Finding trace of a matrix $A $ such that $A^3=A$

Question

If $A$ is a real $n \times n$ matrix satisfying $A^3 = A,$ then Trace of $A$ is always

1) $n$

2) 0

3) $−n$

4) an integer in the set $\{−n,−(n − 1), \dots ,−1, 0, 1, \dots , n\}.$ how to solve this. Thanks in advance.

Answer 1

Suppose $A$ has eigenvalue $\lambda$. Then $A^3$ has eigenvalue $\lambda^3$. However since $A^3=A$, so $\lambda^3=\lambda$ for any eigenvalue $\lambda$ of $A$. Hence the only possible eigenvalues of $A$ are $0,1,-1$. Now take a diagonal $n\times n$ matrix with $k$ of its diagonal elements equal to $1$, and $m$ of its diagonal elements equal to $-1$, where $k+m\leq n$ and the rest being $0$. Then $A^3=A$ and the trace of the matrix is $k-m$. Hence the trace can be anything from the set $\{-n,-(n-1),\cdots,-1,0,1,\cdots,n\}$.

P.S. - Trace of a matrix is the sum of its eigenvalues repeated according to their multiplicities.