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It is well known that every isometry of $\mathbb{R}^n$ has the form $$x\to f(x)=Ax+a \,\, , $$ with $a\in\mathbb{R}^n$ and $A$ and $n\times n$ orthogonal matrix. How can we compute the infimum of the distance between $f(x)$ and $x$ $$inf_{x\in\mathbb{R}^n}\, \|Ax+a-x\| \,\, ?$$

If $1$ is not an eigenvalue of $A$ then the answer is zero, since we can take $x=-(A-I)^{-1}a\,\, $. My guess is that, if $1$ is an eigenvalue of $A$ then the answer is $\|a\|$. Any idea how to prove it? Or is the answer completely different?

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    Your guess not correct: it can be $0$ even though $a\ne 0$. Consider $n=2$ and $$f(x,y)=(-x+2,y)$$ for which every point in the for $(1,y)$ is a fixed point. In general, it is still $0$ as long as $a\in\operatorname{im}(A-I)$. – 2017-02-09
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    @Sassatelli, Yes, that's rigth. So, let's rule out that case too. What about if $1$ is an eigenvalue of $A$ and $a\not\in im(A-I)$? – 2017-02-09

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