The direct sum of two subspaces and of a vector space is another subspace whose elements can be written uniquely as sums of one vector of and one vector of .

Let us start by defining sums of subspaces.

Definition Let be a linear space and and two subspaces of . The sum of and is the set

In other words, the sum of two subspaces is the set that contains all the vectors that can be written as sums of two vectors, one picked from and the other taken from .

Example Let be the space of all column vectors. Let be the subspace of all vectors that can be written as scalar multiples of the vectorthat is, contains all vectors of the formwhere can be any scalar. Similarly, define as the subspace that contains all vectors of the form where can be any scalar. Then, the sum is the set that contains all vectors of the formwhere the scalars and can be chosen arbitrarily.

Remember that a subspace is a subset of a vector space such that any linear combination of vectors of the subspace belongs to the subspace itself.

Proposition Let be a linear space and and two subspaces of . Then, the sum is a subspace of .

Proof

Arbitrarily choose two vectors . By the definition of sum of two subspaces, there exist vectors and such thatandNow, consider a linear combination of and with scalar coefficients and :Since and are subspaces, we have thatandTherefore, the linear combination can be written as a sum of a vector of and a vector of . Thus,We have proved that any linear combination of elements of belongs to itself. In other words, is a linear subspace, which is what we needed to prove.

Sums of more than two subspaces are defined analogously to sums of two subspaces.

Definition Let be a linear space. Let be subspaces of . The sum of is the set

Also in the case of more than two subspaces, their sum is a subspace.

The direct sum is a special kind of sum.

Definition Let be a linear space. Let be subspaces of . The sum is called direct sum and is denoted byif and only if are linearly independent whenever and for .

In other words, in a direct sum, non-zero vectors taken from the different subspaces being summed must be linearly independent.

Example The sum defined in the previous example was a direct sum because the two vectorsare linearly independent if and .

The usual summation symbol can be used to denote the sum of subspaces :

Interestingly, there is a symbol also for direct sums:

The next proposition shows a necessary and sufficient condition for a sum to be direct. It could be used as an alternative definition of direct sum.

Proposition Let be a linear space. Let be subspaces of . The sum is a direct sum if and only if the unique way to obtainis to set , ..., .

Proof

Let us prove the "only if" part, starting from the hypothesis that is a direct sum. By contradiction, suppose there exist vectors for such thatand at least one of the vectors is different from zero. We can assume without loss of generality that the first vectors are different from zero (otherwise we can re-number them). Then, we have thatThus, there is a linear combination (with coefficients all equal to ) of that gives the zero vector as a result. This implies that are linearly dependent. Arbitrarily choose non-zero vectors for . A fortiori, the set is a set of linearly dependent non-zero vectors each chosen from a different subspace . This is not possible because in a direct sum such sets cannot exist (by the definition of direct sum). Thus, we have arrived at a contradiction of the assumption that and at least one of the vectors is different from zero. Therefore, the unique way to obtain the zero vector as a result of the sum is to pick , ..., . Let us prove the "if" part, starting from the hypothesis that the unique way to obtainwith is to pick , ..., . Take non-zero vectors for . By contradiction, suppose they are linearly dependent. Then, there exist scalars not all equal to zero such thatSince are subspaces, for all . Moreover, some of the vectors are non-zero. This contradicts the initial hypothesis (the unique way to obtain the zero vector as the result of a sum of vectors coming from different subspaces is to pick all of them equal to zero). Thus, must be linearly independent. As a consequence is a direct sum.

Here is a necessary condition that is often easy to verify in applications.

Proposition Let be a linear space. Let be subspaces of . If is a direct sum, then for .

Proof

Assume that a vector belongs to the intersection for . Then, , and (because is a subspace). Moreover, we have thatwhich contradicts the hypothesis that is a direct sum (because the only way to obtain zero is to pick all zeros from the subspaces involved in the direct sum). Hence, we conclude that only the zero vector can belong to for .

If there are only two subspaces involved in the sum, the necessary condition becomes also sufficient.

Proposition Let be a linear space. Let and be subspaces of . The sum is a direct sum if and only if

Proof

The "only if" part has been already proved in the more general proposition above. Let us prove the "if" part, starting from the hypothesis that . Suppose that , and Then,which implies that . Therefore, Thus, the only way to obtain the zero vector as a result of the sum in equation (1) is to pick the zero vector as a summand from both subspaces. Hence, is a direct sum.

Here is an example.

Example Let be the space of all column vectors having real entries. Let be the subspace containing all vectors of the formwhere and can be any real numbers. Let be the subspace of all vectorswhere , and can be any real numbers satisfying . Is a direct sum? No because, for example, the vectorbelongs both to and to . As a consequence, the intersection contains other vectors besides the zero vector, which implies that is not a direct sum.

The most important fact about direct sums is that vectors can be represented uniquely as sums of elements taken from the subspaces.

Proposition Let be a linear space. Let be subspaces of . The sum is a direct sum if and only if, for any vector belonging to the sum, there exists a unique set of vectors such that

Proof

Let us prove the "only if" part, starting from the hypothesis that the sum is direct. By contradiction, suppose that there exists a different set of vectors such that Then,and at least one of the summands is not zero, which is impossible because the sum is direct. Thus, there does not exist another representation of the vector . Let us prove the "if" part, starting from the hypothesis that each vector has a unique representation of the kind shown in the proposition. SupposeBut we also haveSince the representation of the sum is unique, then . Thus, the only way to obtain zero is as a sum of zero vectors. Hence, the sum is direct.

Also this proposition could be taken as a definition of direct sum, as done, for example by Axler (1997).

Below you can find some exercises with explained solutions.

Let be the space of all column vectors having real entries. Let be the subspace containing all vectors of the formwhere , and can be any real numbers satisfying . Let be the subspace of all vectorswhere , and can be any real numbers satisfying .

Is a direct sum?

Solution

Let us find whether the intersection of and contains other vectors besides the zero vector. A vector belonging to the intersection should contemporaneously satisfy the two equationsIf we add the first equation to the second, we obtainso that the system becomesThe two equations are the same (multiply the second by ). Then, the only other equation to be satisfied isThe system formed by equations (1) and (2) is solved, for example, by , , . Thus, the vectorbelongs to the intersection and is not a direct sum.

Axler, S. (1997) Linear algebra done right, Second Edition, Springer Science & Business Media.

Please cite as:

Taboga, Marco (2017). "Direct sum", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/direct-sum.

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