Eigenvalues being physical observables I think I'm comfortable with the PDE solutions to the Schrodinger equation. But as soon as we start  putting these values in a matrix (in dirac notation), I lose my understanding and everything becomes plug-and-chug math magic. 
I'm wondering if anyone has an understanding as to why eigenvalues of the eigenstates of a matrix correspond to physical observables. That is, how can we show, using wave mechanics, that the eigenvalues to our eigenfunctions in our PDE can correspond to eigenvalues to our eigenvectors in a matrix? And can we use this understanding to get a better idea of what's going on when we look at the eigenvalues of eigenvectors of a rotation of our matrix?
 A: What you are asking is called, in mathematical terms, spectral theorem. I don't know how much you are interested in details, but any self-adjoint operator $A$ (linear partial differential operator on a Hilbert space) can be written as
$$A=\int_{\sigma(A)}\lambda dP_\lambda\;,$$
where $\sigma(A)$ is the spectrum of $A$ and $dP_\lambda$ the spectral measure (projection valued measure). If the operator has purely discrete spectrum (with finite multiplicities), i.e. the operator is either compact or with compact resolvent, this reduces to the maybe more familiar form:
$$A=\sum_i \lambda_i \lvert \phi_{\lambda_i}\rangle\langle\phi_{\lambda_i}\rvert$$
where $\lambda_i$ are the eigenvalues (repeated if they have multiplicity $>1$), and $\phi_{\lambda_i}$ the corresponding eigenvector.
As you see, there is a natural identification between an operator and its eigenvalues/eigenvectors; and the Dirac notation $\langle\psi \rvert A\lvert\phi\rangle$ is just a way of writing the scalar product between $A\phi$ and $\psi$. Using the decomposition above, we obtain
$$\langle\psi \rvert A\lvert\phi\rangle=\sum_i \lambda_i \langle\psi,\phi_{\lambda_i}\rangle\langle\phi_{\lambda_i},\phi\rangle$$
Keep in mind that we can formally construct the "matrix associated to $A$" to be the (infinite) matrix with elements $M_{ij}=\langle e_i \rvert A\lvert e_j \rangle$, with $\{e_i\}_{i\in\mathbb{N}}$ an orthonormal basis of the Hilbert space. This matrix is diagonal if and only if we chose as a basis the eigenvectors of the operator $A$, and the diagonal elements are the corresponding eigenvalues. Obviously this naïve picture fails if the operator has also continuous spectrum.
A: A measurement is not a primitive in physics. Rather, a measurement is a physical process that takes place according to the same laws of physics as any other physical process. Those same laws apply to the measurement apparatus, to the person doing the measurement and to the records he makes of the measurement. What distinguishes a measurement from any other kind of physical process? One property that is necessary for a particular process to count as a measurement is that it is possible to copy the result. That is, it has to be possible for the result to be present in one system before being copied and in more than one system afterward.
So what sort of operators represent results that can be copied in this way? According to quantum mechanics, systems evolve unitarily. Any unitary operator can be written in the form:
$$
U = \Sigma_a e^{i\phi_a}|a\rangle\langle a|,
$$
where the $|a\rangle$ form an orthonormal set. To copy a result this operator would have to leave the operator being copied unchanged and the only operators it leaves unchanged are normal operators. The normal operators that represent whether something happens or not are projectors, so the values attached to those projectors, eigenvalues, represent possible measurement outcomes.
For a more detailed discussion see
http://arxiv.org/abs/quant-ph/0703160.
