Friday, February 3, 2023
HomeMathematicsAlgebraInner Product Space

# Inner Product Space

## 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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