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Let's say we have a subspace $V$, that is a subset of $\mathbb{R}^n$. Does $V + V^{\perp}$ always span $\mathbb{R}^n$?

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The phrase "linearly independent subspace" doesn't make sense. A set of vectors may be linearly independent (or not), but not a subspace, since the zero vector is contained in any subspace. –  Shaun Ault Jan 2 '13 at 15:49
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Yes, it happens. I'll give you an hint. Set an orthogonal base of $V$, complete it orthogonally and look at the matrix. –  Ivan Jan 2 '13 at 15:51
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I'd make that $+$ into a $\oplus$ to indicate a direct sum of the perp spaces. –  JohnD Jan 2 '13 at 16:02
    
Although you are asking about $\Bbb R$, it's worth noting that this is not true for finite fields. There are, for example, proper subspaces such that $V=V^\perp$ using the "usual" inner product, so that $V+V^\perp$ is not the whole space. –  rschwieb Jan 2 '13 at 17:41

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What you mean is a linear subspace, not a linearly independent subspace, which does not make any sense.
Proof: Take a basis for $V$, $B_V=(v_1,...,v_k)$. The condition $w\in V^\perp$ is equivalent to $\langle v_i,w\rangle=0$ for $i=1,...,k$. Hence you have a system of $k$ linear equations. Since $B_V$ are linearly independent, those equations are independent as well. Hence the space of solutions to this system, which is exactly $V^\perp$, will be of dimension $n-k$. Take a basis $B_{V^\perp}=(u_1,...,u_{n-k})$. Check that $(v_1,...,v_k,u_1,...,u_{n-k})$ is linearly independent and thus is a basis for $\mathbb{R}^n$. So $V+V^\perp=\mathbb{R}^n$

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Yes, every subspace of a normed linear space with finite dimension is closed. It is also true that if $X$ is a closed, linear subspace of a Hilbert space $H$ (in this case $\mathbb{R}^n$), then $H = X \bigoplus X^{\perp}$, where $\bigoplus$ denotes the direct sum of $X$ and $X^{\perp}$ (i.e., $X \cap X^{\perp} = \{0\}$ and $h = x + x'$, for all $h \in H$ with $x \in X$ and $x' \in X^{\perp}$).

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For $w\in \mathbb R^n$, select $v\in V$ that minimizes $|w-v|$ (why is that possible?). Then $w-v\in V^\perp$ (why?).

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I don't really understand what you mean? Does it have anything to do with the zero vector being in both subspaces? –  JohnPhteven Jan 2 '13 at 16:07

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