#math/linear-algebra A linear map from a vector space to itself is called an operator. Suppose $V$ is finite dimensional and $T\in \mathcal{L}(V)$. Then the following is true $ T \text{ is invertible }\iff T \text{ is injective} \iff T\text{ is surjective}. $ > [!From V to W] > If $T\in \mathcal{L}(V,W)$ where $\text{dim V}=\text{dim W}<\infty$ then the theorem also holds true. ## Proof Suppose $T\in \mathcal{L}(V)$ is invertible, then by definition, it is both injective and surjective. Suppose $T\in \mathcal{L}(V)$ is injective. Since $T$ is invertible $\text{null }T=\{ 0 \}$. Thus, by the [[3B3 - Fundamental Theorem of Linear Maps|fundamental theorem of linear maps]] $ \begin{align} \text{dim }V & =\text{dim range }T + \text{dim null } T \\ & =\text{dim range }T. \end{align} $ Then, by the [[3A3 - Linear Maps and Basis of Domain Theorem|basis of domain theorem of linear maps]], it must be that $\text{range }T=V.$ Thus, $T$ is surjective and therefore injective. Suppose $T\in \mathcal{L}(V)$ is surjective. Then $\text{range T}=V$ and $\text{dim range }T=\text{dim }V$. And by the fundamental theorem of linear maps, $\text{dim null }T=0$ and $\text{null }T=\{ 0 \}$. Thus, $T$ is injective and therefore injective.