I have to determine the order of convergence of this iteration specification but I don't know how to do this in $\mathbb{R}^n$. (So far, I've only done this in $\mathbb{R}$). Any help is highly appreciated (and sorry for my bad english).
$x^{(k+1)} = \begin{pmatrix} 0 & -1/2 & -1/4\\ -1/2 & 0 & 0\\ 0 & -1/2 & 0 \end {pmatrix}x^{(k)}+\begin{pmatrix} 3/4 \\ -2 \\ -1/2 \end{pmatrix}$ for $k \geq 0$
$x^{(k)} = \begin{pmatrix} x_1^{(k)} \\ x_2^{(k)} \\ x_3^{(k)} \end{pmatrix}$
On
It works the same in $\Bbb R^n$ as in $\Bbb R$. Find the limit, which will be a vector. Add a different epsilon to each component of the vector, apply the iteration, and see how much closer to the limit the new value is. You will have three orders of convergence, one for each component. Report the poorest as the overall value.
Your iteration is special case of the stationary iteration $$x_{n+1} = Gx_n + f$$ which can occasionally be used to solve the linear system $$x = Gx +f.$$ The initial guess $x_0$ must be selected by the user, but $x_0 = 0$ is a perfectly acceptable choice. If $\|G\|_2<1$, then $I-G$ is nonsingular, and the sequence $\{x_n\}_{n=0}^\infty$ is convergent and $$x_n \rightarrow x = (I-G)^{-1} f, \quad n \rightarrow \infty, \quad n \in \mathbb{N}$$ regardless of the choice of $x_0$. If $x_0 = 0$, then by induction on $n$ you can establish that $$x_n = \sum_{j=0}^{n-1} G^j f.$$ It follows, that $$\|x-x_n\|_2 \leq \left \| \sum_{j=n}^\infty G^j f \right\|_2 \leq \|G\|_2^n\|x\|_2 = \epsilon_n.$$ It follows that $x_n \rightarrow x$ at least linearly as $n \rightarrow \infty$ and $n \in \mathbb{N}$ because $$ \frac{\epsilon_{n+1}}{\epsilon_n} = \frac{ \|G\|_2^{n+1} }{\|G\|_2^{n}} = \|G\|_2 \rightarrow \|G\|_2 < 1.$$ You are free to replace the $2$-norm with any other norm induced by a vector norm. In particular, you can use the $\infty$-norm for which your particular matrix $G$ satisfies $\|G\|_\infty = \frac{3}{4}$.
In any case, the order of convergence is at least $p=1$.