__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.