The Linear Space L(X, Y)

# The Linear Space L(X, Y)

Recall from the Linear Operators on Linear Spaces page that if $X$ and $Y$ are linear spaces, then a linear operator from $X$ to $Y$ is a function $T : X \to Y$ with the following properties:

- $T(x + y) = T(x) + T(y)$ for all $x, y \in X$.

- $T(\lambda x) = \lambda T(x)$ for all $x \in X$ and for all $\lambda \in \mathbb{C}$.

The set of linear operators from $X$ to $Y$ is denoted $\mathcal L(X, Y)$.

Definition: Let $X$ and $Y$ be linear spaces. We define the operation of Addition of Linear Operators for all $S, T \in \mathcal L(X, Y)$ by $(S + T)(x) = S(x) + T(x)$ for all $x \in X$. We define the operator of Scalar Multiplication of Linear Operators for all $T \in \mathcal L(X, Y)$ and for all $\lambda \in \mathbb{C}$ by $(\lambda T)(x) = \lambda T(x)$. |

We will now show that $\mathcal L(X, Y)$ with the operations defined above is itself a linear space.

Theorem 1: Let $X$ and $Y$ be linear operators. Then $\mathcal L(X, Y)$ with the operations of function addition and scalar multiplication is a linear space. |

**Proof:**Let $S, T, U \in \mathcal L(X, Y)$ and let $\alpha, \beta \in \mathbb{C}$. Then:

**Associativity of Addition:**

\begin{align} \quad \quad [S + (T + U)](x) = S(x) + (T + U)(x) = S(x) + [T(x) + U(x)] = [S(x) + T(x)] + U(x) = (S + T)(x) + U(x) = [(S + T) + U](x) \end{align}

**Commutativity of Addition:**

\begin{align} \quad (S + T)(x) = S(x) + T(x) = T(x) + S(x) = (T + S)(x) \end{align}

**Existence of a Zero Linear Operator:**Let $0 : X \to Y$ be defined for all $x \in X$ by $0(x) = 0$. Then:

\begin{align} \quad (S + 0)(x) = S(x) + 0(x) = S(x) + 0 = S(x) \end{align}

(4)
\begin{align} \quad (0 + S)(x) = 0(x) + S(x) = 0 + S(x) = S(x) \end{align}

**Existence of Additive Inverses:**For $S \in \mathcal L(X, Y)$, define $-S : X \to Y$ for all $x \in X$ by $(-S)(x) = -S(x)$. Then:

\begin{align} \quad (S + (-S))(x) = S(x) + (-S)(x) = S(x) - S(x) = 0 \end{align}

(6)
\begin{align} \quad ((-S) + S)(x) = (-S)(x) + S(x) = -S(x) + S(x) = 0 \end{align}

**Associativity of Scalar Multiplication:**

\begin{align} \quad \alpha (\beta S(x)) = \alpha \beta S(x) = (\alpha \beta) S(x) \end{align}

**Existence of a Multiplicative Identity:**For $1 \in \mathbb{C}$ we have that:

\begin{align} \quad 1 \cdot (S(x)) = S(x) = S(x) \cdot 1 \end{align}

**Distributivity:**

\begin{align} \quad \alpha (S + T)(x) = \alpha (S(x) + T(x)) = \alpha S(x) + \alpha T(x) \end{align}

(10)
\begin{align} \quad (\alpha + \beta)S(x) = \alpha S(x) + \beta S(x) \end{align}

- Therefore $\mathcal L(X, Y)$ is a vector space with these operations. $\blacksquare$