Let V and W be vector spaces and T: V --> W be linear. Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearly independent if and only if T(S) is linearly independent.

Relevant Theorems:

(1) T: V --> W is linear if T(x+y) = T(x) + T(y) and T(cx) = cT(x)

(2) nullspace (or kernel) N(T) = {x in V : T(x) = 0}

(3) range (or image) R(T) = {T(x) : x in V}

(4) Let V and W be vector spaces and T: V --> W be linear. Then N(T) and R(T) are subspaces of V and W respectively

(5) Let V and W be vector spaces and T: V --> W be linear. If B = {v1, v2, v3, .... , vn} is a basis for V, then

R(T) = span{ T(B) = span{ T(v1), T(v2), T(v3), ..... , T(vn) } }

(6) If N(T) and R(T) are finite-dimensional, then nullity(T) = dim[ N(T) ], and rank(T) = dim[ R(T) ]

(7) Dimension Theorem: Let V and W be vector spaces and T: V --> W be linear. If V is finite-dimensional, then

nullity(T) + rank(T) = dim(V)

(8) Let V and W be vector spaces, and let T: V --> W be linear. Then T is one-to-one if and only if N(T) = {0}, and T is onto if and only if dim{R(T)} = dim(W)

(9) Let V and W be vector spaces of equal (finite) dimension, and T: V --> W be linear. Then the following are equivalent:

(i) T is one-to-one

(ii) T is onto

(iii) rank(T) = dim(V)

(10) Let V and W be vector spaces over F, and suppose that {v1, v2, ... , vn} is a basis for V. For w1, w2, ... , wn in W, there exists exactly one linear transformation T: V --> W such that T(vi) = wi for i = 1, 2, ... , n.

(11) Let V and W be vector spaces, and suppose that V has a finite basis {v1, v2, ... , vn}. If U, T: V --> W are linear and U(vi) = T(vi) for i = 1, 2, ... , n, then U = T

Note: I CANNOT use the following (a previous result): "T is one-to-one if and only if T carries linearly independent subsets of V onto linearly independent subsets of W".

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