If rank$(A) = 2$, then $A^2 \neq 0_3$ 
Let $A$ be a real $3 \times 3 $ matrix such that rank$(A) = 2$.
Prove that $A^2 \neq 0_3$. where $0_3$ represents the null matrix of order $3$.

I am looking for a solution involving only basic manipulation using matrices. I already have a better solution using the range and the nullity of $A$.
Thank you in advance!
Edit. No Sylvester's inequality, Jordan form or range+nullity / linear transformations. At most use the definition of the rank as the dimension of the column/row space.

If the rank of $A$ is $2$, then there exists a subspace spanned by the non-null vector $v_1$ such that:
$$Av_1 = 0,$$
i.e. $v_1 \in \ker(A).$ On the other hand, there must be at least another non-null vector $v_2$ which is an eigenvector of $A$ for an eigenvalue $\lambda_2 \neq 0$, i.e.
$$Av_2 = \lambda_2 v_2.$$
Of course, $v_1 \in \ker(A^2).$ Moreover:
$$A^2v_2 = A(Av_2) = A(\lambda_2 v_2) = \lambda_2 Av_2 = \lambda_2^2 v_2.$$
This means that $\ker(A^2) \neq \mathbb{R}^3.$ Therefore, $A^2 \neq 0_3$.
