 Solution

Yes. The matrix is a cyclic shift operator on three coordinates. Since $e_1,A{e}_1,A^2{e}_1$ 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 $A{v}_{\lambda }=\lambda v_{\lambda }$ when ${\lambda }^3=1$. These three eigenvectors form an eigenbasis. This is a special case of diagonalization of a cyclic shift over a field.