Let $A$ be an $m\times n$ matrix $(\alpha_{jk};\;j=1,...m,k=1,...,n).$ As we know, $$[Bx]_j = \sum_{k=1}^n\alpha_{jk}x_k,\;\;\;\;\;j=1,...,m,\;\;\;x=(x_1,...,x_n),$$ defines a bounded linear operator $B$ from $\mathbb R^n$ to $\mathbb R^m.$
Show: if both spaces are equipped with the sum-norm, then $$||B|| = \max_{k=1}^n\sum_{j=1}^m|\alpha_{jk}|.$$ Hint: for $\geq$ consider $Be^j$ where $e^j$ is the $j^{th}$ canonical basis vector with all components being zero except the $j^{th}$ component which is one.
$\textbf{MY ATTEMPT:}\\$
we have that $||B|| = \sup_{x\in X}||Bx||,$ for $||x||=1.$ Then, $$||x|| = \sum_{k=1}^n|x_k|.$$ $$||B|| = ||Bx||_\infty = \sum_{j=1}^m|[Bx]_j| = \sum_{j=1}^m\left|\sum_{k=1}^n\alpha_{jk}x_k\right|\leq\sum_{j=1}^m\sum_{k=1}^n |\alpha_{jk}x_k|\leq $$ $$(...)\leq \left(\max_{k=1}^n\sum_{j=1}^m|\alpha_{jk}|\right)\left(\sum_{k=1}^n|x_k|\right) = \max_{k=1}^n\sum_{j=1}^m|\alpha_{jk}|.$$ So $||B|| \leq \max_{k=1}^n\sum_{k=1}^m|\alpha_{jk}|.$
I am having trouble proving that $\max_{k=1}^n\sum_{k=1}^m|\alpha_{jk}|\leq ||B||.$
You need to use the hint.
Let $1\leq k\leq n$, and let $e_k$ be the $k$th element in the standard basis, i.e. all entries of $e_k$ are $0$, except for the $k$th, which is equal to $1$. Clearly, $\|e_k\|=1$, when using the sum norm. Note that $$Be_k=\left(\begin{array}{c}\alpha_{1k}\\\vdots\\\alpha_{mk}\end{array}\right),$$thus $$\|Be_k\|=\sum_{j=1}^m|\alpha_{jk}|$$and $$\|B\|\geq\sum_{j=1}^m|\alpha_{jk}|.$$