For $x\in \mathbb{R}^{n}$ and $x\neq 0$. Show that $sup\frac{\left \| Ax \right \|_{1}}{\left \| x \right \|_{1}}=\left \| A \right \|_{C}$, where $\left \| A \right \|_{C}$ is the maximum absolute column sum of the matrix.
I know that there is a previous answer Why is the matrix norm $||A||_1$ maximum absolute column sum of the matrix. , but I can't really understand it because I'm new in Numerical Aalysis.Any proof or advise would be really helpful.
Denote by $a_j$ the $j$-th column of $A$ and by $A_j$ its sum in absolute value. Then for $x\in\mathbb R^n$ with $\|x\|_1 = 1$ we have \begin{align*} \|Ax\|_1 &= \left\|\sum_jx_ja_j\right\|_1 = \sum_i\left|\left\langle\sum_jx_ja_j,e_i\right\rangle\right| = \sum_i\left|\sum_j x_ja_{ij}\right|\\ &\le \sum_i\sum_j |x_j||a_{ij}| = \sum_j |x_j|\sum_i|a_{ij}| = \sum_j |x_j|A_j\\ &\le \max_jA_j\sum_j|x_j| = \max_jA_j. \end{align*} Now, also $\|Ae_i\|_1 = \|a_i\|_1 = A_i$. This proves the claim.
Let $e_{j}$ be the $j^{th}$ standard basis vector (i.e. 1 in position $j$, and 0 elsewhere). First note that $Ae_{j}=c_{j},$ where $c_{j}$ is the $j^{th}$ column of $A$. Consequently, $\frac{\|Ae_{j}\|_{1}}{\|e_{j}\|_{1}}=\|c_{j}\|_{1}$ . Also, note that $\frac{\|Ax\|_{1}}{\|x\|_{1}}$ is scale invariant, and therefore it suffices to show this for $x$ such that $\|x\|_1=1$.
Let $j^{*}$ be the index of the column of $A$ with maximum 1 norm. Now, any vector $x\in \mathbb{R}^n$ with \|x\|_1 =1$ can be written as
$$x = \sum_{i=1} ^{n} \alpha_i e_i$$ where $$\sum_{i=1} ^{n} |\alpha_i| = 1$$
Note that
$$Ax = \sum_{j=1 } ^n \alpha_j A e_j = \sum _{j=1} ^n \alpha_j c_j $$
Also note that the one norm of $Ax$ is
$$\|Ax\|_1 \le \sum _{j=1} ^n \|\alpha_j c_j\|_1 = \sum _{j=1} ^n |\alpha_j| \cdot \|c_j\|_1 \leq \|c_{j^{*}}\|_1$$
The last inequality follows from the fact that the $\alpha$'s are a "convex combination" which yields the desired result.
