# Orthogonal complement and subspaces

I have the following multiple response question here.

Suppose $$V$$ is a subspace of $$\mathbb{R}^n$$ and $$S$$ is a set of vectors in $$V$$ whose span is the subspace $$W$$ of $$V$$. Select the correct statements.

1. The orthogonal complement of $$S$$ in $$V$$ is the same as the orthogonal complement $$W$$ in $$V$$.
2. Every vector of $$S$$ belongs to the orthogonal complement of $$S$$ in $$V$$.
3. If $$\mathbf{u}$$ is a vector in $$V$$ which belongs to both $$W$$ and its orthogonal complement in $$V$$, then $$\mathbf{u}=\mathbf{0}$$.
4. If $$\mathbf{u}$$ is a vector in $$V$$ which belongs to both $$W$$ and its orthogonal complement in $$V$$, then $$\mathbf{u}$$ belongs to $$S$$.
5. If $$\mathbf{u}$$ is a vector in $$V$$ which belongs to both $$W$$ and its orthogonal complement in $$V$$, then $$|u|=1$$.
6. If $$V$$ has dimension $$10$$, and $$S$$ contains $$7$$ vectors, then the dimension of $$W^\perp$$ is at least $$3$$.
7. If $$V$$ has dimension $$10$$, and $$S$$ contains $$7$$ vectors, then the dimension of $$W^\perp$$ is at most $$3$$.
8. If $$V$$ has dimension $$10$$, and $$S$$ contains $$7$$ vectors, then the dimension of $$W^\perp$$ is always $$3$$.

The correct answer should be $$1$$,$$3$$ and $$6$$ but could someone explain why?

I understand $$3$$ but I don't understand why $$1$$ and $$6$$ are true. Could someone explain why the other statements are false?

For instance, I thought $$8$$ would be right instead of $$6$$ since I thought by the rank-nullity theorem, the nullity must ALWAYS be $$3$$.

By definition $$S^\perp=\{\bar{x} \in V | \bar{x}\cdot \bar{s} = 0, \forall \bar{s} \in S\}$$, where I'm assuming $$(\cdot)$$ is the regular dot product.
Let $$S$$ be a set of vectors in $$V$$.
Then $$\bar{x}\in S^\perp$$ implies $$\bar{x}\cdot\bar{s} =0$$ for all $$\bar{s}\in S$$, and since $$W=\text{span}(S)$$, $$W$$ is the set of all linear combinations of the vectors in $$S$$, thus for $$\bar{w} \in W$$, $$\bar{w} = a_1\bar{s_1} +\cdots + a_n\bar{s_n}$$ for $$a_1,...,a_n \in \mathbb{R}$$, and $$\bar{x}\cdot\bar{w} =a_1(\bar{x}\cdot\bar{s_1})+\cdots+a_n(\bar{x}\cdot\bar{s_n}) = a_1(0)+\cdots+a_n(0)=0$$.
Similarly $$W^\perp = \{\bar{x} \in V|\bar{x}\cdot\bar{w}=0, \forall \bar{w} \in W\}$$, thus $$\bar{x}\in W^\perp$$ because every $$\bar{w} \in W$$ is a linear combination of the vectors in $$S$$ and $$\bar{x}\cdot\bar{s} =0$$ for all $$\bar{s}\in S$$. $$S^\perp \subseteq W^\perp$$.
Every vector $$\bar{s} \in S$$ is an element of $$W$$ thus if $$\bar{x} \in W^\perp$$ then in particular $$\bar{x}\cdot\bar{s} =0$$ for all $$\bar{s} \in S$$, thus $$\bar{x}\in S^\perp$$. $$W^\perp \subseteq S^\perp$$.
In conclusion $$S^\perp = W^\perp$$. As for $$\#6,7,8$$ $$\dim(S^\perp)+\dim(S) = \dim(V) =10$$, however we cannot be certain that $$\dim(S)=7$$ because the vectors in $$S$$ may not be linearly independent, thus at best we know that $$\dim(S) \leq 7$$, and so $$\dim(S^\perp)\geq 10-7=3$$. For example say $$S$$ had $$7$$ vectors but when finding a basis for $$S$$ we find that the basis only has $$5$$ vectors, then $$\dim(S) =5\leq 7$$ and $$\dim(S^\perp)=5 \geq 3$$.