Linear Maps and Spaces of Linear Maps

Linearity and preservation. We suppose that $$V$$ and $$W$$ are vector spaces over the same field $$F$,$ and $\tau$ is a function from $$V$$ to $$W$.$ The function $\tau$ is said to be linear if it preserves the vector-space structures of $$V$$ and $$W$.$

Definition 3.1.1. Let $$V$$ and $$W$$ be vector spaces over the same field $$F$.$ A function $\tau\colon V\to W$ is called a linear map from $$V$$ to $$W$$ if for all $u,v\in V$ and $a\in F$ the following conditions hold:

$$\tau(u+v)=\tau(u) +\tau(v)$;$
$$\tau(av) = a\tau(v)$.$

The set of all linear maps from $$V$$ to $$W$$ is denoted by $$\operatorname{Hom}(V,W)$.$ A linear map from $$V$$ to $$V$$ is called a linear operator on $$V$.$ The identity operator $\operatorname{id}$ on $$V$$ is a linear operator on $$V$$ such that $\operatorname{id}(v)=v$ for all $$v\in V$.$ The set of all linear operators on $$V$$ is denoted by $$\operatorname{End}(V)$.$ A linear operator on a real vector space is called a real operator.

In real analysis, differentiation and integration provide us with two of the most important examples of linear maps.

Example 3.1.2.

Let $C^{\infty}$ denote the set of all real-valued functions defined on $\mathbb R$ that have derivatives of every order. Then $C^{\infty}$ is a vector space. Define $D\colon C^{\infty}\to C^{\infty}$ by $$Df=f'$.$ Then, the operator $$D$$ is linear.
Let $$C([0,1])$$ be a vector space of continuous real-valued functions defined on $$[0,1]$.$ Define
\[I:C([0,1])\to C^1([0,1]),\qquad f\mapsto If,\]
where
\[(If)(x)=\int_0^x f(t)\,dt.\]
Then the linearity of $$I$$ follows from that of integration.

A coordinate map, defined in Definition 2.6.6, is another good example of linear maps.

Proposition 3.1.3. Let $$B$$ be a basis for a vector space $$V$$ over a field $$F$.$ Then the coordinate map $\phi_B\colon V\to F^{(B)}$ is linear.

Proof. Let $$u,v\in V$,$ let $$a\in F$,$ and let

\[S = \operatorname{supp}(c_{u+v}^B) \cup \operatorname{supp}(c_{u}^B) \cup \operatorname{supp}(c_{v}^B).\]

Then $$S$$ is finite and linearly independent. Since

\[\sum_{b\in \operatorname{supp}(c_{u+v}^B)}c_{u+v}^B(b)b = \sum_{b\in \operatorname{supp}(c_{u}^B)}c_{u}^B(b)b + \sum_{b\in \operatorname{supp}(c_{v}^B)}c_{v}^B(b)b,\]

we have

\[\sum_{b\in S}(c_{u+v}^B(b)-c_{u}^B(b)-c_{v}^B(b))b = 0_V.\]

Since $$S$$ is linearly independent, Theorem 2.5.6 yields

\[c_{u+v}^B(b) = c_{u}^B(b) + c_{v}^B(b)\]

for all $$b\in S$,$ and this equality holds for all $$b\in B$.$ It follows that

\[c_{u+v}^B = c_{u}^B + c_{v}^B.\]

Similarly, let

\[T = \operatorname{supp}(c_{av}^B) \cup \operatorname{supp}(c_{v}^B).\]

Then $$T$$ is finite and linearly independent. Since

\[\sum_{b\in \operatorname{supp}(c_{av}^B)}c_{av}^B(b)b = a\sum_{b\in \operatorname{supp}(c_{v}^B)}c_{v}^B(b)b,\]

we have

\[\sum_{b\in T}(c_{av}^B(b)-ac_{v}^B(b))b = 0_V.\]

Since $$T$$ is linearly independent, Theorem 2.5.6 yields

\[c_{av}^B(b) = ac_{v}^B(b)\]

for all $$b\in S$,$ and this equality holds for all $$b\in B$.$ It follows that

\[c_{av}^B = ac_{v}^B.\]

We have shown that $\phi_B(u+v)=\phi_B(u)+\phi_B(v)$ and $$\phi_B(av)=a\phi_B(v)$.$ Therefore $\phi_B$ is linear. $\square$

Proposition 3.1.4. Let $$V$$ and $$W$$ be vector spaces over $$F$,$ and let $\tau\colon V \to W$ be a function.

If $\tau$ is linear, then $$\tau(0_V)=0_W$.$
If $\tau$ is linear, then $\tau(-v)=-\tau(v)$ for all $$v\in V$.$
The function $\tau$ is linear if and only if, for all $u,v\in V$ and $$a\in F$,$ we have
\[\tau(au+v)=a\tau(u)+\tau(v).\]

Proof. (1) We have $$v+0_V=v$$ for all $$v\in V$.$ Since $\tau$ is linear, $$\tau(v) + \tau(0_V) = \tau(v)$.$ Since the zero vector in $$W$$ is unique, $$\tau(0_V)=0_W$.$ (2) Since $$v+(-v)=0_V$,$ we have $\tau(v)+\tau(-v)=0_W$ by (1). Since the inverse of a vector is unique, $$\tau(-v)=-\tau(v)$.$ (3) If $\tau$ is linear, then $$\tau(au+v)=\tau(au) + \tau(v)=a\tau(u) + \tau(v)$.$ We now prove the converse. Setting $$a=1$,$ we obtain $$\tau(u+v)=\tau(u)+\tau(v)$.$ Hence $$\tau(0_V)=0_W$.$ Therefore, setting $$v=0_V$$ gives $$\tau(au)=a\tau(u)$,$ so $\tau$ is linear. $\square$

Theorem 3.1.5. A function $\tau\colon V\to W$ is linear if and only if for any finite set $$I$$ and indexed families $\{v_i\}_{i\in I}$ in $$V$$ and $\{a_i\}_{i\in I}$ in $$F$,$

\[\tau\lowparen{\sum_{i\in I}a_iv_i} = \sum_{i\in I}a_i\tau(v_i)\]

Proof. Suppose that $\tau$ is linear. We proceed by induction on $$n=|I|$.$ If $$n=0$,$ then

\[\tau\lowparen{\sum_{i\in I}a_iv_i} = \tau(0_V) = 0_W = \sum_{i\in I}a_i\tau(v_i).\]

We now suppose that the statements holds for $$n=k$.$ Let $$|I|=k+1$$ and choose $$i'\in I$.$ Define $$I' = I\setminus\{i'\}$.$ Since $$|I'|=k$,$ by induction hypothesis and linearity,

\[\begin{align} \tau\lowparen{\sum_{i\in I}a_iv_i} &= \tau\lowparen{a_{i'}v_{i'} +\sum_{i\in I'}a_iv_i}\\ &= \tau(a_{i'}v_{i'}) + \tau\lowparen{\sum_{i\in I'}a_iv_i}\\ &= a_{i'}\tau(v_{i'}) + \sum_{i\in I'}a_i\tau(v_i)\\ &= \sum_{i\in I}a_i\tau(v_i). \end{align}\]

The backward implication is trivial. $\square$

Proposition 3.1.6. Let $$V$$ and $$W$$ be vector spaces, and let $$\tau\in \operatorname{Hom}(V,W)$.$

If $$U$$ is a subspace of $$W$,$ then $\tau^{-1}(U)$ is a subspace of $$V$.$
If $$U$$ is a subspace of $$V$,$ then $\tau(U)$ is a subspace of $$W$.$

Proof. (1) Since $\tau(0_V)=0_W$ and $$0_W\in U$,$ $$0_V\in \tau^{-1}(U)$.$ Let $$u,v\in \tau^{-1}(U)$.$ Then we have $\tau(u)\in U$ and $$\tau(v)\in U$.$ Since $$U$$ is a subspace, $$\tau(u)+\tau(v)\in U$.$ Since linearity gives $$\tau(u+v)=\tau(u)+\tau(v)$,$ $$u+v\in \tau^{-1}(U)$.$ Similarly, closedness of $$U$$ under scalar multiplication and linearity of $\tau$ gives $$av\in \tau^{-1}(U)$,$ where $$a\in F$.$ By Theorem 2.2.2, $\tau^{-1}(U)$ is a subspace of $$V$.$ (2) Since $$0_V\in U$,$ $0_W\in \tau(U)$ by Proposition 3.1.4(1). Let $u,v\in \tau(U)$ and let $s\in \tau^{-1}(\{u\})$ and $$t\in \tau^{-1}(\{v\})$.$ Since $$s+t\in U$,$ linearity gives $$u+v=\tau(s+t)\in \tau(U)$.$ Similarly, since $$as\in U$,$ we have $$au=\tau(as)\in \tau(U)$,$ where $$a\in F$.$ By Theorem 2.2.2, $\tau(U)$ is a subspace of $$W$.$ $\square$

Proposition 3.1.7. Let $$B$$ be a basis for $$V$.$ For any function $$f\colon B\to W$,$ there is a unique map $\tau\in \operatorname{Hom}(V,W)$ such that $\tau(b)=f(b)$ for all $$b\in B$.$

Proof. Let $v\in V$ and let $c_v^B\colon B\to F$ be the representation of $$v$$ in terms of $$B$.$ Define

\[\tau\colon V\longrightarrow W,\qquad v\longmapsto \sum_{b\in \operatorname{supp}(c_v^B)}c_v^B(b)f(b).\]

Since $$B$$ is linearly independent, $v\in B$ has the unique representation, such that $$c_v^B(v) = 1$$ and $$\operatorname{supp}(c_v^B) = \{v\}$.$ Hence $\tau(b)=f(b)$ for all $$b\in B$.$ We now show the linearity of $$\tau$.$ Let

\[S=\operatorname{supp}(c_u^B)\cup\operatorname{supp}(c_v^B).\]

Since $c_{au+v}^B = ac_u^B + c_v^B$ by Proposition 3.1.3, we have $$\operatorname{supp}(c_{au+v}^B)\subseteq S$.$ Therefore

\[\begin{aligned} \tau(au+v) &=\sum_{b\in\operatorname{supp}(c_{au+v}^B)} c_{au+v}^B(b)f(b)\\ &=\sum_{b\in S}c_{au+v}^B(b)f(b)\\ &=\sum_{b\in S}\left(ac_u^B(b)+c_v^B(b)\right)f(b)\\ &=a\sum_{b\in S}c_u^B(b)f(b) +\sum_{b\in S}c_v^B(b)f(b)\\ &=a\sum_{b\in\operatorname{supp}(c_u^B)}c_u^B(b)f(b) +\sum_{b\in\operatorname{supp}(c_v^B)}c_v^B(b)f(b)\\ &=a\tau(u)+\tau(v). \end{aligned}\]

Hence $\tau$ is linear. Suppose that $\tau$ is a linear map from $$V$$ to $$W$$ such that $\tau(b)=f(b)$ for all $$b\in B$.$ Let $v\in V$ and let $c_v^B\colon B\to F$ be the representation of $$v$$ in terms of $$B$.$ It follows from Theorem 3.1.5 that

\[\begin{align} \tau(v) &= \tau\lowparen{\sum_{b\in \operatorname{supp}(c_v^B)}c_v^B(b)b}\\ &= \sum_{b\in \operatorname{supp}(c_v^B)}c_v^B(b)\tau(b)\\ &= \sum_{b\in \operatorname{supp}(c_v^B)}c_v^B(b)f(b). \end{align}\]

Therefore, the condition $\tau(b)=f(b)$ for all $b\in B$ determines the unique map $$\tau\in \operatorname{Hom}(V,W)$.$ $\square$

Theorem 3.1.8. The set $\operatorname{Hom}(V,W)$ is a vector space under ordinary addition of functions and scalar multiplication of functions by elements of $$F$.$

Proof. Let $u,v\in V$ and $$a,b\in F$.$ Since

\[\begin{align} (\tau + \sigma)(au+v) &= \tau(au + v) + \sigma(au + v)\\ &= (a\tau(u) + \tau(v)) + (a\sigma(u) + \sigma(v))\\ &= a(\tau(u)+\sigma(v)) + (\tau(u) + \sigma(v))\\ &= a(\tau + \sigma)(u) + (\tau + \sigma)(v) \end{align}\]

and

\[\begin{align} (a\tau)(bu + v) &= a\tau(bu + v)\\ &= a(b\tau(u) + \tau(v))\\ &= b(a\tau(u)) + a\tau(v)\\ &= b(a\tau)(u) + (a\tau)(v), \end{align}\]

the two operations are closed. Associativity and commutativity of addition follow from those in $$W$.$ The zero map is the additive identity and $-\tau\in \operatorname{Hom}(V,W)$ defined by $(-\tau)(v) = -\tau(v)$ for all $v\in V$ is the additive inverse of $$\tau\in \operatorname{Hom}(V,W)$.$ Hence $\operatorname{Hom}(V,W)$ is an abelian group. Since scalar multiplication axioms follows from the corresponding axioms in $$W$,$ $\operatorname{Hom}(V,W)$ is a vector space. $\square$

Proposition 3.1.9. If $\sigma \in \operatorname{Hom}(U,V)$ and $$\tau\in \operatorname{Hom}(V,W)$,$ then the composition $\tau\sigma$ is in $$\operatorname{Hom}(U,W)$.$

Proof. Let $u,v\in U$ and $$a\in F$.$ Then we have

\[\begin{align} (\tau\sigma)(au+v) &= \tau(\sigma(au+v))\\ &= \tau(a\sigma(u) + \sigma(v))\\ &= a\tau(\sigma(u)) + \tau(\sigma(v))\\ &= a(\tau\sigma)(u) + (\tau\sigma)(v). \end{align}\]

Therefore, $$\tau\sigma\in \operatorname{Hom}(U,W)$.$ $\square$

Theorem 3.1.10. The vector space $\operatorname{End}(V)$ is an algebra, where multiplication is composition of functions. The identity operator $\operatorname{id}_V$ is the multiplicative identity.

Proof. The closedness of $\operatorname{End}(V)$ under multiplication follows from Proposition 3.1.9. Since composition of functions is associative, multiplication is associative. We now let $\rho,\sigma,\tau\in \operatorname{End}(V)$ and $$v\in V$.$ Since $\rho$ is linear,

\[\begin{align} (\rho(\sigma+\tau))(v) &= \rho(\sigma(v)+\tau(v))\\ &= \rho(\sigma(v)) + \rho(\tau(v))\\ &= (\rho\sigma)(v) + (\rho\tau)(v)\\ &= (\rho\sigma + \rho\tau)(v). \end{align}\]

Therefore $$\rho(\sigma+ \tau) = \rho\sigma + \rho\tau$.$ Similarly, we have $$(\rho + \sigma)\tau = \rho\tau + \sigma\tau$.$ Therefore, multiplication is distributive. By definition 1.2.1, $\operatorname{End}(V)$ is a ring under the addition and multiplication. Furthermore, we have

\[\begin{align} ((a\sigma)\tau)(v) &= (a\sigma)(\tau(v))\\ &= a\sigma(\tau(v))\\ &= a(\sigma\tau)(v)\\ &= (a(\sigma\tau))(v). \end{align}\]

Therefore $$(a\sigma)\tau = a(\sigma\tau)$.$ In the same way, we obtain $$\sigma(a\tau) = a(\sigma\tau)$.$ Hence multiplication is compatible with scalar multiplication. Consequently, $\operatorname{End}(V)$ is an algebra by Definition 2.1.7. Since $(\operatorname{id}\tau)(v) = \operatorname{id}(\tau(v)) = \tau(v)$ and $$(\tau\operatorname{id})(v) = \tau(\operatorname{id}(v)) = \tau(v)$,$ the identity operator $\operatorname{id}$ is the multiplicative identity of $$\operatorname{End}(V)$.$ $\square$