The primary decomposition theorem allows us to use the minimal polynomial of a matrix to write a vector space as a direct sum of invariant subspaces. It is key to understanding and interpreting the information provided by the minimal polynomial.
Let be the space of vectors. Let be a matrix.
Remember that a subspace is said to be invariant under the linear transformation defined by if and only if for any . In other words, is invariant if its elements are transformed by into other elements of .
The minimal polynomial of is the monic annihilating polynomial (i.e., and the leading coefficient of is ) that divides every annihilating polynomial of .
The minimal polynomial is the lowest-degree polynomial in the set of annihilating polynomials and it has the property thatwhere are the distinct eigenvalues of and for any , where is the algebraic multiplicity of .
We can write the minimal polynomial in terms of aswhere is the identity matrix.
In this lecture we are going to prove the Primary Decomposition Theorem, which states that the vector space can be written aswhere denotes the null space of the matrix , that is,and denotes a direct sum.
Thus, by the definition of direct sum, we will be able to uniquely write each vector aswhere for . Moreover, the vectors are guaranteed to be linearly independent whenever they are non-zero.
Furthermore, we will show that the null spaces are invariant.
The next proposition paves the way to the Primary Decomposition Theorem.
Proposition Let be the space of vectors. Let be a matrix. Let and be two polynomials such that: 1) they are relatively prime (i.e., their greatest common divisor is ); 2) they are not identically equal to zero; 3) is an annihilating polynomial. Then,and and are invariant under the linear transformation defined by .
Since and are relatively prime and not identically equal to zero, by Bezout's theorem there exists two polynomials and such thatThus,Arbitrarily choose any and post-multiply both sides of the previous equation by :where we have definedThe previous two equations can be pre-multiplied by and so as to obtain which imply that Since was arbitrary, we havethat is, any can be written as the sum of an element of and an element of . In order to show that the sum is direct, we are going to prove that the representationis unique. Suppose there is another representationwhere and . Subtracting equation (2) from equation (3), we obtainPost-multiplying equation (1) by , we get where: in step we have used the fact that and belong to ; in step we have used equation (4); in step we have used the fact that and belong to . Thus, . In a completely analogous manner, we can prove that . As a consequence, the representation as a sum is unique, which implies thatMoreover, by the properties of matrix polynomials, the null spaces and are invariant under the linear transformation defined by .
A simple example follows.
Example Consider the space of all vectors and a matrix whose characteristic polynomial isIn other words, has an eigenvalue with algebraic multiplicity and another eigenvalue with algebraic multiplicity . The two polynomialssatisfy all the assumptions of the previous proposition because, by the Cayley-Hamilton theorem, the characteristic polynomial is an annihilating polynomial. Then,where the two spaces and are invariant.
Here is the Primary Decomposition Theorem.
Proposition Let be the space of vectors. Let be a matrix. Let be the minimal polynomial of : where are the distinct eigenvalues of . Then,where the spaces , ..., are invariant under the linear transformation defined by .
This proposition is proved by applying recursively the previous proposition. We start with the two polynomials They are relatively prime, non-zero, and their product is an annihilating polynomial since it is equal to the minimal polynomial. Therefore,We now restrict our attention to the space . We define the two polynomialsThey are relatively prime, non-zero, and their product is, by definition, an annihilating polynomial on the space . Therefore,andWe then decompose, with the same technique, the spaces , ..., until we obtain
Thus, the space can be written as a direct sum of invariant spaces of the formwhich contain vectors such thatThese vectors are called generalized eigenvectors and the spaces are called generalized eigenspaces.
In the special case in which , thenthat is, is an eigenvector of associated to the eigenvalue and is the eigenspace of .
The Primary Decomposition Theorem provides the key to understanding the information provided by the minimal polynomial.
It is natural to make a comparison with the characteristic polynomialwhich reveals the eigenvalues and their algebraic multiplicities , but contains no information on the eigenspaces.
The minimal polynomialalso reveals the eigenvalues, but not their algebraic multiplicities; however, it provides an additional piece of information: by looking at the exponents , we can understand whether the eigenspaces are "large enough" to contain a basis of eigenvectors for the space . As explained above and in the next section, if , then the space can be written as a direct sum of eigenspaces, which implies that the eigenvectors of are sufficient to form a basis for . On the contrary, if an exponent is greater than , then the eigenspace associated to is not "large enough" and we are not able to find a basis of eigenvectors for (the eigenvalue is defective). The remedy is to resort to "larger" generalized eigenspaces. What do we mean by larger? In the lecture on matrix powers, we have learned that the larger the power of a matrix is, the larger its null space is. Therefore, if an eigenspacedoes not contain enough linearly independent vectors (to be used in the construction of a basis for ), then we look at larger powers () of the matrix , so as to obtain larger null spaces called generalized eigenspaces.
Thus, the minimal polynomial reveals by how much we need to enlarge the (generalized) eigenspaces in order to be able to form a basis for the space .
Example Let be a matrix. Suppose that the characteristic polynomial of isand the minimal polynomial of isThe eigenvalue has algebraic multiplicity equal to (inferred from the characteristic polynomial), it is not defective (inferred from the exponent equal to in the minimal polynomial) and, as a consequence, its geometric multiplicity is equal to . The eigenvalue has algebraic multiplicity equal to (inferred from the characteristic polynomial), it is defective (inferred from the exponent equal to in the minimal polynomial) and, as a consequence, its geometric multiplicity must be less than (hence ).
Remember that a matrix is diagonalizable if and only if
it is similar to a diagonal matrix;
its eigenvalues are not defective (their algebraic and geometric multiplicities coincide);
there exists a basis of eigenvectors for .
Thanks to the Primary Decomposition Theorem, we can provide another criterion for the diagonalizability of a matrix.
Proposition Let be a matrix. Let be the minimal polynomial of : where are the distinct eigenvalues of . Then, is diagonalizable if and only if
Let us prove the "if" part, starting from the assumption that for every . Let be the space of vectors. Then,In other words, is the direct sum of the eigenspaces of . Pick any vector . Then, we can writewhere belongs to the eigenspace for each . We can choose a basis for each eigenspace and form the unionwhich is a set of linearly independent vectors and a basis for because is a direct sum. The vector can be re-written in terms of the basis by writing each in terms of the basis . Since this is true for an arbitrary vector , we have that is a basis of eigenvectors of for the space . Hence, is diagonalizable. Let us now prove the "only if" part, starting from the assumption that is diagonalizable. As proved in the lecture on the minimal polynomial, for every (otherwise is not an annihilating polynomial). We are going to prove that, when is diagonalizable, becomes an annihilating polynomial by setting (hence it is the minimal polynomial because any other annihilating polynomial has higher degree). Since is similar to a diagonal matrix (having the eigenvalues on its main diagonal), we need to check whether is an annihilating polynomial of (two similar matrices have the same annihilating polynomials). Butbecause all the diagonal elements of are products involving factors of the form , at least one of which is such that . This proves that is the minimal polynomial of and has the desired property ().
Below you can find some exercises with explained solutions.
Let be a matrix. Suppose that the minimal polynomial of is
Is diagonalizable?
The matrix is not diagonalizable because the exponent of the linear factor is greater than .
Let be a matrix. Suppose that the characteristic polynomial of isand the minimal polynomial of isWhat is the geometric multiplicity of the eigenvalue ?
From the characteristic polynomial, we infer that the algebraic multiplicity of is . From the minimal polynomial, we infer that it is not defective. Hence, its geometric multiplicity equals .
Let be the space of vectors. Let be a matrix. Let be the characteristic polynomial of : where are the distinct eigenvalues of and are their respective algebraic multiplicities. Can you modify the Primary Decomposition Theorem so as to write as a sum of null spaces by using the characteristic polynomial instead of the minimal polynomial?
The recursive decomposition used in the proof of the Primary Decomposition Theorem does not depend on any specific property possessed by the minimal polynomial and not by the characteristic one. For example, we can start the decomposition with They are relatively prime, non-zero, and their product is an annihilating polynomial since it is equal to the characteristic polynomial. Therefore,We can then decompose and keep decomposing the null spaces until we obtain
Please cite as:
Taboga, Marco (2017). "Primary decomposition theorem", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/primary-decomposition-theorem.
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