Given that \(U\bot V\) that is \((u, v) = 0\) for any \(u\in U\) and \(v\in V\). Let

\(au+bv=0\)

\(\Rightarrow(au+bv,u)=0 \Rightarrow a(u,v)+b(u,v)=0 \Rightarrow a(u,u)=0\) (since u is a non zero vector) \(\Rightarrow a=0\)

Again

\(au+bv=0 \Rightarrow (au+bv,v)=0 \Rightarrow a(u,v)+b(u,v)=0 \Rightarrow b(v,v)=0\) (since v is a non zero vector) \(\Rightarrow b=0\)

Therefore \(\displaystyle{a}{u}+{b}{v}={0}\to{a}={b}={0}\). Hence \(\{u,v\}\) is linearly independent.