Prove: A is invertible IFF 0 is not an eigenvalue of A

Question

I would appreciate feedback on the correctness of this proof, as well as alternative proofs.

Prove: A is invertible IFF 0 is not an eigenvalue of A.

Proof:If $A$ is invertible, letting $λ=0$ yields $Ax-λx=Ax=0$ which implies that $x=0$. So, $0$ is not an eigenvalue. If $0$ is not an eigenvalue then there is no non-zero $x$ such that $Ax=0$, and because there is only the trivial solution, A must be invertible.

Answer 1

You might want to explore the fact that $det(A) = p(0)$, where $p(x)$ is the characteristic polynomial of $A$.

Answer 2

If $A$ is invertible, then $A$ is injective, which implies that $Null(A)=\{0\}$. Thus, the only solution to $Av=0=0\cdot v$ is $v=0$. Hence, $0$ is not an eigenvalue.

Conversely, if $0$ is not an eigenvalue, then $A=A-0I$ must be injective (otherwise, 0 is an eigenvalue. Theorem 5.6 in Axler), which implies $A$ must be invertible since $A$ is a linear operator.