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?
[Math] Why is the norm $1$ of matrix $A$ is equal to the maximum sum of column
functional-analysislinear algebranormed-spacesnumerical linear algebra
Related Solutions
Since it looks like you are interested in the matrix norm which is induced by a particular vector norm (take a look here) , there is a nice geometric interpretation for $\|A\|$:
$$\|A\|:=\max_{\|x\|=1}\|Ax\|,$$ that is, if you start with any unit vector, compute its image under the transformation $A$, and compute the norm of that image (which is a vector, so its computed via a vector norm), then $\|A\|$ is the largest such resulting value. In other words, $\|A\|$ is the maximum "stretch" that $A$ "does" to a unit vector.
As an example, suppose $A=\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$, so $A:\mathbb{R^2}\to\mathbb{R}^2$, and we will consider $\mathbb{R}^2$ with the 2-norm. Then the matrix norm induced by the (vector) 2-norm described above is summarized graphically with this figure:
Note the unit vectors on the left and then some representative images under $A$. The length of the longest such image is $\|A\|$ (induced by this vector norm).
In $\mathbb{R}^n$, the closed unit ball $\overline{B}$ is compact. The function $f(x) = \|Ax\|$ is continuous, and continuous functions attain their maximum on a compact set, ie, there exists $x_0 \in \overline{B}$ such that $f(x_0) = \max_{x \in \overline{B}} f(x)$. In particular, $\|A x_0 \| = \max_{x \in \overline{B}} \|Ax\|$.
The induced norm is defined as $\|A\| = \max_{x \in \overline{B}} \|A x\|$, and the above remark shows that this is attained for some $x_0$.
Best Answer
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.