Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 26/refcontrol
For further investigations of linear mappings, and, in particular, of trigonalizable mappings, we have first to discuss an important law in the polynomial ring over a field, the Lemma of Bezout.
- The Lemma of Bezout
Recall that a polynomial divides a polynomial , if there exists an polynomial such that
This corresponds to the divisibility concept of the integers . From there, also the concept of a common divisor should be known.
Let denote polynomialsMDLD/polynomials (1) over a fieldMDLD/field . We say that a polynomial is a common divisor of the given polynomials if dividesMDLD/divides (polynomial ring 1)
every .Let denote polynomialsMDLD/polynomials (1) over a fieldMDLD/field . We say that a polynomial is a greatest common divisor of the given polynomials, if is a common divisorMDLD/common divisor (polynomial ring 1) of the , and if has among all common divisors of the maximal
degree.MDLD/degree (polynomial)A greatest common divisor is not uniquely determined, since with also for some constant is also a greatest common divisor. However, if we restrict to normed polynomials, then the greatest common divisor is unique.
PolynomialsMDLD/Polynomials (1) over a fieldMDLD/field are called coprime, if they have, with the exception of constants , no
common divisor.MDLD/common divisor (polynomial ring 1)Let be a fieldMDLD/field and let denote polynomialsMDLD/polynomials (1) over . Let be a greatest common divisorMDLD/greatest common divisor (polynomial ring) of the . Then there exists a representation
with
.We consider the set of all linear combinations
This is an idealMDLD/ideal of , as can be checked directly. Due to Theorem 20.10 , this ideal is a principal ideal,MDLD/principal ideal hence,
with a certain polynomial . This is a common divisor of the . Because of , we have
that is is a factor of every . A similar reasoning shows
for all , and, therefore, also
Hence,
By the condition, has the maximal degree among all common divisors. Therefore, is a constant. Thus, we have
and, in particular, . Therefore, is a linear combination of the .
Let be a fieldMDLD/field and let denote coprime polynomialsMDLD/coprime polynomials over . Then there exists a representation
with
.This follows directly from Theorem 26.4 .
For given polynomials , we can determine explicitly their greatest common divisor and a representation
as stated in Theorem 26.4 . For this, we restrict ourselves to the case . Let the degree of be as large as the degree of . The division with remainder yields
with a remaining polynomial, whose degree is smaller than the degree of , or which is . The main point is that the ideals
are identical and, thus, the greatest common divisor of and and of and coincide. Now we perform again the division with remainder, dividing by with remainder , and again the ideal coincides with the starting ideal. In this way, we obtain a sequence of remaining polynomials
with the property that two adjacent polynomials generate the same ideal. Hence, (the last remaining polynomial different from ) is the greatest common divisor of and . We can find a representation of as a linear combination of the , by working back this algorithm using the equations which describe the divisions with remainder.
The method described in the previous remark is called Euclidean algorithm. It holds accordingly also for the integers.
== Example Example 26.7
change==
We want to determine the greatest common divisorMDLD/greatest common divisor (polynomial) for the polynomials and in . For this, we perform the division with remainder,MDLD/division with remainder (polynomial ring) and we obtain
Thus the two polynomials are coprime.MDLD/coprime (polynomial) A representation of the is
We mention the Lemma of Bezout for the integers.
Every set of integers has a greatest common divisor , and this can be expressed as a linear combination of the , that is, there exist integers such that
Proof
- Generalized eigenspaces
A trigonalizable mapping can be described by an upper triangular matrix. We want to understand whether there are even easier matrices to describe such a mapping. This will be done in two steps. In this lecture, we describe a trigonalizable mapping as a direct sum of mappings on so-called generalized eigenspaces. In the next two lectures, we will discuss endomorphisms on such generalized eigenspaces.
For a linear mappingMDLD/linear mapping on a -vector spaceMDLD/vector space and an eigenvalue ,
is called generalized eigenspace
of for this eigenvalue.If is finite-dimensional, then the chain
becomes stationary, that is, there exists some such that
Generalized eigenspaces are, due to Exercise 26.28 , invariant under the linear mapping. By definition, we have
and for diagonalizable we have equality, see Exercise 26.22 . We want to understand trigonalizable mappings via their generalized eigenspaces.
Let be a linear mappingMDLD/linear mapping on a finite-dimensionalMDLD/finite-dimensional -vector spaceMDLD/vector space , and let
be a factorization of the characteristic polynomialMDLD/characteristic polynomial in coprimeMDLD/coprime (polynomial) polynomials . Then we have the direct sum decompositionMDLD/direct sum decomposition
where these linear subspaces are -invariant.MDLD/invariant (linear)
The restriction of onto is bijective.Due to the Lemma of Bezout, there exist polynomials such that
Set and . Let . Due to the Theorem of Cayley-Hamilton , we have
Therefore, the image of belongs to the kernel of and vice versa. From
we can read off that the left-hand summand belongs to and the right-hand summand belongs to . Therefore, we have a sum decomposition, which is direct, since implies . For the -invarianceMDLD/invariance (endomorphism) of these spaces, see Exercise 26.28 . For , we have
that is, we have . Therefore, the restriction of to the kernel of is surjective, thus bijective.
We consider the permutation matrixMDLD/permutation matrix
over , the characteristic polynomialMDLD/characteristic polynomial is
where the two factors are coprime.MDLD/coprime (polynomial) We want to check Lemma 26.10 in this example. We have
with
and
with
We have
Moreover, we have
and
From this, we can read off that the restriction of to is bijective. The representation of the from Example 26.7 yields the matrix equation
Let
be an endomorphismMDLD/endomorphism on the finite-dimensionalMDLD/finite-dimensional -vector spaceMDLD/vector space , and let . Then the dimensionMDLD/dimension (vs) of the generalized eigenspaceMDLD/generalized eigenspace equals the algebraic multiplicityMDLD/algebraic multiplicity
of .We write the characteristic polynomialMDLD/characteristic polynomial of as
where does not occur in as a linear factor, that is, is the algebraic multiplicity of . Then, and are coprime,MDLD/coprime (polynomial) and, due to Lemma 26.10 , we have the decompositon
and
is a bijection. Moreover,
where the inclusion is clear, and the other inclusion follows from the fact that higher powerse of do not annihilate further elements, by the bijectivity on just mentioned. For the characteristic polynomial, we have, due to the direct sum decomposition according to Lemma 26.10 , the relation
where is the characteristic polynomial of and is the characteristic polynomial of . Since restricted to is the zero mapping, the minimal polynomial of and, hence, also the characteristic polynomial are some power of , say
where
In particular, , as is a divisor of . Assume that . Then is a zero of and is an eigenvalue of . But this is a contradiction to the fact that is a bijection on this space.
For a linear mappingMDLD/linear mapping on a finite-dimensionalMDLD/finite-dimensional -vector spaceMDLD/vector space and two eigenvaluesMDLD/eigenvalues , the corresponding generalized eigenspaceMDLD/generalized eigenspace have trivial intersection, that is
The characteristic polynomial of has the form
where neither nor is a zero of . Because of Theorem 26.12 , applied to , we have
Because of , this implies immediately
Let
be a trigonalizableMDLD/trigonalizable -endomorphismMDLD/endomorphism on the finite-dimensionalMDLD/finite-dimensional -vector spaceMDLD/vector space . Then is the direct sumMDLD/direct sum (vs) of the generalized eigenspaces,MDLD/generalized eigenspaces that is,
where are the different eigenvaluesMDLD/eigenvalues of , and is the direct sumMDLD/direct sum (mapping) of the restrictions
Let
be the characteristic polynomial,MDLD/characteristic polynomial which splits into linear factors according to Theorem 25.10 , where the are different. We do induction over . For , there is only one eigenvalue and only one generalized eigenspace. Due to Corollary 24.3 , the minimal polynomial is of the form and thus . Suppose that the statement is already proven for smaller . We set and . We are then in the situation of Lemma 26.10 and Theorem 26.12 . Therefore, we have a direct sum decomposition into -invariantMDLD/invariant (endomorphism) linear subspaces
The characteristic polynomial is, according to Lemma 26.10 , the product of the characteristic polynomials of the restrictions to the spaces. Because of Theorem 26.12 , the polynomial is the characteristic polynomial of the restriction to the first generalized eigenspace, hence, is the characteristic polynomial of the restriction to . In particular, this restriction is also trigonalizable. By the induction hypothesis, is the direct sum of the generalized eigenspaces for . Altogether, this implies the direct sum decomposition of and of .
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