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Complementary subspace

by , PhD

Two subspaces of a vector space are said to be complementary if their direct sum gives the entire vector space as a result.

Table of Contents

Preliminaries

Let $S$ be a linear space and $S_{1}$ and $S_{2}$ two subspaces of $S$.

Remember that the sum $S_{1}+S_{2}$ is the subspace that contains all the vectors that can be written as sums of a vector from $S_{1}$ and another vector from $S_{2}$:[eq1]

Moreover, $S_{1}+S_{2}$ is said to be a direct sum, and it is denoted by $S_{1}oplus S_{2}$, if and only if one of the following equivalent conditions is met:

Definition

We are now ready to provide a definition of complementary subspace.

Definition Let $S$ be a linear space. Let $S_{1}$ and $S_{2}$ be two subspaces of $S$. $S_{1}$ is said to be complementary to $S_{2}$ if and only if[eq4]

Complementarity, as defined above, is clearly symmetric. If $S_{1}$ is complementary to $S_{2}$, then $S_{2}$ is complementary to $S_{1}$ and we can simply say that $S_{1}$ and $S_{2}$ are complementary.

Thus, when $S_{1}$ and $S_{2}$ are complements, there is a unique way to "decompose" a vector $sin S$ into a component $s_{1}in S$ and another component $s_{2}in S_{2}$, where the decomposition is[eq5]

Let us see a simple example.

Example Let $S$ be the space of all $2 imes 1$ column vectors. Let $S_{1}$ be the space spanned by the vector[eq6]that is, $S_{1}$ contains all scalar multiples of $e_{1}$. Let $S_{2}$ be the space spanned by the vector[eq7]No non-zero vector of $S_{2}$ is a scalar multiple of a vector of $S_{1}$. Therefore, [eq8] and the sum $S_{1}+S_{2}$ is a direct sum. Moreover, any vector[eq9]can be written as[eq10]where [eq11] and [eq12]. Thus, [eq13]which means that $S_{1}$ and $S_{2}$ are complementary.

The complementary subspace is not necessarily unique

A complementary subspace is not necessarily unique.

Example Let $S$, $S_{1}$ and $S_{2}$ be as in the previous example. Let $S_{3}$ be the space spanned by the vector[eq14]No non-zero vector of $S_{3}$ is a scalar multiple of a vector of $S_{1}$. Therefore, [eq15] and the sum $S_{1}+S_{3}$ is a direct sum. Moreover, any vector[eq16]can be written as[eq17]where [eq18] and [eq19]. Thus, [eq20]and $S_{2} eq S_{3}$, which shows that there is no unique complementary subspace of $S_{1}$.

Characterization in terms of bases

The following proposition characterizes complementary subspaces in terms of their bases.

Proposition Let $S$ be a finite-dimensional linear space. Let $S_{1}$ and $S_{2}$ be two subspaces of $S$. Then, $S_{1}$ and $S_{2}$ are complementary if and only if, for any basis [eq21] of $S_{1}$ and any basis [eq22] of $S_{2}$, their union $Bcup C$ is a basis for $S$.

Proof

Let us prove the "only if" part, starting from the hypothesis that $S_{1}$ and $S_{2}$ are complementary. Any vector $sin S$ can be written uniquely as[eq23]where $s_{1}in S_{1}$ and $s_{2}in S_{2}$. Moreover, by the uniqueness of the representation in terms of a basis, [eq24]where the scalars [eq25] are unique. Similarly,[eq26]Thus $s$ has a unique representation in terms of $Bcup C$:[eq27]We have established that $Bcup C$ spans $S$. We still need to prove that it is linearly independent in order to prove that it is a basis. Suppose that[eq28]Since [eq29], it must be that[eq30]and[eq31]Since $B$ and $C$, being bases, are linearly independent sets, the two equations above imply [eq32] and[eq33]Thus, the only linear combination of the vectors of the set $Bcup C$ giving the zero vector as a result has all coefficients equal to zero. This means that $Bcup C$ is linearly independent. Hence, $Bcup C$ is a basis for $S$. We can now prove the "if" part, starting from the hypothesis that, for any bases [eq34] of $S_{1}$ and [eq35] of $S_{2}$, the union $Bcup C$ is a basis for $S$. Choose the two bases arbitrarily (they are guaranteed to exist because we have assumed that $S$ is finite-dimensional). The fact that $Bcup C$ is a basis for $S$ implies that any vector $sin S$ can be uniquely written as a linear combination [eq36]Since $B$ is a basis for $S_{1}$, we have that[eq37]and[eq38]Therefore, any vector $sin S$ can be uniquely represented as[eq39]where $s_{1}in S_{1}$ and $s_{2}in S_{2}$. Thus, [eq29].

How to cite

Please cite as:

Taboga, Marco (2021). "Complementary subspace", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/complementary-subspace.

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