so lately I've been working with The PDE: $2u_{xx}+3u_{yy}+2\sqrt{2}u_{xy}=0$, I know so far that this can be written as $\begin{pmatrix}\partial_{x}&\partial_{y}\end{pmatrix} \begin{pmatrix}2&\sqrt{2}\\\ \sqrt{2}&3\end{pmatrix}\begin{pmatrix}\partial_{x}\\ \partial_{y}\end{pmatrix}[u]=0$ and that the coordinate transformation $\begin{pmatrix}X\\ Y\end{pmatrix}=\begin{pmatrix}-\sqrt{\frac{2}{3}}&\frac{1}{\sqrt{3}}\\\ \frac{1}{\sqrt{3}}&\sqrt{\frac{2}{3}}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}$ reduces the PDE to its canonical form $u_{XX}+4u_{YY}=0$ or $\begin{pmatrix}\partial_{X}&\partial_{Y}\end{pmatrix} \begin{pmatrix}1&0\\\ 0&4\end{pmatrix}\begin{pmatrix}\partial_{X}\\ \partial_{Y}\end{pmatrix}[u]=0$, what my question is now is how I might perform the coordinate change $\zeta=Y+2iX,\eta =Y-2iX$ using this same differential operator notation? since I could have written that $\begin{pmatrix} \zeta \\ \eta \end{pmatrix}=\begin{pmatrix}2i&1\\-2i&1 \end{pmatrix}\begin{pmatrix}X\\Y\end{pmatrix}$
Ive tried applying the same method and have gotten that $\begin{pmatrix} X \\ Y \end{pmatrix}=\begin{pmatrix}-\frac{i}{4}&\frac{i}{4}\\\frac{1}{2}&\frac{1}{2} \end{pmatrix}\begin{pmatrix}\zeta \\\eta \end{pmatrix}\to \begin{pmatrix} \partial_{X} \\ \partial_{Y} \end{pmatrix}=\begin{pmatrix}-\frac{i}{4}&\frac{i}{4}\\\frac{1}{2}&\frac{1}{2} \end{pmatrix}\begin{pmatrix}\partial_{\zeta} \\\partial_{\eta} \end{pmatrix}\to\begin{pmatrix} \partial_{X} \\ \partial_{Y} \end{pmatrix}^T=\begin{pmatrix}\partial_{\zeta} \\\partial_{\eta} \end{pmatrix}^T\begin{pmatrix}-\frac{i}{4}&\frac{i}{4}\\\frac{1}{2}&\frac{1}{2} \end{pmatrix}^T\to \begin{pmatrix} \partial_{X} & \partial_{Y} \end{pmatrix}=\begin{pmatrix}\partial_{\zeta} &\partial_{\eta} \end{pmatrix}\begin{pmatrix}-\frac{i}{4}&\frac{1}{2}\\\frac{i}{4}&\frac{1}{2} \end{pmatrix}\to \begin{pmatrix}\partial_{\zeta} &\partial_{\eta} \end{pmatrix}\begin{pmatrix}-\frac{i}{4}&\frac{1}{2}\\\frac{i}{4}&\frac{1}{2} \end{pmatrix}\begin{pmatrix}1&0\\\ 0&4\end{pmatrix}\begin{pmatrix}-\frac{i}{4}&\frac{i}{4}\\\frac{1}{2}&\frac{1}{2} \end{pmatrix}\begin{pmatrix}\partial_{\zeta} \\\partial_{\eta} \end{pmatrix}[u]=0\to \begin{pmatrix} \partial _{\zeta } & \partial _{\eta } \end{pmatrix}\begin{pmatrix} \frac{15}{16} & \frac{17}{16}\\ \frac{17}{16} & \frac{15}{16} \end{pmatrix}\begin{pmatrix} \partial _{\zeta }\\ \partial _{\eta } \end{pmatrix}[ u] =0 \to \frac{15}{16}u_{\zeta \zeta }+\frac{17}{8}u_{\zeta \eta}+\frac{15}{16}u_{\eta \eta}=0$
but I'm supposed to get $u_{\zeta \eta}=0$ at the end, and this is what I get when I use the chain rule (i.e the standard way of performing coordinate transformations, at least standard in the sense that, that's all I can find on the internet lol)
any help would be appreciated, but to be clear I would like to strictly stick to this differential operator notation, Thanks in advance!