The range null-space decomposition is the representation of a vector space as the direct sum of the range and the null space of a certain power of a given matrix.
Let us revise the concepts of range and null space of a matrix, which are discussed in detail in the lecture on the four fundamental subspaces.
Suppose that is a matrix and is the space of all column vectors.
The matrix defines a linear map such that for any .
The range (or column space) of is the subspacethat is, the set of all values taken by the map as its argument varies over the domain .
The null space of is the subspaceformed by all the elements of that are mapped into the zero vector.
Let us consider the powers . In the lecture on matrix powers we have proved that and its dimension decrease as we increase the power , while and its dimension simultaneously increase. However, at a certain point the two subspaces stabilize: there exists an integer such thatfor any integer .
The smallest non-negative integer for which the above equalities hold is often called the index of , although sometimes the term has different meanings in the linear algebra literature.
The range null-space decomposition, which we are going to prove and discuss below, asserts thatwhere is the index of and the symbol denotes a direct sum.
Remember that the sum of the two subspacesis direct (and we use the instead of the symbol to denote it) if and only ifin which case the representation of any vector as a sum of a vector and a vector is unique.
Furthermore, when the direct sum is equal to the whole space , as in the case of the range null-space decomposition, we say that the two spaces and are complementary.
The first result we are going to prove is that the intersection of and contains only the zero vector.
Proposition Let be a matrix. Let be any non-negative integer such that and . Then,
Let be the space of all column vectors. Let . Since we haveMoreover, since , there exists such that We can pre-multiply both sides of the equation by , so as to getThus, . But because of the stabilization of null spaces. As a consequence, , which implies thatSince was chosen arbitrarily, we can conclude that the intersection contains only the zero vector.
After having revised all the concepts involved in the range null-space decomposition, we are now ready to state it as a proposition.
Proposition Let be the space of all column vectors. Let be a matrix. Let be any non-negative integer such that and . Then,
We are going to use the following result, which we proved in the lecture on complementary subspaces: and are complementary if and only if is a basis for whenever is a basis for and is a basis for . Arbitrarily choose two such bases and . Suppose that is a set of linearly dependent vectors. Then there are scalars , not all equal to zero, such thatorwhere the last inequality descends from the fact that and are sets of linearly independent vectors and the scalars are not all zero. ButandThis is impossible because, as demonstrated above, contains only the zero vector. Thus, we have proved by contradiction that is a set of linearly independent vectors. Moreover, by the rank-nullity theorem, . Thus, is a set of linearly independent vectors. In other words, it is a basis for . Since and were chosen arbitrarily among the bases of and respectively, it descends that and are complementary subspaces, that is,
When the matrix is full-rank, the range null-space decomposition is trivial: since the product of two full-rank matrices is full-rank, is full-rank for any non-negative integer . In other words, the columns of are always linearly independent, and, as a consequence, they span all of . Thus,for any , which makes the decomposition trivial.
Note that (where is the identity matrix) and . As a consequence, the index of a full-rank matrix is .
Everything that we prove for square matrices can be applied to linear operators on finite-dimensional spaces. In fact, in the finite-dimensional case, each square matrix defines an operator and each operator is associated to a square matrix.
Remember that raising a matrix to the -th integer power is the same as composing its associated operator times with itself.
Thus, we can state the range null-space decomposition for linear operators as follows.
Proposition Let be a finite-dimensional vector space. Let be a linear operator. Then, there exists a non-negative integer such that
Below you can find some exercises with explained solutions.
Find the index of and determine the range null-space decomposition of .
We haveThe second power of isfrom which we can clearly see thatThus, the index of is and
Please cite as:
Taboga, Marco (2017). "Range null-space decomposition", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/range-null-space-decomposition.
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