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. 

Suppose $S$ is a six-dimensional subspace of $\mathbb{R}^9$.
a) What are the possible dimensions of subspace orthogonal to S?
Before we get to the answer, let's review the definition of an orthogonal subspace. It is a subspace whose intersection $S$ is exactly the zero vector. Now, how many dimensions could such a subspace have, given that $\dim S = 6$? Certainly the zero vector alone qualifies, which is a subspace of dimension zero. Now, I claim that any such orthogonal subspace could not exceed three dimensions. If it did, it would necessarily contain a nontrivial overlap with $S$. 
Let a basis for $S \subset \mathbb{R}^9$ and let $U \subset \mathbb{R}^9$ be a four dimensional subspace. Let $\{b_1, ..., b_9\}$ form a basis for $\mathbb{R}^9$. Then we can pick six of these to form a basis for $S$ and four of these to form a basis for $U$. But this necessarily implies that $U$ and $S$ have a nontrivial intersection! Using this same reasoning, you should be able to understand that the possible dimensions for a subspace orthogonal to $S$ are $0, 1, 2, 3$.

b) What are the possible dimensions of the orthogonal complement $S^{\perp}$ of S?
The orthogonal complement of a set is everything within the stated universe that is not in $S$ itself, plus the zero vector. More formally, it is the set that completes the universal space via direct sum with $S$. Since $S$ is six dimensional, and our universe is $\mathbb{R}^9$, this implies that the orthogonal complement must have three dimensions.
