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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.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.