Abstract Nonsense

Crushing one theorem at a time

Interesting Theorem Regarding Linear Transformations

Point of post: In this post we discuss an interesting result which tells us precisely when a mapping from one vector space to another is a linear transformation, namely if and only if the graph (to be defined below) is a subspace of the direct sum of the two vector spaces.


As someone who has done the majority of their mathematical work in topology I can say I am well acquainted with the innocuous concept of the graph of a mapping playing an important role in the theory of structure preserving maps. There is the Closed Graph Theorem in functional analysis, the fact that a function from f:X\to Y where X is Hausdorff and Y is compact is continuous if and only if the graph is closed in X\times Y with the product topology ,etc. That said, I had no idea, until now, that there is a simple but satisfying analogue for linear transformations. Namely, if \mathscr{V} and \mathscr{W} are F-spaces and T:\mathscr{V}\to\mathscr{W} we may define the graph, denotes \Gamma_T, to be the set \left\{\left(v,T(v)\right):v\in\mathscr{V}\right\}. Then, T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right) if and only if \Gamma_T is a subspace of \mathscr{V}\boxplus\mathscr{W}.

So, that being said, all I have left to say is the proof of this interesting theorem:

Theorem: Let \mathscr{V} and \mathscr{W} be F-spaces and T:\mathscr{V}\to\mathscr{W}. Then, T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right) if and only if \Gamma_T is a subspace of \mathscr{V}\boxplus\mathscr{W}.

Proof: If T is a linear transformation then we evidently see that

\alpha\left(v,T(v)\right)+\beta\left(v',T(v')\right)=\left(\alpha v+\beta v',\alpha T(v)+\beta T(v')\right)=\left(\alpha v+\beta v',T\left(\alpha v+\beta v'\right)\right)

so that \Gamma_T is indeed a subspace.


Conversely, if \Gamma_T is a subspace of \mathscr{V}\boxplus\mathscr{W}, v,v'\in\mathscr{V} and \alpha,\beta\in F then we note that since \left(v,T(v)\right),\left(v',T\left(v'\right)\right)\in\Gamma_T and \alpha,beta\in F that \alpha(v,T(v))+\beta (v',T(v'))\in\Gamma_T. But since the first coordinate of \alpha(v,T(v))+\beta (v',T(v')) is \alpha v+\beta v' and \alpha(v,T(v))+\beta(v',T(v'))\in\Gamma_T it follows that the second coordinate must equal T\left(\alpha v+\beta v'\right). But, by definition we have that

\alpha(v,T(v))+\beta(v',T(v'))=\left(\alpha v+\beta v',\alpha T(v)+\beta T(v')\right)

and thus it follows that

T\left(\alpha v+\beta v'\right)=\alpha T(v)+\beta T(v')

from where the conclusion follows. \blacksquare



1. Golan, Jonathan S. The Linear Algebra a Beginning Graduate Student Ought to Know. Dordrecht: Springer, 2007. Print.


November 22, 2010 - Posted by | Algebra, Halmos, Linear Algebra | , , ,

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