1

For any matrix $A \in \mathbb{R}^{m,n}$ and any $u \in \mathbb{R}^{m}$ and $v \in \mathbb{R}^n$, how to proof the following inequality?

$$ |u^\top A v | \leq \|A\|_{op} \|u\|_1 \|v\|_1 $$

In general, where can I find reference for matrix norm inequalities? Thanks!

cccfran
  • 23

1 Answers1

1

By a Cauchy-Schwarz inequality we have that $$|(A^tu)\cdot v| \le \| A^tu \|_2 \|v \|_2.$$ Now the operator norm is defined as $$ \|A^t\|_{op} := \sup_{x \ne 0 } \frac{\|A^tx\|_2}{\|x\|_2},$$ therefore $\|A^tu\|_2 \le \|A^t\|_{op}\|u\|_2$. Combining this with the inequality $\|x\|_2 \le \|x\|_1$ - which is $\sum_{i=1}^n x_i^2 \le \left(\sum_{i=1}^n |x_i|\right)^2$ - we have, $$|u^t A v| = |(A^tu)\cdot v| \le \| A^t \|_{op} \| u \|_1 \|v \|_1$$ and the result follows from $\|A^t\|_{op} = \|A\|_{op}$.