A proof of Cartan-Dieudonné's theorem As an assignment, we were asked to prove a theorem by Cartan and Dieudonné in the following form (the Euclidean space $\mathbb{E}^n$ is meant to be the usual $\mathbb{R}^n$ endowed with the usual inner product):
Every isometry of the Euclidean space $\mathbb{E}^n$ is the product of (at most) $n$ reflections about hyperplanes.
I proved it in the following way and I'd like to know whether it's a valid proof or not. I'd also like to know about other proofs (I'm sure there are...) :
We use induction on $n$. Let's assume the proposition for $n-1$. Let $\phi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ be an isometry $\neq id$, and $\{e_1,\ldots, e_n\}$ be the canonical basis of $\mathbb{E}^n$. Consider the reflection $\sigma$ about the hyperplane orthogonal to the vector $e_1-\phi(e_1)$, which we assume non-zero, without loss of generality since $\phi\neq id$.
Clearly $\sigma(e_1)=\phi(e_1)$, so $e_1$ is ok. Since $\{\phi(e_2),\ldots, \phi(e_n)\}$ and $\{\sigma(e_2),\ldots, \sigma(e_n)\}$ are orthormal bases of the subspace $\langle \sigma(e_1)\rangle ^{\perp}$, there is an isometry $\psi: \mathbb{E}^n\rightarrow \mathbb{E}^n$ such that $\psi(\sigma(e_1))=\sigma(e_1)$ and $\psi(\sigma(e_j))=\phi(e_j)$, ($j\neq 1$). Since $\psi$ induces an isometry on $\langle \sigma(e_1)\rangle ^{\perp}$, by the inductive hypothesis, it can be written as a product of reflections about subspaces of $\langle \sigma(e_1)\rangle ^{\perp}$ and dimension $n-2$. Each of these subspaces can be extended to an hyperplane of $\mathbb{E}^n$ so that $\sigma(e_1)$ belongs to that and hence remains unchanged under those reflections. $\square$
 A: While this is a very old question, I find both the proofs, one proposed in the question itself and one by Stillwell cited in the answer by Drew Armstrong, seriously lacking from a formal point of view, even though the basic argument (the same in both proofs) can be made to work well.
Typos aside, here are a few defects. Proofs by induction need (to mention) a base case. There is a fundamental difference between a Euclidean vector space of dimension$~n$ and $\Bbb R^n$ equipped with the standard inner product (although the latter is a (standard) instance of the former): a hyperplane in an $n$-dimensional Euclidean vector space is an $n-1$ dimensional Euclidean vector space, but a hyperplane in $\Bbb R^n$ by no stretch of imagination is $\Bbb R^{n-1}$, unless one wants to entertain the idea that a same set can be subset of another set in many different ways (injective maps were invented to overcome that difficulty). So if one is going to do induction via hyperplanes, it pays to state the theorem for general Euclidean vector spaces rather than just for the spaces $\Bbb R^n$. An essential point is that a reflection of a hyperplane is not a reflection of the whole space, although it can be uniquely extended to one; one needs to prove that the desired properties carry over during this extension (OP does this correctly, Stillwell completely ignores this). Finally, in such a proof by induction one should not exclude the identity from the statement to be proved, since the induction works towards the identity, and would therefore fatally hit upon a case where the induction hypothesis cannot be applied. This is a problem in the OP answer; the Stillwell proof does not exclude the identity form the statement, but simply excludes it from its proof. In both cases the implicit convention is that the identity gives a product of no reflections at all, but this is never said aloud.
But the main obstruction to a nice proof by induction is too weak a statement of the theorem; we need some induction loading. N'en déplaise à Cartan et Dieudonné, the theorem should give the precise number of reflections needed, which is the codimension of the subspace $\ker(\phi-I)$ of vectors fixed by $n$ (clearly this is the minimal number possible, as a product of $k$ reflections fixes at least the intersection of their hyperplanes, which has codimension at most$~k$).
Theorem. Let $\phi$ be an automorphism (orthogonal endomorphism) of an Euclidean vector space $E$ of dimension$~n$; put $d=\dim(\ker(\phi-I_E))$. Then $\phi$ can be written as a product of $n-d$ orthogonal (hyperplane) reflections.
We can prove this by induction on $n-d$. The base case is $n-d=0$, so $\phi=I_E$ which is a product of $0$ reflections. For the inductive case assume $n-d>0$, and the theorem proved for smaller values of $n-d$. Since $\ker(\phi-I_E)\neq E$, one can choose a vector $v\notin\ker(\phi-I_E)$; after doing so, put $H=(\phi(v)-v)^\perp$, a hyperplane, $\sigma_H$ the orthogonal reflection with respect to$~H$, and $\phi'=\sigma_H\circ\phi$. We have $\phi'(v)=v$. Also $\ker(\phi-I_E)\subseteq H$ (since $(w\mid\phi(v)-v)=0$ when $\phi(w)=w)$), so $\ker(\phi'-I_E)\supseteq\ker(\phi-I_E)+\langle v\rangle\supset\ker(\phi-I_E)$. Since on the other hand $\ker(\phi-I_E)\supseteq\ker(\phi'-I_E)\cap H$, it follows that $\dim(\ker(\phi'-I_E))=\dim(\ker(\phi-I_E))+1=d+1$. This shows that the induction hypothesis applies to $\phi'$, and expresses it as a product of $n-d-1$ reflections. Then $\phi=\sigma_H\circ\phi'$ is written as a product of $n-d$ reflections $\square$.
A: There is a proof on page 37 of John Stillwell's Naive Lie Theory.
