# Abstract Nonsense

## 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 -

## 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