6.3: Ranges

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.

1. 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 $T$.
2. The identity map $I:V\to V$ is surjective.
3. 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 $T$.
4. 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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