Page 346 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 31: Finite Dimensional Spectral Theory

If for each i, p x = 0 for each i. Notes
i i
Hence p x + p x + … + p x = 0
i 2 m
(p + p + … + p ) x = 0
1 2 m
Ix = 0

x = 0, a contradiction to the fact that x 0. Hence must be equal to for some i. This in the
i
spectral resolution (1) of T, the scalar are the precisely the eigenvalue of T.
i
If the spectral resolution is not unique.
Let T = Q + Q + … + Q … (3)
1 1 2 2 n n
be another revolution of T. Then , , … is the same set of eigenvalues of T written in
1 2 n
different order. Hence writing the eigenvalues in the same order as in (1) and renaming the
projections, we can write (3) as
T = Q + Q + … + Q … (4)
1 1 2 2 m m
To prove uniqueness, we shall show that p in (1) and Q in (4) are some.
i i
2
Using the fact p = p , p p = 0 i j, we can have
i i i j
0
T = I = p + p + … + p
1 2 m
1
T = p + p + … + p
1 1 2 2 m m
T = p + … + 2 p
2
2
1 1 m m
n
n
and T = p + … + n p for any positive integer n. … (5)
1 1 m m
Now if g (t) is a polynomial with complex coefficient in the complex variable t, we can write
g (T) as
g (T) = g ( ) p + g ( ) p + … + g ( ) p
1 1 2 2 m m
m
= g ( ) p j … (by 5)
j
j 1
Let be a polynomial such that ( ) = 1 and ( ) = 0
i i i i j
if i j
Taking in place of g, we get
i
m m
(T) = ( ) p p p
i i j j ij j i
j 1 j 1
Hence for each i, let p = (T) which is a polynomial in T. The proof is complete if we show the
i i
existence of over the field of complex number.
i
(t ) (t )(t )...(t )
Now (t) 1 i 1 i 1 m
i
( )....( )( ) ( )
i 1 i i 1 i i 1 i m
satisfies our requirements i.e. ( ) = 1 and ( ) = 0 if i j
i i i j
Repeating the above discussion for Q ’s we get in a similar manner Q = (T) for each i.
i i i
p = Q for each i.
i i
This completes the proof of the theorem.

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