 Solution

The adjoint also leaves $E$ invariant, with $L^{\ast }a=5a-15b$ and $L^{\ast }b=3a-9b$. By adjoint in weighted coordinates, its matrix in the same basis is $\left(\begin{array}{cc}5& 3\\ -15& -9\end{array}\right)$. Solving the two eigenvector equations givesThe basis $a,b$ is not orthonormal, so one must not simply transpose the previous numerical matrix without accounting for its Gram matrix. Directly transposing the integral kernel avoids that error.