Cartan–Dieudonné factorization over a real quadratic space

If $n=0$, $O(V,B)=\{I\}$ and the empty product proves the claim. If $n=1$, choose an anisotropic basis vector $u$ by [F1]. Every $A\in O(V,B)$ sends $u$ to $\pm u$, so $A=I$ or $A=r_u$; these have reflection lengths zero and one, respectively, and the parity agrees with the determinant. We now prove the induction step for $n\ge 2$.

Suppose $n\ge 2$ and inductively assume the theorem in smaller dimensions. If $K$ contains an anisotropic $u$, then $A$ fixes $u$ and preserves the nondegenerate space $u^{\perp }$. The induction hypothesis factors $A{\mathrm{\mid }}_{u^{\perp }}$ into at most $n-1$ reflections with normals in $u^{\perp }$; extending these reflections by the identity on $\mathbb{R}u$ gives the same reflections in $V$ and their product equals $A$.

Suppose instead $W$ is not totally isotropic. Then $w\mapsto Q(w)$ and $w\mapsto Q((A-I)w)$ are nonzero real polynomials: the first by nondegeneracy and the second because some vector in $W$ is anisotropic. Their product is a nonzero polynomial because the real polynomial ring is an integral domain. A nonzero polynomial on ${\mathbb{R}}^n$ does not vanish everywhere: for $n=1$ it has finitely many roots; for $n>1$, write $P(x,t)={\sum }_j{p}_j(x)t^j$ with a nonzero coefficient $p_j$, use induction to choose $x$ with $p_j(x)\mathrm{\ne }0$, and then choose $t$ outside the finite root set of the nonzero one-variable polynomial $P(x,t)$. Choose $w$ where the product is nonzero and put $u=(A-I)w$. Then both $w$ and $u$ are anisotropic.

It remains that both $K$ and $W$ are totally isotropic. By [F1], $W=K^{\perp }$. Since $K$ and $W$ are both totally isotropic, [F1] gives $\dim K\le n\mathrm{/}2$ and $\dim W\le n\mathrm{/}2$. But $\dim W=n-\dim K$, so both dimensions equal $n\mathrm{/}2$; the inclusion $K\subseteq K^{\perp }=W$ then gives $K=W=K^{\perp }$ and $n=2\dim K$. The map $A$ is the identity on $K$, and because $(A-I)V=W=K$ it is also the identity on $V\mathrm{/}K$. In a basis extending a basis of $K$, its matrix is block triangular with identity diagonal blocks, so $\det A=1$.

Symmetry and orthogonality give $B((A+I)w,u)=B(Aw,Aw)-B(Aw,w)+B(w,Aw)-B(w,w)=0$. Thus $(A+I)w$ is fixed by $r_u$ and $u=(A-I)w$ is negated, so $r_{u}Aw=w$. The orthogonal map $r_{u}A$ has the anisotropic fixed vector $w$, and the argument of step 1.2 factors it with at most $n-1$ reflections. Prepending $r_u$ factors $A$ with at most $n$.

Since $n$ is even in step 1.4, choose anisotropic $u$ and put $A_1=r_{u}A$. Then $\det A_1=-1$, so $A_1$ cannot be in the isotropic case of step 1.4. Steps 1.2 and 2.1 give a factorization of $A_1$ of length at most $n$. Each reflection has determinant $-1$, so this length is odd; because $n$ is even it is at most $n-1$. The identity $A=r_u{A}_1$ now gives at most $n$ reflections.

Every reflection has determinant $-1$, hence any reflection factorization of $A$ has parity determined by $\det A$. The constructed length is at most $n$ in all cases, and its parity is even precisely for determinant $1$. This proves the assertion, including the isotropic kernel and moved-space case.