Consider the autonomous linear discrete–time system $x(k+1) = Ax(k)$ with $A = \begin{bmatrix} 0.5 & 0 & 0 \\ 2 & -0.6 & 0 \\ -2 & -1 & 1.2 \end{bmatrix}$ and set $S = \{x \in \mathbb{R}^3 : ||x||_\infty \leq 1 \}$ Verify if the set $S$ is invariant for the system.
I know that I have to compute $||A||_\infty = 4.2$ and have to use the induced matrix norm. But I don't know exactly how I then proof that $S$ is invariant.
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and for
$A = \begin{bmatrix} 0.5 & 0 & 0 \\ 0.4 & -0.6 & 0 \\ -0.8 & -0.1 & 0.1 \end{bmatrix}$ ?
Unless I've misunderstood the question, I don't think the set is invariant:
$$\begin{bmatrix}0\\ 0\\ 1.2\end{bmatrix} = A\begin{bmatrix}0\\ 0\\ 1\end{bmatrix}$$
but
$$\begin{bmatrix}0\\ 0\\ 1\end{bmatrix}\in S,\quad\quad \begin{bmatrix}0\\ 0\\ 1.2\end{bmatrix}\not\in S.$$
The induced infinity norm of a matrix can be computed as the maximum absolute row sum of the matrix, that is
$$||A||_{\infty} = \max_{i}\sum_{j}|a_{ij}|$$
so in your case $||A||_{\infty}=1$.
By the definition of the induced norm, if $y=Ax$, then
$$||y||_\infty \leq ||A||_{\infty}||x||_{\infty}.$$
So if $x(t_0)\in S$
$$||x(t_0+1)||_{\infty}\leq ||A||_{\infty}||x(t_0)||_{\infty}\leq 1,$$
thus $x(t_0+1)\in S$. Forward invariance, that is $x(t_0)\in S\Rightarrow x(t)\in S$ for all $t\geq t_0$, follows by induction. Note however that $S$ is not backwards invariant, that is $x(t_0)\in S\Rightarrow x(t)\in S$ for all $t\leq t_0$, (and thus not invariant in the usual sense, which requires both forward and backward invariance):
$$\begin{bmatrix}0.75\\ 0\\ 0\end{bmatrix} = A\begin{bmatrix}1.5\\ 0\\ 0\end{bmatrix}$$
but
$$\begin{bmatrix}0.75\\ 0\\ 0\end{bmatrix}\in S,\quad\quad \begin{bmatrix}1.5\\ 0\\ 0\end{bmatrix}\not\in S.$$