proposition. Full-matrix Muon and Shampoo preserve the orthogonal gauge [ftip-00BA]
✍️source

    Fix one factor, write \(n\in \{d_1,d_2\}\) for its row dimension, and let \(G_t,M_t\in \mathbb R^{n\times k}\) be its gradient and Muon momentum buffer. For \(0\leq \beta <1\), \(M_{-1}=0\), and the source's linear momentum update \(M_t=\beta M_{t-1}+G_t\), Muon's direction is \(\Delta _t=\operatorname {msign}(M_t)\). Under real arithmetic this stateful update and the source's damped Shampoo update are right-equivariant. The Muon claim includes the stated finite Newton--Schulz approximations; the Shampoo claim uses its complete left and right accumulators.

    Proof. 

      By Lemma [ftip-00B1], the primed gradient is \(G'_t=G_tQ\). Starting from \(M'_{-1}=M_{-1}Q=0\), induction through the linear momentum update gives \(M'_t=M_tQ\). If \(M_t=A\Sigma B^{\mathsf T}\) is a compact singular-value decomposition, then \(M_tQ=A\Sigma (Q^{\mathsf T}B)^{\mathsf T}\), so \(\operatorname {msign}(M_tQ)=\operatorname {msign}(M_t)Q\). For \(\epsilon _{\rm NS}>0\), the Newton--Schulz initialization \(X_0=M_t/(\lVert M_t\rVert _F+\epsilon _{\rm NS})\) is covariant because its Frobenius normalization is invariant; each polynomial iterate formed from \(X_j\) and \(X_jX_j^{\mathsf T}\) preserves the same covariance. For Shampoo, the left accumulator is invariant while the right accumulator transforms by \(R_t\mapsto Q^{\mathsf T} R_t Q\). Orthogonal functional calculus conjugates the right matrix function, so the transformed update is the original update right-multiplied by \(Q\).

      Proposition 4.2(3)--(4) and its proof are in Appendix B.3 of The loss does not see the basis, but Adam does[singh2026lossbasis]. Variable splitting, coordinatewise clipping, mixed update rules, finite precision, and different damping conventions lie outside this statement. The Muon part is stateful; it does not replace the momentum buffer by the current gradient or invoke the memoryless classification of Definition [ftip-00B7].