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Let $S = \{1,\cos x, \sin x,...,\cos nx, \sin nx\}$ and $H = Span_\mathbb{c}(S)$. I'm asked to find an ordering of $S$ so that the matrix of $\frac{d}{dx}: H \to H$ in that basis has a simple form.

Isn't the ordered basis just $\{1,\cos x, \sin x,...,\cos nx, \sin nx\}$? I can't think of any other basis, however, I don't think the matrix in this basis will be simple.

Consider the $n=1$ case, the matrix will be $$ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 &0 \\ \end{matrix} $$

and it gets ugly when $n$ gets bigger.

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  • $\begingroup$ When $n =2$, consider the matrix for the basis $\{ 1, \cos x, \cos 2x, \sin x, \sin 2x \}$. $\endgroup$ – user7440 Sep 12 at 4:27
  • $\begingroup$ You end up with the matrix $$ \pmatrix{0\\&0&1\\&-1&0\\&&&0&2\\&&&-2&0\\&&&&&\ddots\\&&&&&&0&n\\&&&&&&-n&0} $$ where the unwritten entries are $0$. I would say that this matrix follows a relatively simple pattern. In particular it is block-diagonal, which is nice. $\endgroup$ – Omnomnomnom Sep 12 at 15:20
  • $\begingroup$ Using the other commenter's approach gives you the block-matrix $$ \pmatrix{0&D\\-D&0} $$ where $D$ is the diagonal matrix with entries $0,1,\dots,n$. $\endgroup$ – Omnomnomnom Sep 12 at 15:23

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