1. Let V and W be vector spaces and T: V rightarrow a one-to-one linear transformation. Prove or
give a counter example to each of the following If {v1,,vk} is a basis of V then {T(vk)} is a basis
of W. If {T(v1),, T(vk)} is a basis of W then {v1,, vk} is a basis of V.
Solution
a.)
False.
Let V=R 2 , W=R 3 .
v 1 = (1,0), v 2 = (0,1) [a basis for V]
T(x,y) = (x,y,0) [clearly one-to-one]
T(v 1 ) = (1,0,0), T(v 2 ) = (0,1,0) [clearly NOT a basis for W]
What caused this to fail? dim(V) does not equal dim(W). Since it is 1-1, we knew that dim(V) <=
dim(W), but not that they were equal, which is needed for the desired result.
b.)
True.
Suppose c 1 v 1 + ... + c k v k = 0. Apply T to both sides:
T(c 1 v 1 + ... + c k v k ) = T(0)
c 1 T(v 1 ) + ... + c k T(v k ) = 0 [since T is linear]
c 1 = c 2 = ... = c k = 0 [since the T(v i ) are linearly independent]
Therefore the v i are linearly independent.
Let v be any element of V. Consider T(v) in W. Since the T(v i ) form a basis, there exist d 1 ,...,d
k such that:
d 1 T(v 1 ) + ... + d k T(v k ) = T(v)
T(d 1 v 1 + ... + d k v k ) = T(v) [since T is linear]
d 1 v 1 + ... + d k v k = v [since T is one-to-one!]
Therefore v is a linear combination of the v i . Hence the v i span V.

2. Since the v i span V and are linearly independent, they from a basis for V.