= Solution
The <characteristic polynomial> is $\det(\lambda I-A)=\lambda^2-\lambda-1$. Its distinct <eigenvalues> are $\alpha=(1+\sqrt5)/2$ and $\beta=(1-\sqrt5)/2$. For either <eigenvalue> $\lambda$, the <vector> $(\lambda,1)^\top$ is an <eigenvector>, since $A(\lambda,1)^\top=(\lambda+1,\lambda)^\top=\lambda(\lambda,1)^\top$.
Choose
$$
\boxed{S=\begin{pmatrix}\alpha&\beta\\1&1\end{pmatrix},\qquad D=\begin{pmatrix}\alpha&0\\0&\beta\end{pmatrix}.}
$$
The <determinant> of $S$ is $\alpha-\beta=\sqrt5\ne0$, so
$$
S^{-1}=\frac1{\sqrt5}\begin{pmatrix}1&-\beta\\-1&\alpha\end{pmatrix}.
$$
Its columns being <eigenvectors> gives $AS=SD$, and left multiplication by $S^{-1}$ proves $S^{-1}AS=D$.
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