Given a linear system $A\textbf{x}=\textbf{b}$, we can express it into the easier-to-solve right-preconditioned form:

$$ AM^{-1}\textbf{y}=\textbf{b}, \quad \textbf{y}= M\textbf{x} $$

On the other hand, the left-preconditioned system is:

$$ M^{-1}A\textbf{x}=M^{-1}\textbf{b} $$

From the literature we know it's easy to show that the left-preconditioned linear system above is equivalent to a fixed-point linear iteration as:

$$ \textbf{x}^{(k+1)}=(I-M^{-1}A)\textbf{x}^{(k)}+M^{-1}\textbf{b} $$

Or equivalently:

$$ \textbf{x}^{(k+1)}=\textbf{x}^{(k)} + M^{-1}\textbf{r}^{(k)} $$

My question is: how can I get an update scheme like those starting from the right-preconditioning formulation?


1 Answer 1


Starting with

$AM^{-1} \mathbf{y} = \mathbf{b}$, where $\mathbf{y} \equiv M \mathbf{x}$,

we can manipulate to

$\mathbf{y} - (I-AM^{-1})\mathbf{y} = \mathbf{b}$

Replacing one instance of $\mathbf{y}$ by $\mathbf{y}^k$ and the other by $\mathbf{y}^{k+1}$ yields the update equation

$\mathbf{y}^{k+1} = (I-AM^{-1})\mathbf{y}^k + \mathbf{b}$

which can also be written as

$\mathbf{y}^{k+1} = \mathbf{y}^k + (\mathbf{b}-AM^{-1}\mathbf{y}^k)$


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