Is the Symmetric Difference of Finite-Dimensional Subspaces Well-Defined?

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The discussion centers on the well-defined nature of the symmetric difference of finite-dimensional subspaces U and V within a finite-dimensional space W. Specifically, the example provided involves U = Span{x, y} and V = Span{y, z} in R^3, leading to a symmetric difference of {x, z} and a query about the uniqueness of Span{x, z}. The author explores the implications of dimension and orthonormal bases, concluding that Span(B \ Bi) is uniquely defined under certain conditions. The discussion also touches on the complexities of defining inner products and norms in spaces over the field Q.

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Talisman
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Suppose U and V are finite-dimensional subspaces of some finite-dimensional space W (over the field Q in my actual case, but probably irrelevant)

I've got a subspace in mind that I can't quite define well. I'll start with a concrete example:

Consider R^3 with orthonormal basis {x, y, z}. Let U = Span{x, y} and V = Span{y, z}. The symmetric difference of {x, y} and {y, z} is {x, z}, and the strange thing I want is Span{x, z}. I resorted to using an "obvious" basis set here, but I want to know if the thing I'm looking for is well-defined. What if I haven't picked a basis; what if it's not a normed space; what if it's not even an inner product space; etc.

A bit more formally: Let U \cap V = I have dimension k > 0. Then I can pick an orthonormal basis for I, Bi = {b1, ..., bk}, and then choose an orthonormal basis for U + V = Span(U \cup V) (call it simply B) that has Bi as a subset. Now Span(B \ Bi) is uniquely defined (I think).

Is that well-defined, or is my intuition wrong? If it is, is there a cleaner way of specifying it?

Even if this is true, my problems have only just begun: the W I have in mind is not an inner product space; it's R over Q, with U and V being subspaces of some n-dimensional subspace S.

S is isomorphic to Q^n, so I can define an inner product and norm, but this seems like a painful way to go about things.
 
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The orthocomplement of Bi in U+V ...

In general, the orthocomplement of K in H (K a subspace of H) is the set of all elements of H orthogonal to K. But you probably need real scalars, not rational.
 
Thanks g_edgar! The following then seems like it should hold, although my proof may be wrong.

Let U \cap V = I have dimension k > 0. Let O be the orthocomplement of I in W = U + V, O_{U} be the orthocomplement of I in U, and O_{V} be the orthocomplement of I in V.

Then O = O_{U} + O_{V}.

Seems straightforward: If U has dimension u, V has dimension v, and I has dimension i, then O is a subspace of W with dimension u+v-2i. Meanwhile, O_u has dimension u-i, and O_v has dimension v-i. Since O_u and O_v are linearly independent subspaces of O and have dimensions summing to u+v-2i, O_u + O_v = O.

Maybe?
 
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