Page 339 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes Since x is an eigenvector of T, corresponding to the eigenvalue 0 and Tx = x.
0 x 0
Hence T ( x) = T(x) = ( x)
Therefore corresponding to an eigenvalue there are more than one eigenvectors.

2. If x is an eigenvector of T, then x cannot correspond to more than one eigenvalue
of T.
If possible let , be two eigenvalues of T, ( ) for eigenvector x. Then
1 2 1 2
Tx = x and Tx = x
1 2
x = x
1 2
( – )x = 0
1 2
– = 0 ( x 0)
1 2
=
l 2
3. Let be an eigenvalue of an operator T on H. If M is the set consisting of all
eigenvectors of T corresponding to together with the vector 0, then M is a non-
zero closed linear subspace of H invariant under T.
By definition x M Tx = x … (1)
By hypothesis 0 M and 0 vector satisfies (1).

M = {x H : Tx = x} = {x H : (T – I) x = 0}
Since T and I are continuous, M is the null space of continuous transformation T – I.
Hence M is closed.

Next we show that if x M , then Tx M . If x M then Tx = x.
Since M is a linear subspace of H, x M x = Tx M .
M is invariant under T.

Definition: Eigenspace

The closed subspace M is called the eigenspace of T corresponding to the eigenvalue .
From property (3), we have proved that each eigenspace of T is a non-zero linear subspace of H
invariant under T.

Note It is not necessary for an operator T on a Hilbert space H to possess an eigenvalue.

Example: Consider the Hilbert space  and the operator T on  defined by
2 2
T (x , x , …) = (0, x , x , …)
1 2 1 2
Let be a eigenvalue of T. Then a non zero vector

y = (y , y , …) in  such that Ty = y.
1 2 2

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