1. First, we need to define the 1- norm of the matrix. For matrix A with n rows and m columns, its 1 norm is defined as the maximum value of the sum of the absolute values of each element of all column vectors, that is:
║A║ 1 = max{ ∑|aij| }，j= 1，2，...，m
2. Next, we need to prove that the above formula is equal to Max {∑| AI 1 |, ∑| AI2 |, ..., ∑| ain |}. For each column vector Ai, we can expand it into an n-dimensional column vector a=[a 1, a2, ..., an]T, where ai represents the i-th element of the vector Ai.
3. Because the column vectors of a matrix have nothing to do with column linearity, we can express each column vector as the sum of other column vectors through linear combination. For example, for the first column vector A 1, we have:
A1= a1e1+a2e2+...+A Ning,
Where e 1, e2, ..., en represent the unit vector of the nth dimension.
4. According to the absolute inequality, we can get:
|ai 1| + |ai2| +...+ |ain| ≤ |a 1| + |a2| +...+ |an|
5. From this, we can get:
∑| AIJ | = max {∑| AIJ | }≤∑| ai 1 |+∑| ai2 |+...+ ∑|ain|，
j= 1，2，...，m
6. On the other hand, for each column vector Ai, we also have:
∑|aij| ≤ ║A║ 1，
j= 1，2，...，m
That is, the 1- norm of a matrix is the upper bound of the sum of the absolute values of all column vectors.
7. Therefore, we draw the following conclusions:
max{ ∑|ai 1|，∑|ai2|，……，∑| ain | }≤║a║ 1≤∑| ai 1 |+∑| ai2 |+...+ ∑|ain|，
j= 1，2，...，m
8. According to the above conclusion, we can prove that the formula of 1- norm of matrix is ║ a ║ 1 = max {∑| AI 1 |, ∑| AI2 |, ..., ∑| ain |}.