When is a sum direct? We begin with a useful reformulation of the property.
Proposition2.2. 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 \(\bw 1ik\).
Proof. Suppose that \(\plst {V}1k\) is direct and let \(\plst {v}1k=0\), with each \(v_i\in V_i\). We can also write \(0=0+\dots +0\) so that the uniqueness in the direct sum property
forces each \(v_i=0\).
Conversely, if the “zero sum” property holds, suppose that, for some \(v\in \plst {V}1k\), we have
and each \(v_i-w_i\in V_i\) so the zero sum property gives \(v_i=w_i\). We conclude that the sum is direct. □
For the case of two summands this gives a very simple way to decide if a sum is direct:
Proposition2.3. Let \(V_1,V_2\leq V\). Then \(V_1+V_2\) is direct if and only if \(V_1\cap V_2=\set 0\).
Proof. Suppose first that \(V_1+V_2\) is direct and let \(v\in V_1\cap V_2\). Then
\(\seteqnumber{0}{2.}{0}\)
\begin{equation*}
0=v+(-v)
\end{equation*}
and \(v\in V_{1}\), \(-v\in V_2\) so that \(v=-v=0\) by Theorem 2.2.
Conversely, suppose that \(V_1\cap V_2=\set 0\) and that \(v_1+v_2=0\), with \(v_i\in V_i\), \(i=1,2\). Then \(v_1=-v_2\in V_1\cap V_2=\set 0\) so that \(v_1=v_2=0\). Thus \(V_1+V_2\) is direct by Theorem 2.2. □
The special case \(V=V_1+V_2\) is important and deserves some terminology:
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 \(V_2\)).
Warning. This notion of the complement of the subspace \(V_1\) has nothing at all to do with the set-theoretic complement \(V\setminus V_1\) which is never a subspace.
Remarks.
(1) From Theorem 2.3, we see that \(V=V_1\oplus V_2\) if and only if \(V=V_1+V_2\) and \(V_1\cap V_2=\set 0\). Many people take these latter properties as the definition of
internal direct sum.
(2) There is a related notion of external direct sum that we will not discuss.
Figure 2.2: \(\R ^3\) as a direct sum of a line and a plane
When there are many summands, the condition that a sum be direct is a little more involved:
Proposition2.4. Let \(\lst {V}1k\leq V\), \(k\geq 2\). Then the sum \(\plst {V}1k\) is direct if and only if, for each
\(1\leq i\leq k\), \(V_i\cap (\sum _{j\neq i}V_j)=\set {0}\).
Proof. This is an exercise in imitating the proof of Theorem 2.3. □
Remark. This is a much stronger condition than simply asking that each \(V_i\cap V_j=\set 0\), for \(i\neq j\).
2.2.1 Induction from two summands
A convenient way to analyse direct sums with many summands is to induct from the two summand case. For this, we need:
Lemma2.5. 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.
Proof. Suppose first that \(\plst {V}1k\) is direct. We use Theorem 2.2 to see that \(\plst {V}1{k-1}\) is direct: let \(\plst {v}1{k-1}=0\) with
each \(v_i\in V_i\), \(\bw 1i{k-1}\). Write this as \(\plst {v}1k=0\) where \(v_k=0\in V_k\) and deduce that each \(v_i=0\), \(\bw 1i{k-1}\).
Again, if \(v=\plst {v}1{k-1}\in \plst {V}1{k-1}\) and \(v_k\in V_k\) with \(v+v_k=0\), then we have \(\plst {v}1k=0\) so that each \(v_i=0\) whence \(v=0\) also. Now Theorem 2.2 tells us that \((\plst
{V}1{k-1})+V_k\) is direct.
Conversely, suppose that both \(\plst {V}1{k-1}\) and \((\plst {V}1{k-1})+V_k\) are direct and that \(\plst {v}1k=0\), with each \(v_i\in V_i\). Let \(v=\plst {v}1{k-1}\in \plst {V}1{k-1}\) so that \(v+v_k=0\). Now Theorem 2.2 and the directness of \((\plst {V}1{k-1})+V_k\) tell us that \(v=v_k=0\). Thus \(\plst {v}1{k-1}=0\) and a final application of Theorem 2.2 yields \(v_i=0\), \(\bw 1i{k-1}\) since \(\plst
{V}1{k-1}\) is direct. □
2.2.2 Direct sums, bases and dimension
When a sum is direct, bases of the summands fit together to give a basis of the sum:
Proposition2.6. 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.
Proof. Clearly \(\cB _1\cB _2\) spans \(V_1+V_2\) and so will be a basis exactly when it is linearly independent.
Suppose that \(V_1+V_2\) is direct and that we have a linear relation \(\sum _{i=1}^k\lambda _{i}v_i+\sum _{j=1}^l\mu _jw_j=0\). Then Theorem 2.2 yields
so that all the \(\lambda _i\) and \(\mu _j\) vanish since \(\cB _1\) and \(\cB _2\) are linearly independent. We conclude that \(\cB _1\cB _2\) is linearly independent and so a basis.
Conversely, if \(\cB _1\cB _2\) is a basis and \(v+w=0\) with \(v\in V_1\) and \(w\in V_2\), write \(v=\sum _{i=1}^k\lambda _{i}v_i\) and \(w=\sum _{j=1}^l\mu _jw_j\) to get a linear relation \(\sum _{i=1}^k\lambda _{i}v_i+\sum _{j=1}^l\mu _jw_j=0\). By
linear independence of \(\cB _1\cB _2\), all \(\lambda _i,\mu _j\) vanish so that \(v=w=0\). Thus \(V_1+V_2\) is direct by Theorem 2.2. □
Again, this along with Theorem 2.5 and induction on \(k\) yields the many-summand version:
Corollary2.7. 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 \(\plst {V}1k\).
Proof. Our induction hypothesis at \(k\) is that \(\plst {V}1k\) is direct if and only if \(\cB _1\ldots \cB _k\) is a basis for \(\plst {V}1k\). This is vacuous at \(k=1\) so let us suppose it
is true for \(k\) and examine the case \(k+1\).
First suppose that \(\plst {V}1{k+1}\) is direct so that \(\plst {V}1k\) and \((\plst {V}1k)+V_{k+1}\) are direct by Theorem 2.5. The induction hypothesis applies to both of these so that, first, \(\cB _1\ldots \cB _k\) is a
basis of \(\plst {V}1k\) and then \((\cB _1\ldots \cB _k)\cB _{k+1}=\cB _1\ldots \cB _{k+1}\) is a basis of \((\plst {V}1k)+V_{k+1}=\plst {V}1{k+1}\).
Conversely, if \(\cB _1\ldots \cB _{k+1}\) is a basis of \(\plst {V}1{k+1}\), \(\cB _1\ldots \cB _k\) is linearly independent and so a basis of \(\plst {V}1k\). By the induction hypothesis, we learn that \(\plst {V}1k\) is direct. Similarly, we see that \((\plst
{V}1k)+V_{k+1}\) is direct whence, by Theorem 2.5, \(\plst {V}1{k+1}\) is direct.
This establishes the induction hypothesis at \(k+1\) and so the result is proved. □
From this we see that dimensions add over direct sums:
Corollary2.8. Let \(\lst {V}1k\leq V\) be subspaces of a finite-dimensional vector space \(V\) with \(\plst {V}1k\)
direct. Then
Exercise.2 Prove the converse of Theorem 2.8: if \(\dim \plst {V}1k=\plst {\dim V}1k\), then the sum is direct.
2 Question 2 on sheet 2.
2.2.3 Complements
For finite-dimensional vector spaces, any subspace has a complement:
Proposition2.9 (Complements exist). Let \(U\leq V\), a finite-dimensional
vector space. Then there is a complement to \(U\).
Proof. Let \(\cB _1:\lst {v}1k\) be a basis for \(U\) and so a linearly independent list of vectors in \(V\). By Theorem 1.1, we can extend the list to
get a basis \(\cB :\lst {v}1n\) of \(V\). Set \(W=\Span {\lst {v}{k+1}n}\leq V\): this is a complement to \(U\).
Indeed, \(\cB _2:\lst {v}{k+1}n\) is a basis for \(W\) and \(\cB =\cB _1\cB _2\) so that \(V=U\oplus W\) by Theorem 2.6. □
In fact, as Figure 2.3 illustrates, there are many complements to a given subspace.
Figure 2.3: Each dashed line is a complement to the undashed subspace.
