Let tall matrices $\mathbf{G}, \mathbf{U} \in \mathbb{R}^{m \times (m-k)}$ be full-column rank, where matrix $\mathbf{U}$ is semi-orthogonal, i.e., $\mathbf{U}^T \mathbf{U} = \mathbf{I}_{m-k}$. The columns of $\mathbf{G}$ and $\mathbf{U}$ span the same subspace, $\mathbb R^{m -k}$. Let $\mathbf{S} \in \mathbb{R}^{(m-k) \times (m-k)}$ be a change-of-basis matrix such that:
$$ \mathbf{G}\mathbf{S} = \mathbf{U}. $$
Can I express a solution of the previous equation as $\mathbf{\bar{S}}= \mathbf{G}^\dagger \mathbf{U}$, where $\mathbf{G}^\dagger$ is the pseudo-inverse of $\mathbf{G}^\dagger$?
Does something change if all the matrix are complex (i.e. $\mathbf{G} \in \mathbb{C}^{m \times (m-k)}$, $\mathbf{U} \in \mathbb{C}^{m \times m-k}$ and $\mathbf{S} \in \mathbb{C}^{(m-k) \times (m-k)}$)?