Questions about contraction

Question

Let $X$ be a metric space and $f:X\to X$. What's true and what's false?

a. If $f$ is bijective and has a unique fixed point, then $f^{-1}:X \to X$ also has a unique fixed point.

b. If $f$ is bijective, then $f$ is a contraction iff $f^{-1}$ is a contraction.

c. $f:\mathbb {R}^2\to \mathbb {R}^2$, $[f]=\begin {pmatrix}\frac {1}{2}&1\\0&\frac {1}{2}\end {pmatrix}$ is a contraction.

I think that a and b are true and I don't really understand c. Am I right about a and b and could someone explain c please?

Answer 1

a) is true

b) is not true: take $f:\mathbb{R}\to \mathbb{R}$, $x\mapsto x/2$. It is a contraction, but $f^{-1}:x\to 2x$ is an expansion.

c) is not true because the image of the vector $(1,1)$ of length $\sqrt{2}$ is $(1/2, 3/2)$ of bigger length $\sqrt{10}/2>\sqrt{2}$.

Answer 2

Clearly $a)$ is true, due the fact that if $x$ and $y$, $x\not=y$ are fixed points of $f^{-1}$, then $f^{-1}(x) = x$ and $f^{-1}(y)=y$. Applying $f$ both sides of these two equations is a contradiction because $f$ would have 2 fixed points.

$b)$ is not true, JCAA answered.

$c)$ $\begin{pmatrix}1/2 & 1\\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} = \begin{pmatrix} x/2+y \\ y/2\end{pmatrix}$, given $z=(1,1)$ and $w=(0,1)$ $$\Vert f(z) - f(w)\Vert = \sqrt{(3/2)^2 + (1/2)^2} = \sqrt{10/4} >1$$ $$\Vert z-w\Vert = 1$$ So $\Vert z - w \Vert < \Vert f(z) - f(w) \Vert$, then $f$ isn't a contraction.