Suppose \(A\) is a \(3\times3\) matrix with real entries and eigenvalues \(-1\), \(0\), and \(1\). Prove that \(A^3=A\).
Solution
1. The characteristic polynomial of \(A\) is \(p_A(x)=x^3-x\).
\(p_A(x)=\det(xI-A)\) is monic of degree \(3\), and its roots in \(\mathbb C\) are the eigenvalues of \(A\). The three eigenvalues \(-1,0,1\) are distinct, so they are all the roots, each simple, and \(p_A(x)=(x+1)x(x-1)=x^3-x\).
2. \(A^3=A\).
By the Cayley–Hamilton theorem \(p_A(A)=0\), that is, \(A^3-A=0\) by 1.