Vector Space of Linear Transformations

Why

Can we think of linear maps as vectors?

Definitions

Suppose $V$ and $W$ are some vector spaces over a field $\F $. Denote the linear maps from $V$ to $W$ by $\mathcal{L} (V, W)$ as usual.

Addition. Given $S, T \in \mathcal{L} (V, W)$ the sum of $S$ and $T$ is the linear map $R \in \mathcal{L} (V, W)$ defined by

\[ Rv = Sv + Tv \quad \text{for all } v \in V \]

Scalar multiplication. Given $S \in \mathcal{L} (V, W)$ the (scalar) product of $\lambda $ and $T$ is the linear map $Q \in \mathcal{L} (V, W)$ defined by

\[ Qv = \lambda Tv \quad \text{for all } v \in V \]

Suppose $V$ and $W$ are two vector spaces over the same field $\F $. Then $\mathcal{L} (V, W)$ is a vector space over the field $\F $ with respect to the operations of addition and scalar multiplication just defined.

The additive identity of the vector space $\mathcal{L} (V, W)$ is the zero map $0 \in \mathcal{L} (V, w)$.

Notation

Given $S, T \in \mathcal{L} (V, W)$ the sum of $S$ and $T$ and $\lambda \in \F $, we denote the sum of $S$ and $T$ by $S + T$. Hence,

\[ (S + T)(v) = Sv + Tv \quad \text{for all } v \in V \]

We denote the product of $\lambda $ and $T$ by $\lambda T$. Hence,

\[ (\lambda T)(v) = \lambda (Tv) \quad \text{for all } v \in V \]