2.2 Direct sums

Let \(\lst {V}1k\leq V\). Any \(v\in \plst {V}1k\) can be written

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

with each \(v_i\in V_i\). We distinguish the case where the \(v_i\) are unique.

  • 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 \(\plst {V}1k\).

(A direct sum.)

Figure 2.1: \(\R ^2=V_1\oplus V_2\)

  • Example. Define \(V_1,V_2\leq \F ^3\) by

    \begin{align*} V_1&=\set {(x_1,x_2,0)\st x_1,x_2\in \F }\\ V_2&=\set {(0,0,x_3)\st x_3\in \F }. \end{align*} Then \(\F ^3=V_1\oplus V_2\).

When is a sum direct? We consider the case of two summands first where there is a very simple answer.

  • Proposition 2.2. Let \(V_1,V_2\leq V\). Then \(V_1+V_2\) is direct if and only if \(V_1\cap V_2=\set 0\).

  • Proof. First suppose that \(V_1+V_2\) is direct and let \(v\in V_1\cap V_2\). Then we can write \(v\) in two ways:

    \begin{align*} v&=v_1+0\\&=0+v_2, \end{align*} with \(v=v_1=v_2\). The uniqueness of the decomposition now forces \(v=0\).

    For the converse, suppose that \(V_1\cap V_2=\set 0\) and that \(v\in V_1+V_2\) can be written

    \begin{equation*} v=v_1+v_2=w_1+w_2 \end{equation*}

    with \(v_i,w_i\in V_i\), \(i=1,2\). Then

    \begin{equation*} (v_1-w_1)=(w_2-v_2) \end{equation*}

    with the left hand in \(V_1\), the right in \(V_2\) and so both in \(V_1\cap V_2\) from which we immediately get \(v_i=w_i\), \(i=1,2\) so that \(V_1+V_2\) is direct.  □

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 Proposition 2.2, 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.

(A direct sum of a line and a plane.)

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:

  • 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 \(1\leq i\leq k\), \(V_i\cap (\sum _{j\neq i}V_j)=\set {0}\).

  • Proof. This is an exercise in imitating the proof of Proposition 2.2.  □

  • 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 Direct sums and projections

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

  • Exercise.1 If \(\pi \) is a projection, \(V=\ker \pi \oplus \im \pi \).

1 Question 3 on sheet 2.

In fact, all direct sums arise this way:

  • 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, \(\id _V=\pi _1+\pi _2\).

  • Proof. Item (c) tells us how to define the \(\pi _i\) so we do this first: for \(v\in V\), we have that \(v=v_1+v_2\) for unique \(v_i\in V_i\), \(i=1,2\). So we define \(\pi _i(v)\) to be

    \begin{equation*} \pi _i(v):=v_i, \end{equation*}

    for \(i=1,2\).

    Our first task is to prove that the \(\pi _i\) are linear: for \(v,w\in V\) and \(\lambda \in \F \), we have

    \begin{align*} v&=\pi _1(v)+ \pi _2(v)\\ w&=\pi _1(w)+ \pi _2(w)\\ v+\lambda w&=\pi _1(v+\lambda w)+ \pi _2(v+\lambda w). \end{align*} However, the first two equalities also give

    \begin{equation*} v+\lambda w=(\pi _1(v)+\lambda \pi _1(w))+(\pi _2(v)+\lambda \pi _2(w)) \end{equation*}

    so the uniqueness in the third equality gives

    \begin{equation*} \pi _i(v+\lambda w)=\pi _i(v)+\lambda \pi _i(w), \end{equation*}

    \(i=1,2\), so that both \(\pi _i\) are linear.

    By definition, \(\im \pi _i\leq V_i\). For the converse, note that, for \(v_1\in V_{1}\), we have \(v_1=v_1+0\), with \(0\in V_2\), so that \(\pi _{1}(v_1)=v_1\). In particular, \(V_1\leq \im \pi _1\) so that \(\im \pi _1=V_1\). Moreover, taking \(v_1=\pi _1(v)\), we get \(\pi _1(\pi _1(v))=\pi _1(v)\), for any \(v\in V\) so that \(\pi _1\) is a projection. Similarly \(\pi _2\) is a projection and (a) holds.

    Finally, \(v=\pi _{1}(v)+\pi _2(v)\in \ker \pi _1\) if and only if \(v=\pi _2(v)\), or, as we have just seen, \(v\in V_2\). Thus \(\ker \pi _1=V_2\) and similarly \(\ker \pi _2=V_1\) settling (b).  □

As a corollary, we see that dimensions add in direct sums:

  • Proposition 2.5. Let \(V=V_1\oplus V_2\) with \(V\) finite-dimensional. Then

    \begin{equation*} \dim V=\dim V_1+\dim V_2. \end{equation*}

  • Proof. We apply the rank-nullity theorem to \(\pi _1\):

    \begin{align*} \dim V&=\dim \im \pi _1+\dim \ker \pi _{1}\\ &=\dim V_1+\dim V_2. \end{align*}  □

2.2.2 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:

  • 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.

  • Proof. Suppose first that \(\plst {V}1k\) is direct. Then any \(v\in \plst {V}1{k-1}\) can be written

    \begin{equation*} v=\plst {v}1{k-1}+0 \end{equation*}

    for unique \(v_i\in V_i\), \(\bw 1i{k-1}\) so that \(\plst {V}1{k-1}\) is direct. Moreover, any \(v\in (\plst {V}1{k-1})+V_k=\plst {V}1k\) can be written

    \begin{equation*} v=\plst {v}1k=(\plst {v}1{k-1})+v_k \end{equation*}

    with, in particular, unique \(v_k\in V_k\) so that \((\plst {V}1{k-1})+V_k\) is direct.

    For the converse, suppose that \(\plst {V}1{k-1}\) and \((\plst {V}1{k-1})+V_k\) are both direct. Then any \(v\in \plst {V}1k\) can be written \(v=w+v_k\) for unique \(w\in \plst {V}1{k-1}\) and \(v_k\in V_k\). Also, there is a unique way to write \(w\) as

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

    with \(v_i\in V_i\), \(\bw 1i{k-1}\). Putting this together, we get

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

    for unique \(v_i\in V_i\), \(\bw 1ik\) so that \(\plst {V}1k\) is direct.  □

Here is a sample application:

  • Corollary 2.7. 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*}

  • Proof. We induct on \(k\) using Proposition 2.5 and Lemma 2.6 in the induction step. In more detail: the induction hypothesis is that the formula holds for \(k\) summands. The base case reads

    \begin{equation*} \dim V_1=\dim V_1 \end{equation*}

    which trivially holds. For the induction step, suppose that the formula holds for any \(k-1\) summands. Then, if \(\plst {V}1k\) is direct, Lemma 2.6 says that \(\plst {V}1{k-1}\) is direct and then the induction hypothesis says that \(\dim \oplst {V}1{k-1}=\plst {\dim V}1{k-1}\). Now Lemma 2.6 says that

    \begin{equation*} \oplst {V}1k=(\oplst {V}1{k-1})\oplus V_k \end{equation*}

    so that Proposition 2.5 applies to give

    \begin{equation*} \dim \oplst {V}1k=(\plst {\dim V}1{k-1})+\dim V_k. \end{equation*}

     □

2.2.3 Direct sums and bases

Proposition 2.5 suggests that there is a relation between the bases of \(V_1,V_2\) and the basis of \(V_1\oplus V_2\). This is indeed the case:

  • 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 concatenation2 \(\cB _1\cB _2:\lst {v}1k,\lst {w}1l\) is a basis of \(V_1+V_2\).

2 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

    \begin{equation*} \sum _{i=1}^k\lambda _{i}v_i=-\sum _{j=1}^l\mu _jw_j\in V_1\cap V_2 \end{equation*}

    which last is the zero subspace by Proposition 2.2. Thus

    \begin{equation*} \sum _{i=1}^k\lambda _{i}v_i=\sum _{j=1}^l\mu _jw_j=0 \end{equation*}

    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 \(v\in V_1\cap V_2\), we can write \(v\) in two ways: \(v=\sum _{i=1}^k\lambda _iv_i\), for some \(\lst \lambda 1k\in \F \), since \(v\in V_1\) and, similarly, \(v=\sum _{j=1}^l\mu _jw_j\). We therefore have a linear relation \(\sum _{i=1}^k\lambda _{i}v_i-\sum _{j=1}^l\mu _jw_j=0\) and so, by linear independence of \(\cB _1\cB _2\), all \(\lambda _i,\mu _j\) vanish so that \(v=0\). Thus \(V_1\cap V_2=\set 0\) and \(V_1+V_2\) is direct by Proposition 2.2.  □

Again, this along with Lemma 2.6 and induction on \(k\) yields the many-summand version:

  • 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

For finite-dimensional vector spaces, any subspace has a complement:

  • Proposition 2.10 (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 Proposition 1.3, 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 Proposition 2.8.  □

In fact, as Figure 2.3 illustrates, there are many complements to a given subspace.

(A subspace has many complements.)

Figure 2.3: Each dashed line is a complement to the undashed subspace.

Here is an application:

  • 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 restriction3 of \(\Phi \) to \(U\) is \(\phi \): \(\Phi \restr {U}=\phi \). Otherwise said: for all \(u\in U\)

    \begin{equation*} \Phi (u)=\phi (u). \end{equation*}

3 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 \(a\in A\).

  • Proof. By Proposition 2.10, \(U\) has a complement and so, by Proposition 2.4, there is a projection \(\pi :V\to V\) with image \(U\).

    Set \(\Phi =\phi \circ \pi :V\to W\). This is a linear map and

    \begin{equation*} \Phi \restr {U}=\phi \circ \pi \restr {U}=\phi \end{equation*}

    since, for \(u=\pi (v)\in \im \pi =U\), \(\pi (u)=\pi (\pi (v))=\pi (v)=u\).  □