QUESTION 2

Question 1

INVERTIBLE OPERATOR

Answer 1

T∈B(X,Y) if
1. T:X->Y is bijection
2. T^(-1)∈B(Y,X)

Question 2

CHARACTERISATION INVERSION

Answer 2

T∈B(X,Y) invertible
⟺
∃S∈B(Y,X): ST=I_X and TS=I_Y

Question 3

INVERSION BY GEOMETRIC SERIES THEOREM

Answer 3

if X is banach and T∈B(X), then
Σ||T^k|| < ∞
⟹
(I-T)^(-1) = ΣT^k∈B(X)

Question 4

RESOLVENT SET

Answer 4

ρ(T) = {λ∈K : (T-λ)^(-1) ∈ B(X)}

i.e. all λ such that (T-λ) is invertible and (T-λ)^(-1) is bounded

Question 5

λ ∈ ρ(T )

Answer 5

if (T − λ)^(−1) is bounded

Question 6

SPECTRUM

Answer 6

δ(T) = K\ρ(T)

Question 7

EIGENVALUE of T∈B(X)

Answer 7

λ∈K if ∃x≠0: (T-λ)x=0

Question 8

EIGENSPACE of T∈B(X)

Answer 8

ker(T-λ)

Question 9

POINT SPECTRUM of T∈B(X)

Answer 9

δ_p(T)={eigenvalues of T}

Question 10

APPROXIMATE EIGENVALUE of T

Answer 10

λ∈K if ∃(xn): ||xn||=1, ∀n∈N
and (T-λ)xn -> 0

Question 10

SPECTRAL MAPPING THEOREM

Answer 10

assume X is banach over K=C and T∈B(X)
for any polynomial p:K->K we have
δ(p(T)) = {p(λ) : λ∈δ(T)}

Question 10

Tx= λx
⟹ x=0

Answer 10

then no eigenvalues

Question 11

SPECTRUM OF IDENTITY OPERATOR

Answer 11

σ(I)={1}