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Unit 23: Hilbert Spaces: The Definition and Some Simple Properties
(i) ( x + y, z) = (x, z) + (y, z), x, y, z X and , C. (Linearity property) Notes
(ii) (x, y) = (y, x) (Conjugate symmetry)
(iii) (x, x) 0, (x, x) = 0 x = 0 (Non-negativity)
A complex inner product space X is a linear space over C with an inner product defined on it.
Notes
1. We can also define inner product by replacing C by R in the above definition. In that
case, we get a real inner product space.
2. It should be noted that in the above definition (x, y) does not denote the ordered pair
of the vectors x and y. But it denotes the inner product of the vectors x and y.
Theorem 1: If X is a complex inner product space then
(a) ( x – y, z) = (x, z) – (y, z)
(b) (x, y + z) = (x, y) + (x, z)
(c) (x, y – z) = (x, y) – (x, z)
(d) (x, 0) and (0, x) = 0 for every x X.
Proof: (a) ( x – y, z) = ( x + (– ) y, z)
= (x, z) + (– ) (y, z)
= (x, z) – (y, z).
(b) (x, y + z) = ( y z, x) ( y, x) ( z, x)
= (y, x) (z,x)
= (x, y) (x,z)
(c) (x, y – z) = (x, y + (– ) z) = (x, y) ( )(x,z)
= (x, y) (x,z)
(d) (0, x) = (0 , x) = 0 ( , x) = 0, where is the zero
element of x and (x, 0) (0, x) 0 0 .
Further note that (x, y + z) = (x, |y + 1|z) = 1 (x, y) + 1 (x, z)
Hence (x, y + z) = (x, y) + (x, z).
This completes the proof of the theorem.
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