Suppose that I have a program that can solve the system of linear equation $Ay=x$ efficiently when $A$ is a Hermitian matrix. I have a system of equations which is $(I+bH)y=x$, where $I$ is the identity matrix and $b$ is a complex scalar and $H$ is the hermitian matrix. Obviously $(I+bH)$ is not hermitian and neither is $(I/b+H)$. Can I reformulate this problem in terms of the Hermitian problem: $Ay=x$ ?


1 Answer 1


Introduce a variable $z=(I+bH)^*x$. Where $(I+bH)^*$ denotes the conjugate transpose of $(I+bH)$. Then solve $(I+bH)^*(I+bH)y=z$ using your method for Hermitian matrices since $(I+bH)^*(I+bH)$ is Hermitian.

$y$ doesn't necessarily solve your original problem, but if it doesn't, then it is a least squares solution to your equation if $(I+bH)^*(I+bH)$ is not singular.

If $(I+bH)^*(I+bH)$ is singular then $y=(I+bH)^+(I+bH)^{+*}z$ is a least squares solution to your original equation, where $A^+$ denotes the Moore-Penrose pseudoinverse of A.

Assuming your algorithm is coded correctly though, it should handle both of these cases already.

  • $\begingroup$ This does not solve the problem since you have shifted the problem to $z=(I+bH)^*x$ even if $z$ is computed correctly. You recover the original problem when obtaining $x$. $\endgroup$
    – user13838
    Commented Dec 2, 2011 at 15:08
  • $\begingroup$ It was my understanding that y is the variable vector and x is some known vector. This is the usual use of the notation, and since the OP did not specify, then I assumed it was the case. If it is not the case that x represents some known vector, then it is not really clear what the poster is asking. In any case: For any known x we can calculate the corresponding y(provided it exists) using the method I described $\endgroup$ Commented Dec 10, 2011 at 20:18
  • $\begingroup$ @percusse or to rephrase: I am assuming we are looking for some relation between y and x. I was assuming we are looking for y as a function of x, since the other direction is trivial. We define this relationship then in terms of it's values given a particular x. This is the only reasonable interpretation of the poster's notation that I could think of, and my solution solves this. $\endgroup$ Commented Dec 10, 2011 at 20:49
  • $\begingroup$ I understand your solution but it is still using a intermediate step where the linear equation $z=(I+bH)^*x$ with a non-hermitian matrix. This is what OP is trying to by-pass, that is solving a linear system with a non-Hermitian A matrix. $\endgroup$
    – user13838
    Commented Dec 11, 2011 at 20:11
  • $\begingroup$ @percusse You apparently don't understand my solution then. We don't have to solve the system $z=(I+bH)^*x$ . x is a given. We only have to apply the transformation to x so that we can obtain z. We never have to find x in terms of z, so there is no solving going on. The only system we have to solve is the Hermitian one. $\endgroup$ Commented Dec 13, 2011 at 9:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .