> [!NOTE] > Suppose that $A, B \in L\left(\mathbb{R}^n, \mathbb{R}^k\right)$ satisfy $\|A x\| \geqslant \alpha\|x\| \text { for some } \alpha>0$and $\|B\|_{o p}<\alpha$. Then $A+B$ is still injective. ###### Proof Applying [[Euclidean spaces are normed spaces|triangle inequality]] ($\lVert x \rVert\leq \lVert x+y \rVert +\lVert y \rVert\implies \lVert x+y \rVert\geq \lVert x \rVert-\lVert y \rVert$), $\|(A+B) x\| \geqslant\|A x\|-\|B x\| \geqslant \alpha\|x\|-\|B\|_{o p}\|x\|=\delta\|x\|,$ where $\delta:=\alpha-\|B\|_{o p}>0$. Therefore $(A+B) x=0 \Rightarrow x=0$, which proves that $A+B$ is injective. $\square$ **Remark**: This proposition can be interpreted as saying that if (5.13) holds then the open ball $\mathbb{B}(A, \alpha) \subset$ $L\left(\mathbb{R}^n, \mathbb{R}^k\right)^4$ is contained in the set of injective linear transformations in $L\left(\mathbb{R}^n, \mathbb{R}^k\right)$. So, a larger value of $\alpha$ in (5.13) indicates that $A$ is able to withstand perturbations by 'larger' (as measured by the operator norm) linear transformations while maintaining injectivity.