Degress of freedom in defining planes in vector spaces

Question

Let $V$ be a finite dimensional vector space with dimension $n$. Let us call any $n-1$ dimensional subspace a (hyper)plane. Then, one needs $n-1$ independent vectors to define a plane in $V$. Suppose now that we endow the space the structure of an inner product. Then the same plane is defined only by a single vector, since a vector and its perpendicular space are direct sum that equals the space $V$. How is this, that giving the space a structure decreases the number of degrees of freedom you have when defining such a plane? (Choosing a plane was of course - arbitrary)

Answer 1

Rouge-Capelli theorem relates number of homogeneous linear equations in a system with dimension of its space of solutions. That is, if $W$ is subspace of $V$, $\dim V = n$ and $\dim W = r$ then $W$ can be described by $n-r$ homogeneous equations, so $n-1$ dimensional plane in $n$ dimensional space can be described by one linear equation.

Let $V=F^n$, where $F$ is a field and $\langle\cdot,\cdot\rangle$ be the standard scalar product. Then if you choose a vector $\alpha=(a_1,..,a_n)$ perpendicular to a plane, say $\Pi$, then $\Pi=\left\{(x_1,...,x_n)\in F^n|a_1x_1+...+a_nx_n=0\right\}$, that is, $\Pi$ is described by a single linear equation. Introducing structure of scalar product doesn't change degrees of freedom in defining subspaces, it only gives you a fancy way of finding system of equations describing given space.