Here is interesting linear algebra fact: let $A$ be an $n \times n$ matrix and $u$ be a vector such that $u^{\top}A = \lambda u^{\top}$. Then for any matrix $B$, $u^{\top}((A-B)(\lambda I - B)^{-1}) = u^{\top}$.

The proof is just basic algebra: $u^{\top}(A-B)(\lambda I - B)^{-1} = (\lambda u^{\top} - u^{\top}B)(\lambda I - B)^{-1} = u^{\top}(\lambda I - B)(\lambda I - B)^{-1} = u^{\top}$.

Why care about this? Let's imagine that $A$ is a (not necessarily symmetric) stochastic matrix, so $1^{\top}A = 1^{\top}$. Let $A-B$ be a low-rank approximation to $A$ (so $A-B$ consists of all the large singular values, and $B$ consists of all the small singular values). Unfortunately since $A$ is not symmetric, this low-rank approximation doesn't preserve the eigenvalues of $A$ and so we need not have $1^{\top}(A-B) = 1^{\top}$. The $(I-B)^{-1}$ can be thought of as a "correction" term such that the resulting matrix is still low-rank, but we've preserved one of the eigenvectors of $A$.