We extend our definition of linear transformation to general vector spaces in a straightforward way:[br][br][u]Definition[/u]: Let [math]V[/math] and [math]W[/math] be two vector spaces and [math]T:V\to W[/math] be a linear transformation from [math]V[/math] to [math]W[/math] if[br][br][math]T(u+v)=T(u)+T(v)[/math] for any [math]u,v[/math] in [math]V[/math][br][math]T(cu)=cT(u)[/math] for any real number [math]c[/math] and [math]u[/math] in [math]V[/math][br][br]Here are some examples of linear transformations:[br][br][u]Example 1[/u]: Let [math]T:M_{m\times n}\to M_{n\times m}[/math] such that [math]T(A)=A^T[/math] for any [math]A[/math] in [math]M_{m\times n}[/math]. [br][br][u]Example 2[/u]: Let [math]D:\mathbb{P}\to\mathbb{P}[/math] such that [math] D(p(t))=\frac{d(p(t))}{dt}[/math] for any [math]p(t)[/math] in [math]\mathbb{P}[/math].[br][br][u]Example 3[/u]: Let [math]F[/math] be the vector space of all continuous functions from [math][a,b][/math] to [math]\mathbb{R}[/math]. We define [math]\phi:F\to F[/math] as follows:[br][br][math]\phi(f)(x)=\int_a^x f(t) \ dt[/math] for any [math]f\in F[/math] and [math]x\in [a,b][/math][br][br][br][u]Example 4[/u]: Let [math]L:\mathbb{S}\to \mathbb{S}[/math] such that [math]L\left(\left(a_n\right)\right)=\left(a_{n+1}\right)[/math]. [math]L[/math] is called the [b]left shift operator[/b] of sequences.[br][br][br]Compositions of two linear transformations is again a linear transformation. The definitions of injective, surjective, and bijective linear transformations are essentially the same as before. Bijective linear transformations are also called [b]isomorphisms[/b]. Two vector spaces are said to be [b]isomorphic[/b] if there exists an isomorphism between them.[br][br][br][br]

Given a linear transformation, we define the following two important subspaces:[br][br][u]Definition[/u]: Let [math]T:V\to W[/math] be a linear transformation. Then [math]\left\{u\in V \ | \ T(u)=0\right\}[/math] is called the [b]kernel[/b] of [math]T[/math], denoted by [math]\text{Ker}(T)[/math]. And [math]\left\{v\in W \ | \ T(u)=v \ \text{for some} \ u\in V\right\}[/math] is called the [b]Image[/b] of [math]T[/math], denoted by [math]\text{Im}(T)[/math].[br][br][math]\text{Ker}(T)[/math] is a subspace of [math]V[/math] and [math]\text{Im}(T)[/math] is a subspace of [math]W[/math]. (Why?)[br][br]By definition, a linear transformation [math]T[/math] is surjective if and only if [math]\text{Im}(T)=W[/math]. And the kernel is useful for detecting whether a linear transformation is injective thanks to the following theorem:[br][br][u]Theorem[/u]: [math]T[/math] is injective if and only if [math]\text{Ker}(T)=\{0\}[/math].[br][br]The proof is essentially the same as the proof of the similar theorem that we had before.[br][br]A useful relation between the dimensions of [math]\text{Ker}(T)[/math] and [math]\text{Im}(T)[/math] is given by the following theorem:[br][br][u]Theorem[/u]: Suppose [math]T:V\to W[/math] is a linear transformation and [math]V[/math] is finite-dimensional. Then [math]\dim(V)=\dim(\text{Ker}(T))+\dim(\text{Im}(T))[/math].[br][br]The following is a useful corollary of the above theorem:[br][br][u]Corollary[/u]: Suppose [math]T:V\to W[/math] is a linear transformation.[br][list=1][*]Suppose [math]\dim(V)=\dim(W)[/math]. Then [math]T[/math] is injective if and only if [math]T[/math] is surjective.[/*][*]If [math]T[/math] is bijective, then [math]\dim(V)=\dim(W)[/math][/*][/list][br][br][math]\dim(\text{Im}(T))[/math] is called the [b]rank[/b] of [math]T[/math] and [math]\dim(\text{Ker}(T))[/math] is called the [b]nullity[/b] of [math]T[/math]. Hence, the above theorem is sometimes called the [b]rank-nullity theoerm[/b].[br][br][br]

Since [math]V[/math] is finite-dimensional, so is [math]\text{Ker}(T)[/math]. Let [math]\dim(\text{Ker}(T))=p\leq \dim(V)=n[/math]. We can find a basis [math]\left\{v_1,v_2,\ldots,v_n\right\}[/math] for [math]V[/math] such that [math]\text{Span}\left\{v_1,v_2,\ldots,v_p\right\}=\text{Ker}(T)[/math] i.e. [math]\left\{v_1,v_2,\ldots,v_p\right\}[/math] is a basis for [math]\text{Ker}(T)[/math]. If [math]p=n[/math], [math]T[/math] is the zero linear transformation and [math]\text{Im}(T)=\{0\}[/math]. Hence the theorem is true. [br][br]Assume [math]p<n[/math], we consider [math]\left\{T(v_{p+1}),\ldots,T(v_n)\right\}[/math]. We will show that it is a basis for [math]\text{Im}(T)[/math]:[br][br]For any [math]w[/math] in [math]\text{Im}(T)[/math], there exists [math]v[/math] in [math]V[/math] such that [math]T(v)=w[/math]. Write [math]v=c_1v_1+c_2v_2+\cdots+c_pv_p+c_{p+1}v_{p+1}+\cdots+c_nv_n[/math]. Then[br][br][math]\begin{eqnarray}w=T(v)&=&c_1T(v_1)+c_2T(v_2)+\cdots+c_pT(v_p)+c_{p+1}T(v_{p+1})+\cdots+c_nT(v_n)\\&=&c_{p+1}T(v_{p+1})+\cdots+c_nT(v_n)\end{eqnarray}[/math][br]since [math]T\left(v_i\right)=0[/math] for [math]i=1,\ldots,p[/math]. Hence [math]\text{Span}\left\{T(v_{p+1}),\ldots,T(v_n)\right\}=\text{Im}(T)[/math]. [br][br]Then we need to show that [math]\left\{T(v_{p+1}),\ldots,T(v_n)\right\}[/math] is a linearly independent set. Suppose[br][br][math]a_{p+1}T(v_{p+1})+\cdots+a_nT(v_n)=0[/math] for some real numbers [math]a_{p+1},\ldots,a_n[/math].[br][br][br]Then we have [math]T\left(a_{p+1}v_{p+1}+\cdots+a_nv_n\right)=0[/math], which means [math]a_{p+1}v_{p+1}+\cdots+a_nv_n[/math] is in [math]\text{Ker}(T)[/math]. Therefore,[br][math]a_{p+1}v_{p+1}+\cdots+a_nv_n=a_1v_1+a_2v_2+\cdots+a_pv_p[/math] for some real numbers [math]a_1,a_2,\ldots,a_p[/math]. Rewrite it as follows:[br][br][math]a_1v_1+a_2v_2+\cdots+a_pv_p-a_{p+1}v_{p+1}-\cdots-a_nv_n=0[/math][br][br]Then [math]a_i=0[/math] for [math]i=1,\ldots,n[/math] because [math]\left\{v_1,v_2,\ldots,v_n\right\}[/math] is a linearly independent set. In particular, [math]a_{p+1}=\cdots=a_n=0[/math]. Hence [math]\left\{T(v_{p+1}),\ldots,T(v_n)\right\}[/math] is a linearly independent set.[br][br]Therefore, [math]\dim(\text{Im}(T))=n-p[/math].[br][br][br][br]