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Let $V$ be a real vector space of finite dimension and let $\langle \cdot, \cdot \rangle$ be a non-degenerate symmetric bilinear form on $V$. Let $U, W \subseteq V$ be linear subspaces such that the bilinear forms $\langle \cdot, \cdot \rangle \vert_U$ and $\langle \cdot, \cdot \rangle \vert_W$ are isometric and non-degenerate.

It is quite easy to prove that there exists an isometry $f$ of $V$ such that $f(U) = W$. I would like to know whether there are some sufficient conditions such that there exists an isometry $f$ of $V$ such that $f(U) = W$ and $f(W) = U$.

Thanks to all!

EDIT. Since I have not received any answer, I want to say that I'm interested in a very particular case: $V = \mathbb{R}^{n+1}$, $\langle \cdot , \cdot \rangle$ is the Lorentzian scalar product on $\mathbb{R}^{n+1}$, i.e. $\langle x, y \rangle = \sum_{i=1}^n x_i y_i - x_{n+1} y_{n+1}$, and $U$ and $W$ are $2$-dimensional subspaces of $\mathbb R^{n+1}$ such that $\langle \cdot, \cdot \rangle \vert_U$ and $\langle \cdot, \cdot \rangle \vert_W$ have signature $(1,1)$.

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    Wait, what is the isometry of $\mathbb R^3$ under the usual dot product that maps the two-dimensional subspace $U = \{(x,y,0)\}$ to a one-dimensional subspace $W = \{(x,0,0)\}$?2012-03-11
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    He/she is supposing that there exists an isometry between $U$ and $W$, therefore $U$ and $W$ have the same dimension.2012-03-11
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    How about a function that just swaps $U\leftrightarrow W$ and doesn't do anything to any subspaces perpendicular to $U + W$? Maybe write down bases for $U$, $W$, extend to $V$, and then define $f$ on the basis?2012-03-11
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    @Neal: if the symmetric form isn't positive, the orthogonal complement of a subspace F isn't necessarily a complementary subspace of F.2012-03-11

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