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Chapter 2 Sums and quotients
We will discuss various ways of building new vector spaces out of old ones.
2.1 Sums of subspaces
\(\plst {V}1k\) is the smallest subspace of \(V\) that contains each \(V_i\). More precisely:
-
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
\(\seteqnumber{0}{2.}{0}\)\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 \(v_i+\lambda w_i\in V_i\). □