Linear subspace/Dual space/1/Example

We consider the linear subspace

${\displaystyle {}U=\langle {\begin{pmatrix}8\\6\\5\end{pmatrix}},{\begin{pmatrix}4\\7\\-3\end{pmatrix}}\rangle \subseteq \mathbb {R} ^{3}\,.}$

The orthogonal space of ${\displaystyle {}U}$ consists of all linear forms

${\displaystyle f\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} }$

with ${\displaystyle {}f{\begin{pmatrix}8\\6\\5\end{pmatrix}}=0}$ and ${\displaystyle {}f{\begin{pmatrix}4\\7\\-3\end{pmatrix}}=0}$. Because a linear form is described by a row matrix ${\displaystyle {}\left(a,\,b,\,c\right)}$ with respect to the standard basis, we are dealing with the solution set of the linear system

${\displaystyle {}8a+6b+5c=0\,,}$

${\displaystyle {}4a+7b-3c=0\,.}$

The solution space is

${\displaystyle {}U^{\perp }={\left\{s\left({\frac {17}{4}},\,1,\,-8\right)\mid s\in K\right\}}\,.}$