Is a matrix $A$ with an eigenvalue of $0$ invertible?

Just wanted some input to see if my proof is satisfactory or if it needs some cleaning up.

Here is what I have.

Proof

Suppose $A$ is square matrix and invertible and, for the sake of contradiction, let $0$ be an eigenvalue. Consider, $(A-\lambda I)\cdot v = 0$ with $\lambda=0 $ $$\Rightarrow (A- 0\cdot I)v=0$$
$$\Rightarrow(A-0)v=0$$
$$\Rightarrow Av=0$$

We know $A$ is an invertible and in order for $Av = 0$, $v = 0$, but $v$ must be non-trivial such that $\det(A-\lambda I) = 0$. Here lies our contradiction. Hence, $0$ cannot be an eigenvalue.

Revised Proof

Suppose $A$ is square matrix and has an eigenvalue of $0$. For the sake of contradiction, lets assume $A$ is invertible.

Consider, $Av = \lambda v$, with $\lambda = 0$ means there exists a non-zero $v$ such that $Av = 0$. This implies $Av = 0v \Rightarrow Av = 0$

For an invertible matrix $A$, $Av = 0$ implies $v = 0$. So, $Av = 0 = A\cdot 0$. Since $v$ cannot be $0$,this means $A$ must not have been one-to-one. Hence, our contradiction, $A$ must not be invertible.

Your proof is correct. In fact, a square matrix $A$ is invertible if and only if $0$ is not an eigenvalue of $A$. (You can replace all logical implications in your proof by logical equivalences.)

Hope this helps!

This looks okay. Initially, we have that $Av=\lambda v$ for eigenvalues $\lambda$ of $A$. Since $\lambda=0$, we have that $Av=0$. Now assume that $A^{-1}$ exists.

Now by multiplying on the left by $A^{-1}$, we get $v=0$. This is a contradiction, since $v$ cannot be the zero vector. So, $A^{-1}$ does not exist.

If $\lambda_1,\dotsc,\lambda_n$ are the (not necessarily distinct) eigenvalues of an $n\times n$ matrix $A$, then $$ \det(A)=\lambda_1\dotsb\lambda_n\tag{1} $$ A nice proof of this fact can be found here.

Now, $A$ is invertible if and only if $\det(A)\neq0$. Hence $(1)$ implies $A$ is invertible if and only if $0$ is not an eigenvalue of $A$.