Define the matrix norm subordinate to or induced by vector norms as:
$$
\|A\|_{p}:=\max_{x\neq0}\frac{\|Ax\|_{p}}{\|x\|_{p}}=\max_{\|x\|_{p}=1}\|Ax\|_{p}
$$
I am on a quest to show that $\|A\|_{1}$ is the maximum sum obtained by summing elements of a specific column $j$ in an $m\times n$ matrix $A$.
Proof. By definition, we have:
$$
\|A\|_{1}:=\max_{x\neq0}\frac{\|Ax\|_{1}}{\|x\|_{1}}=\max_{\|x\|_{1}=1}\|Ax\|_{1}
$$
Then we have that:
\begin{align*}
\|Ax\|_{1}&=\sum_{i=1}^{m}|A(i,:)*x|\\
&=\sum_{i=1}^{m}\left|\sum_{j=1}^{n}a_{i,j}x_{j}\right|\\
&\leqslant\sum_{i=1}^{m}\sum_{j=1}^{n}|a_{i,j}||x_{j}|\\
&=\sum_{j=1}^{n}|x_{j}|\sum_{i=1}^{m}|a_{i,j}|\\
&\leqslant\sum_{j=1}^{n}|x_{j}|\max_{j}\sum_{i=1}^{m}|a_{i,j}|\\
&=\|x\|_{1}\max_{j}\sum_{i=1}^{m}|a_{i,j}|
\end{align*}
this implies:
$$
\frac{\|Ax\|_{1}}{\|x\|_{1}}\leqslant\max_{j}\sum_{i=1}^{m}|a_{i,j}|
$$
I am stuggling with the equality part. Here is how I tackled it.
There exists $x\in\mathbb{R}^{n}$ such that $\|x\|_{1}=1$. For this purpose, consider $x:=e_{j_{0}}$ such that:
$$
\|Ax\|_{1}=\max_{j}\sum_{i=1}^{m}|a_{i,j}|
$$
Then one may observe that:
$$
\max_{j}\sum_{i=1}^{m}|a_{i,j}|=\sum_{i=1}^{m}|a_{i,j_{0}}|
$$
and
$$
Ax=Ae_{j_{0}}=A(:,j_{0})
$$
$$
\blacksquare
$$
Is my assertion for the equality part correct?
Best Answer
You are very close, but I think your reasoning may be improved.
So you want to find some $x\in \mathbb{R}^n$ with $\|x\|_1 = 1$ such that $\displaystyle{\|Ax\|_1= \max_{j} \sum_i |a_{ij}|}$. Let $j_0 \in \{1,2, \ldots, n \}$ such that $$\max_{j} \sum_i |a_{ij}| = \sum_i |a_{ij_0}|$$ Take $x = e_{j_0}$. Then $\|x\|_1 = 1$ and $\displaystyle{\|Ax\|_1} = \|A_{j_0} \|_1 = \sum_i |a_{ij_0}|$, where $A_{j_0}$ is the $j$-th column of $A$.