Perturbation Theory - Exact Solutions and Good Quantum States I'm having a problem with the following question:

Problem:
  Consider the unperturbed, degenerate Hamiltonian $H_0=\bigg(\begin{matrix} E &0\\
0& E\end{matrix}\bigg)$. Add the perturbation $H_1=\bigg(\begin{matrix} 0 &\delta\\
\delta& 0\end{matrix}\bigg)$, where $\delta<<E$, to form the full hamiltonian $H$.
a) Find the exact eigenstates and eigenvectors for $H$.
b) Use these to perform non-degenerate perturbation theory to first order. Why is this okay?
c) Consider the basis vectors $v_1=\frac{1}{\sqrt2}\bigg(\begin{matrix}1\\i\end{matrix}\bigg)$ and $v_2=\frac{1}{\sqrt2}\bigg(\begin{matrix}1\\-i\end{matrix}\bigg)$. Verify that these are eigenvectors
  of $H_0$. Use these to perform degenerate perturbation theory with the perturbation $H_1$. What are the
  resulting energy splittings? What are the “good" quantum states for the perturbed Hamiltonian?
  Compare these to the exact solution from part (a)

What I did/tried:
Part A is easy, just simple linear algebra: $u_1=\frac{1}{\sqrt{2}}\bigg(\begin{matrix}1\\1\end{matrix}\bigg)$ and $u_2=\frac{1}{\sqrt{2}}\bigg(\begin{matrix}1\\-1\end{matrix}\bigg)$, with $\lambda_1=E+\delta$, $\lambda_2=E-\delta$. 
Part B is fine as well. One gets that the energy shift for the first eigenvector in a) is $\delta$, and for the one in b), $-\delta$. But I'm unsure if I get the 'why is this okay'. Is it because the eigenvalues in a) were different?
For Part C, I'm able to check that the proposed vectors are indeed eigenvectors of $H_0$. But, performing degenerate perturbation theory, I get the following $W$ matrix, $W_{ij}=v_i^TH_1v_j$, $\bigg(\begin{matrix} 2\delta i & 0\\ 0&-2\delta i\end{matrix}\bigg)$, so that $E_{\pm} = \frac{1}{2}\big[W_{11}+W_{22} \pm \sqrt{(W_{11}-W_{22})^2+4|W_{12}|^2} \big]=\pm \delta i$, which is complex... I'm also a bit confused whether I get what the good states would be, from $W$ it seems that $v_1, v_2$ would be these good states? If it weren't for the $i$, the shift would be the same as in a.
So basically, I'm uncomfortable with the notion of good states here, the value I got for the shift in C and the question on why it was okay to do the non degenerate perturbation theory in b.
 A: 
Ques. 1: Why is it okay to apply nondegenerate perturbation theory in b)?

It's okay because $u_1$ and $u_2$ are "good states" here. This means they diagonalize* $H_1$. That is if you express $H_1$ in the $  \{u_1,u_2 \}$ basis, the diagonal elements are zero
$$H_1=\bigg(\begin{matrix} \delta &0\\
0& -\delta\end{matrix}\bigg)$$

Ques. 2:  the value I got for the shift in C?

The correct expression is $W_{ij}=(v_i^{*})^{T} H_1v_j$. You forgot to complex conjugate $v_i$. When applying the correct expression you will get
$$W=\bigg(\begin{matrix} 0 &i\delta\\
-i\delta& 0\end{matrix}\bigg)$$
and you get $E_{\pm}=\pm \delta$ as expected. Here $\{v_1,v_2 \}$ are bad states since the diagonal elements $W_{12}$ and $W_{21}$ are not zero, so you're forced to use nondegenerate perturbation theory.
*Caveat: The more precise definition of good states is this. Suppose you have $N$ eigenstates $\{\psi_1,\psi_2,...,\psi_N \}$ of $H_0$ where $N$ could be infinite. Among them, there are $n$ degenerate eigenstates $\{\psi_k,\psi_{k+1},...,\psi_{k+(n-1)} \}$ with same eigenvalue $E$. These degenerate states are said to be good states if they diagonalize your perturbation $H_1$ (i.e., $W_{ij}=0$ whenever $i \ne j$ where $i,j\in\{k,k+1,...,k+(n-1) \})$. You don't care if the set of all eigenstates $\{\psi_1,\psi_2,...,\psi_N \}$ diagonalize $H_1$, you only worry about the particular subset of degenrate eigenstates $\{\psi_k,\psi_{k+1},...,\psi_{k+(n-1)} \}$. Good states allow you to directly apply nondegenrate perturbation theory to find energy corrections.
Your example is very special since here $N=n=2$, so the set of all eigenstates is exactly the set of degenerate eigenstates of $H_0$.
To illustrate, suppose you have 3-level state (i.e., $N=3$) with two-fold degeneracy $n=2$. So it has three eigenstates$\{\psi_1,\psi_{2},\psi_{3} \}$ and suppose $\psi_{1}$ and $\psi_{2}$ are degenerate, then $H_0$ expressed in $\{\psi_1,\psi_{2},\psi_{3} \}$ basis will look like this
$$H_0=\Bigg(\begin{matrix} E &0 &0\\
0& E &0\\
0& 0 &E_3 \end{matrix}\Bigg)$$
$\{\psi_1,\psi_{2} \}$ are good states if they diagonalize $H_1$
$$W=\bigg(\begin{matrix} W_{11} &0\\
0& W_{22}\end{matrix}\bigg)$$
You don't care if the whole set of eigenstates diagonalize $H_1$ which could look like this
$$H_1=\Bigg(\begin{matrix} W_{11} &0 &W_{13}\\
0& W_{22} &W_{23}\\
W_{31}& W_{32} & W_{33} \end{matrix}\Bigg)$$
You only care that $H_1$ is diagonal in the degenerate subspace to be able to apply nondegenerate perturbation theory.
