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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) ​​\( \|T_\alpha\|=\|\alpha\| , \forall \ \alpha\in U \)​​Â
(3) ​\( T \)​ preserves isometry.
     i.e. ​\( \|T_\alpha-T_\beta\|=\|\alpha-\beta\| , \forall \ \alpha, \beta \in U \)​Â
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) ​\( \|T(x)\|=\|x\| \)​Â
(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 ​\( V \)​.Â
Proof:
First we will prove that (1) ​\( \implies \)​ (2)Â
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)Â
Assume that ​\( \|T_x\|=\|x\| , \forall \ x \in V \)​Â
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.Â
​\( \implies \|T_x+T_y\|^2=\|x+y\|^2 \)​
​\( \implies \langle T_x+T_y, T_x+T_y \rangle=\langle x+y, x+y \rangle \)​  By def. of norm.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 ​\( x, y \in V \)​ ​\( \implies \|T_x\|^2=\|x\|^2 \)​ & ​\( \|T_y\|^2=\|y\|^2 \)​
​\( \therefore 2\langle T_x, T_y \rangle=2\langle x, y \rangle \)​
​\( \therefore \langle T_x, T_y \rangle=\langle x, y \rangle \)​
Thus, ​\( \langle T_x, T_y \rangle=\langle x, y \rangle, \forall \ x, y\in V \)​
​\( \implies \)​ ​\( T \)​ is orthogonal.Â
Now, to prove that (1) ​\( \implies \)​ (3)Â
 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 ​\( V \)​.Â
As ​\( \{e_i\}_{i=1}^{n} \)​ is orthogonal
​\( \implies \langle e_i, e_j \rangle= 1 \)​  , if i=j.Â
          =0,  if ​\( i\ne j \)​
and ​\( \{e_i\}_{i=1}^{n} \)​ basis of\( V \implies dim V=n \).Â
Now, ​\( \langle T_{e_i}, T_{e_j}\rangle=\langle e_i, e_j \rangle \)​, ​\( \because T \)​ is orthogonal.
          =1, if i=jÂ
          =0, if ​\( i\ne j \)​
​\( \therefore \langle T_{e_i}, T_{e_j}\rangle=1 \)​, if i=jÂ
             =0, if ​\( i\ne j \)​
​\( \therefore \{T_{e_i}\}_{i=1}^{n} \)​ is orthonormal.Â
Every orthonormal set is linearly independent ​\( \implies \{T_{e_i}\}_{i=1}^{n} \)​ is linearly independent.
As dim V=n= No. of elements in ​\( \{T_{e_i}\}_{i=1}^{n} \)​,
\( \implies \{T_{e_i}\}_{i=1}^{n} \)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.
​\( \therefore \{T_{e_i}\}_{i=1}^{n} \)​ 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.Â
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)
​\( T_x=\sum_{i=1}^{n}{x_iT_{e_i}} \)​ & ​\( T_y=\sum_{j=1}^{n}{y_jT_{e_j}} \)​Â
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,
​\( \langle T_x, T_y \rangle=\langle \sum{x_iT_{e_i}}, \sum{y_jT_{e_j}} \rangle \)​Â
         ​\( =\sum{x_iy_j}\langle T_{e_i}, T_{e_j} \rangle \)​
         ​\( =\sum_{i,j=1}^{n}{x_iy_j} \)​
​\( \therefore \langle T_x, T_y \rangle=\langle x, y \rangle,    \forall \ x, y \in V \)​
Since, (1)​\( \iff \)​(2) and (1)​\( \iff \)​(3)Â
​\( \therefore \)​ above all statements are equivalent.Â
Â
Also Read:Â Cayley Hamilton Theorem
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