
# 6.1: Deﬁnition and elementary properties


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 map $$I:V\to V$$ 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

$\label{eq:T} T(v)=T(a_1 v_1 + \cdots + a_n v_n) = a_1T(v_1)+\cdots+a_nT(v_n) =a_1 w_1 + \cdots + a_n w_n, \tag{6.1.3}$

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

$\begin{equation*} (ST)p(z)=z^2 p'(z) \quad \text{but} \quad (TS)p(z) = z^2 p'(z)+2zp(z). \end{equation*}$