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}[1]{_{|#1}}\)

Chapter 2 Sums and quotients

We will discuss various ways of building new vector spaces out of old ones.

  • Convention. In this chapter, all vector spaces are over the same field \(\F \) unless we say otherwise.

2.1 Sums of subspaces

  • Definition. Let \(\lst {V}1k\leq V\). The sum \(\plst {V}1k\) is the set

    \begin{equation*} \plst {V}1k:=\set {\plst {v}1k\st v_i\in V_i, 1\leq i\leq k}. \end{equation*}

\(\plst {V}1k\) is the smallest subspace of \(V\) that contains each \(V_i\). More precisely:

  • Proposition 2.1. Let \(\lst {V}1k\leq V\). Then

    • (1) \(\plst {V}1k\leq V\).

    • (2) If \(W\leq V\) and \(\lst {V}1k\leq W\) then \(\lst {V}1k\leq \plst {V}1k\leq W\).

  • Proof. It suffices to prove (2) since (1) then follows by taking \(W=V\).

    For (2), first note that \(\plst {V}1k\) is a subset of \(W\): if \(v_i\in V_i\) then \(v_i\in W\) so that \(\plst {v}1k\in W\) since \(W\) is closed under addition.

    Now observe that each \(V_i\leq \plst {V}1k\) since we can write any \(v_i\in V_i\) as \(0+\dots +v_i+\dots +0\in \plst {V}1k\). In particular, \(0\in \plst {V}1k\).

    Finally, we show that \(\plst {V}1k\) is a subspace. If \(\plst {v}1k,\plst {w}1k\in \plst {V}1k\), with \(v_i,w_i\in V_i\), for all \(i\), and \(\lambda \in \F \) then

    \begin{equation*} (\plst {v}1k)+\lambda (\plst {w}1k)=\plus {(v_1+\lambda w_1)}{(v_k+\lambda w_k)}\in \plst {V}1k \end{equation*}

    since each \(v_i+\lambda w_i\in V_i\). □

  • Remark. The union \(\bigcup _{i=1}^kV_{i}\) is almost never a subspace of \(V\) so we use sums as a substitute for unions in Linear Algebra.