Two subspaces of a vector space are said to be complementary if their direct sum gives the entire vector space as a result.
Let be a linear space and and two subspaces of .
Remember that the sum is the subspace that contains all the vectors that can be written as sums of a vector from and another vector from :
Moreover, is said to be a direct sum, and it is denoted by , if and only if one of the following equivalent conditions is met:
;
two non-zero vectors and are linearly independent whenever and ;
the only way to choose and such that is to choose ;
for any , there exists a unique couple and such that .
We are now ready to provide a definition of complementary subspace.
Definition Let be a linear space. Let and be two subspaces of . is said to be complementary to if and only if
Complementarity, as defined above, is clearly symmetric. If is complementary to , then is complementary to and we can simply say that and are complementary.
Thus, when and are complements, there is a unique way to "decompose" a vector into a component and another component , where the decomposition is
Let us see a simple example.
Example Let be the space of all column vectors. Let be the space spanned by the vectorthat is, contains all scalar multiples of . Let be the space spanned by the vectorNo non-zero vector of is a scalar multiple of a vector of . Therefore, and the sum is a direct sum. Moreover, any vectorcan be written aswhere and . Thus, which means that and are complementary.
A complementary subspace is not necessarily unique.
Example Let , and be as in the previous example. Let be the space spanned by the vectorNo non-zero vector of is a scalar multiple of a vector of . Therefore, and the sum is a direct sum. Moreover, any vectorcan be written aswhere and . Thus, and , which shows that there is no unique complementary subspace of .
The following proposition characterizes complementary subspaces in terms of their bases.
Proposition Let be a finite-dimensional linear space. Let and be two subspaces of . Then, and are complementary if and only if, for any basis of and any basis of , their union is a basis for .
Let us prove the "only if" part, starting from the hypothesis that and are complementary. Any vector can be written uniquely aswhere and . Moreover, by the uniqueness of the representation in terms of a basis, where the scalars are unique. Similarly,Thus has a unique representation in terms of :We have established that spans . We still need to prove that it is linearly independent in order to prove that it is a basis. Suppose thatSince , it must be thatandSince and , being bases, are linearly independent sets, the two equations above imply andThus, the only linear combination of the vectors of the set giving the zero vector as a result has all coefficients equal to zero. This means that is linearly independent. Hence, is a basis for . We can now prove the "if" part, starting from the hypothesis that, for any bases of and of , the union is a basis for . Choose the two bases arbitrarily (they are guaranteed to exist because we have assumed that is finite-dimensional). The fact that is a basis for implies that any vector can be uniquely written as a linear combination Since is a basis for , we have thatandTherefore, any vector can be uniquely represented aswhere and . Thus, .
Please cite as:
Taboga, Marco (2021). "Complementary subspace", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/complementary-subspace.
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