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.
Also Read: Cayley Hamilton Theorem
