Prove that a linear operator $T$ on a finite-dimensional vector space is invertible if and only if zero is not an eigenvalue of $T$.
Definition: Let $T$ be a linear operator on a vector space $V$. A nonzero vector $v \in V$ in an eigenvector of $T$ if there exists a scalar $\lambda$ such that $T(v)= \lambda v$. The scalar $\lambda$ is called an eigenvalue.
Let $A$ be in $M_{n,n}(F)$. A nonzero vector $v\in F^n$ is an eigenvector of $A$ if $v$ is an eigenvector of $L_{A}$. The scalar $\lambda$ is called the eigenvalue of $A$.
Proof: $\Rightarrow$ Let $T$ be a finite linear operator, and $T(v)=Av=\lambda v$ for $A$ to be a $M_{n,n}(F)$ matrix. If $T(v)$ is invertible, then $T(T^{-1})=(Av)(Av)^{-1}=(\lambda v)(\lambda v)^{-1}=I_n$. That means $\lambda v$ is nonzero.
$\Leftarrow$ If zero is not an eigenvalue of $T$, that means $\lambda v=Av \neq0$, then $det(Av)$ $\neq$ $0$. Hence $T$ is invertible.
I know this is a crappy work, but this is all I can think about.