Theorem 2.2

If $T: V \rightarrow W$ is linear and $\beta=\{v_1,...,v_n\}$ is basis for $V$, then $$ \begin{align*} R(T) = span(T(\beta)) = span(\{T(v_1),...,T(v_n)\} \end{align*} $$

Proof:

To show that $R(T) = span(T(\beta))$ we will show that $span(T(\beta)) \subseteq R(T)$ and $R(T) \subseteq span(T(\beta))$.

$span(T(\beta)) \subseteq R(T)$: By definition, we know that for any $v \in \beta$, $T(v) \in R(T)$ so $T(\beta) \subseteq R(T)$. Since $R(T)$ is a subspace and $R(T)$ contains the set $T(\beta)$, then by theorem 1.5, it must contain the span of this set as well and so

$$ \begin{align*} span(T(\beta)) = span(\{T(v_1),...,T(v_n)\}) \subseteq R(T) \end{align*} $$

$R(T) \subseteq span(T(\beta))$: Let $w \in R(T)$. We know that $w = T(v)$ for some $v \in V$. Since $\beta$ is a basis, then we can express $v$ as a linear combination of the elements in $\beta$ such that

$$ \begin{align*} v = a_1v_1 + ... + a_nv_n, \end{align*} $$

for some scalars $a_1, ..., a_n$. Since $T$ is linear then we can see that

$$ \begin{align*} w = T(v) &= T(a_1v_1 + ... + a_nv_n) \\ &= a_1T(v_1) + ... + a_nT(v_n). \end{align*} $$

So $w$ is a linear combination of the elements of $T(\beta)$. This means that $w$ is also in $span(T(\beta))$ by the definition of a span and so $R(T) \subseteq span(T(\beta))$ as we wanted to show. $\blacksquare$

References:

Linear Algebra 5th Edition