Finding the dimension of a subspace in $\mathbb R^n$ Given $S_{u}^\bot:=\{v\in \mathbb R^n:v \cdot u=0\}$, I've proved that $S_{u}^\bot$ is a subspace of $\mathbb R^n$. The next part of the question asks me to prove that $S_{u}^\bot$ has dimension of $n-1$.
I assume that it has something to do with the number of elements in the basis of the condition inside $S_{u}^\bot$ but do not know how to tackle. I also solved some other parts of the same question but I don't think that they're related to this part particularly.

For any vector $s \in V$, $s - (s \cdot u)u \in S_u^\perp$, by definition of $S_u^\perp$. In particular, $s = (s\cdot u)u + v$ for some $v \in S_u^\perp$.
This shows that every vector can be written as a linear combination of a vector in the span of $u$, and in $S_u^\perp$. In other words, $S_u^\perp + \overline{\{u\}}= V$.
However, as the subspaces are disjoint (why?) the sum is infact a direct sum, so by the usual relation between the dimensions of two vector spaces and that of their direct sum, one can conclude.
