 Solution

Over the complex numbers, the eigenvaluesare distinct. Thus distinct eigenvalues imply diagonalizability, and $A$ is diagonalizable over $\mathbb{C}$, in fact already over $\mathbb{R}$.

An explicit eigenbasis isFor either quadratic root $\lambda$, the identity ${\lambda }^2=\lambda +1$ verifies $A{v}_{\lambda }=\lambda v_{\lambda }$. Taking these three columns as $P$ gives $P^{-1}AP=\mathrm{diag}(1,{\lambda }_{+},{\lambda }_{-})$. Their distinct eigenvalues ensure linear independence.