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Hello: I need help with this problem:

Let $V=(V,b)$ be a finite-dimensional vector space equipped with $b$ a symmetric and positive definite bilineal form. And let $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

For a matrix $A \in \mathrm{O}(V)$, let $\mathrm{O}_*(V)$ the subset of $\mathrm{O}(V)$ such that be the matrix $P_A:=\frac{A-JAJ}{2}$ is invertible, where $J$ is a complex structure (a matrix such that $J^2=-1$).

Let $n=\dim \ker(P_A)$. For every $j \in \{1,…,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r(Je_j)=-e_j$ and $r_j(v)=v$ for and $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dots r_n \in \mathrm{O}(V).$$

I need to prove that $$RA \in \mathrm{SO}_*(V),$$ where similarly as $\mathrm{O}(V)$: $\mathrm{SO}_*(V)$ is the subset of $\mathrm{SO}(V)$ such that $P_B:=\frac{B-JBJ}{2}$ is invertible.

I already proved that $RA \in \mathrm{SO}(V)$; the only thing that I haven’t been able to figure out is to prove that $\frac{1}{2}(RA-JRAJ)$ is invertible, since $n$ can be even or odd.

Also, $P_{r_j}$ is not invertible since $\det(r_j)=-1$.

¿What is good and optimized approach to deal with the product of reflections $$R=r_1r_2\cdots r_n?$$

In summary: I’m trying to prove that $RA$ (where $R$ the product of reflections $r_1r_2\dots r_n$) is a ortogonal matrix with $\det(RA)=+1$ and that the matrix $\frac{1}{2}(J-J(RA)J)$ is invertible.

Any help will be greatly appreciated. Thanks :).

UPDATE: Two things:

1. I made a typo, $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$, not on $\ker(P_A)$.

2. On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $R$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

And that's it, I think the author's argument has many gaps or things that I'm not getting :(. He says he's following this article, but I have read any multiply times and I don't see anything like what I'm trying to prove (or at least eith this notation).

Let $V=(V,b)$ be a finite-dimensional vector space equipped with $b$ a symmetric and positive definite bilinear form. And let $\{e_1,\dotsc,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

For a matrix $A \in \mathrm{O}(V)$, let $\mathrm{O}_*(V)$ be the subset of $\mathrm{O}(V)$ such that the matrix $P_A:=\frac{A-JAJ}{2}$ is invertible, where $J$ is a complex structure (a matrix such that $J^2=-1$).

Let $n=\dim \ker(P_A)$. For every $j \in \{1,\dotsc,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r_j(Je_j)=-e_j$ and $r_j(v)=v$ for any $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dotsm r_n \in \mathrm{O}(V).$$

I need to prove that $$RA \in \mathrm{SO}_*(V),$$ where similarly as $\mathrm{O}(V)$: $\mathrm{SO}_*(V)$ is the subset of $\mathrm{SO}(V)$ such that $P_B:=\frac{B-JBJ}{2}$ is invertible.

I already proved that $RA \in \mathrm{SO}(V)$; the only thing that I haven’t been able to figure out is to prove that $\frac{1}{2}(RA-JRAJ)$ is invertible, since $n$ can be even or odd.

Also, $P_{r_j}$ is not invertible since $\det(r_j)=-1$.

What is a good and optimized approach to deal with the product of reflections $$R=r_1r_2\dotsm r_n?$$

In summary: I’m trying to prove that $RA$ (where $R$ is the product of reflections $r_1r_2\dotsm r_n$) is an orthogonal matrix with $\det(RA)=+1$ and that the matrix $\frac{1}{2}(J-J(RA)J)$ is invertible.

Any help will be greatly appreciated.

UPDATE: Two things:

1. I made a typo, $\{e_1,\dotsc,e_n\}$ is an orthonormal basis for the subspace $\ker((P_A)^t)$, not $\ker(P_A)$.

2. On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dotsc, n \}$) has the identity restricted to the subspace orthogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dotsm r_n$, then the operator $R$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

And that's it, I think the author's argument has many gaps or things that I'm not getting. He says he's following Ruijsenaars - On Bogoliubov Transformations. II. The General Case, but I have read it multiple times and I don't see anything like what I'm trying to prove (or at least with this notation).

Let $V=(V,b)$ be a finite dimensional vector space equipped with $b$ a symmetric and positive definite bilinear form. And let $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

For a matrix $A \in \mathrm{O}(V)$, let $\mathrm{O}_*(V)$ the subset of $\mathrm{O}(V)$ such that be the matrix $P_A:=\frac{A-JAJ}{2}$ is invertible, where $J$ is a complex structure (a matrix such that $J^2=-1$).

Let $n=\mathrm{dim} \mathrm{ker}(P_A)$. For every $j \in \{1,…,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r(Je_j)=-e_j$ and $r_j(v)=v$ for and $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dots r_n \in \mathrm{O}(V).$$

I need to prove that $$RA \in \mathrm{SO}_*(V),$$ where similarly as $\mathrm{O}(V)$: $\mathrm{SO}_*(V)$ is the subset of $\mathrm{SO}(V)$ such that $P_B:=\frac{B-JBJ}{2}$ is invertible.

I already proved that $RA \in \mathrm{SO}(V)$; the only thing that I haven’t been able to figure out is to prove that $\frac{1}{2}(RA-JRAJ)$ is invertible, since $n$ can be even or odd.

Also, $P_{r_j}$ is not invertible since $\det(r_j)=-1$.

¿What is good and optimized approach to deal with the product of reflections $$R=r_1r_2\cdots r_n?$$

In summary: I’m trying to prove that $RA$ (where $R$ the product of reflections $r_1r_2\dots r_n$) is a ortogonal matrix with $\det(RA)=+1$ and that the matrix $\frac{1}{2}(J-J(RA)J)$ is invertible.

Any help will be greatly appreciated. Thanks :).

EDIT: Two things:

1-) I made a typo, $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$, not on $\ker(P_A)$.

2-) On the main reference I'm using, the author establishes the following:

*This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $r$ is in block form, its lower right corner being the identity on $(\ker(P_A)^t^ Hello: I need help with this problem:

Let $V=(V,b)$ be a finite dimensional vector space equipped with $b$ a symmetric and positive definite bilineal form. And let $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

Let $n=\mathrm{dim} \mathrm{ker}(P_A)$. For every $j \in \{1,…,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r(Je_j)=-e_j$ and $r_j(v)=v$ for and $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dots r_n \in \mathrm{O}(V).$$

EDIT: Two things:

1-) I made a typo, $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$, not on $\ker(P_A)$.

2-) On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $r$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

And that's it, I think the author's argument has many gaps or things that I'm not getting :(. He says he's following this article, but I have read any multiply times and I don't see anything like what I'm trying to prove.

Let $V=(V,b)$ be a finite-dimensional vector space equipped with $b$ a symmetric and positive definite bilineal form. And let $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

Let $n=\dim \ker(P_A)$. For every $j \in \{1,…,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r(Je_j)=-e_j$ and $r_j(v)=v$ for and $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dots r_n \in \mathrm{O}(V).$$

UPDATE: Two things:

1. I made a typo, $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$, not on $\ker(P_A)$.

2. On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $R$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

And that's it, I think the author's argument has many gaps or things that I'm not getting :(. He says he's following this article, but I have read any multiply times and I don't see anything like what I'm trying to prove (or at least eith this notation).

This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $r$ is in block form, its lower right corner being the identity on $\ker((P_A)^t)^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

*This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $r$ is in block form, its lower right corner being the identity on $(\ker(P_A)^t^ Hello: I need help with this problem:

Let $V=(V,b)$ be a finite dimensional vector space equipped with $b$ a symmetric and positive definite bilineal form. And let $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

For a matrix $A \in \mathrm{O}(V)$, let $\mathrm{O}_*(V)$ the subset of $\mathrm{O}(V)$ such that be the matrix $P_A:=\frac{A-JAJ}{2}$ is invertible, where $J$ is a complex structure (a matrix such that $J^2=-1$).

Let $n=\mathrm{dim} \mathrm{ker}(P_A)$. For every $j \in \{1,…,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r(Je_j)=-e_j$ and $r_j(v)=v$ for and $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dots r_n \in \mathrm{O}(V).$$

I need to prove that $$RA \in \mathrm{SO}_*(V),$$ where similarly as $\mathrm{O}(V)$: $\mathrm{SO}_*(V)$ is the subset of $\mathrm{SO}(V)$ such that $P_B:=\frac{B-JBJ}{2}$ is invertible.

I already proved that $RA \in \mathrm{SO}(V)$; the only thing that I haven’t been able to figure out is to prove that $\frac{1}{2}(RA-JRAJ)$ is invertible, since $n$ can be even or odd.

Also, $P_{r_j}$ is not invertible since $\det(r_j)=-1$.

¿What is good and optimized approach to deal with the product of reflections $$R=r_1r_2\cdots r_n?$$

In summary: I’m trying to prove that $RA$ (where $R$ the product of reflections $r_1r_2\dots r_n$) is a ortogonal matrix with $\det(RA)=+1$ and that the matrix $\frac{1}{2}(J-J(RA)J)$ is invertible.

Any help will be greatly appreciated. Thanks :).

EDIT: Two things:

1-) I made a typo, $\{e_1,…,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$, not on $\ker(P_A)$.

2-) On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dots, n \}$) has the identity restricted to the subspace ortogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dots r_n$, then the operator $r$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.