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Euclidian space en U is a linear subspace

  • Thread starter damabo
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Homework Statement



show that (U[itex]\bot[/itex])[itex]\bot[/itex]=U, if (ℝ,V,+,[,.,]) an Euclidian space en U is a linear subspace of V.

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The Attempt at a Solution



suppose [itex]\beta={u_1,...,u_k}[/itex] is an orthonormal basis of U.
pick u in U. Then [itex]u=x_1u_1+...+x_ku_k[/itex] for certain x_1,...,x_k in ℝ.

pick u'_1,...,u'_k as orthonormal basisvectors of U[itex]\bot[/itex], where u'_i[itex]\bot[/itex]u_i for all i. and consequently take basisvectors u''_1,...,u''_k for U[itex]\bot\bot[/itex] that are orthogonal to the previous. then it follows that u''=x_1u''_1+...+x_ku''_k = x_1u_1+...+x_ku_k = u.
this means that U[itex]\subset[/itex]U[itex]\bot\bot[/itex] and also that both have the same dimensions, and consequently that they are the same subspaces (since [itex]span(u_1,...,u_k)=span(u''_1,...,u''_k)=ℝ^k [/itex]).
 
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Answers and Replies

  • #2
HallsofIvy
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I am confused as to what you mean by U⊥. I would expect that to mean "the orthogonal complement of U" but that requires that U be a subset of some larger space. If you mean "orthogonal complement, what larger space are you assuming?
 
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see the problem statement: [itex]U\bot[/itex] is indeed the orthogonal complement and V is the Euclidian space of which U is a subspace. [itex]U\bot\bot[/itex] is thus the orthogonal complement of the orthogonla complement.
 
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