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?

  • $\begingroup$ 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. $\endgroup$ – hardmath Oct 5 '18 at 16:24

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.


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