I can't remember a name for this "matrix decomposition" (which isn't one really, but I would still like to work with it).
Given a set of $2N$ $D\times 1$ column vectors $u_i,v_i$, $i=1,\dots,N$, one can form a $D\times D$ matrix $$M=1+\sum u_i v_i^T$$ where $1$ is the $D\times D$ identity matrix. I have no particular restriction on $N$ compared to $D$ so for $N$ large enough, you overparametrize the matrix.
I've been trying to calculate the determinant of $M$ explicitly (for all N), in terms of the individual scalar products $(u_i^T,v_i$), but it's taking me a little longer than I thought it would, and it's irritating me that I can't find the very name of this "decomposition" (I haven't studied maths in English, usually I get by, but not this time).
First, note that we can write $$ \sum_{i} u_i v_i^T = \pmatrix{u_1 & \cdots & u_k} \pmatrix{v_1 & \cdots & v_k}^T = UV^T. $$ With that, the "Sylvester" matrix identity gives us $$ \det(M) = \det(1_D + UV^T) = \det(1_N + V^TU). % M^{-1} = (I + UV^T)^{-1} = I - U \left(I + V^TU \right)^{-1} V^T. $$ Note that $V^TU$ is your matrix of scalar products. In particular, we have $$ [V^TU]_{ij} = v_i^T u_j. $$