What is the range of a linear transformation? And when is that transformation onto?

For a *linear* transformation $T \in
\mathcal{L} (V, W)$, we refer to $\range(T)$ as
the range space (or
image space) of $T$.
The language is justified by the following
proposition.

Suppose $T \in \mathcal{L} (V, W)$.
Then $\range T$ is a subspace of $W$.

We will show that $\range(T)$ both (a) contains
$0 \in W$ and (b) is closed under vector
addition and scalar multiplication.
*vector
addition*.
*scalar
multiplication*.

For any linear map we have $T(0) = 0$, since scalar multipliciation commutes with the application of $T$. (We can express the origin $0 \in W$ as the scalar $0$ times the origin.) Hence, the origin in $V$ is mapped to the origin in $W$ and so $0 \in W$.

To see that $W$ is closed under vector addition, let $w_1, w_2 \in \range T \subset W$ Then there exists $v_1, v_2 \in V$ so that

\[ Tv_1 = w_1 \text{ and } Tv_2 = w_2 \]

We conclude\[ w_1 + w_2 = Tv_1 + Tv_2 = T(v_1 + v_2) \]

We have written the vector $w_1 + w_2$ as the image of the vector $v_1 + v_2$, and so $w_1 + w_2 \in \range T$. Since $w_1$ and $w_2$ were arbitrary, we conclude that $W$ is closed underSimilarly, to see that $W$ is closed under scalar multiplication, let $w \in \range T$ and $\lambda \in \F $. Here $\F $ denotes the field over $W$ and $V$. Then there exists $v \in V$ satisfying $w = Tv$. We claim

\[ \lambda w = \lambda T(v) = T(\lambda v) \]

The second equality holds because $T$ commutes with scalar multiplication. Since $w$ and $\lambda $ were arbitrary, we conclude that $W$ is closed under