Orthogonal Bases for W and W$^\perp$ in relation to $R^n$

Question

W is a linear subspace in $R^n$. If $B_1$ is an orthogonal basis for W and $B_2$ is an orthogonal basis for $W^\perp$, how I prove that the union of $B_1$ and $B_2$ forms an orthogonal basis for all of $R^n$? I know that I have to split it up into 2 sections - first must prove that it is a basis, and then that it is an orthogonal set. Any help with this?

Answer 1

If $v$ in $\mathbb R^n$, let $w$ be the orthogonal projection of $v$ on $W$ and let $w'=v-w$. Then $w'\in W^\perp$. Since you can write $w$ as a linear combination of elements of $B_1$, since you can write $w'$ as a linear combination of elements of $B_2$, and since $v=w+w'$, you can write $v$ as a linear combination of elements of $B_1\cup B_2$. So, $B_1\cup B_2$ spans the whole space.

Now, if $B_1=\{u_1,\ldots,u_k\}$, if $B_2=\{u_{k+1},\ldots,u_n\}$ and if$$v=\overbrace{\alpha_1u_1+\cdots+\alpha_ku_k}^{\phantom W\in W}+\overbrace{\alpha_{k+1}u_{k+1}+\cdots+\alpha_nu_k}^{\phantom{W^\perp}\in W^\perp}$$then, since there is only one way of writing $v$ as the sum of an element of $W$ with an element of $W^\perp$ and since $B_1$ and $B_2$ are basis, the $\alpha_m$'s are unique.

We are assuming that the $u_m$'s with $m\leqslant k$ are orthogonal and also that the $u_k$'s with $m>k$ are orthogonal. Finally, since the $u_k$'s with $m\leqslant k$ belong to $W$ and the $u_k$'s with $m>k$ belong to $W^\perp$, $u_m$ and $u_p$ are orthogonal if $m\leqslant k$ and $p>k$. So, $B_1\cup B_2$ is an orthogonal basis.