Introuction to Inner Product Space 

  Definition: #

        Let \(\pmb{ V}\) be vector space over a field \(\pmb{ F}\). An \(\fcolorbox{blue}{white}{\pmb{Inner Product}}\) on \(\pmb{ V}\) is a mapping from \(\pmb{ V \times V \to F}\), which is denoted by \( \pmb{\lang x,y\rang~\forall~ x,~y\in V }\), that satisfies the following conditions:
 a) \( \pmb{\lang x+y,z\rang=\lang x,z\rang+\lang y,z\rang}\) \( \pmb{\forall~ x,~y,~z\in V }\)
 b) \( \pmb{\lang cx,y\rang=c\lang x,y\rang}\) \( \pmb{\forall~ x,~y\in V }\) and \( \pmb{\forall~ c\in F }\)
 c) \( \pmb{\overline{\lang x,y\rang}=\lang y,x\rang}\) \( \pmb{\forall~ x,~y\in V }\) where \( \pmb{\overline{\lang x,y\rang}}\) denotes the \(\fcolorbox{blue}{white}{\pmb{complex conjugate}}\) of \( \pmb{\lang x,y\rang}\)
 d) \( \pmb{\lang x,x\rang > 0}\) if \( \pmb{x\ne 0 }\)
 And \(\pmb{ V}\) is said to be \(\fcolorbox{blue}{white}{\pmb{Inner Product Space}}\) with inner product \( \pmb{\lang ~,~\rang }\).

  Definition: #

        Let \(\pmb{ V}\) be vector space over the field \(\pmb{ \mathcal{C}}\). An \(\fcolorbox{blue}{white}{\pmb{Inner Product}}\) on \(\pmb{ V}\) is a mapping from \(\pmb{ V \times V \to \mathcal{C}}\), which is denoted by \( \pmb{\lang x,y\rang~\forall~ x,~y\in V }\), that satisfies the following conditions:
 a) \( \pmb{\lang x+y,z\rang=\lang x,z\rang+\lang y,z\rang}\) \( \pmb{\forall~ x,~y,~z\in V }\)
 b) \( \pmb{\lang cx,y\rang=c\lang x,y\rang}\) \( \pmb{\forall~ x,~y\in V }\) and \( \pmb{\forall~ c\in \mathcal{C} }\)
 c) \( \pmb{\overline{\lang x,y\rang}=\lang y,x\rang}\) \( \pmb{\forall~ x,~y\in V }\) where \( \pmb{\overline{\lang x,y\rang}}\) denotes the \(\fcolorbox{blue}{white}{\pmb{complex conjugate}}\) of \( \pmb{\lang x,y\rang}\)
 d) \( \pmb{\lang x,x\rang > 0}\) if \( \pmb{x\ne 0 }\)
 And \(\pmb{ V}\) is said to be \(\fcolorbox{blue}{white}{\pmb{Complex Inner Product Space}}\) with inner product \( \pmb{\lang ~,~\rang }\).

  Example:
        Let us consider the space \(\pmb{ V=\mathcal{C}^{2}}\) and \(\pmb{ F=\mathcal{C} }\).
        Let \( \pmb{\lang x,y\rang=x_{1}\overline{y_{1}}+x_{2}\overline{y_{2}} }\) where \( \pmb{\forall~ x,~y \in \mathcal{C}^{2} }\) such that \( \pmb{x=(x_{1},x_{2}) }\) and \( \pmb{y=(y_{1},y_{2}) }\) and \( \pmb{x_{1},x_{2},y_{1},y_{2}\in \mathcal{C}} \)

Let \( \pmb{a_{i},b_{i},c_{i} \in \mathcal{C},~i=1,2 }\) and let \( \pmb{d \in \mathcal{C} }\) then \( \pmb{a,b,c \in \mathcal{C}^{2} }\) such that \( \pmb{a=(a_{1},a_{2}) }\), \( \pmb{b=(b_{1},b_{2}) }\) and \( \pmb{c=(c_{1},c_{2}) }\) where \( \pmb{a_{i},b_{i},c_{i} \in \mathcal{C},~i=1,2 }\).

 a)   \begin{align*} & \pmb{\lang a+c,b\rang}\\ &= \pmb{\lang (a_{1},a_{2})+(c_{1},c_{2}),(b_{1},b_{2})\rang}\\ &= \pmb{\lang (a_{1}+c_{1},a_{2}+c_{2}),(b_{1},b_{2})\rang}\\ &=\pmb{\left(a_{1}+c_{1}\right)\overline{b_{1}}+\left(a_{2}+c_{2}\right)\overline{b_{2}} }\\ &=\pmb{\left(a_{1}\overline{b_{1}}+a_{2}\overline{b_{2}}\right)+ \left(c_{1}\overline{b_{1}}+c_{2}\overline{b_{2}}\right) }\\ &= \pmb{\lang a,b\rang+\lang c,b \rang} \end{align*}  Therefore \( \pmb{\lang a+c,b\rang=\lang a,b\rang+\lang c,b \rang}\) \( \pmb{\forall~ a,b,c \in \mathcal{R}^{2} }\).

 b)   \begin{align*} & \pmb{\lang da,b\rang}\\ &= \pmb{\lang d(a_{1},a_{2}) ,b\rang}\\ &= \pmb{\lang (da_{1},da_{2}) ,b\rang}\\ &=\pmb{\left(da_{1}\right)\overline{b_{1}}+\left(da_{2}\right)\overline{b_{2}} }\\ &=\pmb{d\left(a_{1}\overline{b_{1}}+a_{2}\overline{b_{2}}\right)}\\ & =\pmb{d\lang a,b\rang} \end{align*}  Therefore \( \pmb{\lang da,b\rang=d\lang a,b\rang}\) \( \pmb{\forall~ a,b \in \mathcal{R}^{2} }\) and \( \pmb{\forall~d \in \mathcal{R}}\).

 c)   \begin{align*} & \pmb{\overline{\lang a,b\rang}}\\ &=\pmb{\overline{\left(a_{1}\overline{b_{1}}+a_{2}\overline{b_{2}}\right) }}\\ &=\pmb{\left(\overline{a_{1}\overline{b_{1}}}+\overline{a_{2}\overline{b_{2}}}\right) }\\ &=\pmb{\left(b_{1}\overline{a_{1}}+b_{2}\overline{a_{2}}\right) }\\ &= \pmb{\lang b,a\rang}\\ \end{align*}  Therefore \( \pmb{\lang a,b\rang=\lang b,a\rang}\) \( \pmb{\forall~ a,b \in \mathcal{C}^{2} }\).

 d)   Let \(\pmb{a\ne 0} \) i.e., \(\pmb{(a_{1},a_{2})\ne (0,0)} \) implies \(\pmb{a_{1}\ne 0} \) and \(\pmb{a_{2}\ne 0} \) therefore \begin{align*} & \pmb{\lang a,a\rang}\\ &=\pmb{\left(a_{1}a_{1}+a_{2}a_{2}\right) }\\ &=\pmb{\left(a_{1}^{2}+a_{2}^{2}\right) }\\ &\gt 0 \end{align*}  Therefore \( \pmb{\lang a,a\rang\gt 0}\) when \( \pmb{ a\ne 0 }\).

 Hence \(\pmb{ \mathcal{C}^{2}}\) is inner product space.

  Definition: #

        Let \(\pmb{ V}\) be vector space over the field \(\pmb{ \mathcal{R}}\). An \(\fcolorbox{blue}{white}{\pmb{Inner Product}}\) on \(\pmb{ V}\) is a mapping from \(\pmb{ V \times V \to \mathcal{R}}\), which is denoted by \( \pmb{\lang x,y\rang~\forall~ x,~y\in V }\), that satisfies the following conditions:
 a) \( \pmb{\lang x+y,z\rang=\lang x,z\rang+\lang y,z\rang}\) \( \pmb{\forall~ x,~y,~z\in V }\)
 b) \( \pmb{\lang cx,y\rang=c\lang x,y\rang}\) \( \pmb{\forall~ x,~y\in V }\) and \( \pmb{\forall~ c\in \mathcal{R} }\)
 c) \( \pmb{\lang x,y\rang=\lang y,x\rang}\) \( \pmb{\forall~ x,~y\in V }\)
 d) \( \pmb{\lang x,x\rang > 0}\) if \( \pmb{x\ne 0 }\)
 And \(\pmb{ V}\) is said to be \(\fcolorbox{blue}{white}{\pmb{Real Inner Product Space}}\) with inner product \( \pmb{\lang ~,~\rang }\).

  Example:
        Let us consider the space \(\pmb{ V=\mathcal{R}^{2}}\) and \(\pmb{ F=\mathcal{R} }\).
        Let \( \pmb{\lang x,y\rang=x_{1}y_{1}+x_{2}y_{2} }\) where \( \pmb{\forall~ x,~y \in \mathcal{R}^{2} }\) such that \( \pmb{x=(x_{1},x_{2}) }\) and \( \pmb{y=(y_{1},y_{2}) }\) and \( \pmb{x_{1},x_{2},y_{1},y_{2}\in \mathcal{R}} \)

Let \( \pmb{a_{i},b_{i},c_{i} \in \mathcal{R},~i=1,2 }\) and let \( \pmb{d \in \mathcal{R} }\) then \( \pmb{a,b,c \in \mathcal{R}^{2} }\) such that \( \pmb{a=(a_{1},a_{2}) }\), \( \pmb{b=(b_{1},b_{2}) }\) and \( \pmb{c=(c_{1},c_{2}) }\) where \( \pmb{a_{i},b_{i},c_{i} \in \mathcal{R},~i=1,2 }\).

 a)   \begin{align*} & \pmb{\lang a+c,b\rang}\\ &= \pmb{\lang (a_{1},a_{2})+(c_{1},c_{2}),(b_{1},b_{2})\rang}\\ &= \pmb{\lang (a_{1}+c_{1},a_{2}+c_{2}),(b_{1},b_{2})\rang}\\ &=\pmb{\left(a_{1}+c_{1}\right)b_{1}+\left(a_{2}+c_{2}\right)b_{2} }\\ &=\pmb{\left(a_{1}b_{1}+a_{2}b_{2}\right)+ \left(c_{1}b_{1}+c_{2}b_{2}\right) }\\ &= \pmb{\lang a,b\rang+\lang c,b \rang} \end{align*}  Therefore \( \pmb{\lang a+c,b\rang=\lang a,b\rang+\lang c,b \rang}\) \( \pmb{\forall~ a,b,c \in \mathcal{R}^{2} }\).

 b)   \begin{align*} & \pmb{\lang da,b\rang}\\ &= \pmb{\lang d(a_{1},a_{2}) ,b\rang}\\ &= \pmb{\lang (da_{1},da_{2}) ,b\rang}\\ &=\pmb{\left(da_{1}\right)b_{1}+\left(da_{2}\right)b_{2} }\\ &=\pmb{d\left(a_{1}b_{1}+a_{2}b_{2}\right)}\\ & =\pmb{d\lang a,b\rang} \end{align*}  Therefore \( \pmb{\lang da,b\rang=d\lang a,b\rang}\) \( \pmb{\forall~ a,b \in \mathcal{R}^{2} }\) and \( \pmb{\forall~d \in \mathcal{R}}\).

 c)   \begin{align*} & \pmb{\lang a,b\rang}\\ &=\pmb{\left(a_{1}b_{1}+a_{2}b_{2}\right) }\\ &=\pmb{\left(b_{1}a_{1}+b_{2}a_{2}\right) }\\ &= \pmb{\lang b,a\rang}\\ \end{align*}  Therefore \( \pmb{\lang a,b\rang=\lang b,a\rang}\) \( \pmb{\forall~ a,b \in \mathcal{R}^{2} }\).

 d)   Let \(\pmb{a\ne 0} \) i.e., \(\pmb{(a_{1},a_{2})\ne (0,0)} \) implies \(\pmb{a_{1}\ne 0} \) and \(\pmb{a_{2}\ne 0} \) therefore \begin{align*} & \pmb{\lang a,a\rang}\\ &=\pmb{\left(a_{1}a_{1}+a_{2}a_{2}\right) }\\ &=\pmb{\left(a_{1}^{2}+a_{2}^{2}\right) }\\ &\gt 0 \end{align*}  Therefore \( \pmb{\lang a,a\rang\gt 0}\) when \( \pmb{ a\ne 0 }\).

 Hence \(\pmb{ \mathcal{R}^{2}}\) is inner product space.