Norm of Vector #
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
\(\pmb{||x||=\sqrt{\lang x,x \rang}}\) #
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 \(\pmb{||x||=\sqrt{x_{1}^{2}+x_{2}^{2}+…+x_{n}^{2}}}\) .
Properties of Norm #
Statement-1:
Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then prove that
\( \pmb{||c\centerdot x||=|c|\centerdot ||x||}\) \( \pmb{\forall~ x\in V } \) and \( \pmb{ c\in F } \) #
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
\( \pmb{|| x||=0}\) if and only if \( \pmb{x=0 } \) #
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 \( \pmb{x=0 } \) } \).
Cauchy Schwarz Inequality #
Statement-3:
Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then
\( \pmb{ |\lang x,y \rang| \le ||x||\centerdot ||y|| }\) \( \pmb{\forall~ x,~y\in V } \) #
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*}
Triangle Inequality #
Statement-4:
Let \(\pmb{ V}\) be inner product space over a field \(\pmb{ F}\). Then
\( \pmb{ ||x+y|| \le ||x||+||y|| }\) \( \pmb{\forall~ x,~y\in V } \) #
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}{ \( \pmb{ ||x+y|| \le ||x||+||y|| }\) } \) \( \pmb{\forall~ x,~y\in V } \).