$\det(aI-T) = 0$ implies $a$ is an eigenvalue of $T$ In Hoffman and Kunze's linear algebra, right after the definition of an eigenvalue for an endomorphism $T: V \to V$ (i.e. $a\in F$ is an eigenvalue if there exists $\alpha\in V$, $\alpha \neq 0$, such that $T\alpha = a\cdot \alpha$) there is a theorem that states the equivalence of three conditions:


$a$ is an eigenvalue of $T$
$aI-T$ is singular
$\det (aI-T) = 0$


I understand $1 \Rightarrow 2 \Rightarrow 3$, but I'm unable to understand the argument that underlies 3 $\Rightarrow$ 1, the book states (and I paraphrase):


  If the space $V$ is of finite dimension, then $aI - T$ is non-injective just when its determinant is 0


I know that, if $aI - T$ is non-injective, the existence of an $\alpha$ such that $(aI - T)(\alpha) = 0$ (and, therefore, $T\alpha = a\cdot \alpha$) follows, but I don't understand the relation between $\det(aI-T)$ and the non-injectivity of $aI-T$. Any hints?
$\det(aI-T)\neq 0$ is equivalent to $aI-T$  is invertible, hence $(aI-T)\cdot\alpha=0$ implies $\alpha=0$, which means injectivity for linear mappings.
A matrix $M$ has $\det M = 0$ if and only if that matrix is not invertible.  Thinking of $M$ as a linear transformation, that it's not invertible means it either fails to be injective or it fails to be surjective.  But the rank-nullity theorem will tell you that because the dimensions of the domain and codomain are equal, failing to be surjective implies that it also fails to be injective.  So either way it's not injective.