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# 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.

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