Does the first-order energy correction in the degenerate case equals to the eigenvalues of the perturbation matrix?

According to Griffiths, the degenerate perturbation theory says that the first-order corrections to the energies are the eigenvalues of the perturbation matrix. Griffiths solves for the eigenvalues in the unperturbed energy eigenbasis. But, because the eigenvalues of a matrix are independent of the choice of basis, the eigenvalues are just the eigenvalues of the perturbation matrix $$\delta H$$ on any basis.

However, I encounter a problem where $$H_0=\frac{1}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \quad \text{and} \quad \delta H= \begin{pmatrix} 0 & 0 & \frac{1}{2} & \frac{1}{4} \\ 0 & 0 & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & 0 & 0 \end{pmatrix}.$$

The eigenvalues of the perturbation matrix $$\delta H$$ equals to $$-\frac{3}{4}, \frac{3}{4}, -\frac{1}{4}, \frac{1}{4}$$. But the solution says the first order corrections to the energies are $$-\frac{1}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{1}{4}$$. Why they are different?

The relevant matrix that you need to diagonalize is not $$\delta H$$ itself, but its projection onto each one of the degenerate subspaces of $$H_0$$. Since the $$E=1$$ subspace is generated by $$v_1=(1,0,0,0),v_2=(0,0,0,1)$$ the matrix you need to diagonalize is then $$\delta H_1 =\pmatrix{0 &\frac{1}{4}\\\frac{1}{4}&0}$$ (and similarly for $$-1$$) which will give the right corrections.