Addition Multiples and Compositions of Linear Maps Examples 1

# Addition Multiples and Compositions of Linear Maps Examples 1

Recall from the Addition Multiples and Compositions of Linear Maps page that if $S, T \in \mathcal L(V, W)$ then we can define a new linear map, $S + T : V \to W$ by $(S + T)(v) = S(v) + T(v)$ for every vector $v \in V$. Furthermore, for $a \in \mathbb{F}$ we can also defined a linear map $aT : V \to W$ by $(aT)(v) = aT(v)$ for every vector $v \in V$. We called the linear map $S + T$ to be the sum of the linear maps $S$ and $T$, and we called the linear map $aT$ to be $a$ times the linear map $T$.

We also looked at compositions of linear maps. If $T \in \mathcal (U, V)$ and $S \in \mathcal (V, W)$ then we can define the composition (product) of $S$ and $T$ as a linear map $ST : U \to W$ defined by $(ST)(v) = S(T(v))$ for all vectors $v \in V$.

We will now look at some examples regarding these linear maps.

## Example 1

Let $V$ and $W$ be vector spaces. Recall that $\mathcal L (V, W)$ represents the set of all linear maps from $V$ to $W$. Let $R, S, T \in \mathcal L (V, W)$ and define addition on $\mathcal L (V, W)$ by $(S + T)(v) = S(v) + T(v)$ and for all $a \in \mathbb{F}$ define scalar multiplication on $\mathcal L (V, W)$ by $(aT)(v) = aT(v)$. Show that $\mathcal L (V, W)$ is a vector space.

Recall from the Vector Spaces page that to show that a set is a vector space, we must show that all ten of the axioms hold.

Clearly addition is commutative since $(S + T)(v) = S(v) + T(v) = T(v) + S(v) = (T + S)(v)$.

We also have addition is associative since $((R + S) + T)(v) = (R + S)(v) + T(v) = R(v) + S(v) + T(v) = R(v) + (S + T)(v) = (R + (S + T))(v)$.

The zero vector for $\mathcal L (V, W)$ is the zero map $0(v) = 0$ for all $v \in V$.

For each linear map $T \in \mathcal L (V, W)$, the additivity inverse is the linear map $-T$, that is $(T + (-T)) = T(v) + (-T(v)) = T(v) - T(v) = 0$.

For scalars $a, b \in \mathbb{F}$ we have that scalar multiplication is associative, that is $a(bT(v)) = a(b(T(v)) = (ab)T(v)$.

The multiplicative identity is $1$ since $(1T)(v) = 1T(v) = T(v)$.

Multiplication is also distributive over vector addition since $(a(S + T))(v) = a[S(v) + T(v)] = aS(v) + aT(v) = (aS)(v) + (aT)(v)$.

Furthermore, multiplication is also distributive over field addition since $((a + b)T)(v) = (a + b)T(v) = aT(v) + bT(v) = (aT)(v) + (bT)(v)$.

We have already seen on the Addition, Multiples, and Compositions of Linear Maps pages that $\mathcal L (V, W)$ is closed under vector addition and scalar multiplication - that is, if $S, T \in \mathcal L (V, W)$ then $(S + T) \in \mathcal L (V, W)$ and if $a \in \mathbb{F}$ and $T \in \mathcal L (V, W)$ then $(aT) \in \mathcal L (V, W)$.

Therefore all ten axioms holds and so $\mathcal L (V, W)$ is a vector space.

## Example 2

Let $T : \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $T(x, y) = (x, y, x + y)$ and let $S : \mathbb{R}^3 \to \mathbb{R}^4$ be defined by $S(x, y, z) = (x + y, y + z, z + x, z)$. Define the linear map $ST : \mathbb{R}^2 \to \mathbb{R}^4$.

Take a vector $(x, y) \in \mathbb{R}^2$. Note that $ST(x, y) = S(T(x, y))$ and so:

(1)
\begin{align} \quad ST(x, y) = S(T(x,y)) = S(x, y, x+ y) = (x + y, y + (x + y), (x + y) + x, (x + y)) = (x + y, x + 2y, 2x + y, x + y) \end{align}

Therefore $ST : \mathbb{R}^2 \to \mathbb{R}^4$ is the linear map defined by $ST(x, y) = (x + y, x + 2y, 2x + y, x + y)$ for all $(x, y) \in \mathbb{R}^2$.