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Sorry about deleting my answer so quickly but I pushed the preview button while the page was jumping around due to rendering and I missed! Full answer available soon I hope!
While you're at it, once you multiply $(\alpha - 2qcos \Omega t$ by $x$, you can get rid of $\alpha$ or $\beta$ altogether. That would be the Matthieu equation I know and love . . .
@Tony: I'm having a little trouble understanding your question. Are the "subcomponents" orthogonal projections of $x$ onto a set of three (for example) mutually orthogonal, complementary subspaces whose sum is $R^{k}$, the space in which $x$ lives? That's my guess, but if you could clarify it might help folks to answer your question.
In fact, the TM can be set up so that the first thing it does is print the program on the blank tape, then transfer control to the UTM. So Garabed's case includes Gerhard's.
If I am not mistaken, the requirement that all eigenvalues of $M$ are positive implies $det(M)$ is positive and hence $M$ has maximal rank, since the rows must be linearly independent. So what gives? How can $M$ have low rank? Or am I missing something?
If I am not mistaken, the construction of the harmonic conjugate of a given harmonic function works locally, so simple connectedness only becomes an issue on a larger scale.
