Properties of Inner Products

 Statement-1:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that

 Proof:   Let \( \pmb{ x,~y,~z\in V }\), then \begin{align*} \pmb{\lang x,y+z\rang}& = \pmb{\overline{\lang y+z,x\rang}~since~\overline{\lang x,y\rang}=\lang y,x\rang}\\ &= \pmb{\overline{\lang y,x\rang+\lang z,x\rang}~since~\lang x+y,z\rang=\lang x,z\rang+\lang y,z\rang}\\ &= \pmb{\overline{\lang y,x\rang}} + \pmb{\overline{\lang z,x\rang}~since~\overline{a+b}=\overline{a}+\overline{b}}\\ &= \pmb{\lang x,y\rang + \lang x,z\rang ~since~\overline{\lang x,y\rang}=\lang y,x\rang} \end{align*}  Therefore \( \fcolorbox{blue}{white}{\(\pmb{\lang x,y+z\rang=\lang x,y\rang+\lang x,z\rang}\)} \) \( \pmb{\forall~ x,~y,~z\in V }\).

 Statement-2:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that

 Proof:   Let \( \pmb{ x,~y\in V }\) and \(\pmb{c\in F }\), then \begin{align*} \pmb{\lang x,cy\rang} & = \pmb{\overline{\lang cy,x\rang}~since~\overline{\lang x,y\rang}=\lang y,x\rang}\\ & = \pmb{\overline{c\lang y,x\rang}~since~\lang cx,y\rang=c\lang x,y\rang}\\ & =\pmb{\overline{c}\overline{\lang y,x\rang}~since~\overline{ab}=\overline{a}\overline{b} }\\ & =\pmb{\overline{c}\lang x,y\rang~since~\overline{\lang x,y\rang}=\lang y,x\rang} \end{align*}  Therefore \( \fcolorbox{blue}{white}{\(\pmb{\lang x,cy\rang=\overline{c}\lang x,y\rang}\)} \) \( \pmb{\forall~ x,~y\in V }\) and \(\pmb{\forall~ c\in F }\).

 Statement-3:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that

 Proof:    Let \( \pmb{ x\in V }\) and \(\pmb{c\in F }\) then \begin{align*} \pmb{\lang x,0\rang} & = \pmb{\lang x,0y\rang~since~ 0=0y~\forall~ y\in V}\\ & = \pmb{\overline{0}\lang x,y\rang~since~\lang x,cy\rang=\overline{c}\lang x,y\rang} \\ & = \pmb{0\lang x,y\rang ~since~\overline{0}=0}\\ & = \pmb{0} \end{align*}  Therefore \( \fcolorbox{blue}{white}{\(\pmb{\lang x,0\rang=0}\) }\) \( \pmb{\forall~ x\in V }\). Again
\begin{align*} \pmb{\lang 0,x\rang} & = \pmb{\lang 0y,x\rang~since~ 0=0y~\forall~ y\in V}\\ & = \pmb{0\lang y,x\rang~since~\lang cx,y\rang=c\lang x,y\rang}\\ & = \pmb{0} \end{align*}  Therefore \( \fcolorbox{blue}{white}{\(\pmb{\lang 0,x\rang=0}\) }\) \( \pmb{\forall~ x\in V }\).

 Statement-4:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that

 Proof:  Let \(\pmb{\lang x,x\rang=0} \)
   To proof \( \pmb{x=0 }\)
 If possible let \( \pmb{x\ne 0 }\) then \(\pmb{\lang x,x\rang \gt 0} \) but by hypothesis \(\pmb{\lang x,x\rang=0} \). A contradiction.
 Therefore our assumption is wrong. Then \( \pmb{x=0 }\).
 Conversely let \( \pmb{x=0 }\)
   To proof \(\pmb{\lang x,x\rang=0} \)
 Since we have \(\pmb{\lang x,0\rang=0}\) ( \pmb{\forall~ x\in V }\) then \(\pmb{\lang x,x\rang=0} \).
 Therefore \( \fcolorbox{blue}{white}{\(\pmb{\lang x,0\rang=0}\) iff \( \pmb{x=0 }\) }\).

 Statement-5:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that

 Proof:   We have \( \pmb{\forall~ x\in V }\) \begin{align*} & \pmb{\lang x,y\rang = \lang x,z\rang} \\ \implies & \pmb{\lang x,y\rang – \lang x,z\rang=0}\\ \implies & \pmb{\lang x,y\rang +(-1) \lang x,z\rang=0}\\ \implies & \pmb{\lang x,y\rang +(\overline{-1}) \lang x,z\rang=0}\\ \implies & \pmb{\lang x,y\rang + \lang x,(-1)z\rang=0~since~\lang x,cy\rang=\overline{c}\lang x,y\rang}\\ \implies & \pmb{\lang x,y\rang + \lang x,-z\rang=0}\\ \implies & \pmb{\lang x,y-z\rang=0~since~\lang x,y+z\rang=\lang x,y\rang+\lang x,z\rang} \end{align*}  Therefore \(\pmb{\lang x,y-z\rang=0}\) \( \pmb{\forall~ x\in V }\) implies \( \pmb{y-z=0 }\) i.e., \(\pmb{ y=z} \).
 Hence \( \fcolorbox{blue}{white}{ \( \pmb{\lang x,y\rang=\lang x,z\rang}\) \( \pmb{\forall~ x\in V \implies y=z }\) }\).