# Range of a Linear Map

Definition: If $T \in \mathcal L (V, W)$ then the Range of the linear transformation $T$ is the subset of $W$ defined as $\mathrm{range} (T) = \{ T(v) : v \in V \}$, that is, the set of all vectors $T(v) \in W$ to which are mapped to from vectors $v \in V$. |

Before we look at some examples of ranges of vector spaces, we will first establish that the range of a linear transformation can never be equal to the empty set. This should intuitively make sense. If $T : V \to W$ is a linear transformation, then since $V \neq \emptyset$ (since $V$ is a linear transformation) then there must contain at least one element in $V$, and so this element is mapped to some vector in $W$. We will verify this with the following lemma.

Lemma 1: If $T \in \mathcal L (V, W)$ then the range of $T$ contains at least one element from $W$, that is $\mathrm{range} (T) \neq \emptyset$. |

**Proof:**Since $V$ and $W$ are vector spaces, then we have that both $V$ and $W$ are nonempty sets. Furthermore, both $V$ and $W$ contain their respective additive identities $0_V$ and $0_W$. From the Null Space of a Linear Map page, we know that $T(0_V) = 0_W$ and so $0_W \in \mathrm{range} (T)$, so $\mathrm{range} (T) \neq \emptyset$. $\blacksquare$

Now the following lemma will tell us that the $\mathrm{range} (T)$ is a subspace of $W$.

Lemma 2: If $T \in \mathcal L (V, W)$ then the subset $\mathrm{range} (T)$ is a subspace of $W$. |

**Proof:**Since $\mathrm{range} (T) \subseteq W$, all we must do is verify that $\mathrm{range} (T)$ is closed under addition, closed under scalar multiplication, and contains the zero vector of $W$.

- Let $w, y \in W$ and $a \in \mathbb{F}$.

- Since $w, y \in W$ we have that there exists vectors $u, v \in V$ such that $w = T(u)$ and $y = T(v)$. Therefore $w + y = T(u) + T(v)$ and since $T$ is a linear transformation then $T(u) + T(v) = T(u + v)$. Therefore $w + y = T(u + v)$ and so $(w + y) \in \mathrm{range} (T)$ so $\mathrm{range} (T)$ is closed under addition.

- Once again, since $w \in W$, there exists a vector $u \in V$ such that $w = T(u)$. So $aw = aT(u)$ and since $T$ is a linear tranformation then $aT(u) = T(au)$. Therefore $aw = T(au)$ and so $(au) \in \mathrm{range} (T)$ so $\mathrm{range} (T)$ is closed under scalar multiplication.

- From Lemma 1, we have that $0_W = \mathrm{range} (T)$, and so $\mathrm{range} (T)$ contains the zero vector of $W$. Therefore $\mathrm{range} (T)$ is a subspace of $W$. $\blacksquare$

*Notice that lemmas 1 and 2 above are analogous that of lemmas 1 and 2 from the Null Space of a Linear Map page. It is important to note that if $T : V \to W$ is a linear map from the vector spaces $V$ to $W$, then both the null space of $T$ and the range of $T$ are nonempty and the null space of $T$ is a subspace of $V$, while the range of $T$ is a subspace of $W$.*

We will now look at some examples of ranges of linear transformations.

## The Range of the Zero Map

If $0 \in \mathcal L (V, W)$ represents the zero map, then $\mathrm{range} (T) = \{ 0 \}$ since every vector $v \in V$ is mapped to $0_W \in W$.

## The Range of the Identity Map

If $I \in \mathcal L (V, V)$ represents the identity map, then $\mathrm{range} (T) = V$ since every vector $v \in V$ is mapped to itself, so the range contains all vectors from $V$.

## The Range of the Left Shift Operator

If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the left shift operator, then $\mathrm{range} (T) = \mathbb{F}^{\infty}$ since any sequence $(x_2, x_3, ...) \in \mathbb{F}^{\infty}$ is mapped from the sequence $(x_1, x_2, ...) \in \mathbb{F}^{\infty}$.

## The Range of the Right Shift Operator

If $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ represents the right shift operator, then $\mathrm{range} (T) = \{ (0, x_1, x_2, ...) : x_1, x_2, ... \in \mathbb{F} \}$. We note that any sequence $(x_1, x_2, ...)$ where $x_1 \neq 0$ cannot be in the range of $T$ since the first term of any sequence under $T$ will be zero.