# 6.1: Deﬁnition and elementary properties

- Page ID
- 271

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Throughout this chapter, \(V \) and \(W \) denote vector spaces over \(\mathbb{F} \). We are going to study functions from \(V \) into \(W \) that have the special properties given in the following definition.

**Definition 6.1.1.** A function \(T:V\to W \) is called **linear** if

\[T(u+v) = T(u) + T(v), ~~ \rm{~for ~all~} u,v\in V , \tag{6.1.1}\]

\[T(av) = aT(v), ~~ \rm{~for~ all~} a\in \mathbb{F} \rm{~and~} v\in V . \tag{6.1.2}\]

The set of all linear maps from \(V \) to \(W \) is denoted by \(\mathcal{L}(V,W) \). We also write \(Tv \) for \(T(v) \). Moreover, if \(V = W \), then we write \(\mathcal{L}(V,V) = \mathcal{L}(V) \) and call \(T\) in \(\mathcal{L}(V) \) a **linear operator** on \(V \).

**Example 6.1.2. **

1. The** zero map** \(0:V\to W \) mapping every element \(v\in V \) to \(0\in W \) is linear.

2. The **identity ma**p \(I:V\to V \) defined as \(Iv=v \) is linear.

3. Let \(T:\mathbb{F}[z] \to \mathbb{F}[z] \) be the **differentiation map** defined as \(Tp(z)=p'(z) \).

Then, for two polynomials \(p(z),q(z)\in\mathbb{F}[z] \), we have

\[ T(p(z)+q(z)) = (p(z)+q(z))' =p'(z)+q'(z)=T(p(z))+T(q(z)). \]

Similarly, for a polynomial \(p(z)\in \mathbb{F}[z] \) and a scalar \(a\in \mathbb{F} \), we have

\[ T(ap(z))=(ap(z))'=ap'(z)=aT(p(z)). \]

Hence \(T \) is linear.

4. Let \(T:\mathbb{R}^2\to \mathbb{R}^2 \) be the map given by \(T(x,y)=(x-2y,3x+y) \). Then, for \((x,y),(x',y')\in \mathbb{R}^2 \), we have

\begin{equation*}

\begin{split}

T((x,y)+(x',y')) &= T(x+x',y+y') = (x+x'-2(y+y'),3(x+x')+y+y')\\

&= (x-2y,3x+y) + (x'-2y',3x'+y') = T(x,y) + T(x',y').

\end{split}

\end{equation*}

Similarly, for \((x,y)\) in \(\mathbb{R}^2 \) and \(a\) in \(\mathbb{F}\), we have

\[ T(a(x,y)) = T(ax,ay) = (ax-2ay,3ax+ay) = a(x-2y,3x+y) = aT(x,y). \]

Hence \(T\) is linear. More generally, any map \(T: \mathbb{F}^n \to \mathbb{F}^m \) defined by

\[ T(x_1,\ldots,x_n) = (a_{11}x_1+\cdots +a_{1n} x_n, \ldots,a_{m1}x_1+\cdots+a_{mn}x_n) \]

with \(a_{ij}\in\mathbb{F} \) is linear.

5. Not all functions are linear! For example, the exponential function \(f(x)=e^x \) is not linear since \(e^{2x} \neq 2 e^x \) in general. Also, the function \(f:\mathbb{F} \to \mathbb{F} \) given by \(f(x)=x-1 \) is not linear since \(f(x+y)=(x+y)-1 \neq (x-1)+(y-1)=f(x)+f(y) \).

An important result is that linear maps are already completely determined if their values on basis vectors are specified.

**Theorem 6.1.3. ** *Let* \((v_1,\ldots,v_n) \) *be a basis of *\(V \) *and* \((w_1,\ldots,w_n) \) *be an arbitrary list of vectors in* \(W \). *Then there exists a unique linear map*

\[ T:V\to W \quad \text{such that \(T(v_i)=w_i, \, \forall \, i = 1, 2, \ldots, n \).} \]

*Proof. * First we verify that there is at most one linear map \(T \) with \(T(v_i)=w_i \). Take any \(v\in V \). Since \((v_1,\ldots,v_n) \) is a basis of \(V \) there are unique scalars \(a_1,\ldots,a_n\) in \(\mathbb{F} \) such that \(v = a_1 v_1 + \cdots + a_n v_n \). By linearity, we have

\[ \begin{equation} \label{eq:T}

T(v)=T(a_1 v_1 + \cdots + a_n v_n) = a_1T(v_1)+\cdots+a_nT(v_n)

=a_1 w_1 + \cdots + a_n w_n, \tag{6.1.3}

\end{equation} \]

and hence \(T(v) \) is completely determined. To show existence, use Equation (6.1.3) to define \(T \). It remains to show that this \(T \) is linear and that \(T(v_i)=w_i \). These two conditions are not hard to show and are left to the reader.

The set of linear maps \(\mathcal{L}(V,W) \) is itself a vector space. For \(S,T\in \mathcal{L}(V,W) \) addition is defined as

\[ \begin{equation*}

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

\end{equation*} \]

For \(a\in \mathbb{F} \) and \(T\in \mathcal{L}(V,W) \), scalar multiplication is defined as

\[ \begin{equation*}

(aT)(v) = a(Tv), \quad \text{for all \(v\in V \).}

\end{equation*} \]

You should verify that \(S+T \) and \(aT \) are indeed linear maps and that all properties of a vector space are satisfied.

In addition to the operations of vector addition and scalar multiplication, we can also define the **composition of linear maps**. Let \(V,U,W \) be vector spaces over \(\mathbb{F} \). Then, for \(S\in\mathcal{L}(U,V) \) and \(T\in \mathcal{L}(V,W) \), we define \( T\circ S\in\mathcal{L}(U,W)\) by

\begin{equation*}

(T\circ S)(u) = T(S(u)), \quad \text{for all \(u\in U \).}

\end{equation*}

The map \(T\circ S \) is often also called the** product** of \(T \) and \(S \) denoted by \(TS \). It has the following properties:

1.** Associativity:** \((T_1 T_2)T_3=T_1(T_2 T_3) \), for all \(T_1\in \mathcal{L}(V_1,V_0) \), \(T_2 \in \mathcal{L}(V_2,V_1) \) and \(T_3\in\mathcal{L}(V_3,V_2) \).

2.** Identity:** \(TI=IT=T \), where \(T\in \mathcal{L}(V,W) \) and where \(I \) in \(TI \) is the identity map in \(\mathcal{L}(V,V) \) whereas the \(I \) in \(IT \) is the identity map in \(\mathcal{L}(W,W) \).

3.**Distributivity:** \((T_1+T_2)S=T_1S+T_2S \) and \(T(S_1+S_2)=TS_1+TS_2 \), where \(S,S_1,S_2\in\mathcal{L}(U,V) \) and \(T,T_1,T_2\in\mathcal{L}(V,W) \).

Note that the product of linear maps is not always commutative. For example, if we take \(T\in\mathcal{L}(\mathbb{F}[z],\mathbb{F}[z]) \) to be the differentiation map \(Tp(z)=p'(z) \) and \(S\in\mathcal{L}(\mathbb{F}[z],\mathbb{F}[z])\) to be the map \(Sp(z)=z^2p(z) \), then

\[ \begin{equation*}

(ST)p(z)=z^2 p'(z) \quad \text{but} \quad (TS)p(z) = z^2 p'(z)+2zp(z).

\end{equation*}\]

## Contributors

## Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.