Suppose I have some matrix $G \in \mathbb{C}^{2 \times 4}$. Then the columns of $G$ are the Gram vectors of the positive semidefinite (psd) matrix $G^H G$ (Hermitian transpose of $G$ times itself). Now consider the matrix $$ P := G^H G +I.$$ Here, $I$ is the identity matrix of size 4 by 4.
Note that rank$(P)=4$, and that $P$ is psd. So there must exist some matrix $Z \in \mathbb{C}^{4 \times 4}$ such that $P = Z^H Z$. Can this matrix $Z$ be expressed in terms of $G$? I already have that $$P = \widetilde{G}^H \widetilde{G}, \text{ for } \widetilde{G} = \begin{bmatrix} G \\ \, I \end{bmatrix} \in \mathbb{C}^{6 \times 4}$$ But $\widetilde{G}$ has 6 rows, instead of 4. So I'm thinking there must exist some unitary matrix $R \in \mathbb{C}^{6 \times 6}$ such that $$R \widetilde{G} = \begin{bmatrix} Z \\ \mathbf{0}_{2 \times 4}\end{bmatrix},$$ but how can you find this $R$? And does a closed form exist?