Relationship between Fourier coefficients, eigenvalues, and the spectrum of a ring for dummies? As the question title suggests, what is an explanation for dummies of the relationship between Fourier coefficients, eigenvalues, and the spectrum of a ring?
 A: The Fourier series of a periodic function can be thought of as its decomposition into eigenfunctions for the translation operator $f(x) \mapsto f(x + t)$ on periodic functions. Say we're talking about $\mathbb{C}$-valued functions with period $2 \pi$: then the eigenfunctions are $e^{inx}, n \in \mathbb{Z}$ with eigenvalues $e^{int}$ (for translation by $t$). 
The connection to the spectrum of a ring is a little more indirect. Suppose $T : V \to V$ is a linear operator on a finite-dimensional complex vector space. It generates a commutative subring of the ring $\text{End}(V)$ of endomorphisms of $V$ isomorphic to $\mathbb{C}[T]/m(T)$ where $m$ is the minimal polynomial of $T$. The spectrum of this ring, in the sense of algebraic geometry, has one closed point for each eigenvalue of $T$ (exercise). 
A: This is a basic question about the overloaded usage of the word "spectrum," as all three terms are called spectra. Only the eigenvalues of a linear operator and the spectrum of a ring are directly related; Fourier coefficients are separate. As such, we are only going to explain that connection.
I am not sure if there is an "explanation for dummies," in the sense that we can explain how the Fourier transform works via observable behavior of sound and light waves. However, there is definitely a way to use undergraduate linear algebra to explain this, and we will try to do so.$$\textbf{eigenvalues} \leftrightarrow \textbf{spectrum of a ring}$$Let $V$ be a vector space over a field $k$. The set of eigenvalues of linear operator, e.g. a matrix, $T: V \to V$ is called the spectrum of $T$, denoted $\text{Spec}\,T$. Note that we include the multiplicities of each eigenvalue, e.g. the number of times it is an eigenvalue. The spectrum is computed by finding the roots of the characteristic polynomial, $\chi(x)$.
The spectrum of a ring $R$, $\text{Spec}\,R$ is defined as the set of prime ideals of $R$.
How are these two related? Consider the set of polynomials of $T$ with coefficients in $k$,$$k[T] = \{a_0 + \ldots + a_nT^n : a_i \in k,\,n \in \mathbb{N}\}.$$Now, recall that the Cayley-Hamilton theorem says that every linear operator satisfies its characteristic polynomial, e..g $\chi(T) = 0$. If we "enforce" this constraint in $k[T]$, e.g. taking the quotient of $k[T]$ by the idea generated by $\chi(T)$, we get a new ring $R' = k[T]/(\chi(T))$. It turns out that the prime ideals of $R'$ are simply generated by the polynomials $(x - \lambda_1)$, $\ldots$, $(x - \lambda_n)$, where $\lambda_i$ are the associated eigenvalues. With a little more work, we can show that we need the multiplicities, e.g. we can not quotient by the minimal polynomial. See here for a longer discussion of this and other algebraic geometric concerns.
