Let $T: L^2(\mathbb R^d) \to L^2(\mathbb R^d)$ be defined by $\displaystyle (Tf)(x) = \int_{\mathbb R^d}f(y)K(|x-y|)dy$ where $K(|z|) = \frac{1}{1+|z|^\alpha}$ where $z \in \mathbb R^d$ and $\alpha > \frac{d}{2}$.
Show that the spectrum of $T$ (the set of eigenvectors) spans $L^2(\mathbb R^d)$
What I tried (false proof):
If we can show $T$ is a Hilbert-Schmidt integral operator, we can use the spectral theorem.
Notice that $\displaystyle \int \frac{1}{(1+|z|^\alpha)^2}dz = \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz + \int_{|z| > 1} \frac{1}{(1+|z|^\alpha)^2}dz \leq \\ \displaystyle \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz + \int_{|z| > 1} \frac{1}{(|z|^\alpha)^2}dz$.
$\displaystyle \int_{|z| \leq 1} \frac{1}{(1+|z|^\alpha)^2}dz \leq \int_{|z| \leq 1}1dz = \pi < \infty$ so this part is not problematic. Let's analyze the other part of the integral.
Define $A_k = \{z: 2^k < |z| \leq 2^{k+1}\}$. Then we have $\displaystyle \{z: |z| > 1\} = \bigcup_{k=0}^{\infty}A_k$, and because $A_k$ is simply every element of $A_0$ but multiplied by $2^k$, from dilation property of measures $m(A_k) = (2^k)^d m(A_0) = 3\cdot \pi \cdot 2^{kd}$
Finally then, \begin{align*} \int_{|z| > 1} \frac{1}{(|z|^\alpha)^2}dz &= \sum_{k=0}^{\infty}\int_{A_k}\frac{1}{(|z|^\alpha)^2}dz \leq \sum_{k=0}^{\infty}\int_{A_k}\frac{1}{(2^k)^{2\alpha}}dz \\ &= \sum_{k=0}^{\infty}2^{-2k\alpha}m(A_k) = 3\pi \sum_{k=0}^{\infty}2^{kd-2k\alpha} \\ &< \infty\end{align*}
This sum converges because $2\alpha > d$. Hence, $K(|z|) \in L^2(\mathbb R^d)$.
Now the false part
Because $K(|z|) \in L^2(\mathbb R^d)$, we can say that $K(|x-y|) \in L^2(\mathbb R^d \times \mathbb R^d)$. [Not really. This is actually false. Infact, $\iint |K(|x-y|)|^2dxdy = \infty]$
So we can say that $T$ is hilbert schmidt, the fact that it's self adjoint is trivial, and by the spectral theorem we are done.
As I said, there's a problem with this proof. Is there a way to make it correct? Is the problem statement even correct?
As remarked by DisintegratingByParts in the comments, the spectrum contains the eigenvalues (and possibly other spectral values), not the eigenvectors. As suggested by the OP, I'll read the question as: Are the eigenfunctions of $T$ total in $L^2(\mathbb{R}^d)$?
The answer is no, at least for $d=1$ and $\alpha=2$. In this case, the eigenfunction equation $Tf=\lambda f$ corresponds under Fourier transform to $$ e^{-|\xi|}\hat f(\xi)=\lambda \hat f(\xi). $$ This equation clearly does not have a non-trivial solution in $L^2$.
I don't expect the answer to be different for other values of $d$ of $\alpha$, but I don't have a proof (there don't seem to be any explicit formulas for the Fourier transform of $K$ in this case). The only multiplication operators with a total set of eigenfunctions are (possibly infinite) linear combinations of indicator functions, so it remains to show that the Fourier transform of $K$ cannot take this form. This seems at least plausible, since you have some decay at $\infty$.