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How might we show that the isometry given by $Ax+b$ where $A$ is an orthogonal matrix, $b, x$ are vectors in $R^2$ must have either a fixed point or a fixed line? For example, wouldn't a rotation followed by a translation not fix anything?

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A rotation followed by a translation is in fact equal to a rotation around a point different from the origin:

Rotation around a point $c$:

$$A(x - c) + c = Ax + (I - A)c$$

and since $A$ has no eigenvalue equal to 1, the matrix $I - A$ is invertible, and thus we can set $c = (I - A)^{-1}b$.

Letting $x = c$ will give you your fixed point.

So what is left is to show that it also holds for cases where $A$ has 1 as an eigenvalue, which means it's either the identity matrix, or a reflection.

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thank you. How did you see that the rotation-translation map is actually a rotation about another point? –  Lin Feb 9 '12 at 12:49

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