If $W$ and $W^{\perp}$ are subspaces of $V$, both with orthonormal basis, then $V = W \oplus W^{\perp}$

Question

Let $W$ be a subspace of $V$, $W^{\perp}$ it's orthogonal complement in $V$, and say that both have orthonormal basis, and that $V$ is a finite dimensional inner product space, then $V = W \oplus W^{\perp}$.

I can see that $W \cap W^{\perp} = \{0\}$ but can someone show me why $\mathrm{span}\{W,W^{\perp}\}=V$?

Thanks

Answer 1

Take $v\in V$. If $\{w_1,\ldots,w_n\}$ is an orthonormal basis of $W$, then

$$w=\langle v,w_1\rangle w_1+\langle v,w_2\rangle w_2+ \dots + \langle v,w_n\rangle w_n\in W.$$

And $v-w$ is orthogonal to each $w_k$ and therefore $v-w\in W^\perp$. So,$$v=w+(v-w)\in\operatorname{span}\{W,W^\perp\}.$$