I know that the isometries of $\Bbb{E}^n$ are of the following form: $$ T(v) = Rv+t$$ where $R$ is an orthogonal transformation and $t$ is a translation vector. I am trying to prove that this is true. It is easy enough to see that such a transformation is an isometry, but I'm not sure how to prove that all isometries are of this form. All I know is that an isometry $g$ has the property that $\|v\| = \|g(v)\|$, or equivalently that $\|v\|^2 = \|g(v)\|^2$. I am not sure how to work with this to show that all isometries are of the form of $T$ above. Any suggestions on how to approach this would be appreciated.

Edit: Does it make sense to think about what could happen to the standard basis vectors? We know that isometries must preserve the angles between the vectors, so it should follow that after acting on $\Bbb{E}^n$ by an isometry, the images of the basis vectors should remain orthogonal.

Edit 2: Before I read the solution that was posted, let me explain what I've come up with so far based on the comments. Suppose $f$ is an isometry of $\Bbb{E}^n$. Then let $g(v) = f(v) - f(0)$ so that there is no translation. Because $f$ was an isometry and translation is an isometry, $g$ should also be an isometry. Then because isometries preserve distances, any point $v$ that was a distance $r$ from the origin should also have that $g(v)$ is a distance $r$ from the origin. So if we assume for now that $g$ is a linear transformation, (I'll have to return to this), then we know that $\|v\| =\|g(v)\|$, so $g$ must be an orthogonal transformation. Then it follows that $f(v) = g(v) + f(0)$, which is the form that I want.

Other than proving that $g$ is a linear transformation, does this look valid thus far?

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    $\begingroup$ The key is to alter your isometry so that it preserves the origin (replace $T$ by $T - T(0)$). Then it follows that it is linear/orthogonal. Now add back in to re-translate and that gives you your form. $\endgroup$ – Randall Jun 6 '18 at 2:33
  • $\begingroup$ There is some subtlety here. It may take you a bit to sort through it. $\endgroup$ – James S. Cook Jun 6 '18 at 2:38
  • $\begingroup$ That looks good so far. Proving origin fixing, distance preserving maps are necessarily linear is the hard part. The answer given is interesting, I recall doing it a different way which did not use the continuity argument. Instead, I had to use orthonormal basis arguments. $\endgroup$ – James S. Cook Jun 6 '18 at 14:07

By translating, we may assume without loss of generality that $T(0) = 0$. The hard part is showing that $T$ is linear; then the polarization identity tells us that it preserves inner products. I'm sure there are many ways of doing this, but here's mine:

Note that $T$ preserves midpoints, since the midpoint $(v+w)/2$ of $v$ and $w$ is the unique point with distance $\|v-w\|/2$ from each. Explicitly, for any $v$ and $w$ we have $$T\Bigl(\frac{v+w}{2}\Bigr) = \frac{T(v) + T(w)}{2}.$$ Putting $w = 0$ and doubling $v$ gives $T(v) = T(2v)/2$. So, $$T(v+w) = T\Bigl(\frac{2v+2w}{2}\Bigr) = \frac{T(2v)+T(2w)}{2} = \frac{2T(v)+2T(w)}{2} = T(v) + T(w)$$ and $T$ is additive.

By induction with $w = av$, this gives $T(nv) = nT(v)$ for any integer $n$. If $n \ne 0$ it follows that $T(v)/n = T(v/n)$, so $$T(av) = aT(v)$$ for any rational $a$. Since $T$ is continuous this actually holds for any real $a$, and we're done.

Actually there is a more general result called the Mazur–Ulam theorem which doesn't require the norm to be Euclidean.

  • $\begingroup$ neat argument. Definitely not how I saw this done before. $\endgroup$ – James S. Cook Jun 6 '18 at 14:08
  • $\begingroup$ I just fixed an error in the proof (of course I must use $T(0) = 0$ somewhere!). $\endgroup$ – arkeet Jun 6 '18 at 16:48

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