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

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

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 $\text N 1$: Positive Definiteness

\(\ds \norm M_\infty\) \(=\) \(\ds \max_{\begin {split} 
                 1 \mathop \le i \mathop \le m \\
                 1 \mathop \le j \mathop \le n 
                 \end {split} } 
           \size {a_{i j} }\)
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:

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

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

$\Box$


Norm Axiom $\text N 2$: Positive Homogeneity

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

$\Box$


Norm Axiom $\text N 3$: Triangle Inequality

Let $P, Q \in \R^{m \times n}$.

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

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

We have that:

\(\ds \size {p_{i j} + q_{i j} }\) \(\le\) \(\ds \size {p_{i j} } + \size {q_{i j} }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \max_{\begin {split} 
                 & 1 \mathop \le i \mathop \le m\\
                 & 1 \mathop \le j \mathop \le n
                 \end {split} } \size {p_{i j} } + 
           \max_{\begin {split}
                 & 1 \mathop \le i \mathop \le m\\
                 & 1 \mathop \le j \mathop \le n
                 \end {split} } \size {q_{i j} }\)
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:

\(\ds \norm {P + Q}_\infty\) \(=\) \(\ds \max_{\begin {split} 
                 & 1 \mathop \le i \mathop \le m\\
                 & 1 \mathop \le j \mathop \le n
                 \end {split} } \size {p_{i j} + q_{i j} }\)
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.

$\blacksquare$


Sources

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