TITLE: Solving a linear equation given the solution of another
QUESTION [6 upvotes]: Suppose I have a matrix $S$ having a one-dimensional nullspace $\{ e \}$ such that $S + ee^\top$ is a positive definite symmetric matrix.
Now let $b \in Range(S)$ and suppose I solve the equation $(S + ee^\top)x = b$ is there anyway I can derive the solution $x'$ of the equation $Sx' = b$?   I was trying a Sherman Morrison Woodbury type formula, but this fails since the denominator is $0.$
Any help would be appreciated.

REPLY [7 votes]: Suppose $x$ is the solution to $(S+ee^T)x = b$. Since $b\in Range(S)$, we may write $b = Sz$ for some z. Now compute:
$$
e^T(S+ee^T)x = e^Tb = e^TSz
$$
Since S is symmetric, and e is in its nullspace, we have $e^TS = 0$. So the above equation simplifies to $e^Tx = 0$. But this implies
$$
(S+ee^T)x = Sx
$$
So x is a solution to the equation $Sx=b$ as well. As noted above, the solution to $Sx=b$ is not unique; $x + \lambda e$ is also a solution for any real $\lambda$.