# If A is an $n\times n$ square matrix such that $A^3=A$ , then show that $\operatorname{rank}(A) + \operatorname{tr}(A)$ is even

If $$A$$ is an $$n\times n$$ square matrix such that $$A^3=A$$ , then show that $$\operatorname{rank}(A) + \operatorname{tr}(A)$$ is even.

$$\mathbf {My \ attempt}:$$ Actually, I have been trying over this problem which was posed as a multiple choice question in an exam . First of all to show that $$\operatorname{rank}(A) ≥ \operatorname{tr}(A)$$ . Consider, the rank factorization of $$A$$, $$A= PQ$$ where $$P$$ and $$Q$$ are $$n\times r$$ and $$r \times n$$ matrices respectively for $$r=\operatorname{rank} (A$$) , and as $$P$$ and $$Q$$ are left and right invertible, then $$\operatorname{tr}((QP)^2)=\operatorname{rank} A\geq \operatorname{tr}(QP)=\operatorname{tr}(PQ)=tr(A).$$ But, I can't approach for the part stated in the question . Any help is appreciated .

• Can someone please edit the text using Latex or MathJax (as I am completely unable of doing it) ??? Jul 13, 2019 at 2:45
• It may be worth your while to check out math.meta.stackexchange.com/questions/5020/… . Jul 13, 2019 at 3:00

## 1 Answer

Since $$A^3=A$$, the eigenvalues of $$A$$ are from $$\{0,+1,-1\}$$. In addition, the $$A$$ is diagonalizable, as the minimal polynomial of $$A$$ divides $$t^3-t=(t-1)(t+1)t$$, hence it factors into linear factors.

Let me denote there algebraic multiplicities by $$n_0,n_+,n_-$$, respectively. Then the rank of $$A$$ is equal to $$rank(A)=n_++n_-,$$ while the trace is (the sum of the eigenvalues) $$tr(A) = n_+-n_-.$$ So $$rank(A) + tr(A) = n_++n_-+n_+-n_- = 2n_+,$$ which is even.