Let $G = AA^T$ be a gram matrix with $$ A=\begin{pmatrix} | & & | \\ d_1 & \dots & d_k \\ | & & | \end{pmatrix} $$ where $d_i\in\mathbb{R}^n$ and $n>k$. We are now interested in the sum of the identity $\mathbb{I}$ with the gram matrix $G$ $$ \mathbb{I} + G = \mathbb{I} + \sum_{i=1}^k d_i d_i^T. $$
I am looking for $B$ such that $$ \mathbb{I} + G = BB^T $$
Special Case
If the $d_i$ are orthogonal, we can complete them into an orthonormal basis $Q = (\frac{d_1}{\|d_1\|},\dots,\frac{d_n}{\|d_n\|})$ and immediately obtain the eigendecomposition $$ Q^T(\mathbb{I} + G) Q = \mathbb{I} + \text{diag}(\|d_1\|^2,\dots, \|d_k\|^2, 0\dots,0) $$ and thus $$ \mathbb{I} + G = B B^T $$ with $B = \Bigl(\frac{\sqrt{1+\|d_1\|^2}}{\|d_1\|} d_1, \dots, \frac{\sqrt{1+\|d_k\|^2}}{\|d_k\|} d_k, \frac{d_{k+1}}{\|d_{k+1}\|}, \dots, \frac{d_n}{\|d_n\|}\Bigr)$.
In particular for $k=1$ this is always possible.
Now I am asking, whether or not something similar is possible if the $d_i$ are not orthogonal and have to be orthogonalized first. I could not make it work so far.
