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.



*

*The orthogonal complement of $S$ in $V$ is the same as the orthogonal complement $W$ in $V$.

*Every vector of $S$ belongs to the orthogonal complement of $S$ in $V$.

*If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}=\mathbf{0}$.

*If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $\mathbf{u}$ belongs to $S$.

*If $\mathbf{u}$ is a vector in $V$ which belongs to both $W$ and its orthogonal complement in $V$, then $|u|=1$.

*If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at least $3$.

*If $V$ has dimension $10$, and $S$ contains $7$ vectors, then the dimension of $W^\perp$ is at most $3$.

*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 $3$.
 A: 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 $\dim(S^\perp)=5 \geq 3$.
