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__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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