 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:Then $\det P=\phi -\psi =\sqrt{5}\ne 0$, so $P$ is an invertible matrix, withMultiplication gives $PA=\left(\begin{array}{cc}\phi +1& \phi \\ \psi +1& \psi \end{array}\right)=DP$, since ${\phi }^2=\phi +1$ 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 $A=QD{Q}^{-1}$, so the convention must be checked rather than copied mechanically.