Dimensionality of a vector space & number of othogonal vectors I read that the dimensionality $n$ of a vector space is equal to the maximum number $n_\perp$ of mutually orthogonal vectors   in it.
Now suppose we start with an arbitrary vector $|V\rangle$ in the vector space, find all the mutually orthogonal vectors to $|V\rangle$ and form a set $\{|V\rangle,| W \rangle , |X\rangle... \}$, where $\langle V | W\rangle= \langle W |X\rangle=\langle V| X\rangle=...=0$ (i.e. mutually orthogonal),  will the size of this set be equal to $n_\perp$? If yes, why? Is the size of the set we obtained independent of which arbitrary vector $|V\rangle$ we started with?
 A: The answer is yes in both cases.
The thing is that a family of mutually orthogonal vectors is linear independent. Then any maximal mutually orthogonal set $O$ of $n^\bot$ vectors is a maximal linear independent set of vectors, hence a basis. Since the dimension $n$ of a vector space does not depend on a chosen basis we get $n^\bot= n$ for any set $O$ we start with.
This is a rough sketch, there are of cause still details to check!

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*There is no canonical set of all vectors mutually orthogonal to $v$
Think of $\Bbb R^2$ and fix $v=(1,0)$. Then $w=(0,1)$ is orthogonal to $v$, but so are $-w=(0,-1)$ and $2w=(0,2)$. So we have to choose a set of mutually orthogonal vectors such as $\{v,w\}$ or $\{v,-w\}$ or $\{v,2w\}$.


*mutually orthogonal vectors are linearly independent
Let $\{v_1,…,v_n\}$ be mutually orthogonal and $\sum_{k=1}^n \lambda_kv_k=0$. Then for any $i$
$$\begin{align*}
0&=\langle \sum_{k=1}^n \lambda_kv_k,v_i\rangle \\&= \sum_{k=1}^n \lambda_k\langle v_k,v_i\rangle\\&= \lambda_i \langle v_i,v_i\rangle\end{align*}$$
yields $\lambda_i=0$.


*A maximal mutually orthogonal  set is a maximal linear independent set
Given a maximal mutually orthogonal set $O$ assume there is a vector $v$ linear independent to it. Use Gram-Schmidt-orthogonalization to make the set $O\cup \{v\}$ mutually orthogonal. This contradicts the maximality of $O$.
