I have to prove that matrix norm $||A||_\infty$ induced by vector norm $||x||_\infty = \smash{\displaystyle\max_{1 \leq k \leq n}} |x_k|$ where $x_k$ is k-th element of vector can be disribed by equation $||A||_\infty = \smash{\displaystyle\max_{1 \leq i \leq n}} \sum\limits_{j=1}^{n}|a_{i,j}|$.
I can use fact that induced norm $||A||_\alpha = \sup \limits_{x\neq 0} \frac{||Ax||_\alpha}{||x||_\alpha}$.
Oblvius conclusion is that $||A||_\infty = \smash{\displaystyle\max_{1 \leq i \leq n}} |\sum\limits_{j=1}^{n}a_{i,j}x_j| : \smash{\displaystyle\max_{1 \leq i \leq n}}|x_i|$ and i cannot move past that point. I think i am missing some knowledge about x. Can anyone give me a hint how to move past that point?
Best Answer
Notice that since $\|x\|_\infty>0$ you can push the denominator into the max, then into the absolute value, and then apply the distributive property to the sum, i.e. $$\frac{\max_{i}\left|\sum_j a_{ij} x_j\right|}{\|x\|_\infty}=\max_{i}\left|\sum_j a_{ij}\frac{x_j}{\|x\|_\infty}\right|$$
You can then upper bound this value due to the fact that $x_j \le \|x\|_\infty \Rightarrow\frac{x_j}{\|x\|_\infty}\le1$, hence: $$\max_{i}\left|\sum_j a_{ij}\frac{x_j}{\|x\|_\infty}\right|\le \max_{i}\left|\sum_j a_{ij}\right|$$
Now, you only need to to find an $x$ such that this upper bound is achieved. I leave that to you (it's quite straightforward, really).