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:
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.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces