This is true only for injective transformations. For example, the projection of two-dimensional vectors on the horizontal axis maps any two vectors into linearly dependent vectors.
What is the best way to answer part 2?
Prove that the set of images of a linearly dependent set of vectors under a linear transformation is linearly dependent. Let {v1¬, , … , vp} be a set of linearly dependent vectors, and let T: Rn Rm be a linear transformation. Show that {T(v1), …, T(vp)} is linearly dependent. Part 2: Is it also true that the images of a linearly independent set of vectors under a linear transformation is also linearly independent? Explain.
Thanks.
suppose there exist c_{1},c_{2},...,c_{n} not all 0 with:
c_{1}v_{1} + c_{2}v_{2} +...+ c_{n}v_{n} = 0
(that is {v_{1},v_{2},...,v_{n}} is a linearly dependent set), and that T is a linear transformation.
then T(c_{1}v_{1} + c_{2}v_{2} +...+ c_{n}v_{n}) = T(0) = 0.
but since T is linear:
0 = T(c_{1}v_{1} + c_{2}v_{2} +...+ c_{n}v_{n}) = c_{1}T(v_{1}) + c_{2}T(v_{2}) +...+ c_{n}T(v_{n})
which shows that {T(v_{1}),T(v_{2}),...,T(v_{n})} is linearly dependent.
it is NOT true that the image of a linearly independent set under a linear transformation is linearly independent.
for example, if T is the 0-map, T(v) = 0, for all v in V, then even if B = {v_{1},v_{2},...,v_{n}} is a basis,
T(B) = {T(v_{1}),T(v_{2}),...,T(v_{n})} = {0,0,...,0} = {0} (since repeated elements of a SET don't "count extra"),
and {0} is ALWAYS a linearly dependent set.
some equivalent (sufficient) conditions for T(S) to be LI, when S is LI:
a) T is injective
b) det(T) ≠ 0
c) ker(T) = {0}
d) rank(T) = dim(V) (where V is the domain of T)
note these conditions are not necessary, it may well be that T(S) is LI when S is, even if none (thus all) of the above hold. for example, let T be multiplication by the matrix:
[1 0 0]
[0 1 0]
[0 0 0].
then for S = {(1,0,0),(0,1,0)}, T(S) is LI, even though T is singular. however, S' = {(0,1,0),(0,2,2)} is also LI, but T(S) = {(0,1,0),(0,2,0)}, which is NOT linearly independent, since the second vector is a scalar multiple of the first:
we have 2(0,1,0) + (-1)(0,2,0) = (0,0,0) and neither of {2,-1} is 0.