3.3 Vector space homomorphisms

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If $V$ and $W$ are two vector spaces over the same field $\mathbb{K}$, a map $T:V\to W$ is called linear, or a vector space homomorphism from $V$ into $W_{;}$ , if

$$T(au+bv)=aTu+bTv\text{\hspace{1em}(3.1)}$$

for all $a,b\in \mathbb{K}$ and all $u,v\in V$ . The notation $Tu$ on the right-hand side is commonly used in place of $T(u)$ . Vector space homomorphisms play a similar role to group homomorphisms in that they preserve the basic operations of vector addition and scalar multiplication that define a vector space. They are the morphisms of the category of vector spaces.

Since $T(u+0)=Tu=Tu+T0.$ , it follows that the zero vector of $V$ goes to the zero vector of $W$ under a linear map, $T0=0$ . Note, however, that the zero vectors on the two sides of this equation lie in different spaces and are, strictly speaking, different vectors.

A linear map $T:V\to W$ that is one-to-one and onto is called a vector space isomorphism. In this case the inverse map $T^{-1}:W\to V$ must also be linear. for if $u,v\in W$ let

$u^{\prime }=T^{-1}u,v^{\prime }=T^{-1}v$ then

$$\begin{array}{rl}{T}^{-1}(au+bv)& =T^{-1}(aT{u}^{\prime }+bT{v}^{\prime })\\ & =T^{-1}(T(a{u}^{\prime }+b{v}^{\prime }))\\ & ={\mathrm{id}}_V(a{u}^{\prime }+b{v}^{\prime })\\ & =a{u}^{\prime }+b{v}^{\prime }\\ & =a{T}^{-1}u+b{T}^{-1}v.\end{array}$$

Two vector spaces $V$ and $W$ are called isomorphic, written $V\cong W$ , if there exists a vector space isomorphism $T:V\to W$ . Two isomorphic vector spaces are essentially identical in all their properties.

Example 3.10

Consider the set ${\mathcal{P}}_n(x)$ of all real-valued polynomials of degree $\le n.$

$$f(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n.$$

Polynomials of degree $\le n$ can be added and multiplied by scalars in the obvious way,

$$\begin{array}{c}f(x)+g(x)=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+\cdots +(a_n+b_n)x^n,\\ cf(x)=c{a}_0+c{a}_{1}x+c{a}_2{x}^2+\cdots +c{a}_n{x}^n,\end{array}$$

making ${\mathcal{P}}_n(x)$ into a vector space. The map $S:P_n(x)\to {\mathbb{R}}^{n+1}$ defined by

$$S(a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n)=(a_0,a_1,a_2,\dots ,a_n)$$

is one-to-one and onto and clearly preserves basic vector space operations,

$$S(f(x)+g(x))=S(f(x))+S(g(x)),S(af(x))=aS(f(x)).$$

Hence $S$ is a vector space isomorphism, and $P_n(x)\cong {\mathbb{R}}^{n+1}$

The set ${\hat{\mathbb{R}}}^{\infty }$ of all sequences of real numbers $(a_0,a_1,\dots )$ having only a finite number of non-zero members $a_i\ne 0$ is a vector space, using the same rules of vector addition and scalar multiplication given for ${\mathbb{R}}^{\infty }$ in Example 3.6. The elements of ${\hat{\mathbb{R}}}^{\infty }$ are real sequences ofthe form $(a_0,a_1,\dots ,a_m,0,0,0,\dots )$ . Let $\mathcal{P}(x)$ be the set ofall real polynomials, $\mathcal{P}(x)=$ ${\mathcal{P}}_0(x)\cup {\mathcal{P}}_1(x)\cup {\mathcal{P}}_2(x)\cup ..$ . This is clearly a vector space with respect to the standard rules of addition of polynomials and scalar multiplication. The map $S:{\hat{\mathbb{R}}}^{\infty }\to \mathcal{P}$ defined by

$$S:(a_0,a_1,\dots ,a_m,0,0,\dots )\mapsto a_0+a_{1}x+\cdots +a_m{x}^m$$

is an isomorphism.

It is simple to verify that the inclusion maps defined in Section 1.4,

$$i_1:{\hat{\mathbb{R}}}^{\infty }\to {\mathbb{R}}^{\infty }\text{ and }{i}_2:{\mathcal{P}}_n(x)\to \mathcal{P}(x)$$

are vector space homomorphisms.

Let $L(V,W)$ denote the set of all linear maps from $V$ to $W$. If $T$, $S$ are linear maps from $V$ to $W$ , addition $T+S$ and scalar multiplication $aT$ are defined by

$$(T+S)(u)=Tu+Su,(aT)u=aTu.$$

The set $L(V,W)$ is a vector space with respect to these operations. Other common notations for this space are $\mathrm{Hom}(V,W)$ and $\mathrm{Lin}(V,W)$.

Exercise

Verify that $L(V,W)$ satisfies all the axioms of a vector space.

If $T\in L(U,V)$ and $S\in L(V,W)$ , define their product to be the composition map $ST=S\circ T:U\to W;$

$$(ST)u=S(Tu).$$

This map is clearly linear since

$$ST(au+bv)=S(aTu+bTv)=aSTu+bSTv.$$

If $S$ and $T$ are invertible linear maps then so is their product $ST$ , and $(ST{)}^{-1}:W\to U$ satisfies

$$(ST{)}^{-1}=T^{-1}{S}^{-1},\text{\hspace{1em}(3.2)}$$

since

$$T^{-1}{S}^{-1}ST=T^{-1}{\mathrm{id}}_{V}T=T^{-1}T={\mathrm{id}}_U.$$

Linear maps $S:V\to V$ are called linear operators on $V$. They form the vector space $L(V,V)$ . If $S$ is an invertible linear operator on $V$ it is called a linear transformation on $V$. It may be thought of as a vector space isomorphism of $V$ onto itself, or an automorphism of $V$. The linear transformations of $V$ form a group with respect to the product law of composition, called the general linear group on $V$ and denoted $GL(V)$ . The group properties are easily proved:

Closure: if $S$ and $T$ are linear transformations of $V$ then so is $ST$ , since (a) it is a linear map, and (b) it is invertible by Eq. (3.2).

Associativity: this is true of all maps (see Section 1.4).

Unit: the identity map $\mathrm{id}$ is linear and invertible.

Inverse: as shown above, the inverse $T^{-1}$ of any vector space isomorphism $T$ is linear.

Note, however, that $GL(V)$ is not a vector space, since the zero operator that sends every vector in $V$ to the zero vector 0 is not invertible and therefore does not belong to $GL(V)$.

Problems

Problem 3.3

Show that the infinite dimensional vector space ${\mathbb{R}}^{\infty }$ is isomorphic with a proper subspace of itself.

Problem 3.4

On the vector space $\mathcal{P}(x)$ of polynomials with real coefficients over a variable $x$, let $x$ be the operation of multiplying by the polynomial $x$, and let $D$ be the operation of differentiation,

$$x:f(x)\mapsto xf(x),D:f(x)\mapsto \frac{\mathrm{d}f(x)}{\mathrm{d}x}.$$

Show that both of these are linear operators over $\mathcal{P}(x)$ and that $Dx-xD=I$ , where $I$ is the identity operator.