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.
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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