# Euclidian space en U is a linear subspace

1. Sep 22, 2012

### damabo

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

pick u'_1,...,u'_k as orthonormal basisvectors of U$\bot$, where u'_i$\bot$u_i for all i. and consequently take basisvectors u''_1,...,u''_k for U$\bot\bot$ 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$\subset$U$\bot\bot$ and also that both have the same dimensions, and consequently that they are the same subspaces (since $span(u_1,...,u_k)=span(u''_1,...,u''_k)=ℝ^k$).

Last edited: Sep 22, 2012
2. Sep 22, 2012

### HallsofIvy

Staff Emeritus
Re: U$\bot$$\bot$=U

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?

3. Sep 22, 2012

### damabo

Re: U$\bot$$\bot$=U

see the problem statement: $U\bot$ is indeed the orthogonal complement and V is the Euclidian space of which U is a subspace. $U\bot\bot$ is thus the orthogonal complement of the orthogonla complement.

Last edited: Sep 22, 2012