PlanetMath: dense ring of linear transformations

Note that the linear independence of the

's is essential in insuring the existence of a linear transformation

. Otherwise, suppose $0=\sum d_iv_i$ where $d_1\neq 0$ . Pick

's so that they are linearly independent. Then $0=f(\sum d_iv_i)=\sum d_if(v_i)=\sum d_iw_i$ , contradicting the linear independence of the

's.

The notion of “dense” comes from topology: if

is given the discrete topology and $\operatorname{End}_D(V)$ the compact-open topology, then

is dense in $\operatorname{End}_D(V)$ iff

is a dense ring of linear transformations of

Proof. First, assume that

is a dense ring of linear transformations of

. Recall that the compact-open topology on $\operatorname{End}_D(V)$ has subbasis of the form $B(K,U):=\lbrace f\mid f(K)\subseteq U\rbrace$ , where

is open and

is compact in

. Since

is discrete,

is finite. Now, pick a point $g\in \operatorname{End}_D(V)$ and let

$\displaystyle B=\bigcup_{\alpha\in I}\bigcap_{i=1}^{n(\alpha)}B(K_{i\alpha},U_{i\alpha})$

be a neighborhood of

some index set. Then for some $\alpha\in I$ , $g\in \bigcap B(K_{i\alpha},U_{i\alpha})$ . This means that $g(K_{i\alpha})\subseteq U_{i\alpha}$ for all $i=1,\ldots,n(\alpha)$ . Since each $K_{i\alpha}$ is finite, so is $K:=\bigcup K_{i\alpha}$ . After some re-indexing, let $\lbrace v_1,\ldots, v_n\rbrace$ be a maximal linearly independent subset of

. Set

, $j=1,\ldots,n$ . By assumption, there is an $f\in R$ such that

, for all

. For any $v\in K$ ,

is a linear combination of the

's: $v=\sum d_jv_j$ , $d_j\in D$ . Then $f(v)=\sum d_jf(v_j)=\sum d_jg(_j)=g(v)\in U_{i\alpha}$ for some

. This shows that $f(K_{i\alpha})\subseteq U_{i\alpha}$ and we have $f\in \bigcap B(K_{i\alpha},U_{i\alpha})\subseteq B$ .

Conversely, assume that the ring is a dense subset of the space $\operatorname{End}_D(V)$ . Let $v_1,\ldots,v_n$ be linearly independent, and $w_1,\ldots,w_n$ be arbitrary vectors in . Let be the subspace spanned by the 's. Because the 's are linearly independent, there exists a linear transformation such that and for $v\notin W$ . Let $K_i=\lbrace v_i\rbrace$ and $U_i=\lbrace w_i\rbrace$ . Then the 's are compact and the 's are open in the discrete space . Clearly $g\in\lbrace h\mid h(v_i)=w_i\rbrace=B(K_i,U_i)$ for each $i=1,\ldots,n$ . So lies in the neighborhood $B=\cap B(K_i,U_i)\subseteq \operatorname{End}_D(V)$ . Since is dense in $\operatorname{End}_D(V)$ , there is an $f\in R\cap B$ . This implies that for all . $\qedsymbol$