Let T V W be a linear transformation and let U be a subspace

Let T: V-> W be a linear transformation and let U be a subspace of W. Prove that the set T^-1(U) = {v E V| T(v) E U} is a subspace of V. What is T^-1 (U), if U = {0}?

Solution

Given that V and W are two vector spaces and T is linear

T: V -->W

U is a subspace of W

Since U is a subspace for any two elements u1 and u2

T^-1(u1+u2) = T^-1(u1) + T^-1(u2) (since T is linear)

and since u1+u2 is in U

T^-1(u1+u2) will be in the inverse of U.

In other words for two elements T^-1(u1) and T^-1(u2) addition also belongs to T^-1(U)

So addition is closed.

T^-1(cU1) = cT^-1(u1) belongs to  T^-1(U)

Identity for  T^-1(U) will be  T^-1(e) where e is the identity of W

and inverse will be  T^-1(-u1) for  T^-1(u1)

Hence T^-1(U) = {v E V| T(v) E U} is a subspace of V

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If U ={0}, then 0 must be the identity element for W since U is a subspace of W.

So if U contains only identity element 0,

then T^-1(U) = {e} where e is the additive identity for V.