Differentiation as Rotation I am trying to make a connection between linear algebra and the Fourier transform. Functions form a vector space and differentiation is an operator. 
Fourier transforming a function from what i understand is writing the function in the basis of complex exponentials. Differentiation in the Fourier domain is multiplication by the imaginary unit which is also the eigenvalue of the 90 degree rotation matrix of the 2x2 rotational matrix.
Is it possible to say that the the differential operator is diagonalised in the Fourier domain?
and
Is the differential operator the equivalent of the rotation matrix in finite dimensional matrices?
 A: Suppose you have a selfadjoint matrix $A$ on $\mathbb{C}^N$. The spectral theorem in projection form gives you orthogonal projections such that
$$
                      P_k^2 = P_k^{\star} = P_k \\
                      P_j P_k = 0,\;\;\; j\ne k \\
                      I = P_1+P_2+\cdots+P_n \\
                      AP_j = \lambda_j P_j.
$$
Therefore,
$$
                            A = \lambda_1 P_1 + \cdots +\lambda_n P_n.
$$
Any power of $A$ can be computed
$$
                      A^{m} = \lambda_1^m P_1 + \cdots + \lambda_n^m P_n
$$
Other functions of $A$ can be computed such as
$$
                   e^{A} = e^{\lambda_1}P_1 + \cdots e^{\lambda_n}P_n
$$
This abstraction using orthogonal projections is a useful one for transitioning to the general case.
The Fourier transform is a continuous diagonalization of $A=\frac{1}{i}\frac{d}{dt}$. One orthogonal projection is
$$
                   P_{S}f = \mathcal{F}^{-1}(\chi_{S}\mathcal{F}f),
$$
where $\chi_{S}$ is the characteristic function of a measurable subset $S$ of $\mathbb{R}$. You can see that $P_{S}^2f=P_{S}f$ because $\chi_{S}^2 =\chi_{S}$. And, it's not hard to check that $(P_Sf,g)=(f,P_Sg)$. And $P_{\mathbb{R}}=I$ because $P_{\mathbb{R}}f=\mathcal{F}^{-1}\mathcal{F}f$. As a starting point for comparing the matrix case, consider
$$
               I = \sum_{n=-\infty}^{\infty}P_{[\frac{n}{N},\frac{n+1}{N}]}
$$
Using Parseval's identity,
$$
        Af =\frac{1}{i}\frac{d}{dx}P_{[\frac{n}{N},\frac{n+1}{N}]}f \approx \frac{n}{N}P_{[\frac{n}{N},\frac{n+1}{N}]}f
$$
More precisely, using Parseval's identity,
\begin{align}
            \left\|\left(A-\frac{n}{N}I\right)P_{[\frac{n}{N},\frac{n+1}{N}]}f\right\|^2 &=\int_{\frac{n}{N}}^{\frac{n+1}{N}}|s-\frac{n}{N}|^2|\hat{f}(s)|^{2}ds \\
  & \le \frac{1}{N^2}\int_{\frac{n}{N}}^{\frac{n+1}{N}}|\hat{f}(s)|^2ds \\
  & \le \frac{1}{N^2}\|P_{[\frac{n}{N},\frac{n+1}{N}]}f\|^2.
\end{align}
Therefore,
$$
                   AP_{\frac{n}{N},\frac{n+1}{N}}f \approx \frac{n}{N}P_{[\frac{n}{N},\frac{n+1}{N}]}f
$$
So you have an approximate decomposition
$$
               I = \sum_{n=-\infty}^{\infty}P_{[\frac{n}{N},\frac{n+1}{N}]}
 \\
              A \approx \sum_{n=-\infty}^{\infty}\frac{n}{N}P_{[\frac{n}{N},\frac{n+1}{N}]}
$$
You can't quite make everything exact, because $e^{isx}$ is not in the space $L^2$, but you can make the operator approximation as close as you like by taking $N$ large enough, which is the granularity of the decomposition of the real axis. The reason you can't make it exact is because $\int_{n/N}^{(n+1)N}e^{isx}ds$ is in the space and is an approximate eigenvector with approximate eigenvalue $n/N$, but the function $e^{i(n/N)x}$ is not in $L^2$. The sum gives way to an integral in the limit as $N\rightarrow\infty$. The Spectral Theorem for selfadjoint operators makes this exact. If $P(t)f=\mathcal{F}^{-1}(\chi_{(-\infty,t]}\mathcal{F}f)$, then the following is made precise in a Hilbert space sense:
$$
    \frac{1}{i}\frac{d}{dx} = \int_{-\infty}^{\infty}\lambda dP(\lambda).
$$
And, you have
$$
     \|f\|^2 = \int_{-\infty}^{\infty}d_{\lambda}\|P(\lambda)f\|^2.
$$
