Let be a -vector space, an -endomorphism of and a function. Here, we want to make sense of ; this should be another endomorphism of which is somehow related to both and .
Let us make this more precise. For that, let be a subalgebra of the -algebra of endomorphisms of , containing the identity , and let be a -subalgebra of the -algebra of functions , containing the identity and the constant functions. We say that is a functional calculus if satisfies the following conditions:
- for a fixed , the map , is a -algebra homomorphism with and ;
- for a fixed , the map , is -algebra with .
We usually write for if it is clear which is meant.
Note that contains all polynomial functions , i.e. the functions of the type , where is a polynomial. Note that for polynomial functions , the value of for is completely determined by the fact that is an -algebra homomorphism with and , as : if , then
In particular, this gives a canonical functional calculus , where is the image of the canonical map . (In case you are curious, if, and only if, is infinite; in the other case, , where .)
What about functions which are not polynomial? In case is diagonalizable, i.e. has a basis consisting of eigenvectors of , one can define for an arbitrary function by defining as the linear map which maps an eigenvector of with eigenvalue to . If one sets , one obtains a functional calculus .
In Functional Analysis, one is interested in such functional calculi with or , and one obtains ones for holomorphic functions , for continuous functions and even for certain Borel-measureable functions . But for today, we want to stick to the situation of an arbitrary . We will use and , i.e. the canonical functional calculus.
Let . In case the canonical map , is not injective, the unique normed generator of is called the minimal polynomial of and denoted by .
In case , every has a minimal polynomial, as and . In case , certain elements of do have a minimal polynomial; for example, has the minimal polynomal ; other elements of do not possess a minimal polynomial, for example any endomorphism with infinitely many different eigenvalues.
In case is an eigenvalue, let be an corresponding eigenvector and let . Then , and corresponds to . Clearly, , whence .
Conversely, assume that . Write , where and satisfies . As , . Now . In case , is an eigenvalue of (let and choose maximal with ; then ); hence, assume . In that case, we must have . But as is a proper divisor of , this cannot be.
The minimal polynomial is a rather powerful tool. In case it exists, one gets an -invariant decomposition of as follows:
Let be a prime divisor of . Then
is an -invariant subspace of . If is another prime divisor of coprime to , then is -invariant and is an monomorphism.
If is an arbitrary prime polynomial, if, and only if, .
Clearly, . As , this is a subspace of . As is -invariant (as ), it follows that is -invariant as well.
As , is -invariant as well. Let and let be minimal with . As are coprime, there exist with . Then . Therefore, is injective.
Finally, the last statement can be proven in exactly the same way as the previous lemma.
Let , where is a set of pairwise distinct monic prime polynomials and . Then is a direct sum.
Assume that this is not a direct sum. Then there exists , not all zero, such that . Assume that the number of non-zero is minimal under this condition. Let be with , and let satisfy . Then , and . If , then as is injective and so is its -th power. But this is only possible by the minimality assumption of for all , i.e. , contradicting the choice of . Therefore, the sum is a direct sum.
Let be a prime factor of and let be the maximal exponent of appearing in , i.e. and . Then .
Let , and write with ; then and are coprime. We have to show . Clearly lies in the kernel of , as . Let be such that ; as and are coprime, there exist with . Therefore,
Note that one can in fact show that .
Let be a prime factor of . Then there exists a -invariant subspace with .
Write with , and . Then and, in particular, for every . Let ; we can write for some . Now , whence , i.e. . Therefore, .
Let such that . Let ; then with . Now , whence , i.e. . As , we see .
Hence, we get . Finally, note that is clearly -invariant.
Let , where is a set of pairwise distinct monic prime polynomials and . Then .
We show this by induction on . In case , we have , which is only possible if . In that case, the statement is obvious. Hence, assume .
Let and be an -invariant direct complement of . Clearly is injective on , whence implies . Therefore, has strictly less than distinct prime factors, whence . In particular,
whence the claim follows.
Note that this is a generalization of the Jordan decomposition. Note that in fact, is the minimal -decomposition of in case . This completes the task started in my post on such decompositions, namely finding minimal -decompositions in case the characteristic polynomial of (assuming ) does not splits into linear factors.
Assume that has a minimal polynomial of the form , where is prime and . Let and . Then and is nilpotent of index . Moreover, is diagonalizable if, and only if . Finally, is diagonalizable if, and only if, and .
Clearly, as both are elements of . Moreover, , whence is nilpotent of index .
If , for some . In that case, . Conversely, if is diagonalizable, any eigenvalue of must be a zero of . This is only possible if .
Finally, assume that is diagonalizable. Hence, is diagonalizable as well; but the only diagonalizable and nilpotent endomorphism is 0, whence and is diagonalizable, i.e. . Conversely, assume and ; then and is diagonalizable.
Assume that has a minimal polynomial. Then is diagonalizable if, and only if, is squarefree and splits over .
Write with and pairwise distinct, monic prime polynomials . Then by the generalized Jordan decomposition. Hence is diagonalizable if, and only if, is diagonalizable for every . For a fixed , we have that , whence by the previous lemma, is diagonalizable if, and only if, and .
Assume that has a minimal polynomial. Then there exist polynomials such that and , where are the endomorphisms from the previous corollary.
As , it suffices to show the existence of . Write with and pairwise distinct monic primes , and set . We want a polynomial such that . Now the minimal polynomial of is , whence , i.e. it suffices to solve the congruences , . But since , , are pairwise coprime, such an exists by the Chinese Remainder Theorem.
Assume that has a minimal polynomial which is separable (i.e. its prime factors do not have multiple roots in their splitting field). Then there exist unique endomorphisms such that
- and ;
- is nilpotent;
- if is a splitting field of over , is diagonalizable.
By the previous lemma and corollary, there exist polynomials with , such that satisfy the conditions. (Note that , and since is separable, is squarefree and splits into linear factors over . Hence, by the second-previous lemma, is diagonalizable.)
Now let be any two endomorphisms which satisfy the conditions above. As and , all of , , and commute with each other. Hence, we have , and is nilpotent and is diagonalizable. But this is possible if, and only if, , i.e. if and .
Let us now return to the original idea of functional calculus. The generalized Jordan decomposition allows us to do a Taylor expansion in the nilpotent part:
Assume that has a minimal polynomial which is separable. Let be the generalized Jordan decomposition of , and let . Finally, let be the nilpotence index of , i.e. let satisfy . Then
Consider , the rational function field. The Taylor expansion of around is given by . Here, we have in fact . As , commute, we can plug in and and obtain
Now for gives the formula.
Note that in case or , this formula holds also for arbitrary analytic functions . In fact, the function only needs to be analytic on an open set which contains the complex eigenvalues of . The most important example is the exponential function , . The above shows that every possessing a minimal polynomial can be decomposed into a diagonalizable part and a nilpotent part of finite index , and in that case,
Now let . Recall the the characteristic polynomial of is defined as . So far, we have not used Cayley-Hamilton's Theorem. In fact, we can use the above stuff to prove the theorem. For that, we first relate the minimal polynomial to the characteristic polynomial.
If is an irreducible prime, then divides if, and only if, divides .
Let be a splitting field of over , and consider . We have and , whence it suffices to show that if, and only if, for every .
For that, note that if, and only if, is an eigenvalue of . But this is equivalent to not being injective, which in turn is equivalent (as ) to that is not invertible, which is the case if, and only if, , i.e. .
In fact, we can show that divides , which implies the Cayley-Hamilton theorem as . For that, we show that for every prime polynomial , where gives the exponent of in the prime factor decomposition of a non-zero element of .
Assume . If with being nilpotent, where , then .
Let be a splitting field of over , and let , and . It then suffices to show that the statement holds for these -endomorphisms of . Hence, we can assume that splits over . In that case, there exists a basis of such that the representation matrix of with respect to is in upper triangular form. Then , where ranges over the diagonal elements of . Now , whence is in upper triangular form as well. As is nilpotent, the diagonal elements of must all be zero. As , we see that .
Assume that has a minimal polynomial. Let , where are pairwise distinct irreducible polynomials. Set
Let and let with ; then, if , satisfies
as since , we get .
For the converse, first note that is -invariant. Assume has a monic prime factor distinct from . Then ; let . Let be such that and be such that . As and are coprime, there exist polynomials with . Hence,
a contradiction. Hence, all prime factors lie in . Therefore,
which shows the claim.
Assume has a minimal polynomial, and let be a prime polynomial. Let be a field extension of over which splits; write with distinct elements and . Then
Notice that . Hence, the corollary follows from the previous lemma.
Let and be a prime polynomial. Then .
We first show that “” holds. Let be a splitting field of over , and write with pairwise distinct and . We have ; since , it suffices to know that the theorem holds in case , as then and, therefore,
Hence, assume that with . In that case, . Let and write with . Note that divides . But is diagonalizable on with only the eigenvalue , whence . Therefore, .
The above argument shows . If are all distinct prime factors of , we get
as all summands are , the theorem follows.
If and , then .