How to relate the spectrum of a self-adjoint unbounded operator to the spectrum of a compact solution operator Apologies if this is a trivial question. I'm having some trouble getting this straight in my head and would appreciate either an explanation or a pointer to a good reference. 
If this looks too long, here's a

tl;dr Say I have a densely defined, bounded below, symmetric operator $T$ defined in a Hilbert space $H$. After shifting by some $\sigma$ one can show there exists a compact solution operator $K$. Why does the spectrum of $T$ coincide with (a shift of) the spectrum of $K$?

Setup
Let $H$ be a Hilbert space with inner product $(\cdot,\cdot)$. Let $T$ be a densely defined symmetric operator bounded below by some constant $c$, with $D = \operatorname{dom}(T)$. Define the form $\mathfrak{t}$ on $D\times D$ by $\mathfrak{t}(u,v) = (Tu,v)$. By symmetry $\mathfrak{t}$ is sesquilinear. Because $T$ is bounded below, if we choose $\sigma > \max\{1,-c\}$, the form defined on $D\times D$ by 
$$ \mathfrak{t}_\sigma(u,v) = \mathfrak{t}(u,v) + \sigma\cdot(u,v) $$
is a positive definite Hermitian inner product on $D$.
Define by $V$ the completion of $D$ with respect to $\mathfrak{t}_\sigma$. As $\mathfrak{t}_\sigma$ majorizes the norm on $H$ the injection $D\hookrightarrow H$ extends to a bounded embedding $\iota : V\hookrightarrow H$. Now assume that $V$ compactly embeds in $H$. (Here, if $T$ is not self-adjoint, take the Friedrichs extension.)
Define an eigenvalue of $\mathfrak{t}$ to be any $\mu\in\mathbb{C}$ such that there exists some $u\in V$ so for all $v\in V$ we have $\mathfrak{t}(u,v) = \mu(u,v)$. I do not assume eigenvectors of $\mathfrak{t}$ are elements of $D$.
By Riesz there exists a solution operator $S_\sigma: V^*\to V$. There is a bounded map $P:H\to V^*$ given by $u\mapsto (v\mapsto (u,v))$. (This is bounded by an application of Cauchy-Schwarz.) Composing these with the compact injection $\iota$ gives a compact map $K = S_\sigma P\iota:V\to V$. Note that $K$ has the property that for any $u\in V$, 
$$ \mathfrak{t}_\sigma(Ku, v) = (u,v). $$
By the spectral theorem for compact operators, the spectrum of $K$ is discrete accumulating only at zero, and all nonzero elements of the spectrum are eigenvalues of finite multiplicity.
Suppose $u$ is an eigenvalue of $K$ with eigenvector $\mu$. Then for arbitrary $v\in V$,
\begin{align*}
\mathfrak{t}(u,v) &= \mathfrak{t}_\sigma(u,v) - \sigma(u,v) \\
  &= \frac{1}{\mu}\mathfrak{t}_\sigma(Ku, v) - \sigma(u,v) \\
  &= \bigg(\frac{1}{\mu}-\sigma\bigg)(u,v)
\end{align*}
and so we have that $\frac{1 - \mu\sigma}{\mu}$ is an eigenvalue of $\mathfrak{t}$ with eigenvector $u$.
Confusion

Does it follow that
  $$ \mbox{spectrum of } T = \bigg\{ \frac{1-\mu\sigma}{\mu}\ \bigg|\ \mu\in\mbox{spectrum of }K - \{0\} \bigg\}? $$

Thoughts
I've been worrying at this for a couple of days and trying to prove something like the following: If $\lambda$ is in the resolvent set of $T$, then $\frac{1}{\sigma+\lambda}$ is in the resolvent set of $K$. This would immediately imply what I want modulo showing that that eigenvalues of $\mathfrak{t}$ are contained in the spectrum of $T$. (This is not a concern, see the note below.) Some algebra and head-scratching later, nothing has panned out -- it looks promising, but I haven't found the zinger yet. 
Maybe more to the point, I don't have enough intuition for where the algebra ought to go. I've started by assuming that $u$ is a $\mu$-eigenvector of $K$ in $H$ (after extending showing $K$ is symmetric and extending $K$ to a map $H\to H$ by density of $V$ and compactness), assuming that $\frac{1}{\lambda + \sigma}$ is in the resolvent set of $T$, and showing that we cannot have $\mu = \frac{1}{\lambda + \sigma}$, but this approach doesn't seem to rely on the assumption that the domain of $(T-\lambda)^{-1}$  is all of $H$, nor that it is bounded.
I'm aware that this closely follows standard reference materials (Gilbarg-Trudinger Ch 8, Evans ch 6.5). In fact in Evans 6.5, at the end of the proof of Theorem 1, Evans writes, "But observe as well that for $\eta\neq 0$, we have $Sw = \eta w$ if and only if $Lw = \lambda w$ for $\lambda = \frac{1}{\eta}$." You might interpret this question as, "why does the 'only if' follow?"
There are at least two other approaches I'm aware of to showing the spectrum of $T$ is discrete. One uses the Rayleigh quotient, shows there exists a lowest eigenvalue, and proceeds by induction to exhaust $H$ with orthonormal eigenspaces (e.g. Gilbarg-Trudinger, last section of ch 8). The other uses the theory of spectral measures (e.g. ch 13 of Green Rudin). I'd prefer to avoid those approaches if possible because I'd like to understand all three in as much detail as possible, in order to more thoroughly understand how they relate to each other. 
(My intuition is that the exhaustion proof is related to how this proof would go if one unpacked the spectral theorem for compact operators, while the theory of spectral measures is "not homotopic," as it were, to this proof, but that's not yet clear to me.)
Anyway, I'm missing a piece of the puzzle here and I'm not sure where the missing piece fits to look more closely for it.
Notes
In the context of PDEs I'm happy to assume the facts bolded in the setup: $V$ is related to a Sobolev space $W^{1,1}$ so Rellich-Kondrachov comes into play, and elliptic regularity gives that the eigenvectors of $\mathfrak{t}$ are in fact eigenvectors of $T$.
If you've read this far, thanks for making it through my long-winded question :) 
 A: Great question, Neal! I've been thinking a lot about this recently, so I may be able to help you.

The trick here is that the map $K$ should be the same map as used in the construction of the Friedrichs extension.

Let's briefly recap that construction, for details see these nice notes by Paul Garrett, which in turn follow the classic text of Riesz-Nagy.
To construct the Friedrichs extension of a positive operator $T$, we construct the form domain $V$. By your assumption $V$ combactly embeds into $H$. Call this embedding $\iota$. Call the isomorphism given by the Riesz representation theorem $H\leftrightarrow^{R_H}H^*$ Now let's deconstruct the map Garrett calls "$B$":
$$ B: H\xrightarrow{R_H}H^*\xrightarrow{\iota^*} V^* \xrightarrow{R_V} V. $$
This is almost the map you call $K$. One now proves that it is injective, hence is invertible on its range. Call the inverse "$A$". After a little more work, we see the map $A$ IS the Friedrichs extension of $T+\sigma$. So indeed the map $K$ is the inverse of the Friedrichs extension of $T+\sigma$.
Now, you restricted the domain of $K$ to $V$ in order to make it compact. However, by Schauder's theorem (see this MSE answer for proof) because we have assumed $\iota$ is compact, we have that $\iota^*$ is compact. Thus $B$ is compact without first restricting its domain to $V$!
Now we're almost done. Because we constructed the Friedrichs extension of $T$ shifted by a scalar, the map $B$ is in fact a compact resolvent of the original map $T$! So now the theory of compactly resolved unbounded operators comes into play, and we have that the spectrum of $T$ is discrete, accumulates at infinity, and in fact comprises eigenvectors. Ta-da!
As for this result's relationship with the theory of spectral measures and the functional calculus, that's not so clear to me right now. I look forward to helping you figure that out in the future!
