Change-of-basis for operators is a little different than change-of-basis for vectors. If $V$ is a vector space with bases $\mathcal{B}$ and $\mathcal{D}$, then there is a change-of-basis matrix $$P_{\mathcal{DB}} = \begin{pmatrix} \phantom{a} &&& \cr [{\bf b}_1]_\mathcal{D} & [{\bf b}_2]_\mathcal{D} & \cdots & [{\bf b}_n]_\mathcal{D} \cr\phantom{a} &&& \end{pmatrix},$$ such that $$[{\bf v}]_\mathcal{D} = P_{\mathcal{DB}} [{\bf v}]_\mathcal{B}.$$

Now consider an operator $L: V \to V$. The rule for relating $[L]_\mathcal{D}$ to $[L]_\mathcal{B}$ is different! \begin{eqnarray*} [L]_\mathcal{D} & = & P_{\mathcal{DB}} [L]_{\mathcal{B}} P_{\mathcal{BD}} \cr & = & P_{\mathcal{DB}} [L]_{\mathcal{B}} P_{\mathcal{DB}}^{-1} \cr & = & P_{\mathcal{BD}}^{-1} [L]_{\mathcal{B}} P_{\mathcal{BD}} \end{eqnarray*} Do not write "$[L]_\mathcal{D} = P_{\mathcal{DB}} [L]_\mathcal{B}$"! That obvious analogue of the formula for vectors is dead wrong.