# $B=(u_1,u_2,u_3,u_4)$ is basis for $\mathbb R^n$, $C=(w_1,w_2,w_3,w_4)$ is its orthogonal. If $v\in sp\{u_1,u_2\}$ then $v\in sp\{w_1+w_2,w_1-w_2\}$

Let $B=(u_1,u_2,u_3,u_4)$ be an ordered basis for $\mathbb R^n$ and $C=(w_1,w_2,w_3,w_4)$ is its orthogonal basis. If $v\in \operatorname{span}\{u_1,u_2\}$, prove that $v\in \operatorname{span}\{w_1+w_2,w_1-w_2\}$.

I'm confused isn't the orthogonal basis to a space just a subset of the space's null space (because for every vector $w_i$ in $C$ it must hold that $B\cdot w_i=0$).

But the null space of a basis is only the zero vector. So how can this be?

No, its not. An orthogonal basis $w_1, \cdots, w_n$ of a space is a basis of space such that $w_i\cdot w_j=0$ for all $i\neq j$.
Now if $u_1, u_2, u_3, u_4$ is an ordered basis of $\mathbb{R}^4$, then its orthogonal basis (obtained by Gram-Schmidt process) $w_1, w_2, w_3, w_4$ has the first two vectors as $$w_1=u_1, w_2=u_2-\frac{u_2\cdot u_1}{u_1\cdot u_1}w_1.$$ So $u_1, u_2\in\textrm{span}\{w_1, w_2\}.$ But $w_1, w_2\in\textrm{span}\{w_1+w_2, w_1-w_2\}$. So if $v\in\textrm{span}\{u_1, u_2\}$ then $v\in\textrm{span}\{w_1+w_2, w_1-w_2\}$.