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Inner Product Space

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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, if ​\( i\ne j \)
\( \therefore \langle T_{e_i}, T_{e_j}\rangle=1 \)​, 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. 
 
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