Page 342 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 31: Finite Dimensional Spectral Theory

Proof: Notes
(a) T is singular a non-zero vector x H such that Tx = 0 or Tx = 0. Hence T is singular
0 is the eigenvalue of T i.e. 0 (T).

(b) Let T be non-singular and (T).
Hence 0 by (a) so that –1 exists. Since is an eigenvalue of T, a non-zero vector
x H s.t. Tx = x.

Premultiplying by T we get
–1
–1
–1
T Tx = T ( x)
1
T (x) = x for x 0
–1
–1
Hence –1 ( (T ))
) = is an eigenvalue of (T ) = T.
Conversely, if –1 is an eigenvalue of T then ( –1 –1 –1 –1
–1
Hence (T).
(c) Let S = ATA . Then we find S – I.
–1
Now S – I = ATA – A ( I) A –1
–1
= A (T – I) A –1
–1
det (S – I) = det (A(T – I) A )
= det (T – I)
is an eigenvalue of T det (T – I) = 0.
Hence det (T – I) = 0 det (S – I) = 0

S and T have the some eigenvalues so that
–1
(ATA ) = (T).
(d) If = (T), is an eigenvalue of T. Then a non-zero vector x such that Tx = x.
Hence T (Tx) = T ( x) = Tx = x.
2
2
Hence if is an eigenvalue of T, then is an eigenvalue of T . Repeating this we get that
2
n
n
if is an eigenvalue of T, then is an eigenvalue of T for any positive integer n.
m
Let P (t) = a + a t + … + a t , a , a , ……, a are scalars.
0 1 m 0 1 m
m
Then [P (T)]x = (a I + a T + …… + a T )x
0 2 m
= a x + a ( x) + …… + a ( m x)
0 1 m
= [a + a ( ) + …… + a m ]x
0 1 m
Hence P( ) = a + a + …… + a m is an eigenvalue of P (T).
0 1 m
This if (T), then P ( ) (P (T)).
This completes the proof of the theorem.
31.1.6 Spectral Theorem

Statement: Let T be an operator on a finite dimensional Hilbert space H with , , …, as the
1 2 m
eigenvalues of T and with M , M , …, M be then corresponding eigenspaces. If P , P , …, P are
1 2 m 1 2 m
the projections on the spaces, then the following statements are equivalent.

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