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# Inner Product Space

## Inner Product Space:

Let â€‹$$U$$â€‹ and â€‹$$V$$â€‹ be any two inner product spaces over the same field â€‹$$\mathbb{R}$$â€‹ and â€‹$$T:U\to V$$â€‹ be a linear transformation then we say that â€‹$$T$$â€‹ preserves inner product if
(1) â€‹$$T$$â€‹ is orthogonal transformation
Â Â  Â Â  i.e. â€‹$$\langle T_\alpha, T_\beta \rangle=\langle \alpha, \beta \rangle$$â€‹
(2) â€‹â€‹$$\|T_\alpha\|=\|\alpha\| , \forall \ \alpha\in U$$â€‹â€‹Â
(3)Â  â€‹$$T$$â€‹Â preserves isometry.
Â Â  Â Â  Â  i.e. â€‹$$\|T_\alpha-T_\beta\|=\|\alpha-\beta\| , \forall \ \alpha, \beta \in U$$â€‹Â

## Theorem:

Let â€‹$$V$$â€‹ be a m-dimensional inner product space over â€‹$$\mathbb{R}$$â€‹ and â€‹$$T:U\to V$$â€‹ be a linear transformation then show that following statements are equivalent:
(1) â€‹$$T$$â€‹ is orthogonal.
(2) â€‹$$\|T(x)\|=\|x\|$$â€‹Â
(3) â€‹$$\{e_i\}_{i=1}^{n}=\{e_1, e_2, … , e_n\}$$â€‹ is orthogonal basis of â€‹$$V$$â€‹ then â€‹$$\{T_{e_i}\}_{i=1}^{n}$$â€‹ is also orthogonal basis of â€‹$$V$$â€‹.Â

## Proof:

First we will prove that (1) â€‹$$\implies$$â€‹ (2)Â
Assume that â€‹$$T$$â€‹ is orthogonal.
i.e. â€‹$$\langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y\in V$$â€‹
Consider, â€‹$$\|T_x\|=\sqrt{\langle T_x, T_x \rangle}$$â€‹ Â  Â  Â … by definition of norm.
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\sqrt{\langle x, x \rangle}$$â€‹Â  Â  â€‹$$\because T$$â€‹ is orthogonal.
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\|x\|$$â€‹
â€‹$$\therefore \|T_x\|=\|x\| , \forall \ x \in V$$â€‹
i.e. (1)â€‹$$\implies$$â€‹ (2) is proved.
Now we will prove that, (2)â€‹$$\implies$$â€‹(1)Â
Assume that â€‹$$\|T_x\|=\|x\| , \forall \ x \in V$$â€‹Â
To prove that â€‹$$T$$â€‹ is orthogonal.
i.e.â€‹$$\langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y\in V$$â€‹
Since, â€‹$$V$$â€‹ is real inner product space,
â€‹$$\implies x+y \in V$$â€‹
â€‹â€‹$$\implies \|T(x+y)\|=\|x+y\|$$â€‹ â€‹………. â€‹â€‹$$\because x+y \in V$$â€‹
â€‹$$\implies \|T_x+T_y\|=\|x+y\|$$â€‹ ……Â  $$\because T$$â€‹ is linear.Â
â€‹$$\implies \|T_x+T_y\|^2=\|x+y\|^2$$â€‹
â€‹$$\implies \langle T_x+T_y, T_x+T_y \rangle=\langle x+y, x+y \rangle$$â€‹Â  Â By def. of norm.Â

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

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

As â€‹$$x, y \in V$$â€‹ â€‹$$\implies \|T_x\|^2=\|x\|^2$$â€‹ & â€‹$$\|T_y\|^2=\|y\|^2$$â€‹
â€‹$$\therefore 2\langle T_x, T_y \rangle=2\langle x, y \rangle$$â€‹
â€‹$$\therefore \langle T_x, T_y \rangle=\langle x, y \rangle$$â€‹
Thus, â€‹$$\langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y\in V$$â€‹
â€‹$$\implies$$â€‹ â€‹$$T$$â€‹ is orthogonal.Â
Now, to prove that (1) â€‹$$\implies$$â€‹ (3)Â
Â Suppose â€‹$$T$$â€‹ is orthogonal. i.e. â€‹$$\langle T_x, T_y\rangle =\langle x, y\rangle, \forall \ x, y\in V$$â€‹
If â€‹$$\{e_i\}_{i=1}^{n}$$â€‹ is orthonormal basis of â€‹$$V$$â€‹ then we have to prove that â€‹$$\{T_{e_i}\}_{i=1}^{n}$$â€‹ is also an orthonormal basis of â€‹$$V$$â€‹.Â
As â€‹$$\{e_i\}_{i=1}^{n}$$â€‹ is orthogonal
â€‹$$\implies \langle e_i, e_j \rangle= 1$$â€‹ Â , if i=j.Â
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â =0,Â  Â if â€‹$$i\ne j$$â€‹
and â€‹$$\{e_i\}_{i=1}^{n}$$â€‹ basis of$$V \implies dim V=n$$.Â
Now, â€‹$$\langle T_{e_i}, T_{e_j}\rangle=\langle e_i, e_j \rangle$$â€‹, â€‹$$\because T$$â€‹ is orthogonal.
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â =1, if i=jÂ
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â =0, if â€‹$$i\ne j$$â€‹
â€‹$$\therefore \langle T_{e_i}, T_{e_j}\rangle=1$$â€‹, if i=jÂ
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  =0, if â€‹$$i\ne j$$â€‹
â€‹$$\therefore \{T_{e_i}\}_{i=1}^{n}$$â€‹Â is orthonormal.Â
Every orthonormal set is linearly independent â€‹$$\implies \{T_{e_i}\}_{i=1}^{n}$$â€‹ is linearly independent.
As dim V=n= No. of elements in â€‹$$\{T_{e_i}\}_{i=1}^{n}$$â€‹,
$$\implies \{T_{e_i}\}_{i=1}^{n}$$Â is maximal linearly independent subset of V.Â
By theorem which states that every maximal linearly independent subset of vector space forms a basis of that vector space.
â€‹$$\therefore \{T_{e_i}\}_{i=1}^{n}$$â€‹Â is orthonormal basis of V.
Now, to prove that (3) â€‹$$\implies$$â€‹ (1)
Consider â€‹$$\{e_i\}_{i=1}^{n}$$â€‹ is orthonormal basis of V and â€‹$$\{T_{e_i}\}_{i=1}^{n}$$â€‹ is also an orthonormal basis of V.Â
We have to prove that T is orthogonal.
i.e. â€‹$$\langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y \in V$$â€‹
Let x, y be any two arbitrary elements of inner product space V.
â€‹$$\implies$$â€‹ â€‹$$x=\sum_{i=1}^{n}{x_ie_i}$$â€‹ & â€‹$$y=\sum_{j=1}^{n}{y_je_j}$$â€‹ Â (â€‹$$\because \{e_i\}_{i=1}^{n}$$â€‹ is basis of V)
â€‹$$T_x=\sum_{i=1}^{n}{x_iT_{e_i}}$$â€‹ & â€‹$$T_y=\sum_{j=1}^{n}{y_jT_{e_j}}$$â€‹Â
Consider, â€‹$$\langle x, y\rangle=\langle\sum_{i=1}^{n}{x_ie_i}, \sum_{j=1}^{n}{y_je_j} \rangle$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\sum_{i,j=1}^{n}{x_iy_j}\langle e_i, e_j \rangle$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\sum_{i,j=1}^{n}{x_iy_j}$$â€‹
Now,
â€‹$$\langle T_x, T_y \rangle=\langle \sum{x_iT_{e_i}}, \sum{y_jT_{e_j}} \rangle$$â€‹Â
Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\sum{x_iy_j}\langle T_{e_i}, T_{e_j} \rangle$$â€‹
Â  Â  Â  Â  Â  Â  Â  Â  Â  â€‹$$=\sum_{i,j=1}^{n}{x_iy_j}$$â€‹
â€‹$$\therefore \langle T_x, T_y \rangle=\langle x, y \rangle,Â Â Â Â \forall \ x, y \in V$$â€‹
Since, (1)â€‹$$\iff$$â€‹(2) and (1)â€‹$$\iff$$â€‹(3)Â
â€‹$$\therefore$$â€‹Â above all statements are equivalent.Â
Â
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