How can a spectrum object be constructed?

There are so many spectrum objects in mathematics.

• spectrum of a linear operator
• spectrum of a ring
• spectrum of a graph
• spectrum in algebraic topology
• spectrum of a sentence (mathematical logic)

There has to be some universal property uniting all of these spectrum objects. How do you use a universal construction to derive each of the five?

• The spectrum of a graph is the spectrum of a linear operator, and the fact that there is also a spectrum of ring is just a coincidence as far as I know. I doubt there's any concept underlying both cases. Feb 27, 2019 at 10:03
• @Arnaud D. There's also a spectrum in algebraic topology, so I don't think it's a coincidence. Feb 27, 2019 at 10:04
• Sounds like even more of a coincidence to me... Feb 27, 2019 at 10:10
• @ArnaudD Spectrum of a sentence (mathematical logic) Feb 27, 2019 at 10:13
• Spectra in algebraic topology and spectra of sentences are totally unrelated to the other notions of spectra you mentioned. Mar 15, 2019 at 6:47

Let $$k$$ be an algebraically closed field (characteristic $$0$$ if you like), and $$T$$ a linear operator acting on a finite-dimensional $$k$$-vector space. Then $$\operatorname{Spec} T$$ is, by definition, the space of "generalized eigenvalues" (i.e. diagonal entries in the Jordan canonical form of $$T$$) with repetition. But if $$\chi_T(x) \in k[x]$$ is the characteristic polynomial of $$T$$ then the roots of $$\chi_T(x)$$ are precisely the generalized eigenvalues with repetition, so the affine scheme $$\operatorname{Spec} k[x]/(\chi_T(x)) = \operatorname{Spec} T$$ in a natural way: namely, its points are generalized eigenvalues of $$T$$, and we can recover their multiplicities as the dimension of the localization as a $$k$$-algebra. Therefore the spectrum of a ring is a generalization of a spectrum of a linear operator.
Since the "spectrum" of a graph is the spectrum of its adjacency matrix $$M$$, this is obviously a special case of the spectrum of a linear operator. So it is the spectrum of the ring $$\mathbb C[x]/(\chi_M(x))$$.