$Trig_N$ is the complex vector space of trigonometric polynomials of degree $\le$ N, i.e. the space of functions f of the form $f(t) = \sum_{|n|\le N} a_ne^{int} $, $a_n \in C$. Let $D : f \to f'$ be the differentiation operator on $Trig_N$ and $T = D^2 + 2D$. What is the matrix of T relative to the basis ${e^{int}}|n|\le N$ ?
I know that for a normal polynomial space, the basis is $\{1,x,x^2,\ldots,x^k\}$ and the matrix for $D: f \to f'$ is $$\begin{bmatrix}0&1&0&0&\ldots&0\\0&0&2&0&\ldots&0\\0&0&0&3&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\cdots&k\\0&0&0&0&\ldots&0\end{bmatrix}.$$ from
$D(1)=0$, $D(x)=1$,$\cdots$, $D(x^k)=kx^{k-1}$. But in this case the basis is much more complicated and it would be hard to derive the derivative matrix like above, let alone the matrix of $T=D^2+D$. Is there a different way to figure out the associated matrix of T?
Note that, for each $n\in\Bbb N$,$$(D^2+2D)(e^{int})=D(ine^{int})+2ine^{int}=(-n^2+2in)e^{int}.$$Therefore, if your vbasis is $\left\{e^{-iNt},e^{-i(N-1)t},\ldots,e^{iNt}\right\}$, then the matrix that your after is the diagonal matrix such that the entries of the main diagonal are$$-N^2-2iN,-(N-1)^2-2i(N-1),\ldots,-N^2+2iN.$$