Page 345 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes M = {0}.
For if M {0}, then every operator on a non-zero finite dimensional Hilbert space must have an
eigenvalue.

Now M = {0} M = H.
Thus M = M + M + …… + M = H and so M ’s span H.
1 2 m i
This complete the proof of the theorem.

31.1.7 Spectral Resolution of an Operator

Let T be an operator on a Hilbert space H. If there exist distinct complex numbers , , …,
1 2 m
and non-zero pairwise orthogonal projections p , p , …, p such that
1 2 m
T = p + p + … + p
1 1 2 2 m m
and p = p + p + … + p , then the expression
1 2 m
T = p + p + … + p for T is called the spectral resolution for T.
1 1 2 2 m m

Note We note that the spectral theorem coincides with the spectral resolution for a
normal operator on a finite dimensional Hilbert space.

Theorem: The spectral resolution of the normal operator on a finite dimensional non-zero Hilbert
space is unique.

Proof: Let T = p + p + … + p
1 1 2 2 m m
be a spectral resolution of a normal operator on a non-zero finite dimensional Hilbert space H.
Then , , …, are distinct complex numbers and p s are non-zero pairwise orthogonal
1 2 m i
projections such that p + p + … + p = 1. We establish that + , …, are precisely the distinct
1 2 m 1 2 m
eigenvalues of T.
To this end we show first that for each i, is an eigenvalue of T. Since p 0 is a projection, a
i i
non-zero x in the range of p such that p x = x
1 i
Let us consider
Tx = ( p + p + … + p )x
1 1 2 2 m m
2
= ( p p + p p + … + p + … + p p )x
1 1 i 2 2 i i i m m i
2
So p ’s are pairwise orthogonal p p = 0 for i j and p = p , we have Tx = p x = x by p x = x.
i i j i i i i i i
is an eigenvalue of T.
i
Next we show that each eigenvalue of T is an element of the set ( , , …, ). Since T is an
1 2 m
operator on a finite dimensional Hilbert space, T must have an eigenvalue.
If is an eigenvalue of T then Tx = x = Ix.
( p + p + … + p ) x = (p + p + … + p ) x
1 1 2 2 m m 1 2 m
{( – ) p + ( – ) p + … + ( – ) p } x = 0 … (2)
1 1 2 2 m m
2
Since p = p and p p = 0 for i j operating with p throughout (2), we get
i i i j i
( – ) p x = 0 for i = 1, 2, …, m.
i i

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