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Direct sum

by , PhD

The direct sum of two subspaces $S_{1}$ and $S_{2}$ of a vector space is another subspace whose elements can be written uniquely as sums of one vector of $S_{1}$ and one vector of $S_{2}$.

Table of Contents

Sums of subspaces

Let us start by defining sums of subspaces.

Definition Let $S$ be a linear space and $S_{1}$ and $S_{2}$ two subspaces of $S$. The sum of $S_{1}$ and $S_{2}$ is the set[eq1]

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 $S_{1}$ and the other taken from $S_{2}$.

Example Let $S$ be the space of all $3	imes 1$ column vectors. Let $S_{1}$ be the subspace of all vectors that can be written as scalar multiples of the vector[eq2]that is, $S_{1}$ contains all vectors $s_{1}$ of the form[eq3]where $lpha $ can be any scalar. Similarly, define $S_{2}$ as the subspace that contains all vectors $s_{2}$ of the form [eq4]where $eta $ can be any scalar. Then, the sum $S_{1}+S_{2}$ is the set that contains all vectors $s$ of the form[eq5]where the scalars $lpha $ and $eta $ can be chosen arbitrarily.

Sums are subspaces

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 $S$ be a linear space and $S_{1}$ and $S_{2}$ two subspaces of $S$. Then, the sum $S_{1}+S_{2}$ is a subspace of $S$.

Proof

Arbitrarily choose two vectors $s,tin S_{1}+S_{2}$. By the definition of sum of two subspaces, there exist vectors [eq6] and [eq7] such that[eq8]and[eq9]Now, consider a linear combination of $s$ and $t$ with scalar coefficients $sigma $ and $	au $:[eq10]Since $S_{1}$ and $S_{2}$ are subspaces, we have that[eq11]and[eq12]Therefore, the linear combination $sigma s+	au t$ can be written as a sum of a vector of $S_{1}$ and a vector of $S_{2}$. Thus,[eq13]We have proved that any linear combination of elements of $S_{1}+S_{2}$ belongs to $S_{1}+S_{2}$ itself. In other words, $S_{1}+S_{2}$ is a linear subspace, which is what we needed to prove.

More than two summands

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

Definition Let $S$ be a linear space. Let [eq14] be subspaces of $S$. The sum of [eq15] is the set[eq16]

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

Direct sum of subspaces

The direct sum is a special kind of sum.

Definition Let $S$ be a linear space. Let [eq17] be subspaces of $S$. The sum [eq18] is called direct sum and is denoted by[eq19]if and only if [eq20] are linearly independent whenever $s_{j}in S_{j}$ and $s_{j}
eq 0$ for $j=1,ldots ,n$.

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

Example The sum $S_{1}+S_{2}$ defined in the previous example was a direct sum because the two vectors[eq21]are linearly independent if $lpha 
eq 0$ and $eta 
eq 0$.

Notation

The usual summation symbol can be used to denote the sum of n subspaces [eq22]:[eq23]

Interestingly, there is a symbol also for direct sums:[eq24]

Sums giving the zero vector as a result

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 $S$ be a linear space. Let [eq25] be subspaces of $S$. The sum [eq18] is a direct sum if and only if the unique way to obtain[eq27]is to set $s_{1}=0$, ..., $s_{n}=0$.

Proof

Let us prove the "only if" part, starting from the hypothesis that [eq28] is a direct sum. By contradiction, suppose there exist vectors $s_{j}in S_{j}$ for $j=1,ldots ,n$ such that[eq29]and at least one of the vectors is different from zero. We can assume without loss of generality that the first $m$ vectors are different from zero (otherwise we can re-number them). Then, we have that[eq30]Thus, there is a linear combination (with coefficients all equal to 1) of [eq31] that gives the zero vector as a result. This implies that [eq32] are linearly dependent. Arbitrarily choose non-zero vectors $t_{j}in S_{j}$ for $j=m+1,ldots ,n$. A fortiori, the set [eq33]is a set of linearly dependent non-zero vectors each chosen from a different subspace $S_{j}$. 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 [eq34]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 $s_{1}=0$, ..., $s_{n}=0$. Let us prove the "if" part, starting from the hypothesis that the unique way to obtain[eq35]with $s_{j}in S_{j}$ is to pick $s_{1}=0$, ..., $s_{n}=0$. Take n non-zero vectors $t_{j}in S_{j}$ for $j=1,ldots ,n$. By contradiction, suppose they are linearly dependent. Then, there exist scalars [eq36] not all equal to zero such that[eq37]Since [eq38] are subspaces, [eq39] for all $j$. Moreover, some of the vectors $lpha _{j}t_{j}$ 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, [eq40] must be linearly independent. As a consequence [eq41] is a direct sum.

Intersections contain only the zero vector

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

Proposition Let $S$ be a linear space. Let [eq25] be subspaces of $S$. If [eq41] is a direct sum, then [eq44]for $j
eq k$.

Proof

Assume that a vector $s
eq 0$ belongs to the intersection $S_{j}cap S_{k}$ for $j
eq k$. Then, $sin S_{j}$, $sin S_{k}$ and $-sin S_{k}$ (because $S_{k}$ is a subspace). Moreover, we have that[eq45]which contradicts the hypothesis that [eq41] 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 $S_{j}cap S_{k}$ for $j
eq k$.

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

Proposition Let $S$ be a linear space. Let $S_{1}$ and $S_{2}$ be subspaces of $S$. The sum $S_{1}+S_{2}$ is a direct sum if and only if [eq47]

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 [eq48]. Suppose that $s_{1}in S_{1}$, $s_{2}in S_{2}$ and [eq49]Then,[eq50]which implies that [eq51]. Therefore, [eq52]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, $S_{1}+S_{2}$ is a direct sum.

Here is an example.

Example Let $S$ be the space of all $3	imes 1$ column vectors having real entries. Let $S_{1}$ be the subspace containing all vectors of the form[eq53]where $lpha $ and $eta $ can be any real numbers. Let $S_{2}$ be the subspace of all vectors[eq54]where $lpha $, $eta $ and $gamma $ can be any real numbers satisfying [eq55]. Is $S_{1}+S_{2}$ a direct sum? No because, for example, the vector[eq56]belongs both to $S_{1}$ and to $S_{2}$. As a consequence, the intersection $S_{1}cap S_{2}$ contains other vectors besides the zero vector, which implies that $S_{1}+S_{2}$ is not a direct sum.

Uniqueness of representation

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

Proposition Let $S$ be a linear space. Let [eq25] be subspaces of $S$. The sum [eq18] is a direct sum if and only if, for any vector $s$ belonging to the sum, there exists a unique set of vectors [eq59] such that [eq60]

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 [eq61] such that [eq62]Then,[eq63]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 $s$. Let us prove the "if" part, starting from the hypothesis that each vector has a unique representation of the kind shown in the proposition. Suppose[eq64]But we also have[eq65]Since the representation of the sum is unique, then [eq66]. 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).

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let $S$ be the space of all $3	imes 1$ column vectors having real entries. Let $S_{1}$ be the subspace containing all vectors of the form[eq67]where $lpha $, $eta $ and $gamma $ can be any real numbers satisfying [eq68]. Let $S_{2}$ be the subspace of all vectors[eq69]where $lpha $, $eta $ and $gamma $ can be any real numbers satisfying [eq70].

Is $S_{1}+S_{2}$ a direct sum?

Solution

Let us find whether the intersection of $S_{1}$ and $S_{2}$ contains other vectors besides the zero vector. A vector belonging to the intersection should contemporaneously satisfy the two equations[eq71]If we add the first equation to the second, we obtain[eq72]so that the system becomes[eq73]The two equations are the same (multiply the second by $-1$). Then, the only other equation to be satisfied is[eq74]The system formed by equations (1) and (2) is solved, for example, by $lpha =0$, $eta =1$, $gamma =-1$. Thus, the vector[eq75]belongs to the intersection and $S_{1}+S_{2}$ is not a direct sum.

References

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

How to cite

Please cite as:

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

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