> [!NOTE] Theorem (Characterization of Injective Linear Maps via Norm Inequalities) > Let $A\in L(\mathbb{R}^n, \mathbb{R}^k)$ where $L(\mathbb{R}^n, \mathbb{R}^k)$ denotes the set of [[Linear maps|linear maps]] from $\mathbb{R}^n$ to $\mathbb{R}^k$. $A$ is injective iff there exists $\alpha>0$ such that for all $x\in \mathbb{R}^n$ $\lVert Ax \rVert \geq \alpha \lVert x \rVert. $ > > Or equivalently, $\lVert Ax_{n} \rVert/\lVert x_{n} \rVert \not \to 0$, for all $x_{n} \in \mathbb{R}^n \setminus \{ 0 \}$ (if $x=0$ any constant works). ###### Proof of Characterization of Injective Linear Maps via Norm Inequality $(\implies)$ Suppose for all $x\in \mathbb{R}^n,$ $\lVert Ax \rVert \geq \alpha \lVert x \rVert$. If $\lVert Ax \rVert=0$ for some $x\in \mathbb{R}^n$, then $\lVert x \rVert\leq \frac{1}{\alpha} \lVert Ax \rVert = 0\implies x=0$ showing that $\text{Ker }A= \{ 0 \}$ since. Since [[Linear Map is Injective iff Kernel Only Contains Zero]], $A$ is injective. $(\impliedby)$ If A is injective then $\|Ax\|>0$ for every $x\ne0$.  Thus the (continuous) function $f\;:\;S^{n-1}\;\to\;(0,\infty),\qquad f(u)=\|A u\|$attains a _minimum_ $\alpha>0$, by the extreme‐value theorem.  Hence for every nonzero $x$ we have $\|Ax\|\;=\;\|A(\|x\|\,u)\| =\|x\|\;\|A u\|\;\ge\;\|x\|\;\alpha.$ ~~Analysis 3: Conversely, we prove the contrapositive. Suppose that there is a sequence $x_j$ in $\mathbb{R}^n \backslash\{0\}$ such that $\left\|A x_j\right\| /\left\|x_j\right\| \rightarrow 0$ as $j \rightarrow \infty$. Set $u_j:=x_j /\left\|x_j\right\|$. Then for all $j\in \mathbb{N}$, $\left\|u_j\right\|=1$ and $\lVert A u_j \rVert \rightarrow 0$ as $j \rightarrow \infty$.~~ ~~Since the unit sphere is [[Subset of Euclidean space is sequentially compact iff closed and bounded|sequentially compact]] (as it is closed & bounded), there exists a subsequence $u_{j_{\ell}}$ which converges to $u \in S^{n-1}$. But the map $x \mapsto\|A x\|$ is continuous and therefore, $\|A u\|=\lim _{l \rightarrow \infty}\left\|A u_{j_{l}}\right\|=0$, i.e., $u \in \operatorname{ker}(A)$. It follows that $A$ is not injective.~~