contraction in five dimensional Euclidean space

Question

$T(u,v,w,x,y)=(v,0.5w,x,y,u)$

$S(u,v,w,x,y)=(y,u,v,0.5w,x)$

(a) Could anyone tell me whether S and T are contraction maps on any matrix norm/induced vector norm?

(b) Is their composition both way i.e ST and TS, contraction?

(c) What can we say about the Contraction property of (SSSS)(TTTT) i.e $S^4T^4$ and the other-way $T^4S^4$? In other words even if $T$, $S$ are not a contraction, but does there exists $k_1, k_2\in\mathbb N$ such that $T^{k_1}$ is a contraction and $S^{k_2}$ is a contraction?

Thank you for your response.

Answer 1

First, $T$ and $S$ cannot be strict contraction mappings with respect to the same norm. To see this, suppose for contradiction that there exists a norm $\lVert \cdot \rVert$ on $\mathbb{R}^5$ (or $\mathbb{C}^5$, doesn't matter) with respect to which $T$ and $S$ are both strict contractions. This means that $\lVert T(x) \rVert < \lVert x \rVert$ and $\rVert S(x) \rVert < \lVert x \rVert$ for all $x \in \mathbb{R}^5 \setminus \{0\}$.

Now $$\lVert (0,1,0,0,0) \rVert = \lVert S(1,0,0,0,0) \rVert < \lVert (1,0,0,0,0) \rVert = \lVert T(0,1,0,0,0) \rVert < \lVert (0,1,0,0,0) \rVert,$$ which is a contradiction.

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On the other hand, $T$ and $S$ are both non-strict contractions with respect to the $\sup$ norm $\lVert (a,b,c,d,e) \rVert := \sup \{\lvert a \rvert, \dots, \lvert e \rvert\}$. I'll give the proof for $T$; the proof for $S$ is completely analogous.

Let $(a,b,c,d,e) \in \mathbb{R}^5$ be arbitrary. Then

$$\lVert T(a,b,c,d,e) \rVert = \lVert (b,0.5c,d,e,a) \rVert = \sup \{\lvert b \rvert, \lvert 0.5c \rvert, \lvert d \rvert, \lvert e \rvert, \lvert a \rvert\} = \sup \{\lvert a \rvert, \lvert b \rvert, 0.5 \lvert c \rvert, \lvert d \rvert, \lvert e \rvert\} \leq \sup \{\lvert a \rvert, \lvert b \rvert, \lvert c \rvert, \lvert d \rvert, \lvert e \rvert\} = \lVert (a,b,c,d,e) \rVert,$$

as desired.

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In all of the above, I'm using the fact that $d(Tx, Ty) := \lVert Tx - Ty \rVert = \lVert T(x-y) \rVert$, so $d(Tx, Ty) \leq d(x,y)$ for all $x,y$ is equivalent to $\lVert Tx \rVert \leq \lVert x \rVert$ for all $x$ (in one direction, take $y = 0$).

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Edit (b) and (c) seem to contain many different questions with many different answers. I'll try answer as many of them as I can identify.

• Neither $ST$ nor $TS$ can be a strict contraction with respect to any norm. This is because $ST(1,0,0,0,0) = (1,0,0,0,0) = TS(1,0,0,0,0)$, and strict linear contractions cannot have nonzero fixed points.
• $S^4 T^4$ is given by $$(a,b,c,d,e) \mapsto (\frac{1}{4}a, \frac{1}{2}b, \frac{1}{2}c, \frac{1}{4}d, \frac{1}{4}e).$$ This a strict contraction with respect to any $\ell^p$ norm (including the $\sup$ norm). Likewise, $$T^4 S^4 = (a,b,c,d,e) \mapsto (\frac{1}{4}a, \frac{1}{4}b, \frac{1}{2}c, \frac{1}{2}d, \frac{1}{4}e)$$ is a strict contraction with respect to any $\ell^p$ norm.
• $S^5$ and $T^5$ both equal $\frac{1}{2} \operatorname{id}$, so both are strict contractions with respect to any norm whatsoever. Of course, this implies that $S^{5k} T^{5k} = T^{5k} S^{5k} = \frac{1}{2^{k+1}} \operatorname{id}$ is a strict contraction for every positive integer $k$.
• For $k \geq 4$, $S^k T^k$ and $T^k S^k$ will be strict contractions with respect to any $\ell^p$ norm. For $1 \leq k \leq 3$, $S^k T^k$ and $T^k S^k$ have nonzero fixed points, so are not strict contractions with respect to any norm.