Induced norm question. I'd like to show that the induced 1-norm satisfies:
$\|A\|_1=\max_{1 \le j \le n}\sum_{i=1}^n |a_{i,j}|$
I'd appreciate some guidance.
 A: Use the definition of $||A||_1$, which is: $$||A||_1 = \max_{||x||_1=1}{||Ax||_1}$$
So say we have $x\in\mathbb R^n$ (assume $A$ is $n\times n$), with $||x||_1=1$, which means: $$(1) \quad \sum_{i=1}^{n}|x_i| = 1$$ 
Now we want to find the maximum of the set (of real numbers) $\{||Ax||\text{ such that } ||x||_1 =1\}$=$U$. For any number in this set,$$q\in U\Rightarrow q=||Ax||_1 = ||\sum_{j=1}^n x_j\vec{a_j}||_1$$
where $\vec{a_j}$ is the $j$-th column of $A$ (whose columns happen to be vectors in $\mathbb R^n$, of which the $1$-norm can be taken), so by the triangle inequality we have $$q \le\sum_{j=1}^n ||x_j\vec{a_j}||_1 = \sum_{j=1}^n |x_j|\ ||\vec{a_j}||_1$$
Because every $||\vec{a_j}||_1$ has a positive coefficient, the total sum is less than or equal to replacing every $||\vec{a_j}||_1$ with the maximum of them all, $\max_{j}{||\vec{a_j}||_1}$, which we'll call $M$. So we have $$q \le\sum_{j=1}^n |x_j|M = M\sum_{j=1}^n |x_j|\stackrel{\text{(1)}} = M\cdot1 = M = \max_{1\le j\le n}{||\vec{a_j}||_1} = \max_{1\le j\le n}{\sum_{i=1}^n |a_{i,j}|}, \text{ QED.}$$
So far we have proved that for any $q\in U$, $q\le\max_{1\le j\le n}\sum_{i=1}^n |a_{i,j}|$. For a set in $\mathbb R^n$, the maximum is defined as the real number that: 1) is greater than or equal to any number in the set, and 2) is in the set. We've proved that $\max_{1\le j\le n}\sum_{i=1}^n |a_{i,j}|$ satisfies 1). Now, note that $\sum_{i=1}^n |a_{i,j}|=||\sum_{i=1}^n Ae_j||_1$, where $e_j$ is the j-th standard basis vector in $\mathbb R^n$(all zeros and a $1$ in the $j$-th position). Since $||e_j||_1 = 0 +\dotsb+0+1+0+\dotsb+0 = 1$, by definition $\sum_{i=1}^n |a_{i,j}|$ is in $U$ and therefore satisfies 2), and therefore is the maximum of $U$.
