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Within each fixed unitary irreducible representation, choose the selected right singular vectors to be orthonormal. Their corresponding left singular vectors are also orthonormal, even when a singular value is repeated: choose an orthonormal basis in each eigenspace of $T_{\rho }^{\ast }{T}_{\rho }$ and put $B_p=T_{\rho }{A}_p/{\lambda }_p$. Therefore, when ${\rho }_p={\rho }_q$,This is (iv), with the unnormalized Hilbert-Schmidt inner product and the Kronecker delta. No orthogonality of differently shaped matrices is being asserted.

For the unheaded continuation, partition $b\in {\mathbb{C}}^m$ into blocks $b_p\in {\mathbb{C}}^{n_p}$. ThenThe Schur averaging of rectangular matrices formula isfor any $n_p\times n_q$ matrix $M$. The average is an intertwiner; the Schur lemma makes it zero for inequivalent irreducible representations, and a scalar multiple of the identity for the same representative. The trace determines that scalar in the latter case.

Apply this with $M=U(p{)}^{\ast }U(q)$. The Hilbert-Schmidt inner product normalization just proved givesExpanding the squared Euclidean norm now yieldsThus the final identity follows from averaging, although $U^{\ast }U$ itself need not be the identity. The whole construction is the spectral inverse theorem for the matrix-valued U2 quantity.

For chosen inequivalent unitary irreducible representations $\rho ,\sigma$ of a finite group and a $d_{\rho }\times d_{\sigma }$ matrix $M$, the average ${\mathbb{E}}_x\rho (x)M\sigma (x{)}^{\ast }$ is zero when $\rho \ne \sigma$. For $\rho =\sigma$, it is $(\mathrm{tr}M/d_{\rho })I$. This follows from the Schur lemma, since the average intertwines the two group representations. Equivalent representations in different bases require the corresponding intertwiner; the identity-matrix formula assumes literally the same representative.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.