Orthogonal Subspaces I am reading orthogonality in subspaces and ran into confusion by reading this part:  
Suppose S is a six-dimensional subspace of nine-dimensional space $\mathbb R^9$.
a) What are the possible dimensions of subspace orthogonal to $S$? Answer: Sub spaces orthogonal to S can have dimensions $0,1,2,3.$
b) What are the possible dimensions of the orthogonal complement $S^{\perp}$ of $S$? Answer: Complement $S^{\perp}$ is the largest orthogonal subspace with dim $3$.
Where I am having trouble is understanding how the answers make sense to the question, or how the answers are pretty much the answers. In other words, for a, how is the dimensions $0,1,2,3$? But maybe I am not understanding the question. Any assistance with helping me understand the answer would be appreciated. 

The orthogonal complement of $S$ is
$$
S^{\perp}=\{v\in\mathbb{R}^9:\langle v, w\rangle=0,\text{ for all }w\in S\}
$$
(where $\langle v,w\rangle$ denotes the inner product). It's immediate that $S^{\perp}$ is a subspace of $\mathbb{R}^9$.
If a vector $u$ belongs to $S\cap S^{\perp}$, then it must be orthogonal to all vectors in $S$, in particular to $u$ itself, so $u=0$. Therefore
$$
S\cap S^{\perp}=\{0\}.
$$
In particular, if $B$ is a basis of $S$ and $C$ is a basis of $S^\perp$, then $B\cup C$ is linearly independent.
Now, another theorem tells us that
$$
S+S^{\perp}=\mathbb{R}^9
$$
which means that $B\cup C$ is a basis of $\mathbb{R}^{9}$. So $C$ must have three elements, $C=\{v_1,v_2,v_3\}$. Hence
$$
\{0\}\subset\operatorname{Span}\{v_1\}
\subset\operatorname{Span}\{v_1,v_2\}
\subset\operatorname{Span}\{v_1,v_2,v_3\}=S^{\perp}
$$
exhibits subspaces of $S^{\perp}$ of dimensions $0$, $1$, $2$, $3$ respectively. They are orthogonal to $S$ because they are included in $S^\perp$. 
Finally, since $\dim S^{\perp}=3$, no subspace with dimension greater than $3$ can be included in it. And any subspace orthogonal to $S$ is included in $S^{\perp}$ by definition.
