2.2 Direct sums

  • Definition. Let \(\lst {V}1k\leq V\). The sum \(\plst {V}1k\) is direct if each \(v\in \plst {V}1k\) can be written

    \begin{equation*} v=\plst {v}1k \end{equation*}

    in only one way, that is, for unique \(v_i\in V_i\), \(\bw 1ik\).

    In this case, we write \(\oplst {V}1k\) instead of \(\plst {V}1k\).

  • Proposition 2.2. Let \(\lst {V}1k\leq V\). Then \(\plst {V}1k\) is direct if and only if whenever \(\plst {v}1k=0\), with \(v_i\in V_i\), \(\bw 1ik\), then \(v_i=0\), for all \(\bw 1ik\).

  • Proposition 2.3. Let \(V_1,V_2\leq V\). Then \(V_1+V_2\) is direct if and only if \(V_1\cap V_2=\set 0\).

  • Definition. Let \(V_1,V_2\leq V\). \(V\) is the (internal) direct sum of \(V_1\) and \(V_2\) if \(V=V_1\oplus V_2\).

    In this case, say that \(V_2\) is a complement of \(V_1\) (and \(V_1\) is a complement of \(V_2\)).

  • Proposition 2.4. Let \(\lst {V}1k\leq V\), \(k\geq 2\). Then the sum \(\plst {V}1k\) is direct if and only if, for each \(1\leq i\leq k\), \(V_i\cap (\sum _{j\neq i}V_j)=\set {0}\).

2.2.1 Induction from two summands

  • Lemma 2.5. Let \(\lst {V}1k\leq V\). Then \(\plst {V}1k\) is direct if and only if \(\plst {V}1{k-1}\) is direct and \((\plst {V}1{k-1})+V_k\) (two summands) is direct.

2.2.2 Direct sums, bases and dimension

  • Proposition 2.6. Let \(V_1,V_2\leq V\) be subspaces with bases \(\cB _1\colon \lst {v}1k\) and \(\cB _2\colon \lst {w}1l\). Then \(V_1+V_2\) is direct if and only if the concatenation1 \(\cB _1\cB _2\colon \lst {v}1k,\lst {w}1l\) is a basis of \(V_1+V_2\).

1 The concatenation of two lists is simply the list obtained by adjoining all entries in the second list to the first.

  • Corollary 2.7. Let \(\lst {V}1k\leq V\) be finite-dimensional subspaces with \(\cB _i\) a basis of \(V_i\), \(\bw 1ik\). Then \(\plst {V}1k\) is direct if and only if the concatenation \(\cB _1\ldots \cB _k\) is a basis for \(\plst {V}1k\).

  • Corollary 2.8. Let \(\lst {V}1k\leq V\) be subspaces of a finite-dimensional vector space \(V\) with \(\plst {V}1k\) direct. Then

    \begin{equation*} \dim \oplst {V}1k=\plst {\dim V}1k. \end{equation*}

2.2.3 Complements

  • Proposition 2.9 (Complements exist). Let \(U\leq V\), a finite-dimensional vector space. Then there is a complement to \(U\).