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 :)
