# Given orthogonal basis what is the orthogonal complement?

The question states:

Let $q_1,q_2,\dots,q_n$ be an orthogonal basis of $\Bbb R^n$ and let $S = \operatorname{span}\{q_1,q_2,\dots,q_k\}$, where $1 \le k \le n-1$. Show that $S^\perp = \operatorname{span}\{q_{k+1},\dots,q_n\}$.

I am not sure what the answer should really be. I know that if $S=\operatorname{span}\{q_1,\dots,q_k\}$ then $S^\perp$ must be the span of vectors that are orthogonal to each vector in $S$. Other than using this logic I am not sure how to prove $S^\perp = \operatorname{span}\{q_{k+1},\dots,q_n\}$. Should I show that if $S$ is a subset of $\Bbb R^n$ then the union of $S$ and $S^\perp$ creates the basis in $\Bbb R^n$?

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HINTS:

1. Is it true that every vector in $\operatorname{span}\{q_{k+1},\dots,q_n\}$ is orthogonal to every vector in $S$? If so, then you know that $\operatorname{span}\{q_{k+1},\dots,q_n\}\subseteq S^\perp$.

2. Suppose that $v\in\Bbb R^n\setminus\operatorname{span}\{q_{k+1},\dots,q_n\}$. Write $v=a_1q_1+\dots+a_nq_n$ for some scalars $a_1,\dots,a_n$. There must be some non-zero $a_i$ with $i\le k$; why? That implies that $v\notin S^\perp$; why? Finally, why does this imply that $S^\perp\subseteq\operatorname{span}\{q_{k+1},\dots,q_n\}$?

Now put (1) and (2) together to conclude that $\operatorname{span}\{q_{k+1},\dots,q_n\}=S^\perp$.

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You need to show two things: First, you must show that if you take any vector in $\operatorname{span}\{q_{k+1},\ldots,q_n\}$ and any vector in $S$, you get 0. (Hint: write both vectors as linear combinations of the $\{q_i\}$ and use linearity of the dot product.)

Then, you need to show the reverse containment. To do this, take a vector not in $\operatorname{span}\{q_{k+1},\ldots,q_n\}$, and show that you can find a vector in $S$ to dot it with that won't give you 0. (Hint: if $q_1$ shows up with nonzero coefficient when you write the expansion for $x$, then $x\cdot q_1\neq0$).

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