Page 340 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 31: Finite Dimensional Spectral Theory

Now Ty = y T (y , y , …) = (y , y , …) Notes
1 2 1 2
(0, y , y , …) = ( y , y …)
1 2 1 2
y = 0, y = y , ……
1 2 1
Now y is a non-zero vector y 0
1
y = 0 = 0.
1
Then y = y y = 0 and this contradicts the fact that y is a non-zero vector. Therefore T cannot
2 1 1
have an eigenvalue.
31.1.5 Spectrum of an Operator

The set of all eigenvalues of an operator is called the spectrum of T and is denoted by (T).

Theorem 1: If T is an arbitrary operator on a finite dimensional Hilbert space H, then the spectrum
of T namely (T) is a finite subset of the complex plane and the number of points in (T) does
not exceed the dimension n of H.
Proof: First we shall show that an operator T on a finite dimensional Hilbert space h is singular
if and only if there exists a non-zero vector x H such that Tx = 0.
Let a non-zero vector x H s.t. Tx = 0. We can write Tx = 0 as Tx = T0. Since x 0, the two distinct
elements x, 0 H have the same image under T. Therefore T is not one-to-one. Hence T does
–1
not exist. Hence it is singular.
Conversely, let T is singular. Let no non-zero vector such that Tx = 0. This means Tx = 0 x =
0. Then T must be one-to-one. Since H is finite dimensional and T is one-to-one, T is onto, so that
T is a non-singular, contradicting the hypothesis that T is singular. Hence there must be non-
zero vector x s.t. Tx = 0.
Now if T is an operator on a finite dimensional Hilbert space H of dimension n. Then A scalar
(T), if there exists a non-zero vector x such that (T – I)x = 0.

Now (T – I)x = 0 (T – I) is a singular.
(T – I) is singular det (T – I) = 0. Thus (T) satisfies the equation det (T – I) = 0.
Let B be an ordered basis for H. Thus det (T – I)
= det ([T – I] )
B
But det ([T – I] ) = det ([T] – [I] )
B B B
Thus det (T – I) = det ([T] – [ ]).
B ij
So det (T – I) = 0 det ([T] – [ ]) = 0 … (1)
B ij
If [T] = [ ] is a matrix of T then (1) gives
B ij

11 12 1n

22 22 2n … (2)
   
 
n1 nn
The expression of determinant of (2) gives a polynomial equation of degree n in with complex
coefficients in the variable . This equation must have at least one root in the field of complex
number (by fundamental theorem of algebra). Hence every operator T on H has eigenvalue so

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