Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 1.
FREDHOLM THEORY
Define z = x − y, and hence z ∈ T −1 (0). Thus X = T (X) + T −1 (0). In particular,
T (X) is closed by Theorem 1.4.2, so that
X = T (X) ⊕ T −1 (0).
Therefore we can find a projection P ∈ B(X) for which
We thus write
P (X) = T (X) and P −1 (0) = T −1 (0).
T =
[ ] [ ]
T1 0 P (X)
:
0 0 P −1 (0)
→
[ ] P (X)
P −1 ,
(0)
where T 1 is invertible because T 1 := T | T X is 1-1 and onto since T (X) = T 2 (X). If
we put
[ ]
T ′ T
−1
=
1 0
,
0 0
then T T ′ T = T and
[ ] [ ]
T T ′ = T ′ T
−1
T =
1 0 T1 0
=
0 0 0 0
which says that T is simply polar.
[ ] I 0
= P,
0 0
Theorem 1.9.8. If T ∈ B(X) then
T is polar ⇐⇒ T n is simply polar for some n ∈ N
[ ]
T1 0
Proof. (⇒) If T is polar then we can write T =
with T
0 T 1 invertible and
[ ]
2
T
T 2 nilpotent. So T n n
= 1 0
, where n is the nilpotency of T
0 0
2 . If we put S =
[ ]
T
−n
1 0
, then T
0 I
n ST n = T n and ST n = T n S.
(⇐) If T n is simply polar then X = T n (X) ⊕ T −n (0). Observe that since T n is
simply polar we have
T (T n X) = T n+1 (X) ⊇ T 2n (X) = T n (X)
T (T −n (0)) ⊆ T −n+1 (0) ⊆ T −n (0)
Thus we see that T | T n (X) is 1-1 and onto, so that invertible. Thus we may write
( )
T1 0
T =
: T n (X) ⊕ T −n (0) −→ T n (X) ⊕ T −n (0),
0 T 2
where T 1 = T | T n (X) is invertible and T 2 = T | T −n (0) is nilpotent with nilpotency n.
Therefore T is polar.
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