# Addition and Multiples of Linear Maps

We have already looked at a bunch of different linear maps from the Linear Maps page. Since the linear transformations $S, T \in \mathcal L (V, W)$ of the $\mathbb{F}$-vector spaces $V$ and $W$ are functions, we can define addition between these linear maps as $(S + T)(v) = S(v) + T(v)$, and scalar multiplication as $(kT)(v) = kT(v)$ for all $v \in V$ and $k \in \mathbb{F}$. Are these "new" maps linear though? Fortunately they are, and will now verify that $S + T$ and $kT$ are both linear maps.

Theorem 1: If $S, T \in \mathcal L (V, W)$, then $S + T$ is a linear map from $V$ to $W$. |

**Proof:**Let $u, v \in V$ and $a, b \in \mathbb{F}$. Then we have that:

- Thus $(S + T)(au + bv) = a(S + T)(u) + b(S + T)(v)$ and so $S + T$ is a linear map. $\blacksquare$

Furthermore, we can multiply a linear map $T \in \mathcal L(V, W)$ by a scalar $k$. Define $(kT)(v) = kT(v)$ for all $v \in V$. We will also verify that $kT$ is a linear map.

Theorem 2: If $T \in \mathcal L (V, W)$ and $k$ is a scalar, then $kT$ is a linear map from $V$ to $W$. |

**Proof:**Let $u, v \in V$ and $a, b \in \mathbb{F}$. Then we have that:

- Thus $(kT)(au + bv) = a(kT)(u) + b(kT)(v)$ and so $kT$ is a linear map. $\blacksquare$

# Compositions of Linear Maps

Since linear maps are functions, we can also sometimes form compositions of these functions to create compositions of linear maps. Let $T \in \mathcal L (U, V)$ and $S \in \mathcal L (V, W)$ be linear maps, where $U$, $V$, and $W$ are vector spaces over the field $\mathbb{F}$. Then we will define $ST = (S \circ T)(v) = S(T(v))$ for all $u \in U$, that is, we first take a vector $u \in U$, apply the linear transformation $T$ to get $T(v) \in V$, and then apply the linear transformation $S$ to get $S(T(v)) \in W$. Therefore $S \circ T : U \to W$. We will now verify that $S \circ T$ is a linear map from $U$ to $W$.

Theorem 3: If $T \in \mathcal L (U, V)$ and $S \in \mathcal L (V, W)$ then $S \circ T$ is a linear map from $U$ to $W$. |

**Proof:**Let $u, v \in U$ and $a, b \in \mathbb{F}$. Then since both $S$ and $T$ are linear maps, we have that:

- Thus $(S \circ T)(au + bv) = a (S \circ T)(u) + b (S \circ T)(v)$ and so $S \circ T$ is a linear map. $\blacksquare$.

*We should note that often times we will use the notation $ST$ instead of $S \circ T$ to represent the composition of the linear maps $S$ and $T$ as the notation is neater in some circumstances.*