Proving that $\mathbf{W}$+$\mathbf{W^{\perp}}$=$\mathbb{R^{n}}$

Solution 1:

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

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

i.e.,

$a_1 = \ldots = a_k = 0$.

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 $W$.

Solution 2:

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 $W^{\perp\perp}$.)

Solution 3:

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 $\mathbb{R}^n$.