# Prove this rank related problem

Suppose $$A$$ is a matrix such that $$A^2\neq0$$ but$$A^3=0$$.Then prove that $$rank(A)>rank(A^2)$$ and $$rank(A)\neq tr(A)$$.

$$rank(AB)\leq$$min{$$rank A,rank B$$}.Then $$rank(A^2)\leq rank(A)$$.How to prove the reamining part?

• You don't mention the size of $A$ but let's call it an $n\times n$ matrix. Now, what do you know about the rank and nullity of $A$? Of $A^2$? This leads to the strictly inequality of ranks that you are asking for. – hardmath Oct 5 '18 at 16:24

## 2 Answers

If $$A^3=0$$, then $$A$$ is nilpotent. Since $$A$$ is nilpotent, all of its eigenvalues are $$0$$, so its trace is also $$0$$ (because the trace is equal to the sum of the eigenvalues).

Now you just need to prove that the rank is strictly bigger than $$0$$. Can you take it from here?

As A^3=O so A is nilpotent matrix.So eigan values of A are all zeros(0). Hence Trace(A)=0;.......(1) Since trace(A) is nothing but the summation of eigan values.

Now A is non zero matrix because A^2 is not a null matrix,so rank(A)>0;.........(2) Since rank(A)=0 iff A=O.

Hence from above (1) and (2) rank(A)>trac(A).

This comple the proof.