In Reply to: Lecture Notes Set #7 p. 8 posted by Doug on November 24, 2002 at 10:35:18:
>Hi Professor Tao,
>What do you mean by "lambda is en eigenvalue of T is and only only N(T-lambda*Iv) is non-zero, i.e. when T-lambda*Iv is not one-to-one"?
>I don't understand how are the two things related.
>Thanks!
If lambda is an eigenvalue of T, then we must have some non-zero
eigenvector v for which
Tv = lambda v
i.e.
Tv = lambda I_V v
(since I_V v = v for all v in V) and hence
(T - lambda I_V) v = 0
and hence v lies in the null space N(T - lambda I_V) of
T - lambda I_V. Thus N(T - lambda I_V) contains a non-zero vector,
which implies that T-lambda I_V is not one-to-one by Lemma 1 of
the Week 3 notes.
Conversely, if T - lambda I_V is not one-to-one, we can reverse
the above steps and conclude that lambda is an eigenvalue of T.
Terry