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.

Let us write $A=(C_1,C_2,C_3)$ where $C_i$ is the column $i$ of $A.$
Assume $A^2=0.$ That is, we have that $C_1,C_2,C_3$ are two linearly independent solutions of the system $Ax=0.$ But since $A$ is of order $3$ and has rank $2$ this is not possible.
