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Suppose I have two operators, $D \equiv \frac{d}{dt}$ and $D^2 \equiv \frac{d^2}{dt^2}$, represented in matrix form by two different bases $\mathbf{e} = \left \{\cos (wt), \sin (wt) \right \}$ and $\mathbf{e'} = \left \{e^{iwt}, e^{-iwt} \right \}$.

For $\mathbf{e}$:

$$D_{\mathbf{e}} = \begin{pmatrix} 0 &w \\ -w & 0 \end{pmatrix}, \ \ D_{\mathbf{e}}^2 = \begin{pmatrix} -w^2 &0 \\ 0 & -w^2 \end{pmatrix}$$

For $\mathbf{e'}$:

$$D_{\mathbf{e'}} = \begin{pmatrix} iw &0 \\ 0 & -iw \end{pmatrix}, \ \ D_{\mathbf{e'}}^2 = \begin{pmatrix} -w^2 &0 \\ 0 & -w^2 \end{pmatrix}$$

How do I proceed if I wanted to find the transformation matrix between the bases $\mathbf{e}$ and $\mathbf{e'}$?

In my old linear algebra textbook there's a tutorial of how to do it for vectors (column matrices) but I got confused when trying to do it with operators (square matrices).

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The basis vectors are not column matrices (as you noted in your question) but they are not square matrices either. They are the functions in the two bases, that is, $\cos wt$ etc. The transformation matrix from $\def\b#1{{\bf#1}}\b e'$ to $\b e$ has columns which are the functions in $\b e'$, expressed as coordinate vectors with respect to $\b e$. Since the first vector (i.e., function) in $\b e'$ can be written $$e^{iwt}=\cos wt+i\sin wt\ ,$$ the matrix is $$\pmatrix{1&?\cr i&?\cr}$$ and I'm sure you can do the rest for yourself.

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  • $\begingroup$ Is it really that simple? I don't believe it (with all due respect). $\endgroup$ – chili and sea bass Jan 12 '16 at 3:22
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    $\begingroup$ Unless I have misunderstood the question, it is really that simple. Transformation matrices between bases depend only on the bases and have nothing to do with any particular operator. $\endgroup$ – David Jan 12 '16 at 3:44

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