A CMB-motivated rotation that separates the dark-energy equation-of-state plane into a direction the CMB constrains tightly and a direction it leaves degenerate. Below: the step-by-step construction of the matrix, then its application to the DESI DR2 best fit.
The CMB constraint on the comoving distance to last scattering \(D_M(z_\*)\) imposes a near-perfect degeneracy in the \((w_0, w_a)\) plane:
$$ w_a = A\,(w_0 + 1), \qquad A = -3.68, $$and the degeneracy line passes through \(\Lambda\)CDM at \((w_0, w_a) = (-1, 0)\).
Measured relative to \(\Lambda\)CDM we write \(\Delta w_0 = w_0 + 1\) and \(\Delta w_a = w_a\). The parallel direction runs along the degeneracy (left unconstrained by the CMB); the perpendicular direction is the one the CMB pins down:
$$ \vec{v}_\parallel = (1,\, A) = (1,\, -3.68), \qquad \vec{v}_\perp = (-A,\, 1) = (3.68,\, 1). $$The common normalisation factor is
$$ N = \sqrt{1 + A^2} = \sqrt{1 + (-3.68)^2} = \sqrt{14.5424} = 3.813, $$giving the orthonormal basis
$$ \hat{u}_\parallel = \tfrac{1}{N}(1,\, A) = (0.262,\, -0.965), \qquad \hat{u}_\perp = \tfrac{1}{N}(-A,\, 1) = (0.965,\, 0.262). $$Stacking \(\hat{u}_\parallel\) and \(\hat{u}_\perp\) as columns gives the transformation from rotated coordinates \((w_\parallel, w_\perp)\) to physical offsets:
$$ \begin{pmatrix} \Delta w_0 \\ \Delta w_a \end{pmatrix} = \begin{pmatrix} 1/N & -A/N \\ A/N & 1/N \end{pmatrix} \begin{pmatrix} w_\parallel \\ w_\perp \end{pmatrix} = \begin{pmatrix} 0.262 & 0.965 \\ -0.965 & 0.262 \end{pmatrix} \begin{pmatrix} w_\parallel \\ w_\perp \end{pmatrix}. $$Forward (rotated → physical):
$$ w_0 = -1 + \frac{w_\parallel - A\,w_\perp}{N}, \qquad w_a = \frac{A\,w_\parallel + w_\perp}{N}. $$Inverse (physical → rotated):
$$ w_\parallel = \frac{(w_0 + 1) + A\,w_a}{N}, \qquad w_\perp = \frac{-A\,(w_0 + 1) + w_a}{N}. $$The \(w_0\) row carries \(-A\,w_\perp\), matching the rotation matrix in Step 3 (with \(A<0\), this adds a positive \(|A|\,w_\perp\) term).
Imposing the CMB constraint on \(D_M(z_\*)\) sets the perpendicular coordinate to zero and lets the parallel coordinate vary freely along the degeneracy:
$$ w_\perp = 0, \qquad w_\parallel \ \text{free}. $$The relation then reduces to a one-parameter family:
$$ w_0 = -1 + \frac{w_\parallel}{3.813}, \qquad w_a = \frac{-3.68\,w_\parallel}{3.813}. $$From the DESI DR2 + Planck CMB \(w_0w_a\)CDM fit (arXiv:2503.14738, DESI + CMB, Eq. 25):
$$ w_0 = -0.42 \pm 0.21, \qquad w_a = -1.75 \pm 0.58. $$Using the inverse transformation with \(A = -3.68\) and \(N = 3.813\):
$$ w_\parallel = \frac{(w_0 + 1) + A\,w_a}{N} = \frac{0.58 + 6.44}{3.813} = \frac{7.02}{3.813} = 1.841, $$ $$ w_\perp = \frac{-A\,(w_0 + 1) + w_a}{N} = \frac{2.134 - 1.75}{3.813} = \frac{0.384}{3.813} = 0.101. $$The Jacobian of the physical → rotated map is
$$ J = \begin{pmatrix} 1/N & A/N \\ -A/N & 1/N \end{pmatrix} = \begin{pmatrix} 0.262 & -0.965 \\ 0.965 & 0.262 \end{pmatrix}. $$Assuming uncorrelated errors on \((w_0, w_a)\):
$$ \sigma_{w_\parallel}^2 = \left(\tfrac{1}{N}\right)^2 \sigma_{w_0}^2 + \left(\tfrac{A}{N}\right)^2 \sigma_{w_a}^2 = (0.262)^2(0.21)^2 + (0.965)^2(0.58)^2 = 0.00303 + 0.3133 = 0.3163, $$ $$ \sigma_{w_\parallel} = 0.562. $$ $$ \sigma_{w_\perp}^2 = \left(\tfrac{A}{N}\right)^2 \sigma_{w_0}^2 + \left(\tfrac{1}{N}\right)^2 \sigma_{w_a}^2 = (0.965)^2(0.21)^2 + (0.262)^2(0.58)^2 = 0.0410 + 0.0231 = 0.0641, $$ $$ \sigma_{w_\perp} = 0.253. $$DESI DR2 + CMB in rotated coordinates:
$$ w_\parallel = 1.84 \pm 0.56, \qquad w_\perp = 0.10 \pm 0.25. $$The perpendicular coordinate \(w_\perp\) is consistent with zero at the \(0.4\sigma\) level — i.e. the DESI DR2 + CMB best fit sits essentially on the CMB degeneracy line, displaced from \(\Lambda\)CDM almost entirely along the unconstrained parallel direction.