# Solving a linear equation given the solution of another

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.

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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 $\lambda$.
@Laurent. Maybe you wanted to say "$x+\lambda e$ is also a solution for any real $\lambda$"? In fact, these are all the solutions, since $[e]$ is the nullspace of $S$. – a.r. Aug 20 '10 at 13:53