MA20216: Algebra 2A

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
\(\seteqnumber{0}{2.}{0}\)
\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 \(V_1,V_2\leq V\). 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 2.3. 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 Direct sums and projections

Definition. Let \(V\) be a vector space. A linear map \(\pi :V\to V\) is a projection if \(\pi \circ \pi =\pi \).

Proposition 2.4. Let \(V_1,V_2\leq V\) with \(V=V_1\oplus V_2\). Then there are projections \(\pi _1,\pi _2:V\to V\) such that:
- (a) \(\im \pi _i=V_i\), \(i=1,2\);
- (b) \(\ker \pi _1=V_2\), \(\ker \pi _2=V_1\);
- (c) \(v=\pi _1(v)+\pi _2(v)\), for all \(v\in V\). Otherwise said, \(\id _V=\pi _1+\pi _2\).

Proposition 2.5. Let \(V=V_1\oplus V_2\) with \(V\) finite-dimensional. Then
\(\seteqnumber{0}{2.}{0}\)
\begin{equation*} \dim V=\dim V_1+\dim V_2. \end{equation*}

2.2.2 Induction from two summands

Lemma 2.6. 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.

Corollary 2.7. Let \(\lst {V}1k\leq V\) be subspaces of a finite-dimensional vector space \(V\) with \(\plst {V}1k\) direct. Then
\(\seteqnumber{0}{2.}{0}\)
\begin{equation*} \dim \oplst {V}1k=\plst {\dim V}1k. \end{equation*}

2.2.3 Direct sums and bases

Proposition 2.8. Let \(V_1,V_2\leq V\) be finite-dimensional subspaces with bases \(\cB _1:\lst {v}1k\) and \(\cB _2:\lst {w}1l\). Then \(V_1+V_2\) is direct if and only if the concatenation¹ \(\cB _1\cB _2:\lst {v}1k,\lst {w}1l\) 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 2.9. 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 \(\plst {V}1k\).

2.2.4 Complements

Proposition 2.10 (Complements exist). Let \(U\leq V\), a finite-dimensional vector space. Then there is a complement to \(U\).

Proposition 2.11 (Extension of linear maps). Let \(V,W\) be vector spaces with \(V\) finite-dimensional. Let \(U\leq V\) be a subspace and \(\phi :U\to W\) a linear map. Then there is a linear map \(\Phi :V\to W\) such that the restriction² of \(\Phi \) to \(U\) is \(\phi \): \(\Phi \restr {U}=\phi \). Otherwise said: for all \(u\in U\)
\(\seteqnumber{0}{2.}{0}\)
\begin{equation*} \Phi (u)=\phi (u). \end{equation*}

² Recall that if \(f:X\to Y\) is a map of sets and \(A\sub X\) then the restriction of \(f\) to \(A\) is the map \(f\restr {A}:A\to Y\) given by \(f\restr {A}(a)=f(a)\), for all \(a\in A\).