- If the ensemble is invariant under every transformation \( H \mapsto R H R^T \) with \( R \) an orthogonal matrix
- and if all components \( H_{ij}, i \le j \) are independent

I prove here the converse, namely, the probability measure \eqref{eq:20151025e} is invariant under transformations \( H \mapsto R H R^T \).

I first prove that the exponent is invariant and secondly that the absolute value of the Jacobian of the transformation is \(1\).

- The exponent is clearly invariant because a short calculation gives \begin{equation*} \mathrm{tr}(RHR^T) = \mathrm{tr}(R^TRH) = \mathrm{tr}(H) \end{equation*} and similarly \begin{equation*} \mathrm{tr}(RHR^T)^2 = \mathrm{tr}(RHR^TRHR^T) = \mathrm{tr}(R^T RHR^TR H) = \mathrm{tr}(H^2) \end{equation*} where I used the cyclicity of the trace and \( R^T R = 1 \).
- The Jacobian of a linear transformation is equal to the determinant of this linear transformation. I therefore need to know the determinant of the linear transformation \( H \mapsto R H R^T \) on symmetric matrices \( H \). A bit of work shows that this determinant is equal to \( \left( \mathrm{det} R \right) ^{n+1} \). Because \( R \) is orthogonal, it follows that \( \mathrm{det} R = \pm 1 \) and thus \(\left( \mathrm{det} R \right) ^{n+1} = \pm 1\).

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