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Suppose $f$ $-$ isometry of the affine space. $f=Ax+b$ where $A \in O(n)$ and $b$ $-$ is a vector. Proposition. f has a fixed point $\iff b$ is orthogonal to space of eigenvectors with eigenvalue equal to $1$.

Is it right? How can I show this?

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Without loss of generality we may suppose that the matrix $A$ is diagonal with elements $(c_1,\ldots,c_k,\overbrace{1,1,\ldots,1}^{\text{ n-k times}})$ on the diagonal with respect to a basis $(e_1,\ldots,e_k, e_{k+1},\ldots,e_n)$. The subspace $V_1$ spanned by the vectors $(e_1,\ldots,e_k)$ is orthogonal to the subspace $V_2$ spanned by $(e_{k+1},\ldots,e_n)$, the former consisting of eigenvectors corresponding to eigenvalues $\neq 1$, the latter to eigenvectors with eigenvalues $=1$. The equation for a fixed point $x$ gives us $\begin{cases} (c_1-1)x_1 + b_1 & = 0\\ (c_2-1)x_2 + b_2 & = 0\\ \ldots \\ (c_k-1)x_k + b_k & = 0\\ b_{k+1}& = 0\\ \ldots\\b_n & = 0\end{cases}$.

This shows that the system has a solution $\iff$ $b$ lies in the subspace $V_1$ and is thus orthogonal to the subspace $V_2$.

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    What is "the equation for a fixed point"? It's the main difficulty for me. I would say that $Ax=x$ but $A$ acts on the vectors meanwhile $x$ a point. – 2017-02-01
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    It's $Ax+b-x =0$, expressing that $f(x)=x$, so that x is a fixed point. – 2017-02-01
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    @iou Note that in the plane the eigenspace with eigenvalue 1 is the zero subspace so every vector $b$ is "perpendicular" to it, so every isometric affine transformation in the plane has a fixed point. In 3d space the condition means that $b$ should be perpendicular to the axis of rotation of $A$. – 2017-02-01
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    let $f=Ax+b$ be the isometric transformation of $E^n$ and $b$ isn't orthogonal to space of eigenvectors with eigenvalue 1, then $f$ has a invariant line, is it right? – 2017-02-01
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    That's quite a different problem where I have no answer for. I think this is a nice question to pose. – 2017-02-01
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    I'm confused. "Without loss of generality we may suppose that the matrix...", i.e. any matrix of isometric transformation is diagonalizable? – 2017-02-02
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    [here](https://en.wikipedia.org/wiki/Orthogonal_matrix#Matrix_properties) you can find this fact at the end of the paragraph "Matrix properties" where it says : *an orthogonal matrix can always be diagonalized over the complex numbers to exhibit a full set of eigenvalues, all of which must have (complex) modulus 1.* – 2017-02-02