Supremum Norm on Vector Space of Real Matrices is Norm

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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 $(\text N 1)$

\(\ds \norm M_\infty\) \(=\) \(\ds \max_{\begin {split} 1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n \end {split} } \size {a_{ij} }\) Greatest Element is Supremum, Definition of Max Operation
\(\ds \) \(\ge\) \(\ds 0\)

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.

$\Box$


Norm Axiom $(\text N 2)$

\(\ds \norm {\alpha \cdot M}_\infty\) \(=\) \(\ds \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size {\alpha m_{ij} }\) Greatest Element is Supremum
\(\ds \) \(=\) \(\ds \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size \alpha \size {a_{ij} }\) Absolute Value of Product
\(\ds \) \(=\) \(\ds \size \alpha \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size {a_{ij} }\)
\(\ds \) \(=\) \(\ds \size \alpha \norm M_\infty\) Greatest Element is Supremum

$\Box$


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:

\(\ds \size {p_{ij} + q_{ij} }\) \(\le\) \(\ds \size {p_{ij} } + \size {q_{ij} }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size {p_{ij} } + \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size {q_{ij} }\) Definition of Max Operation
\(\ds \) \(=\) \(\ds \norm P_\infty + \norm Q_\infty\) Greatest Element is Supremum, Definition of Supremum Norm

This holds for any $i,j$.

Hence:

\(\ds \norm {P + Q}_\infty\) \(=\) \(\ds \max_{\begin {split} & 1 \mathop \le i \mathop \le m\\ & 1 \mathop \le j \mathop \le n \end {split} } \size {p_{ij} + q_{ij} }\) Greatest Element is Supremum
\(\ds \) \(\le\) \(\ds \norm P_\infty + \norm Q_\infty\)

$\Box$


All norm axioms are seen to be satisfied.

Hence the result.

$\blacksquare$


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