> [!NOTE] > Let $L(\mathbb{R}^n,\mathbb{R}^k)$ denote the set of [[Linear maps|linear maps]] from $\mathbb{R}^n$ to $\mathbb{R}^k$. The operator norm of $A=(a_{ij})\in L(\mathbb{R}^n, \mathbb{R}^k)$, denoted $\lVert \lvert A \rvert\rVert$ or $\lVert A \rVert_{op}$, is defined by $\lVert A \rVert_{op} := \text{sup}_{x \neq 0} \frac{\lVert Ax \rVert}{\lVert x \rVert } $or equivalently, $\lVert A \rVert_{op} := \sup_{\lVert x \rVert =1 } \lVert Ax \rVert $ Remark: since we're taking supremums, we have that for all $x\in \mathbb{R}^n$, $\lVert Ax \rVert \leq \lVert A \rVert_{op}\lVert x \rVert$. The [[Operator Norm]] is indeed a [[Norms]]. ###### Proof Proof. The first two items are elementary and the proofs are left to the reader. For the third item, $\|(A+B) x\|=\|A x+B x\| \leqslant\|A x\|+\|B x\| \leqslant\left(\|A\|_{o p}+\|B\|_{o p}\right)\|x\|$and therefore $\|A+B\|_{o p}=\sup _{x \in \mathbb{R}^n \backslash\{0\}} \frac{\|(A+B) x\|}{\|x\|} \leqslant\|A\|_{o p}+\|B\|_{o p} .$