Prove that the induced matrix norm $||A||_{\infty}$ is equal to its maximum absolute row sum.
This is my first time working through a proof in numerical analysis. Is my workings correct? If not, how do I solve this proof? Any help is greatly appreciated.
Suppose the maximum row is obtained from row $k$ of the matrix $A$. Then choose vector $x$ defined by $x_j = 1$ if $a_{kj} \geq 0$ and $x_j = -1$ if $a_{kj} < 0.$ Then $||x||_{\infty} = 1$ and
$$||A||_{\infty} \geq ||Ax||_{\infty} \geq \left|\sum_{j=1}^n a_{kj}x_j \right|= \sum ^{n}_{j=1} |a_{kj}| = R$$.
Hence here we have a specific vector of length $1$ for which the vector norm $A$ dominates the maximum row sum. Hence,
$$||A||_{\infty} = R $$
Edit:
Let $||x||_{\infty} = 1$ then by the definition $|x_i| \leq 1 $ and $|x_k| =1$ for some $k$. Then
$$||Ax||_{\infty} = max _{i}\left|\sum^n_{j=1}a_{ij}x_j \right| \leq max_{i}\sum^{n}_{j=1} |a_{ij}| = R$$
Hence the max row sum is always greater than or equal to the infinity vector norm of $A$
Best Answer
Based on another answer you should have $$ \| A\|_{\infty} = \max_{\| x\|_{\infty} =1} \| Ax\| = \max_{\| x\|_{\infty} =1} \max_{1 \leq i \leq n}| \bigg\| \sum_{j=1}^{n} a_{ij} x_{i} \bigg\| \tag{1} $$
$$ \max_{\| x\|_{\infty} =1} \max_{1 \leq i \leq n}| \bigg\| \sum_{j=1}^{n} a_{ij} x_{i} \bigg\| = \max_{1 \leq i \leq n} \max_{ \| x\|_{\infty} = 1} \bigg\| \sum_{j=1}^{n} a_{ij} x_{j}\bigg\|\tag{2} $$
$$ \max_{\| x\|_{\infty} =1} \max_{1 \leq i \leq n}| \bigg\| \sum_{j=1}^{n} a_{ij} x_{i} \bigg\| = \max_{1 \leq i \leq n} \sum_{j=1}^{n} |a_{ij}|\tag{3} $$
$$ \| x\|_{1} = \sum_{i=1}^{n} |x_{i}| \tag{4} $$
$$ \|A\|_{\infty} = \max_{1 \leq i \leq n} \| a_{i} \|_{1} \tag{5} $$