Inner Product Space

 Inner Product Space : 

Let $U$ and $V$ be any two inner product spaces over the same field $\mathbb{R}$ and $T:U\to V$be a linear transformation then we say that T preserves inner product if

(1) T is orthogonal transformation

      i.e. $\langle T_\alpha, T_\beta \rangle=\langle \alpha, \beta \rangle$ 

(2) $\|T_\alpha\|=\|\alpha\| , \forall \ \alpha\in U$ 

(3)  T preserves isometry.

        i.e. $\|T_\alpha-T_\beta\|=\|\alpha-\beta\| , \forall \ \alpha, \beta \in U$ 

Theorem :

Let $V$ be a m-dimensional inner product space over $\mathbb{R}$ and $T:U\to V$ be a linear transformation then show that following statements are equivalent:

(1) T is orthogonal.

(2) $\|T(x)\|=\|x\|$ 

(3) $\{e_i\}_{i=1}^{n}=\{e_1, e_2, ... , e_n\}$ is orthogonal basis of $V$ then $\{T_{e_i}\}_{i=1}^{n}$ is also orthogonal basis of $V$. 

Proof :

First we will prove that $(1)\implies (2)$ 

Assume that $T$ is orthogonal.

i.e. $\langle T_x, T_y \rangle=\langle x, y \rangle$, $\forall \ x, y\in V$ 

Consider, $\|T_x\|=\sqrt{\langle T_x, T_x \rangle}$      ... by definition of norm.

                                $=\sqrt{\langle x, x \rangle}$     $\because T$ is orthogonal.

                                $=\|x\|$ 

$\therefore \|T_x\|=\|x\| , \forall \ x \in V$ 

i.e. $(1)\implies (2)$ is proved.

Now we will prove that, $(2)\implies (1)$ 

Assume that $\|T_x\|=\|x\| , \forall \ x \in V$ 

To prove that $T$ is orthogonal.

i.e. $\langle T_x, T_y \rangle=\langle x, y \rangle$, $\forall \ x, y\in V$ 

Since, $V$ is real inner product space,

$\implies x+y \in V$ 

$\implies \|T(x+y)\|=\|x+y\|$    .....      $\because x+y \in V$ 

$\implies \|T_x+T_y\|=\|x+y\|$ ......       $\because$ T is linear. 

$\implies \|T_x+T_y\|^2=\|x+y\|^2$ 

$\implies \langle T_x+T_y, T_x+T_y \rangle=\langle x+y, x+y \rangle$   By def. of norm. 

$\implies \langle T_x, T_x \rangle +2\langle T_x, T_y \rangle + \langle T_y, T_y \rangle=\langle x, x \rangle +2\langle x, y \rangle + \langle y, y \rangle$ 

$\implies \|T_x\|^2+2\langle T_x, T_y \rangle + \|T_y\|^2=\|x\|^2+2\langle x, y \rangle +\|y\|^2$ 

As $x, y \in V \implies \|T_x\|^2=\|x\|^2$ & $\|T_y\|^2=\|y\|^2$ 

$\therefore 2\langle T_x, T_y \rangle=2\langle x, y \rangle$ 

$\therefore \langle T_x, T_y \rangle=\langle x, y \rangle$ 

Thus, $\langle T_x, T_y \rangle=\langle x, y \rangle$, $\forall \ x, y\in V$ 

$\implies$ T is orthogonal. 

Now, to prove that $(1) \implies (3)$ 

 Suppose T is orthogonal. i.e. $\langle T_x, T_y\rangle =\langle x, y\rangle, \forall \ x, y\in V$ 

If $\{e_i\}_{i=1}^{n}$ is orthonormal basis of V then we have to prove that $\{T_{e_i}\}_{i=1}^{n}$ is also an orthonormal basis of V. 

As $\{e_i\}_{i=1}^{n}$ is orthogonal $\implies \langle e_i, e_j \rangle= 1$,   if $i=j$. 

                                                                               $=0$,   if $i\ne j$ 

and $\{e_i\}_{i=1}^{n}$ basis of V $\implies$ dim $V=n$. 

Now, $\langle T_{e_i}, T_{e_j}\rangle=\langle e_i, e_j \rangle$, $\because$ T is orthogonal.

                               $=1$, if $i=j$ 

                               $=0$, $i\ne j$ 

$\therefore \langle T_{e_i}, T_{e_j}\rangle=1$, if $i=j$ 

                          $=0$, if $i\ne j$ 

$\therefore \{T_{e_i}\}_{i=1}^{n}$ is orthonormal. 

Every orthonormal set is linearly independent $\implies \{T_{e_i}\}_{i=1}^{n}$ is linearly independent.

As dim $V=n=$ No. of elements in $\{T_{e_i}\}_{i=1}^{n}$,

$\implies \{T_{e_i}\}_{i=1}^{n}$ is maximal linearly independent subset of V. 

By theorem which states that every maximal linearly independent subset of vector space forms a basis of that vector space.

$\therefore \{T_{e_i}\}_{i=1}^{n}$ is orthonormal basis of V.

Now, to prove that $(3) \implies (1)$

Consider $\{e_i\}_{i=1}^{n}$ is orthonormal basis of V and $\{T_{e_i}\}_{i=1}^{n}$ is also an orthonormal basis of V. 

We have to prove that T is orthogonal.

i.e. $\langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y \in V$ 

Let x, y be any two arbitrary elements of inner product space V.

$\implies x=\sum_{i=1}^{n}{x_ie_i}$ & $y=\sum_{j=1}^{n}{y_je_j}$  ($\because \{e_i\}_{i=1}^{n}$ is basis of V) 

$T_x=\sum_{i=1}^{n}{x_iT_{e_i}}$ & $T_y=\sum_{j=1}^{n}{y_jT_{e_j}}$ 

Consider, $\langle x, y\rangle=\langle\sum_{i=1}^{n}{x_ie_i}, \sum_{j=1}^{n}{y_je_j} \rangle$ 

                                 $=\sum_{i,j=1}^{n}{x_iy_j}\langle e_i, e_j \rangle$ 

                                 $=\sum_{i,j=1}^{n}{x_iy_j}$ 

Now,

$\langle T_x, T_y \rangle=\langle \sum{x_iT_{e_i}}, \sum{y_jT_{e_j}} \rangle$ 

                   $=\sum{x_iy_j}\langle T_{e_i}, T_{e_j} \rangle$ 

                   $=\sum_{i,j=1}^{n}{x_iy_j}$ 

$\therefore \langle T_x, T_y \rangle=\langle x, y \rangle,       \forall \ x, y \in V$ 

Since, $(1)\iff (2)$ and $(1)\iff (3)$ 

$\therefore$ above all statements are equivalent. 

Also Read: Cayley Hamilton Theorem