Given a Hermitian matrix $A$ with a kernel (or nullspace) spanned by vector $u$, i.e. $Au=0$. Then for inhomogeneous equation $Ax=b$, must $(x,u) = 0$?
In other words, for singular Hermitian $A$, does a particular solution to the inhomogeneous problem have to be orthogonal to the homogeneous solution space?
Suppose $x$ is a solution to the inhomogenous. Then $x + cu$ is also a solution for any scalar $c$.
Now assume all particular solutions are orthogonal to u. Then $$\langle u, x \rangle = 0$$ and $$\langle u, x + cu \rangle = \langle u, x \rangle + c||u||^2 = 0.$$ This implies $$c||u||^2 = 0.$$ But $u$ is a nonzero vector, and so it has a positive norm, and we have reached a contradiction. Therefore the assumption was false; not all particular solutions to the inhomogenous are orthogonal to $u$.
