= Solution
The requested order is $A=P^{-1}DP$. To obtain this order directly, put the transposes of the two <eigenvectors> in the rows of $P$, using the fact that $A$ is symmetric:
$$
\boxed{P=\begin{pmatrix}\varphi&1\\\psi&1\end{pmatrix},\qquad D=\begin{pmatrix}\varphi&0\\0&\psi\end{pmatrix}.}
$$
Then $\det P=\varphi-\psi=\sqrt5\ne0$, so $P$ is an <invertible matrix>, with
$$
P^{-1}=\frac1{\sqrt5}\begin{pmatrix}1&-1\\-\psi&\varphi\end{pmatrix}.
$$
Multiplication gives $PA=\bigl(\begin{smallmatrix}\varphi+1&\varphi\\\psi+1&\psi\end{smallmatrix}\bigr)=DP$, since $\varphi^2=\varphi+1$ and $\psi^2=\psi+1$. Multiplying by $P^{-1}$ proves the required <diagonalization>. Putting the <eigenvectors> in columns would instead give the equally standard order $A=QDQ^{-1}$, so the convention must be checked rather than copied mechanically.
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