Supremum Norm on Vector Space of Real Matrices is Norm

Theorem
Supremum Norm forms a norm on the vector space of real matrices.

Proof
Let $M \in \R^{m \times n} : m, n \in \N_{\mathop > 0}$ be a real matrix.

Denote the $\paren {i, j}$-th entry of $M$ by $a_{ij}$.

Note that the set of matrix elements of $M$ is a finite set of real numbers.

We have that:


 * Real Numbers form Ordered Field


 * Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements

Therefore, $M$ has the greatest element.

Norm Axiom $(\text N 1)$
Equality is obtained for $M$ being a zero matrix.

Suppose $\norm M_\infty = 0$.

Then:


 * $\displaystyle \forall i, j : 1 \le i \le m, 1 \le j \le n : \size {a_{ij}} = 0$

In other words, $M$ is a zero matrix.

Norm Axiom $(\text N 3)$
Let $P, Q \in \R^{m \times n}$.

Denote their $\paren {i, j}$-th matrix elements as $p_{ij}$ and $q_{ij}$ respectively.

Fix $i,j \in \N : 1 \le i \le m, 1 \le j \le n$.

We have that:

This holds for any $i,j$.

Hence:

All norm axioms are seen to be satisfied.

Hence the result.