I’m studying Linear Algebra and I stumbled into an interesting problem:
Let $V=(V,b)$ be a finite-dimensional vector space with a symmetric and positive-definite bilinear form $b$. Let $M \in \mathrm{O}(V)$. Also, let $k=\dim \ker (p_M^t)=\dim \ker(p_M)$, where $$p_M:=\frac{1}{2} (M-JMJ).$$ Here $J:V \to V$ is the lineal transformation such that $J^2=-1$ and $J^{-1}=J^t=-J$. Finally, let $\{\epsilon_1, \epsilon_2, …, \epsilon_k\}$ be a orthormormal basis of the kernel of $p_M^t$. $V$ is a complex vector space; if $\dim_{\mathbb{R}}V =2m$ then $\dim_{\mathbb{C}} V=m$ and $\dim \ker(p_M) = \dim(p_M^t) = k \in \{0,1,…,m\}$.
For every $j \in \{1,…,k\}$, define a reflection $r_k$ such that $r_j(\epsilon_j) = -J\epsilon_j$, $r_j(J\epsilon_j) =-\epsilon_j$. Also, each $r_j$ satisfies that $r_j(v)=v$ when $b(v,\epsilon_j)=b(v,J\epsilon_j)=0$.
If $r=r_1r_2\dots r_k$ is an ortogonal matrix on $V$, prove that $\det(rM)=1$.
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My attempt:
We know that if $r$ is a reflection then $\det(r)=-1$ and that $\det(M)=\pm 1$, since is orthogonal. And so we have two cases:
- If $\det(M)=1$ then $$\det(rM)=\det(Mr_1r_2\dots r_k)=1 \cdot (-1)^k,$$ and so if $k$ is even we are done, however, if $k$ is odd then the determinant is $-1$.
- If $\det(M)=-1$ then $$\det(rM)=\det(Mr_1r_2\dots r_k)=-1 \cdot (-1)^{k+1}$$ and so if $k$ is odd we are done, but if $k$ is even then the determinant is $-1$.In both cases I reach a dead end.
I tried using Cartan-Dieudonné theorem, but no dice(since I arrive basically at the same wall I can’t get through. ¿Am I missing something? The conditions about the reflections obviously are important, but I don’t know how to work them in order to prove that $\det(rM)=1$ always.
Please help.
EDIT: Maybe if I express each reflection $r_j$ as a block matrix I can go somewhere, but with that $J$, I don’t know how to proceed….
EDIT 2: Since each $r_j$ swaps $\epsilon_j$ and $J\epsilon_j$ then the matrix representation of each reflection (which is a block matrix) is a permutation matrix. Still, permutation matrices have determinant equal to $\pm 1$. I believe this doesn’t help me anyhow. Am I right?
EDIT 3: I noticed that since $b(v,e_j)=0=b(v,Je_j)$ if $r_j(v)=v$, $v$ is ortogonal to each element of the basis $\{e_1,…,e_k\} $ (and hence to every element of $\ker p_M^t$). Also, since $r_j(e_j+Je_j)=-(e_j+Je_j)$, the (block) matrix representation of $r_j$ is $$\begin{bmatrix} -J & 0\\ 0 & J \end{bmatrix}.$$ Am I right? I think this doesn’t solve my problem, but it’s something… And I also think this works when $k$ is even… :(.
EDIT 4: I noticed that $\ker(P_M)$ is precisely the linear subspace of vectors $v \in V$ such that $MJv = - JMv$, since $P_M= \dfrac{-1}{2}J(MJ + JM)$. So, in particular $MJ\epsilon_j= -JM\epsilon_j$ for all $j \in \{1,…,k\}$, so there must be a way in which $M$ and $r=r_1r_2\dots r_k$ are related somehow, but I haven’t been able to find out what that relation is…
EDIT 5: I had a typo, the set $\{ \varepsilon_1,…,\varepsilon_k \}$ is a basis for $\ker(p_M^t)$, not for $\ker(p_M)$.
I tried using the fact that a real ortogonal matrix $A$ has positive determinant if and only if $\dim\ker(A+I_k)$ is even, where $I_k$ is the identity matrix of order $k$, but again, I don’t get anything helpful, I think…