Prove this rank related problem Suppose $A$ is a matrix such that $A^2\neq0$ but$A^3=0$.Then prove that $rank(A)&gt;rank(A^2)$ 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?
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.