### Range and Column Space

##### 1. Range

Recall the range of a transformation T: XY is all the images under the transformation

rangeT = {T(x): any xX}
= {bY: There is xX, such that T(x) = b}
= {bY: T(x) = b has solutions}.

For a linear transformation, we have

The range of a linear transformation T: VW is a subspace of W.

Proof w, w' ∈ rangeTw = T(v), w' = T(v') for some v, v' ∈ Vw + w' = T(v) + T(v') = T(v + v') ∈ rangeT, where the linearity of T is used in the red equality. The proof of wrangeTcwrangeT is similar.

Example The subset (see this exercise)

{(a + b, 3a - b, 2a + b): a, bR} ⊂ R3

is a subspace because it is the range of the linear transformation

T(a, b) = (a + b, 3a - b, 2a + b): R2R3.

The subset (see this exercise)

{a cost + b sint: a, bR} ⊂ F(R)

is a subspace because it is the range of the linear transformation

C(a, b) = a cost + b sint: R2F(R).

We also recall that a transformation is onto (surjective) if and only if the range is the whole target space. In other words, any element of Y is the image of some element of X.

Example Consider the derivative transformation

D(f) = f': C1(R) → C(R).

By the fundamental theorem of calculus, any continuous function fC(R) has an antiderivative (for example, given by F(t) = ∫0tf(t)dt). This implies that D is onto. However, if we change the target space from C(R) to F(R), then D is no longer onto.

Example Consider the linear transformation

M(f) = (t2+1)f: F(R) → F(R)

of multiplying the function t2+1. For any function fF(R), we may construct another function F(t) = f(t)/(t2+1). Then we have M(F) = f. Thus the transformation M is onto. Note that the key here is that t2 + 1 never vanishes, so that it can be "inverted". If we change the function to t2 - 1, then M is no longer onto.

For the similar reason, if X is an invertible n by n matrix, then the linear transformation

M(X) = AX: M(k, n) → M(m, n)

is onto.