**Theorem on Iterative Method Convergence**

> 
> *If $\vert \vert I - Q^{-1}A \vert \vert <1$ for some subordinate matrix norm, then the sequence produced by $Qx^{(k)} = (Q-A)x^{(k-1)}
> + b$ converges to the solution of $Ax = B$ for any intial vector $x^{(0)}$.*
> 
> 
> 

My question is related to the proof of this theorem. In my textbook they have the following steps in the proof:

Obtain $x^{(k)} = (I-Q^{-1}A)x^{(k-1)} + Q^{-1}b$ from $Qx^{(k)} = (Q-A)x^{(k-1)} + b$

Now obtain the actual solution $x = (I-Q^{-1}A)x+Q^{-1}b$. Subtract these and obtain:

$x^{(k)}-x = (I-Q^{-1}A)(x^{(k-1)}-x)$

The following statements can be found:

$\vert \vert x^{(k)}-x \vert \vert \leq \vert \vert I-Q^{-1}A \vert \vert \cdot \vert \vert x^{(k-1)}-x\vert \vert $

By repeating this step obtain

$\vert \vert x^{(k)}-x \vert \vert \leq \vert \vert I-Q^{-1}A \vert \vert ^{k} \cdot \vert \vert x^{(0)}-x\vert \vert $

Thus if $\vert \vert I - Q^{-1}A \vert \vert <1$ then $\lim\_{k \to \infty}\vert \vert x^{(k)}-x \vert \vert = 0$ for any $x^{0}$

**My question**

I don't understand the last conclusion: 

> 
> Thus if $\vert \vert I - Q^{-1}A \vert \vert <1$ then $\lim\_{k \to \infty}\vert \vert x^{(k)}-x \vert \vert = 0$ for any $x^{(0)}$
> 
> 
> 

My quess is that when you compute a new $x^{(k)}$, you can interpretend this as a $x^{(0)}$ value, and therefore the next value $x^{(k+1)}$ will be closer to x?

Someone who can explain it?