#math/linear-algebra Suppose $T\in \mathcal{L}(V)$. A number $\lambda \in \mathbf{F}$ is called an `eigenvalue` of $T$ if there exists $v\in V$ such that $v\neq{}0$ and $Tv=\lambda v$. $ \begin{array}{c} V\text{ has a one-dimensional}\\ \text{subspace invariant under }T \end{array} \iff T\text{ has an eigenvalue} $ # Eigenvalue Conditions for Finite Spaces Suppose $V$ is finite-dimensional, $T\in \mathcal{L}(V)$, and $\lambda \in \mathbf{F}$. Then the following are equivalent $ \begin{align} \mathbf{(1)} & \quad \lambda \text{ is an eigenvalue of }T; \\ \mathbf{(2)} & \quad T-\lambda I\text{ is not injective};\\ \mathbf{(3)} & \quad T-\lambda I \text{ is not surjective};\\ \mathbf{(4)} & \quad T-\lambda I \text{ is not invertible};\\ \end{align} $ where $I\in \mathcal{L}(V)$ is the identity operator, i.e. $Iv=v$ for all $v\in V$. **Proof:** Let $S=T-\lambda I$. Then if $\mathbf{(1)}$ is true, then there exists some $v\in V$ with $v\neq{}0$ such that $Tv=\lambda v$. Then $S(v)=(T-\lambda I)v=Tv-\lambda v=0$. Thus, since $v\neq{}0$ and $S(0)=0$ , $S=T-\lambda I$ is not injective. For the remainder, by [[3D4 - Invertible Operators]], since $S=T-\lambda I$ is an operator $ S\text{ is injective}\iff S\text{ is surjective}\iff S\text{ is invertible}. $