Norm of Vector in Inner Product Space

 Definition:

 Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\) and \(\pmb{x\in V}\). Then the Norm of the vector \(\pmb{x}\) is denoted by \(\pmb{||x||}\) and is define by

 Example:   Let \(\pmb{ V=F^{n}}\) and \(\pmb{x\in F^{n}}\) such that \(\pmb{x=(x_{1},x_{2},…,x_{n})}\) then the norm of \(\pmb{x}\) is \(\pmb{||x||=\sqrt{x_{1}^{2}+x_{2}^{2}+…+x_{n}^{2}}}\) .

 Statement-1:

 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{||c\centerdot x||}& =\pmb{ \sqrt{\lang c\centerdot x,c\centerdot x \rang}} \\ & =\pmb{ \sqrt{c^{2}\centerdot\lang x,x \rang}} \\ & =\pmb{|c|\centerdot ||x||} \end{align*}  Therefore \( \fcolorbox{blue}{white}{\( \pmb{||c\centerdot x||=|c|\centerdot ||x||}\)}\) \( \pmb{\forall~ x\in V } \) and \( \pmb{ c\in F } \).

 Statement-2:

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

 Proof:   Let \( \pmb{|| x||=0}\) then \begin{align*} & \pmb{|| x|| = 0} \\ \pmb{\implies} & \pmb{ \sqrt{\lang x,x \rang} = 0} \\ \pmb{\implies} & \pmb{ \lang x,x \rang = 0} \\ \pmb{\implies} & \pmb{ x = 0} \\ \end{align*}    Conversely, let \( \pmb{x=0}\) then \begin{align*} & \pmb{ x = 0} \\ \pmb{\implies} & \pmb{ \lang x,x \rang = 0} \\ \pmb{\implies} & \pmb{ \sqrt{\lang x,x \rang} = 0} \\ \pmb{\implies}& \pmb{|| x|| = 0} \\ \end{align*}  Therefore \( \fcolorbox{blue}{white}{ \( \pmb{|| x||=0}\) if and only if \( \pmb{x=0 } \) } \).

 Statement-3:

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

 Proof:   Let \( \pmb{ x,~y\in V } \).
 Case-1:   Let \( \pmb{ y=0 } \) then \( \pmb{ |\lang x,0 \rang|=0} \) and \( \pmb{ ||x||\centerdot ||0||= 0} \).
  Therefore \( \pmb{ |\lang x,y \rang| = ||x||\centerdot ||y|| }\) when \( \pmb{ y=0 } \) .
 Case-2:   Let \( \pmb{ y \ne 0 } \)
 Let \( \pmb{ c\in F} \) such that $$ \pmb{ c=\frac{\lang x,y \rang}{\lang y,y \rang}=\frac{\lang x,y \rang}{||y||^{2}} } $$   Now \begin{align*} &\pmb{ ||x-c\centerdot y||^{2}\ge 0} \\ \pmb{\implies} & \pmb{ \lang x-c\centerdot y,x-c\centerdot y \rang\ge 0} \\ \pmb{\implies} & \pmb{ \lang x-c\centerdot y,x \rang -\overline{c}\lang x-c\centerdot y,y \rang\ge 0} \\ \pmb{\implies} & \pmb{ \lang x,x \rang-\lang c\centerdot y,x \rang -\overline{c}\lang x ,y \rang +\overline{c}\lang c\centerdot y,y \rang\ge 0} \\ \pmb{\implies} & \pmb{ \lang x,x \rang-\lang c\centerdot y,x \rang -\overline{c}\lang x ,y \rang +c\overline{c}\lang y,y \rang\ge 0} \\ \pmb{\implies} & \pmb{ ||x||^{2}-\frac{\lang x,y \rang}{||y||^{2}} \lang y,x \rang -\overline{\frac{\lang x,y \rang}{||y||^{2}}}\lang x ,y \rang +\frac{\lang x,y \rang}{||y||^{2}} \overline{\frac{\lang x,y \rang}{\cancel{||y||^{2}}}} \cancel{||y||^{2}}\ge 0} \\ \pmb{\implies} & \pmb{ ||x||^{2}-\frac{\lang x,y \rang \overline{\lang x,y \rang}}{||y||^{2}} -\cancel{\frac{|\lang x,y \rang|^{2}}{||y||^{2}}} +\cancel{\frac{|\lang x,y \rang|^{2}}{||y||^{2}}} \ge 0} \\ \pmb{\implies} & \pmb{ ||x||^{2} -\frac{|\lang x,y \rang|^{2}}{||y||^{2}} \ge 0} \\ \pmb{\implies} & \pmb{ ||x||^{2}\centerdot ||y||^{2} -|\lang x,y \rang|^{2}\ge 0} \\ \pmb{\implies} & \pmb{ |\lang x,y \rang|^{2} \le ||x||^{2}\centerdot ||y||}^{2} \\ \pmb{\implies} & \pmb{ |\lang x,y \rang| \le ||x||\centerdot ||y||} \\ \end{align*}

 Therefore \( \fcolorbox{blue}{white}{ \( \pmb{ |\lang x,y \rang| \le ||x||\centerdot ||y|| }\) } \) \( \pmb{\forall~ x,~y\in V } \).

 Statement-4:

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

 Proof:   Let \( \pmb{ x,~y\in V } \). \begin{align*} & \pmb{ ||x+y||^{2} } = \pmb{ \lang x+y,x+y \rang} \\ \pmb{\implies}& \pmb{ ||x+y||^{2} } = \pmb{ \lang x+y,x \rang+ \lang x+y,y \rang}\\ \pmb{\implies}& \pmb{ ||x+y||^{2} } = \pmb{ \lang x,x \rang+ \lang y,x \rang+ \lang x,y \rang+ \lang y,y \rang}\\ \pmb{\implies}& \pmb{ ||x+y||^{2} } = \pmb{ ||x||^{2}+\overline{\lang x,y \rang}+ \lang x,y \rang+ ||y||^{2}}\\ \pmb{\implies}&\pmb{ ||x+y||^{2} } = \pmb{ ||x||^{2}+ 2\centerdot Real~Part~of~\lang x,y \rang+ ||y||^{2}}\\ \pmb{\implies}&\pmb{ ||x+y||^{2} } = \pmb{ ||x||^{2}+ 2\centerdot |\lang x,y \rang|+ ||y||^{2}}\\ \pmb{\implies}& \pmb{ ||x+y||^{2} } \le \pmb{ ||x||^{2}+ 2\centerdot ||x||\centerdot||y||+ ||y||^{2}}\\ \pmb{\implies}&\pmb{ ||x+y||^{2} } \le \pmb{ \left(||x||+ ||y||\right)^{2}}\\ \pmb{\implies}&\pmb{ ||x+y|| } \le \pmb{||x||+ ||y||} \end{align*}  Therefore \( \fcolorbox{blue}{white}{ \( \pmb{ ||x+y|| \le ||x||+||y|| }\) } \) \( \pmb{\forall~ x,~y\in V } \).