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I am trying to prove that given a subspace $\mathbf{W}$ in $\mathbb{R^{n}}$, the subspace and its orthogonal complement 'cover' whole of $\mathbb{R^{n}}$ through '+' where we define $\mathbf{W}$+$\mathbf{W^{\perp}}$ as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks!

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up vote 4 down vote accepted

Suppose $W$ has dimension $k$, and start with a basis $v_1,\ldots,v_k$ of $W$. This is a linearly independent subset of $\mathbb{R}^n$, so it may be extended to a basis $v_1,\ldots,v_n$ of $\mathbb{R}^n$.

Apply the Gram-Schmidt process to $v_1,\ldots,v_n$ to obtain an orthonormal basis $w_1,\ldots,w_n$ for $\mathbb{R}^n$ such that $w_1,\ldots,w_k$ is an orthonormal basis for $W$. Now take any vector $z = a_1 w_1 + \ldots + a_k w_k + a_{k+1} w_{k+1} + \ldots + a_n w_n$ of $\mathbb{R}^n$. The conditions for $z \in W^{\perp}$ are precisely

$\langle z,w_1 \rangle = \ldots = \langle z,w_k \rangle= 0$,


$a_1 = \ldots = a_k = 0$.

Thus $W^{\perp}$ is the span of $w_{k+1},\ldots,w_n$ and $\mathbb{R}^n = W \oplus W^{\perp}$.

In the above argument, the key is showing that $\dim W + \dim W^{\perp} = \dim \mathbb{R}^n$ (as Jonas Meyer says in his answer). Another argument for this, valid for any nondegenerate symmetric bilinear form $(u,v) \mapsto B(u,v)$ on a vector space $V$ (not necessarily finite-dimensional) over a field $K$ is given in $\S 4$ of these notes on quadratic forms. Note that a general nondegenerate bilinear form could be isotropic: that is, one may have nonzero vectors $v$ with $B(v,v) = 0$ and thus $Kv \cap (Kv)^{\perp} \neq 0$. But by definition an inner product is anisotropic, which forces $W \cap W^{\perp} = 0$ for all subspaces $W$.

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Can you produce a basis of the right size using vectors in $W$, $W^{\perp}$? There is a standard result that perpendicular vectors are linearly-independent (assume not, then set a non-trivial combination to zero,then dot both sides by the "right" vector to get a contradiction). Start with the fact that $W$, $W^{\perp}$ are both subspaces, then try using perpendicular-> linearly-independent to produce a basis of the right size for $\mathbb{R}^n$. Overall, you have $k$ linearly-independent vectors in a basis for $W$, $(n-k)$ linearly-indep in a basis for $W^{\perp}$, and you can show that their set union is a basis for $\mathbb{R}^n$.

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You can do $W^\perp$ with $W^\perp$. If you want to see how TeX on this site was done, you can right click, then click on "Show Source". – Jonas Meyer Mar 3 '11 at 7:49
I tried editing Kevin's comment to texify those symbols, but apparently my changes are still being reviewed. – Anthony Labarre Mar 3 '11 at 8:58
@Anthony: No, I approved your edit right away, but Kevin was editing at the same time and accidentally undid it. You can look at the edit history by clicking on the time to the right of "edited" in the bottom center of the post. – Jonas Meyer Mar 3 '11 at 12:48
Oh, ok, thanks for the clarification. – Anthony Labarre Mar 3 '11 at 12:59

If you know how to prove that the dimensions of $W$ and $W^\perp$ add up to $n$, then the same argument shows that the dimensions of $W+W^\perp$ and $(W+W^\perp)^\perp$ add up to $n$, and the latter is $0$. (Any vector in $(W+W^\perp)^\perp$ is perpendicular to itself, because it is in both $W^\perp$ and $W^{\perp\perp}$.)

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