Prove that Frobenius norm is consistent

Question

Prove that Frobenius is consistent. Meaning: $$\|Ax\|\le \|A\|\|x\|$$

My try:
$$\|Ax\|^2 = \| \sum_{j=1}^n x_j A_{*j} \|^2 \le \sum_{i,j} (x_j a_{ij})^2 = \sum_{j=1}^n x_j^2 \sum_{i=1}^n a_{ij}^2 =\sum_{j=1}^n x_j^2 \|A\|^2 = \|x\|^2\|A\|^2$$

Hence,

$$\|Ax\| \le \|A\|\|x\|$$

I'd like to know if that's correct.

Thanks!

Answer 1

Watch out, $$\|\sum_{j=1}^nA_{\ast j}x_j\|^2=\sum_{i=1}^n\left(\sum_{j=1}^n A_{ij}x_j\right)^2$$

Clearly, if you use Cauchy-Schwarz's inequality
$$\left(\sum_{j=1}^n A_{ij}x_j\right)^2\leq \left(\sum_{j=1}^n A_{ij}^2\right)\left(\sum_{j=1}^n x_j^2\right),$$ and put it all together, you'll have that $$\|Ax\|^2\leq\sum_{i=1}^n\left(\sum_{j=1}^n A_{ij}^2\right)\left(\sum_{j=1}^n x_j^2\right)=\left(\sum_{j=1}^n x_j^2\right)\left(\sum_{i=1}^n\sum_{j=1}^n A_{ij}^2\right)=\|x\|^2\|A\|^2.$$

Answer 2

No, it isn't correct. The inequality $\| \sum_{j=1}^n x_j A_{*j} \|^2 \le \sum_{i,j} (x_j a_{ij})^2$ isn't true. In case $A$ has size 2, you are essentially saying that $$ (ax+by)^2 + (cx+dy)^2 \le (ax)^2 + (by)^2 + (cx)^2 + (dy)^2. $$ Why is that true?

In addition, somewhere near the middle you have an expression $$\sum_{j=1}^n x_j^2 \sum_{i=1}^n a_{ij}^2.$$ What does that mean? Is it $\sum_{j=1}^n\left(x_j^2\sum_{i=1}^n a_{ij}^2\right)$ or $\left(\sum_{j=1}^nx_j^2\right)\left(\sum_{i=1}^n a_{ij}^2\right)$?

• If it's $\sum_{j=1}^n\left(x_j^2\sum_{i=1}^n a_{ij}^2\right)$, it isn't equal to $\sum_{j=1}^nx_j^2\|A\|^2$.
• If it's $\left(\sum_{j=1}^nx_j^2\right)\left(\sum_{i=1}^n a_{ij}^2\right)$, the latter summation doesn't make sense, as $j$ is unspecified.