$n_S\times n_S$ matrix which only needs to satisfy $$V=U^T U$$ It is evident that $U$ is not uniquely determined. It is only determined up to an orthogonal transformation. We may take advantage of this freedom to do additional orthogonal transformation to further diagonalize the scalar mass matrix while still keeping scalar kinetic terms in their canonically normalized form.

After the transformation Eq.  we obtain $$\begin{aligned}
\mL_{quad}+\mL_{gf}\supset & \frac{1}{2}(\partial_\mu S_i)(\partial^\mu S_i)
-\frac{1}{2}\xi^p\tilde{G}_p^2-\frac{1}{2}((U^{-1})^T\bM_G^2 U^{-1})_{ij}S_i S_j \nonumber \\
& -\frac{1}{2\xi^p}(\partial_\mu Z_p^\mu)^2+\frac{1}{2}(\bM_V^2)_{pq}Z_{p\mu}Z_q^\mu
\label{eq:quad1}\end{aligned}$$ In the above equation $\tilde{G}_p$’s can be viewed as linear combinations of $S_i$’s. It should be noted from a physical perspective that the $n_S$ scalar degrees of freedom with which we started could be divided into two categories (after appropriate linear combinations if needed): unphysical scalars and physical scalars. Specifically, $n_M$ unphysical scalars should exist and serve as unphysical Goldstones to be eaten by $n_M$ gauge bosons to make them massive. The remaining $n_E=n_S-n_M$ scalar degrees of freedom then must be physical scalars. By virtue of this observation, there must exist an orthogonal transformation $$\bar{S}_i=P_{ij}S_j
\label{eq:pt}$$ which diagonalizes the $-\frac{1}{2}((U^{-1})^T\bM_G^2 U^{-1})_{ij}S_i S_j$ term. Then Eq.  becomes $$\begin{aligned}
\mL_{quad}+\mL_{gf}\supset\frac{1}{2}(\partial_\mu\bar{S}_i)(\partial^\mu\bar{S}_i)
-\frac{1}{2}\xi^p\tilde{G}_p^2-\frac{1}{2}\nu_r^2\bar{S}_r^2
-\frac{1}{2\xi^p}(\partial_\mu Z_p^\mu)^2+\frac{1}{2}(\bM_V^2)_{pq}Z_{p\mu}Z_q^\mu
\label{eq:quad2}\end{aligned}$$ The index $r$ ranges from $n_M+1$ to $n_S$ (this 