Why is the norm $1$ of matrix $A$ is equal to the maximum sum of column first, I know that there exists a similar question to mine which is in here, and it is actually very well explained. However, there is just one part that I do not understand. That is the conclusion. First, how did they conclude that norm of Matrix $A$ is less than the maximum column sum of matrix $A$? Then how did they conclude that it did not the inequality turned into equality? 
 A: The one-norm of $A=(a_{ij})$ is defined as the maximum of $\Vert Ax\Vert_{1}$ with respect to all vectors $x$ which are unit in the one-norm.
In other words,
$$
\Vert A\Vert_{1}=\max_{\Vert x\Vert_{1}=1}\Vert Ax\Vert_{1}.
$$
In the answer you link to, the author first shows that for any  vector $x$ satisfying $\Vert x \Vert_1 = 1$,
$$
\Vert Ax\Vert_{1}\leq\max_{j}\sum_{i}|a_{ij}|=\sum_{i}|a_{im}|
$$
where $j=m$ is some index that maximizes the expression on the right hand side of the above inequality.
Next, the author notes that the standard basis vector $e_{m}$ (i.e., the vector whose $m$-th entry is 1 and whose other entries are zero) satisfies
$$
\Vert e_{m}\Vert_{1}=1\qquad\text{and}\qquad\Vert Ae_{m}\Vert_{1}=\sum_{i}|a_{im}|.
$$
Putting this all together,
$$
\sum_{i}|a_{im}|=\Vert Ae_{m}\Vert_{1}\leq\Vert A\Vert_{1}\leq\max_{j}\sum_{i}|a_{ij}|=\sum_{i}|a_{im}|.
$$
Since the leftmost and rightmost expressions are identical, it follows that all of the inequalities above must be equalities.
The desired result follows.
