Linear Mappings between Vector Spaces form Vector Space

Theorem
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.

Let $X, Y$ be vector spaces over the same field $\struct {F, +, \times}$.

Let $\map \LL {X, Y}$ be the set of linear mappings.

Let $x \in X$.

Define pointwise addition $T + S \in \map \LL {X, Y}$ such that:


 * $\forall x \in X: \map {\paren {T + S} } x := \map T x + \map S x$

For $\alpha \in F$, define pointwise scalar multiplication $\alpha \cdot T$ such that:


 * $\forall x \in X: \map {\paren {\alpha \cdot T} } x := \alpha \cdot \map T x$

Let $\mathbf 0: X \to Y$ be the zero mapping.

Then $\struct {\map \LL {X, Y}, +, \, \cdot \,}$ is a vector space over $\struct {F, +, \times}$.

Proof
Let $T, S, P \in \map \LL {X, Y}$ such that:


 * $T, S, P: X \to Y$

Let $\lambda, \mu \in F$.

$\forall x \in X: \map T x, \map S x \in Y$.

By assumption, $Y$ is a vector space.

By, $T + S \in \map \LL {X, Y}$

$\forall x \in X: \map T x, \map S x \in Y$.

By assumption, $Y$ is a vector space.

By, $T + S = S + T$

$\forall x \in X: \map T x, \map S x, \map P x \in Y$.

By assumption, $Y$ is a vector space.

By, $\paren {T + S} + P = T + \paren {S + P}$.

We have that:


 * $\forall x \in X: \map T x \in Y$

By assumption, $Y$ is a vector space.

By :


 * $\forall x \in X \exists \paren {- \map T x}: \map T x + \paren {-\map T x} = \mathbf 0_Y$