Abstract Nonsense

Crushing one theorem at a time

Characterization of Linear Homomorphisms In Terms of Bases


Point of post: In this post we give a characterization of linear homomorphisms which has to do entirely with how the homomorphism acts on a basis.

Motivation

Often it is laborious to check that that a linear transformation is injective or surjective. It turns out that there is a nice characterization of injectivity and surjectivity for finite dimensional spaces which depends entirely on how the the transformation acts on a basis of the domain space.

Theorem: Let \mathscr{V} and \mathscr{W} be finite dimensional F-spaces  and let T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right). Then for any basis \mathscr{B}=\{x_1,\cdots,x_n\} of \mathscr{V} it’s true that:

\begin{aligned}&(1)\quad T\textit{ is a monomorphism if and only if there exists a basis }\mathscr{D}\textit{ for }\mathscr{W}\textit{ for which }T:\mathscr{B}\to\mathscr{D}\textit{ is an injection.}\\ & (2)\quad T\textit{ is an epimorphism if and only if }\textit{ there exists a subset of }T\left(\mathscr{B}\right)\textit{ which is a basis for }\mathscr{W}.\\ &(3)\quad T\in\text{GL}\left(\mathscr{V}\right)\textit{ if and only if }T\left(\mathscr{B}\right)\text{ is a basis for }\mathscr{W}.\end{aligned}

Proof:

(1): First suppose that T is a monomorphism. We claim that T\left(\mathscr{B}\right) is linearly independent. To see this suppose that

\displaystyle \sum_{j=1}^{n}\alpha_j T(x_j)=\bold{0}

then

T\displaystyle \left(\sum_{j=1}^{n}\alpha_j x_j\right)=\bold{0}

but since T is a monomorphism we know from prior discussion that \ker T=\{\bold{0}\} so that the above implies that

\displaystyle \sum_{j=1}^{n}\alpha_j x_j=\bold{0}

and since \mathscr{B} is linearly independent it follows that \alpha_1=\cdots=\alpha_n=0, from where the linear independence of T\left(\mathscr{B}\right) follows. But, this implies that T\left(\mathscr{B}\right) may be extended to a basis as desired.

Conversely, suppose that T\left(\mathscr{B}\right) can be injected into a basis, clearly then T\left(\mathscr{B}\right) may be extended to a basis for \mathscr{W}. Then, we see that T\left(\mathscr{B}\right) is linearly independent. Assume then that \displaystyle \sum_{j=1}^{n}\alpha_j x_j\in\ker T then

\displaystyle T\left(\sum_{j=1}^{n}\alpha_j x_j\right)=\sum_{j=1}^{n}\alpha_j T(x_j)=\bold{0}

and thus by assumption \alpha_1=\cdots=\alpha_j=0, thus \ker T=\{\bold{0}\} and so T is a monomorphism.

(2): Suppose first that T is an epimorphism. We note then that

\text{span }T\left(\mathscr{B}\right)=T\left(\text{span }\mathscr{B}\right)=T\left(\mathscr{V}\right)=\mathscr{W}

and thus T\left(\mathscr{B}\right) spans \mathscr{W}. But, we may then eliminate elements of T\left(\mathscr{B}\right) to assure that the resulting set is linearly independent and still spans, namely we may find a subset of T\left(\mathscr{B}\right) which is a basis for \mathscr{W}.

Conversely,  suppose that T\left(\mathscr{B}\right) contains a subset \mathscr{S} which spans \mathscr{W}. Then, we see that

T\left(\mathscr{V}\right)=T\left(\text{span }\mathscr{B}\right)=\text{span }T\left(\mathscr{B}\right)\supseteq\text{span }\mathscr{S}=\mathscr{W}

from where it follows that T is an epimorphism.

(3): First suppose that T\in\text{GL}\left(\mathscr{V}\right). Then, since T is a monomorphism we have from (1) that T\left(\mathscr{B}\right) is linearly independent and since T is an epimorphism we have from (2) that \text{span }T\left(\mathscr{B}\right)=\mathscr{W}. It easily follows that T\left(\mathscr{B}\right) is a basis for \mathscr{W}.

Conversely, suppose that T\left(\mathscr{B}\right) is a basis. Since T\left(\mathscr{B}\right) may be extended to a basis (it already is one) we see from (1) that T is a monomorphism. Also, since T\left(\mathscr{B}\right) contains a subset which is a basis for \mathscr{W} (namely itself) we see from (2) that T is an epimorphism. Thus, putting these two together implies that T\in\text{GL}\left(\mathscr{V}\right).

\blacksquare

Notice that the methodology in the above proof serves to prove the following equivalent statement:

Corollary: Let \mathscr{V} and \mathscr{W} be finite dimensional F spaces. Then if T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right) we see that

\begin{aligned}&(1)\quad T\textit{ is a monomorphism if and only if }T\left(\mathscr{B}\right)\textit{ is linearly independent.}\\ &(2)\quad T\textit{ is an epimorphism if and only if }\text{span }T\left(\mathscr{B}\right)=\mathscr{W}.\\ &(3)\quad T\in\text{GL}\left(\mathscr{V}\right)\textit{ if and only if }T\left(\mathscr{B}\right)\textit{ is a basis for }\mathscr{W}.\end{aligned}

 

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

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December 2, 2010 - Posted by | Algebra, Halmos, Linear Algebra | , , ,

3 Comments »

  1. […] a) This was proven, in some detail (recalling that an endomorphism on a finite dimensional space is an isomorphism if and only if it’s a monomorphism) here […]

    Pingback by Halmos Section 36: Inverses (Pt. II) « Abstract Nonsense | December 3, 2010 | Reply

  2. […] follows directly from our earlier characterization of isomorphisms since evidently is linear and letting be the usual basis for we see then that which is […]

    Pingback by Matrix Algebra (Pt. I) « Abstract Nonsense | December 13, 2010 | Reply

  3. […] let be the unique element of such that . Since this is a bijection between bases our previous characterization of isomorphisms let’s us conclude that […]

    Pingback by Halmos Sections 37 and 38: Matrices and Matrices of Linear Transformations(Pt. III) « Abstract Nonsense | December 19, 2010 | Reply


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