= Solution
\b[Yes.] The matrix is a <cyclic shift operator> on three coordinates. Since $e_1,Ae_1,A^2e_1$ are the three standard <basis vectors>, no <polynomial> of degree below three annihilates $A$. Also $A^3=I$, so its <minimal polynomial> is $X^3-1$. Over the <complex numbers> this has the distinct roots $1,\omega,\omega^2$, where $\omega=e^{2\pi i/3}$. The <minimal polynomial criterion for diagonalizability> therefore proves <diagonalizability>.
Explicitly, $v_\lambda=(1,\lambda^2,\lambda)^{\mathsf T}$ satisfies $Av_\lambda=\lambda v_\lambda$ when $\lambda^3=1$. These three <eigenvectors> form an <eigenbasis>. This is a special case of <diagonalization of a cyclic shift over a field>.
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