Suppose we have an $m \times n$ matrix $A$ represented in block form as
$$ A = \left[\begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$$
where $$A_{11}$$ is $k\times k$. The Schur complement of $A_{11}$ is $S = A_{22} - A_{21}A_{11}^{-1}A_{12}$. Now suppose $A$ underwent a rank 1 update. That is we computed
$$ B= A + uv^T$$
for $u \in \mathbb{R}^m$, $v \in \mathbb{R}^n$. Partitioning $B$ the same as $A$ we get a new Schur complement
$$ T = B_{22} - B_{21}B_{11}^{-1}B_{12} $$
Now since $B$ is the result of a rank 1 update to $A$, $T$ should also be the result of a rank one update of $S$. At least that's what my advisor says. Does anyone have any idea why this might be true? And better yet do you know of a way to actually compute the rank one update $uv^T$ such that $S + uv^T = T$?