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||.

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