean norm in both cases, the induced matrix norm is called the Frobenius norm is given by the formula $$\|A\|= \sqrt{\sum_{i=1, j=1}^n|a_{ij}|^2}.$$ In this formula, the $|\cdot|$ denotes the modulus of a complex value. Thus, your statement is correct for real matrices and needs to be amended for complex matrices.
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