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\[appendix:equal-spectra\]

Appendix A: Proof of Lemma \[maj\]
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For a MIC $\{E_i\}$ and a post-measurement set $\{\sigma_j\}$, $$\left[\Phi^{-1}\right]_{ij}=\tr E_i\sigma_j\;.$$ The elements of the MIC may be expanded in the SIC basis $$\label{alpha}
    E_i=\sum_k\left[\alpha\right]_{ik}H_k\;,$$ so we may write $$\left[\Phi^{-1}\right]_{ij}=\sum_k\left[\alpha\right]_{ik}\tr H_k\sigma_j=\sum_k\left[\alpha\right]_{ik}p(k|j)\;,$$ where $p(k|j)$ is the probabilistic representation of the state $\sigma_j$ with respect to the SIC $\{H_k\}$. The $\alpha$ matrix must be invertible because it is a transformation between two bases, so the probability vectors can be written $$\label{alphaconditionals}
    p(i|j)=\sum_k\left[\alpha^{-1}\right]_{ik}\left[\Phi^{-1}\right]_{kj}\;.$$ We know that SIC probability vectors satisfy [@Fuchs:2013] $$\sum_ip(i|j)^2\leq\frac{2}{d(d+1)}\quad \forall j\;,$$ so we have $$\label{sicbound}
    \sum_{i}\left(\sum_k\left[\alpha^{-1}\right]_{ik}\left[\Phi^{-1}\right]_{kj}\right)^{\!2}\leq \frac{2}{d(d+1)}\quad \forall j\;.$$ Summing over $j$, we then have $$\sum_{ij}\left(\sum_k\left[\alpha^{-1}\right]_{ik}\left[\Phi^{-1}\right]_{kj}\right)^{\!2}\leq \frac{2d}{d+1}\;.$$ This expression is the sum of the absolute square entries of a matrix, which is equivalent to the square of the Frobenius norm of the matrix: $${\left\lVert\alpha^{-1}\Phi^{-1}\right\rVert}_2^2=\sum_is^2(\alpha^{-1}\Phi^{-1})\leq\frac{2d}{d+1}\;.$$ From [@Horn:1994] 3.1.11, for any square matrix $A$, $$\sum_i|\lambda_i(A)|^2\leq\sum_i\varsigma^2(A)\;,$$ so we have a general bound on the absolute squared spectrum: $$\label{SICupperbound}
    \sum_i|\lambda_i(\alpha^{-1}\Phi^{-1})|^2\leq \frac{2d}{d+1}\;.$$ Eq.  shows that $\alpha^{-1}\Phi^{-1}$ is column-stochastic and thus that one of its eigenvalues is $1$, so we may write: $$\sum_{i>1}|\lambda_i(\alpha^{-1}\Phi^{-1})|^2\leq\frac{2d}{d+1}-1=\frac{d-1}{d+1}\;.$$ Now, 