lemma is of enough interest to present in full.

\[zhu\] Let $\lambda$ be the spectrum of the Gram matrix $g$ of a normalized basis of positive semidefinite operators $\Pi_j$ sorted in nonincreasing order. Then $\lambda\succ\lambda_\textnormal{SIC}$ with equality iff $\Pi_j$ forms a SIC.

By assumption $\text{tr}\Pi_j^2=1$ for all $j$. Since the eigenvalues of $\Pi_j$ are nonnegative, $$1=\text{tr}\Pi_j^2=\sum_i\lambda^2_i(\Pi_j)\leq\sum_i\lambda_i(\Pi_j)=\text{tr}\Pi_j.$$ Define the frame superoperator $$\mathcal{F}=\sum_j\kketbra{\Pi_j}{\Pi_j}\;,$$ where $\kket{A}:=\sum_{ij}[A]_{ij}\ket{i}\ket{j}$. $\mathcal{F}$ has the same spectrum as the Gram matrix $[g]_{ij}=\bbraket{\Pi_i}{\Pi_j}=\tr\Pi_i\Pi_j$. To see this, form a projector out of the state $\sum_i\kket{\Pi_i}\ket{i}$ where $\ket{i}$ is an orthonormal basis in $\mathcal{H}_{d^2}$ and perform partial traces over each subsystem. The results are $g^{\rm T}$ and $\mathcal{F}$, and so, by the Schmidt theorem, the spectra of $\mathcal{F}$ and $g$ are equal: $\lambda(g)=\lambda(\mathcal{F})=\lambda$.

The expectation value of any operator with respect to an arbitrary normalized state is less than or equal to its maximal eigenvalue. Thus, a lower bound on the maximal eigenvalue $\lambda_1$ of $\mathcal{F}$ is given by $$\label{evalbound}
        \lambda_1\geq\frac{1}{d}\bbra{I}\mathcal{F}\kket{I}=\frac{1}{d}\sum_j(\tr\Pi_j)^2\geq d.$$ As our basis is normalized, $\tr g=d^2$, so $\sum_i\lambda_i=d^2$. With this constraint and our bound on the maximal eigenvalue, we have $$\begin{split}
        \lambda&\succ\left(\lambda_1,\frac{d^2-\lambda_1}{d^2-1},\ldots,\frac{d^2-\lambda_1}{d^2-1}\right)\\
        &\succ\left(d,\frac{d}{d+1},\ldots,\frac{d}{d+1}\right)=\lambda_{\rm SIC}.
    \end{split}$$ The second majorization becomes an equality when $\lambda_1=d$. From Eq. , we can see 