Determining the rank of the matrix (Linear algebra) Prove the following statements or provide a counterexample if it is false
If $A$ is a square matrix of order $5$ and $A^{2}=0$ (the zero matrix) then rank(A) is at most $1$.
From the answer given below the statement is false and a counterexample was given. But what im still unclear is that for such questions how would u be able to know whether u need to prove the statement or to disprove it. And also what are the thought process needed to come out with such counterexamples. It seems like the person is able to come out with counterexamples out of nowhere like magic. Could anyone explain. Thanks
 A: What about
$$
  A=\left( {\begin{array}{ccccc}
   0 & 1 & 0 & 0 & 0\\
   0 & 0 & 0 & 0 & 0\\
   0 & 0 & 0 & 1 & 0\\
   0 & 0 & 0 & 0 & 0\\
   0 & 0 & 0 & 0 & 0\\
\end{array} } \right)
$$
Why is your reasoning failing with that $A$?
A: So in order to find the answer to this you need to think around the question.
The claim is that if $A^2=0$ then the image of $A$ must be single dimensional, this is something that immediately appears should be false.
The reasoning for this is that the statement is requiring that the initial space is reduced to 1 dimension (by definition of $\text{rank}(A)=\text{dim}(\text{Im}(T)$) from 5 in order to achieve $A^2=0$. When you think about possible matrices that could achieve this it makes sense that the they could produce a 2-dimensional space and then further reduce it to a null space.
I apologise if I am not making this extremely clear, but in future with this style of question, I suggest trying to begin to lay out a proof. Think about what you would use to prove such a statement and this should help make sense of whether or not it should be true.
