Solution
= Solution
By the definition of an <eigenvalue>, there is a nonzero <eigenvector> $v$ with $Av=\lambda v$. Induction using <matrix multiplication> gives
$$
A^{r+1}v=A(A^rv)=A(\lambda^rv)=\lambda^rAv=\lambda^{r+1}v,
$$
starting from $r=1$. Thus the same nonzero <eigenvector> establishes
$$
\boxed{A^rv=\lambda^rv,\qquad \lambda^r\text{ is an eigenvalue of }A^r\quad(r\geq1).}
$$