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

Then

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 .

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