Computing the eigenvalues of the precision operator $C_0^{-1}=\eta(-\triangle)^p+KI$ 
Consider $L_2(\mathbb{T})$ with the basis $$\phi_{2k}(x)=\sqrt{2}\cos(2\pi k x)\\ \phi_{2k-1}(x)=\sqrt{2}\sin(2\pi k x)$$


for $k\in\mathbb{N}$. The functions $\phi_k$ belong to the domain $H^{2p}$ of the operator $C_0^{-1}=\eta(-\triangle)^p+KI$ where $\triangle$ is the Lapalcian and I the identity,K and $\eta$ are constants.
A simple application of the operator on the basis above yields:
$$C_0^{-1}\phi_{2k-1}=\eta((4\pi^2k^2)^p+k)\phi_{2k}\\C_0^{-1}\phi_{2k}=\eta((4\pi^2k^2)^p+K)\phi_{2k-1} $$


It follows that $C_0$ is the operator on $L_2(\mathbb{T})$ which is diagonalized by the basis of $\phi_k$, with eigenvalues:


$$\lambda_k=\eta \left(\left( 4\pi^2\left[\frac{k}{2}\right]^2\right)^p +K\right)^{-1} $$

I am not understanding how the eigenvalues are computed in order to diagonalize the operator. I cannot visualize a matrix that contains the operator.
Question:
How do I compute the eigenvalues $\lambda_k$?
Thanks in advance!
 A: There seem to be some typos in the above question. Here is how to calculate the eigenvectors of $C_0$ for $C_0^{-1} = \eta(-\Delta)^p + KI$:
At first, your formula with $C_0^{-1}$ applied to $\phi_{2k}$ and $\phi_{2k-1}$ doesn't seem quite right to me. Keep in mind that $-\Delta$ is a 2nd derivative. So $\cos$ becomes $\cos$ again and $\sin$ becomes $\sin$.
By taking second derivatives, you may verify that
\begin{align}
-\Delta \phi_{2k} &= 4 \pi^2 k^2 \phi_{2k}\\ -\Delta \phi_{2k-1} &= 4 \pi^2 k^2 \phi_{2k-1}
\end{align}
So $ \phi_{2k}, \phi_{2k-1} $ are eigenvectors of $ -\Delta $ with the eigenvalues $ 4 \pi^2 k^2 $ (the equation above is the very definition of an eigenvector with an eigenvalue).
Since the $ \phi_{2k}, \phi_{2k-1} $ form an orthonormal basis in $ L_2(\mathbb{T}) $, functions of $ - \Delta $ can be defined by the so-called spectral calculus: we define any function $ f(-\Delta) $ as being the operator which multiplies by $ f(4 \pi^2 k^2) $, i.e.
\begin{align}
f(-\Delta) \phi_{2k} &= f(4 \pi^2 k^2) \phi_{2k}\\ f(-\Delta) \phi_{2k-1} &= f(4 \pi^2 k^2)  \phi_{2k-1}
\end{align}
Now, $C_0^{-1} = C_0^{-1}(-\Delta) = \eta(-\Delta)^p + KI$ is such a function of $ -\Delta $, so
\begin{align}
C_0^{-1} \phi_{2k} &= (\eta(4 \pi^2 k^2)^p+K) \phi_{2k}\\ C_0^{-1} \phi_{2k-1} &= (\eta(4 \pi^2 k^2)^p+K)  \phi_{2k-1}
\end{align}
Similarly, $ C_0 $ acts as a multiplication
\begin{align}
C_0 \phi_{2k} &= (\eta(4 \pi^2 k^2)^p+K)^{-1} \phi_{2k}\\ C_0 \phi_{2k-1} &= (\eta(4 \pi^2 k^2)^p+K)^{-1}  \phi_{2k-1}
\end{align}
That $ C_0 $ acts as a pure multiplication on $ \phi_{2k}, \phi_{2k-1} $ is the same as saying, $ \phi_{2k}, \phi_{2k-1} $ are eigenvectors of $ C_0 $. The multiplication factors are the eigenvalues
$$ \lambda_k = (\eta(4 \pi^2 k^2)^p+K)^{-1}, $$
you are looking for.
It seems like within the result given in the question, the $ \eta $ is in a wrong position and there is an additional $ 1/2 $ within the $k$.
In such a situation, it makes sense to check again, whether either the definition of $ C_0 $ or the result for $ \lambda_k $ in your question is copied correctly.
The matrix visualization is a bit harder: for each choice of a basis, the operator $ C_0 $ becomes an infinitely large matrix (loosely speaking $ \infty \times \infty $). Within the basis consisting of $ \phi_{2k}, \phi_{2k-1} $, this matrix is diagonal and consists of $ 2 \times 2 $ blocks of the form $ \begin{pmatrix} (\eta(4 \pi^2 k^2)^p+K)^{-1} &0\\0&(\eta(4 \pi^2 k^2)^p+K)^{-1} \end{pmatrix} $ with increasing integers $ k \ge 0 $. The constant $K > 0$ is needed, such that all $ (\eta(4 \pi^2 k^2)^p+K) $ are nonzero, so the matrix entries really exist.
However, it is often inconvenient to represent such operators by matrices as they are infinitely large and heavily depend on your basis choice. It makes only sense if you compute with finitely many matrix elements.
