## Properties of a matrix norm

Given an $n \times n$ matrix $A$ over $\mathbb{C}$, define $||A|| = \sum_{i,j} |a_{i,j}|$, where bars denote the complex modulus. Prove the following for all $A,B \in \mathsf{Mat}_n(\mathbb{C})$ and all $\alpha \in \mathbb{C}$.

1. $||A+B|| \leq ||A|| + ||B||$
2. $||AB|| \leq ||A|| \cdot ||B||$
3. $||\alpha A|| = |\alpha| \cdot ||A||$

Say $A = [a_{i,j}]$ and $B = [b_{i,j}]$.

We have $||A+B|| = \sum_{i,j} |a_{i,j} + b_{i,j}|$ $\leq \sum_{i,j} |a_{i,j}| + |b_{i,j}|$ by the triangle inequality. Rearranging, we have $||A+B|| \leq (\sum_{i,j} |a_{i,j}| + \sum_{i,j} |b_{i,j}|$ $= ||A|| + ||B||$ as desired.

Now $||AB|| = \sum_{i,j} |\sum_k a_{i,k}b_{k,j}|$ $\leq$latex \sum_{i,j} \sum_k |a_{i,k}|b_{k,j}|\$ by the triangle inequality. Now $||AB|| \leq \sum_{i,j} \sum_{k,t} |a_{i,k}| |b_{t,j}|$ since all the new terms are positive, and rearranging, we have $||AB|| \leq \sum_{i,k} \sum_{j,t} |a_{i,k}||b_{t,j}|$ $= (\sum_{i,k} |a_{i,k}|)(\sum_{j,t} |b_{t,j}|)$ $= ||A|| \cdot ||B||$.

Finally, we have $||\alpha A|| = \sum_{i,j} |\alpha a_{i,j}|$ $= |\alpha| \sum_{i,j} |a_{i,j}|$ $= |\alpha| \cdot ||A||$.