Consider the following sequence of vectors for any given starting vector $x_0$: $$ x_k = x_{k-1} + B^{-1}(b - Ax_{k-1}) $$ I'm trying to prove that if $C = I- B^{-1}A$ has a spectral radius less than $1$ (i.e., the maximum absolute value of an eigenvalue of $C$ is less than $1$), then $x_k$ converges to the solution $x$ to $Ax = b$.
I proceeded as follows. We can write $x_k = Cx_{k-1} + B^{-1}b$. Now, it is pretty easy to prove that $$ x_k = C^kx_0 + \left(\sum_{j=1}^k C^j\right)B^{-1}b $$ Now, I'm using a pretty shaky argument here, but I'm pretty sure it's provable. Essentially, since $C$ has spectral radius less than $1$, $C^kx_0\to 0$ for any starting vector $x_0$ as $k\to\infty$. Furthermore (on similarly shaky ground), $\sum_{j=1}^k C^j\to (I - C)^{-1}$ as $k\to\infty$ (for the same reason that $C$'s spectral radius is less than $1$). I don't have an appropriate text on me, but I'm pretty sure a result like this is proven in most texts on operator theory.
So, lets assume for a second that my arguments are indeed correct. Then the limit of the sequence should be $x_\infty = (I - C)^{-1}B^{-1}b$, but since $C = I- B^{-1}A$, this is simply $B^{-1}AB^{-1}b$. Hence $$ Ax_\infty = AB^{-1}AB^{-1}b $$
Now, the point of the iterative method is to choose a matrix $B$ that approximates $A^{-1}$, but is easier to calculate than $A^{-1}$. So, in a sense $AB^{-1} = I$.
However, I'm pretty sure the result should be exact, in the sense that $Ax_\infty = b$ identically. So, is there somewhere I'm going awry (other than my shaky arguments)? Am I going in the right direction?
Edit: If I've done my algebra correctly, then the only way the convergence is exact is if $$ x_\infty = A^{-1}(I - B^{-1}A)^{-1}(I - AB^{-1})b $$ I haven't checked if this can be simplified further (i.e. using the Woodbury Matrix identity), but it seems unlikely that this is actually true.
I'm an idiot. $C = I - B^{-1}A$, and so $(I - C)^{-1} = (B^{-1}A)^{-1} = A^{-1}B$, so $$ x_\infty = A^{-1}BB^{-1}b = A^{-1}b $$ as expected. I must have been very tired last night.