I am thinking about the following linear algebra problem: $$ Ax = b $$ where $A$ is an $n$ by $n$ positive semi-definite matrix, in particular, it is rank $n-1$ with null space span$\{e=(1,1,\ldots,1)^\top\}$. We assume that $b$ is in the range of $A$ and we are looking for a solution $x$ in the range of $A$. The solution is unique.
I would like to assume that the Cholesky decomposition is given: $$ A = CC^\top$$ where $C$ is lower triangular matrix. Because $A$ is rank $n-1$, the right bottom most entry of $C$ will satisfy that $C(n,n)=0$.
The new system I get is now $$ CC^\top x = b$$ if we let $z=C^\top x$ and consider $Cz = b$, we can find first $n-1$ entries of $z$ by a lower triangular solve. The last entry of $z$ has to be zero because $C^\top$ is upper triangular and $C^\top(n,n) = 0$.
My problem is how do we do the next step, that is to find $x$ such that $e^\top x = 0$ and that $$ C^\top x = z$$ Of course this can be solved by adding the equation $e^\top x = 0$ into the linear system. But is there a good way where we can exploit the upper triangular property of $C^\top$? If possible, I would also assume that $C$ is sparse, it would be better if we could exploit the sparsity as well.
I found an easy way to do it. I can simply solve the system $$ C^\top y = z$$ By assuming $y(n) = 0$, I can obtain a unique solution $y$ via upper triangular solve. The difference between $x$ and $y$ satisfies that $$ C^\top (x-y) = 0$$ therefore it lives in the null space of $C^\top$, i.e. the span of $e$. That means $$ x-y = c e$$ Notice that $e^\top x = 0$, we must have $$ e^\top(x-y) = -e^\top y = c e^\top e = cn $$ so $c = mean(y)$, we can easily see that $$ x = y - \bar{y} e $$ where $\bar{y}$ is the average of $y$.