# 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

• Why will all of the zero rows except the first be zero after row reduction? Nov 1, 2015 at 15:21
• Hint Do you have Jordan canonical form available? Nov 1, 2015 at 15:21
• By that reasoning you can only conclude that $\rank(A) \le 4$.
– user258700
Nov 1, 2015 at 15:23
• Im unsure about what jordan canoical form is. Could u explain that. Thanks Nov 1, 2015 at 15:34
• In that case probably this is not the intended method for the problem. Every matrix is similar to a more-or-less unique matrix that has a particularly simple form called "Jordan canonical (or normal) form". If one is asking a question about a property (like rank) that is invariant under similarity, often one can answer the question more easily by working only with matrices in this form: en.wikipedia.org/wiki/Jordan_normal_form Nov 1, 2015 at 19:24

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 $A$?

• Okay i know why my reasoning does not work. How do u tackle this question. Could u explain Thanks Nov 1, 2015 at 15:44
• You have a method to write all nilpotent matrices in a canonical way. This is based on a method from Jordan. This is the background I used. Nov 1, 2015 at 15:55
• In fact, up to similarity, this is the only $5 \times 5$ matrix for which $A^2 = 0$ but $\operatorname{rank} A > 1$. Nov 1, 2015 at 19:25

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.