Let X,Y be vector spaces both over R and C. Let T be a linear operator with domain X and Range in Y.?

How do i show that


i. The inverse of T carrying R(T) into D(T) exists if and only if


Tx =0 implies that x=0.


ii. If the inverse eists then it is a linear operator


iii. If the dimension of D(T)=n %26lt; infinity and the inverse of T exists then the dimension of R(T)=dimension of D(T).

Let X,Y be vector spaces both over R and C. Let T be a linear operator with domain X and Range in Y.?
i). Note that the inverse exists iff T is 1-1, since T is onto R(T) by definition.


(=%26gt;) Suppose the inverse exists. Then T must be 1-1, so for any x, y, if Tx = Ty then x = y. In particular, for y = 0 we have that for any x with Tx = T0 = 0 then x = 0.


(%26lt;=) Suppose Tx = 0 =%26gt; x = 0. Then let x, y be any vectors with Tx = Ty. Then T(x-y) = Tx - Ty = 0, so by the assumption x-y = 0 and hence x = y. So T is 1-1 and thus the requested inverse exists.





ii). Let the inverse exist and call it U. Let a = Tx and b = Ty be any vectors in R(T). Hence U(a) = x and U(b) = y. Let k be any scalar.





Now T(x+y) = Tx + Ty (since T is linear) = a + b. So U(a+b) = x+y = U(a) + U(b).


Also, T(kx) = kTx (since T is linear) = ka. So U(ka) = kx = kU(a).


Hence U is linear.





iii). Let {x1, ..., xn} be a basis for D(T) and suppose the inverse exists. Let {y1, ..., yn} be defined by yi = Txi. We claim that {y1, ..., yn} is a basis for R(T).


Proof: Let y ∈ R(T), then y = Tx for some x. Since {x1, ..., xn} is a basis for D(T) we can write x = a1 x1 + ... + an xn for some constants a1, ..., an. Then Tx = a1 Tx1 + ... + an Txn = a1 y1 + ... + an yn. So {y1, ..., yn} spans R(T).


Now suppose b1 y1 + ... + bn yn = 0 for some scalars b1, ..., bn. Then we have b1 Tx1 + ... + bn Txn = 0


=%26gt; T(b1 x1) + ... + T(bn xn) = 0 since T is linear


=%26gt; T(b1 x1 + ... + bn xn) = 0 since T is linear


=%26gt; b1 x1 + ... + bn xn = 0 from (i)


=%26gt; b1, ..., bn = 0 since {x1, ..., xn} is linearly independent.


Hence {y1, ..., yn} is linearly independent.


Since {y1, ..., yn} is linearly independent, is a subset of R(T) and spans R(T), it is a basis for R(T).


Since R(T) has a basis with n elements, it has dimension n.