# 6.3: Ranges

- Page ID
- 273

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**Definition 6.3.1. **Let \(T:V\to W \) be a linear map. The **range** of \(T \), denoted by \(\range(T) \), is the subset of vectors in \(W \) that are in the image of \(T \). I.e.,

\[ \range(T) = \{ Tv \mid v\in V\} = \{ w\in W \mid \rm{~ there~ exists~} v \in V \rm{~ such~ that~} Tv=w\}.\]

**Example 6.3.2. **The range of the differentiation map \(T:\mathbb{F}[z] \to \mathbb{F}[z] \) is \(\range(T) =\mathbb{F}[z]\) since, for every polynomial \(q\in \mathbb{F}[z] \), there is a \(p\in \mathbb{F}[z] \) such that \(p'=q \).

**Example 6.3.3. **The range of the linear map \(T(x,y)=(x-2y,3x+y) \) is \(\mathbb{R}^2 \) since, for any \((z_1,z_2)\in \mathbb{R}^2 \), we have \(T(x,y)=(z_1,z_2) \) if \((x,y)=\frac{1}{7}(z_1+2z_2,-3z_1+z_2) \).

**Proposition 6.3.4.** *Let* \(T:V\to W \) *be a linear map. Then *\(\range(T) \) *is a subspace of* \(W \).

*Proof. *

We need to show that \(0\in \range(T) \) and that \(\range(T) \) is closed under addition and scalar multiplication. We already showed that \(T0=0 \) so that \(0\in \range(T) \).

For closure under addition, let \(w_1,w_2\in \range(T) \). Then there exist \(v_1,v_2\in V\) such that \(Tv_1=w_1 \) and \(Tv_2=w_2 \). Hence

\begin{equation*}

T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2,

\end{equation*}

and so \(w_1+w_2\in \range(T) \).

For closure under scalar multiplication, let \(w\in \range(T) \) and \(a\in \mathbb{F} \). Then there exists a \(v\in V \) such that \(Tv=w \). Thus

\begin{equation*}

T(av)=aTv=aw,

\end{equation*}

and so \(aw \in \range(T) \).

**Definition 6.3.5.** A linear map \(T:V\to W \) is called **surjective** if \(\range(T)=W \). A linear map \(T:V\to W \) is called **bijective **if \(T \) is both injective and surjective.

**Example 6.3.6. **

- The differentiation map \(T:\mathbb{F}[z] \to \mathbb{F}[z] \) is surjective since \(\range(T) = \mathbb{F}[z] \). However, if we restrict ourselves to polynomials of degree at most \(m \), then the differentiation map \(T:\mathbb{F}_m[z] \to \mathbb{F}_m[z] \) is not surjective since polynomials of degree \(m \) are not in thecrange of \(T \).
- The identity map \(I:V\to V \) is surjective.
- The linear map \(T:\mathbb{F}[z] \to \mathbb{F}[z] \) given by \(T(p(z)) = z^2 p(z) \) is not surjective since, for example, there are no linear polynomials in the range of \(T \).
- The linear map \(T(x,y)=(x-2y,3x+y) \) is surjective since \(\range(T)=\mathbb{R}^{2} \), as we calculated in Example 6.3.3.

## Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

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