Let $w:\mathbb{R} \times \mathbb{R}^n \rightarrow Mat(n,\mathbb{R})$ be a smooth function, $R_{ij}$ be a fixed skew-symmetric $n\times n$ real matrix, and $A\in\mathbb{R}$. Consider the equation $$\frac{\partial w}{\partial t} - \sum_{i=1}^n\left(\frac{\partial}{\partial x^i} + \frac{1}{4}\sum_j {R}_{ij} x^j\right)^2 w + Aw = 0.$$
My question is, if one wants to solve for $w$, why is it sufficient to take $R_{ij}$ to be block-diagonal, consisting of $2\times 2$ skew-symmetric blocks? I know that a skew-symmetric matrix can always be block-diagonalised this way, but am having a hard time writing down rigorously what the solution to the equation would be if $R$ was changed to a block-diagonal form, say $P^{-1} RP$.
I'm looking for a proof that shows something like, if $w$ solves the above equation, then $P^{-1}wP$ solves the equation with $R$ replaced by $P^{-1}RP$.
(This question arose from trying to understand the proof of Proposition 12.25, p. 162, of John Roe's book on the Atiyah-Singer index theorem. http://www.maths.ed.ac.uk/~aar/papers/roeindex.pdf)
Thanks for any help.