Is a matrix AA with an eigenvalue of 00 invertible?

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

Here is what I have.

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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.

Answer

Attribution
Source : Link , Question Author : Derrick J. , Answer Author : Amitesh Datta