TITLE: Is a matrix $A$ with an eigenvalue of $0$ invertible?
QUESTION [52 upvotes]: 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.

REPLY [2 votes]: Suposse that 0 is an eigenvalue of A. 
Considering $J$ the Jordan canonical form of A, you have a triangular matrix with the eigenvalues in the diagonal, with the same determinant than $A$.  
It's known that determinant of a triangular matrix is the product of the elements in the diagonal and we've supossed that one of them is 0, then is clear that $Det(A)=Det(J)=0$, so $A$ isn't invertible.