How to compute $||e_m||_1$ and $||Ae_m||_1$ and use it to prove

Question

1. Let m be the index of the column of A with maximum 1 norm. That is, ||a_m||₁ >= ||a_i||₁ for 1 <= i <= n. Let e_m be the standard basis vector that has e_m(m) = 1 and e_m(i) = 0 for i != m. Compute ||e_m||₁ and ||Ae_m||₁.

2. Show that ||A||₁ = max_1<=I<=n||a_i/sub>||1.

For:

1. I know that ||Ax|| <= ||A|| * ||x|| as well as the definitions for the 1-norm of a vector and the induced matrix norm but I don't know how they are useful in solving the question.

2. I know that I need to show that both the >= and <= inequalities hold and I need to use ||Ax||₁ <= max_1<=I<=n||a_i||1.

Please help. Thank you.

Answer 1

Your thoughts for part 1 are not relevant to the question. The point of the question is to observe that $Ae_m = a_m$, which means that $\|Ae_m\| = \|a_m\|$.

This question is helpful for part $2$ because it gives us a lower-bound for $\|A\|_1$. Indeed, $$ \|A\|_1 = \max_{\|x\| = 1}\|Ax\|_1 \leq \|Ae_n\| = \max_{1 \leq i \leq m} \|a_i\|. $$ So, in order to answer question to, we need only show that $\|A\|_1 \leq \max_{1 \leq i \leq m}\|a_i\|$. Equivalently, we need to show that for a vector $x$ satisfying $\|x\|_1=1$, we have $\|Ax\| \leq \max_{1 \leq i \leq m}$.

As a hint towards how to obtain this inequality, begin by using the triangle inequality to note that $$ \left\|Ax \right\|_1 = \left\|\sum_{i=1}^n x_i a_i \right\|_1 \leq \sum_{i=1}^n |x_i|\cdot \|a_i\|_1. $$