Note that the below material is well-known among experts.

Let be any prime not dividing . Then modulo is invertible, and modulo , the system has a unique solution. Moreover, for any integer , the system has a unique solution modulo : this is true since is also invertible modulo – for that, it suffices to check that the determinant of is a unit, which it is since it is coprime to . Moreover, if is a solution to over the integers, then modulo is the unique solution of modulo . Hence, if we choose such that bounds all coefficients of the solution , we can recover a solution to over the integers from a solution to modulo , by chosing the unique preimages in .

This opens the question on how to solve modulo . For that, a Hensel-like lifting technique can be used. (In fact, this follows from Bourbaki’s generalization since the Jacobian of the map , equals .) Assume that we have an which satisfies . We want to find with . Write with . As , and as is divisible by , we obtain the linear system . Hence, it suffices to solve linear systems over the prime field to solve over .

This yields the following algorithm:

1. Choose p := 2.
2. Solve A x = b modulo p.
3. If a unique solution exists:
   1. Set e = 0 and lift x to the integers with coordinates in .
   2. Compute c := A*x - b.
   3. If c = 0, return x.
   4. Solve A y = c/p^e modulo p.
   5. Set x := x + y*p^e and e := e + 1.
   6. Adjust x modulo such that all coefficients are in .
   7. Go back to Step 3.2.

   Else:

   1. Choose the next prime p and go back to Step 2.

The only subprogram we need is a linear systems solver for with square over a finite field, which returns information on the number of solutions. (Note that is not divisible by if and only if there is a unique solution.) If more information is known on the matrix , for example its determinant has been already computed, this information can be used as well.

Let us analyze the running time of this algorithm. Denote by the smallest prime not dividing , and by the time the linear system solver over needs to solve a system of size . Let (resp. resp. ) denote the largest absolute value of an coefficient of (resp. resp. ).

Clearly, the number of iterations is in . In each iteration, one linear system over of size has to be solved, and has to be evaluated. The former takes operations, and the latter involves multiplications and additions of integers of size and , and substractions of integers of size and . For simplicity, assume that . Finally, to compute , we need multipliations of integers of size and , and additions which can be neglected. Clearly, the multiplications can also be neglected, since the evaluation of already is slower.

Let denote the time a multiplication of two numbers of size needs. Then inside the main loop, we need time units, and the main loop alltogether needs time units. Finding needs time units (using the Prime Number Theorem).

Assuming that we use a naive Gaussian algorithm as well as naive multiplication, i.e. and , we obtain a total running time of Using fast multiplication, i.e. (using FFT methods), and fast linear system solving, i.e. , where , we obtain a total running time of

Now let us try to eliminate from this expression. Clearly, the the second part, we can use that . To eliminate from the first part, we need to find an upper bound. For that, let us first stick to , the smallest prime not dividing the integer . (Letting be a -matrix yields ; in general, using this notation.) Now is divisible by , whence for we have . Note that for integral , , where denotes the Chebyshev function. Using known bounds on , we get Therefore, becomes true for for every . This shows that eventually holds for , yielding and, thus, . Using the Leibniz formula, .

Finally, we can use some linear algebra to bound in terms of and . First note that , where denotes the identity matrix and denotes the adjungate matrix of . As , we see that it suffices to bound . Now the coefficients of are determinants of matrices with coefficients coming from , whence . Therefore, when assuming that .

This can be combined into the following theorem:

Theorem.

Assuming that and , the above algorithm needs time units to compute the unique solution of using naive arithmetic in , , and naive Gaussian elimination to solve linear systems over . Using fast linear algebra and fast multiplication, we only need time units for any .

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Edit: in fact, the theorem follows from the more general result stated here. Also see the comments.

Theorem.

Let be a field and . Then the matrix has determinant

Before I present the proofs, I want to say a few words on how this matrix comes up. Consider the linear system , where Assuming that all , one sees that this system has no solution. Instead, one can try to find a vector which minimizes . A well-known fact in Linear Algebra says that this is the case for a unique , and that is the solution to the system . Now and , whence we can describe the unique solution using Cramer’s rule by Now the above theorem gives an easy to evaluate formula for the determinants, namely The solution vector can be evaluated in operations, and in case the are rational numbers, one can hence efficiently compute an exact (i.e. rational) solution.

Proof (First proof).

For this is easy to verify. Hence, assume that and that the statement is true for . Using the multilinearity of the determinant and Lagrange expansion, both for the first row, we see that The same argument applied to the second row of the second determinant, we obtain that the second determinant Since has two identical rows – namely the first two –, its determinant is 0. Moreover, by induction hypothesis, and Plugging this in yields which shows the claim.

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Proof (Second proof).

For this is clear. Hence, assume that and that the statement is true for . We do a Lagrange expansion by the last column. This yields a term , which by induction hypothesis equals The other terms are of the form , where is with the last column and the -th row removed. Note that by cyclically permuting rows to of , we obtain the matrix . The permutation has sign , whence we see that Combining everything so far, we see that what we wanted to show.

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Assuming that it is known that there is, up to scale, only one alternating -linear form (i.e. that ), one can proceed as follows. Given , consider the map , . This is clearly -linear and alternating, whence there exists some such that . Evaluating at the identity matrix gives . Evaluating at gives .

Of course, using the trick similar to the first lemma here, it suffices to show this for to obtain it for any unitary commutative ring, after showing that the determinant is in fact a polynomial equation with integer coefficients (for example, by showing the Leibniz formula).

In case , and are not isomorphic: a basis of has a cardinality strictly larger than the one of . Moreover, the canonical map is still a monomorphism, but no longer surjective. In the case of , one has as well a canonical monomorphism , but it is no longer surjective as well. We want to study the images of the canonical maps and .

We begin with an auxiliary lemma.

Lemma.

Let with . Then there exists some with . Hence, if satisfies for all , then .

Proof.

Choose a -basis of such that there exists some with . Define by . Then and .

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Proposition.

The map is a monomorphism and its image is In particular, if for some , then .

Proof.

Clearly, for , is -linear. Moreover, one quickly sees that is -linear itself. To see that is injective, let with . Now, by the lemma, there exists a with ; this shows that , whence . Therefore, an is injective.
Now, if , , whence . This shows that the image of is contained in the given set. Now assume that satisfies ; say, . Then for all , whence . By the Homomorphism Theorem, there exists a homomorphism such that commutes. Now , whence . As (as ), is an isomorphism and we must have for some . But then, lies in the image of .
Finally, if , we saw that we have for any non-zero , with depending on . Since is injective, this shows that we must have .

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This allows us to show that is surjective if, and only if, .

Corollary.

We have that is surjective if, and only if, .

Proof.

First, if , we see that . Since is injective, it follows that is in fact an isomorphism.
Now assume that . It suffices to construct a hyperplane in with ; this defines an element of which is not contained in the image of by the above proposition. For that, chose a basis of (using Zorn’s lemma). This defines a family of elements of by , . Let be the subspace of generated by the ‘s. If we would have , we could emply Zorn’s lemma a second time to obtain a hyperplane with ; this would prove our claim.
Hence, we have to show that . Note that for , ; in particular, for every , only finitely many of the ‘s are non-zero. Hence, it makes sense to define , . We claim that in case : for that, note that is a linear independent set in , since for every linear combination , we get for every .

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Note that in fact, the proof shows that is isomorphic to a -fold direct product of , while is isomorphic to a -fold direct sum of . In case , these are of the same dimension, but in case , they are not.

Proposition.

The map is a monomorphism and its image is the -vector space of finite dimensional -homomorphisms .

Proof.

One quickly sees that defines an element of , whence is well-defined and its image is contained in . Moreover, one quickly sees that is a homomorphism.
Let with , i.e. with for all . Without loss of generality, we can assume that our representation of satisfies that the ‘s are linearly independent. In that case, implies for all . But since this is true for all , it follows that for all . But then, . Therefore, , whence is injective.
Now let , and let be a basis of . Let be the projections with an for . Set . Then for all since for all ; therefore, . This shows that , whence we have equality.

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Now is a -algebra, whence for , it makes sense to define . We are interested on how can be described in terms of and . This is resolved by the following result:

Proposition.

Let be -vector spaces. The map is -linear and the following diagram commutes:

Proof.

Let , and be the canonical maps. Since these are isomorphisms, we have to show that for , and , we have . For that, let . Then what we had to show.

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In particular, is a -algebra isomorphic to ; it posseses a if, and only if, .

Now consider transposition Clearly, transposition is injective:

Lemma.

The map is -linear and injective.

Proof.

It is clear that is -linear. To see that it is injective, let with . Let ; then for all , whence by the first lemma. But that means .

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We show that transposition restricts to the subspaces of the homomorphism spaces of homomorphisms with finite-dimensional image.

Lemma.

Let . The map is an isomorphism. In particular, .

Proof.

Let . The map , is well-defined and a homomorphism as and . Now let with , i.e. with . But since is defined on , this means that . Hence, is injective.
Now let , i.e. there exists some with . Set ; then and . Therefore, is injective.
Finally, in case , we have and , whence . On the contrary, if , we have , whence .

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Now we have seen that and in a canonical way, and we have the canonical monomorphism . We show that these maps behave nicely with transposition:

Proposition.

The map is the unique homomorphism which makes the diagram commuting.

Proof.

Let and . Then for , and for all and . Hence, , what we had to show.

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Now consider double transposition, i.e. and its finite-dimensional image restriction The above shows that using the canonical isomorphisms and , we can describe double transpotition by the following commuting diagram:

If , we obtain a map

Lemma.

The map is -linear and injective.

Proof.

First, if is fixed, for all , ; hence, . Now, if and , , we have whence is -linear.
To see that is injective, let be such that . Let and ; since for all , we see that , but since this is the case for all we get .

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Note that we have the following diagram: Moreover, using the canonical embeddings and , we can define a map by , and a map by . It turns out that these map make the diagram commute:

Lemma.

The maps , and , are -linear and make the diagram commute. In particular, is injective.

Proof.

That and are -linear is clear. For the lower triangle, let ; we have to show that . For that, let and ; then For the right triangle, let ; we have to show that . For that, let and ; then

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Now note that is injective. We can use this to determine the image of . For example, for , the last equivalence follows from the first proposition. Unfortunately, this criterion does not really helps in practice.

In case anyone knows a better description of the image of or , I’d be happy to know.

Unfortunately, the answer turns out to be “no” for the case (as well as ):

Example.

Let . Consider the matrix as well as the sequence Clearly, . Assume that is open in ; then we must have for almost all . But is in Jordan canonical form, and clearly not diagonalizable; but this means that for all . Therefore, is not open in .

But nonetheless, contains a Zariski-open subset of in case is algebraically closed (which implies that lies dense in ). For that recall that is the characteristic polynomial of .

Proposition.

Consider the set Then and is Zariski-open in . In fact, is the complement of a hypersurface in .

Proof.

Note that in case is squarefree, splits into distinct linear factors since is algebraically closed. Hence, has distinct eigenvalues in and therefore one obtains linearly independent eigenvectors of ; i.e., is diagonalizable over . Therefore, .
Now we show that is a hypersurface in , i.e. there exists a polynomial such that . For that, consider the maps defined by for all . Obviously, these must be polynomials. Next, consider the discriminant of ; this is a polynomial expression in the coefficients of , i.e. in , whose value is zero if, and only if, is squarefree. Therefore, . Finally, is a polynomial, whence is Zariski-open in .

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Note that the situation is different over :

Proposition.

In the standard topology,

Proof.

Assume that has at least one eigenvalue with imaginary part . If is a sequence of matrices with , each must have an eigenvalue with . But then, for infinitely many , we must have (since is open), whence we cannot have for infinitely many . Hence, is not in the closure of .
Now assume that has only real eigenvalues. Then there exist some with in Jordan canonical form. By pertubing the diagonal elements of slightly, we can obtain a sequence of matrices with . But then, and for every .
Note that this implies ; moreover, this also implies . Hence, the first two equalities hold. The third equality is standard.

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Also note that for , as the example (which is diagonalizable over , with eigenvalues ) shows. So what about ? In fact, as in the case of , it turns out that is .

Proposition.

We have

For the proof, we need a little lemma.

Lemma.

Let . Then .

Proof.

Let be an arbitrary polynomial. Write , where and is irreducible and monic, . Now the coefficients of all ‘s (except the highest coefficients) are a finite set in of elements, whence there exists sequences with pairwise distinct such that converges to one coefficent of one . In particular, we can construct monic polynomials with , and for every . Even more, we can make sure that every is irreducible; this enforces that is squarefree, i.e. . Therefore, we found a sequence in converging to , whence .

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Proof (Proof of the Proposition).

Let whose characteristic polynomial can be written as , with not necessarily distinct, but monic and irreducible polynomials . There exists a matrix with where is the companion matrix of ; this is a Frobenius normal form of . Now we can find a sequence of squarefree polynomials such that for every , are pairwise coprime, and that . Then set clearly, . Moreover, the characteristic polynomial of is given by , i.e. it is squarefree by choice of the . Therefore, and .

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The usual solution to this is to complexify : define , and define an action this turns into a -vector space. If one identifies by its image under , , then for all , . Now we are left to extend to . It turns out that there is exactly one choice to extend to a -linear map , i.e. that . Namely, one has to define ; this is obviously -linear, whence it suffices to show that : Now if is a -basis of , it is as well an -basis of ; moreover, . If now is a complex eigenvalue of , then there exists some such that . So is indeed an eigenvalue of . Abusing notation, we say that is an eigenvalue of ; this will always mean that we are talking of . This process is called complexification of and .

But does this generalize? What if is the base field and one has an eigenvalue of the matrix? Can we do the same thing here? And what if and we have an eigenvalue in ? The answer is yes. The idea is as follows. A basis of over is given by , . Hence, we defined , where the first corresponds to 1 and the second to : i.e. should mean . Now has a basis with three elements, so one could define . And for if , , we need an infinite basis and an infinite direct sum.

It would be nice if we could avoid working with bases, both of and of the field extension . This can indeed be done, using the tensor product. We begin with a very abstract defintion.

Definition.

Let be a ring and -modules. A pair , where is a -module and is -bilinear, is said to be a tensor product of and over if the following universal property holds:
If is any -module and is -bilinear, there exists exactly one homomorphism such that .

Theorem.

Tensor products exist and are unique up to unique isomorphism. More precisely, if and are tensor products of and over , there exists exactly one -isomorphism with .

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From now on, we write for and for , , . In case the base is clear, we will drop the subscript.

As we are interested in tensor products of vector spaces over a field, we can be more concrete.

Theorem.

Let and be -vector spaces. Let be a basis of and be a basis of . Then is a basis of . In particular, .

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A different interpretation is that is the set of linear combinations of elements of , where the coefficients are elements of . Hence, we extend the range of the coefficients of elements of from to . Every element of can be written in the form with , , .

Now let be a field extension of . Then is a -vector space, whence we can consider the tensor product . As expected, this turns out to be a -vector space with scalar multiplication , . In case , this definition coincides with the natural -vector space structure of .

Let us consider the special case , . Then is a -basis of ; if is an -basis of , then is an -basis of : every element of can be written in the form with . Moreover, is a -basis of . Compare this with the ad-hoc definition of at the beginning of this post.

Now, let us consider what to do with -linear maps , where and are -vector spaces. We begin with a general result on tensor products.

Theorem.

Let be -modules, , and let be -module homomorphisms. Then there exists exactly one -homomorphism with .

Proof.

Set and define One quickly checks that is bilinear. Hence, by the definition of the tensor product , there exists exactly one -homomorphism with

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Now let us consider -vector spaces , , a -linear map and the identity map . By the theorem, there exists exactly one -linear map with . But since is , using the -vector space structure of , we obtain , i.e. is -linear.

Finally, let be a -basis of and be a -basis of . Then is as well a -basis of and is as well a -basis of , whence we can consider the matrices and . Write ; then . Now Therefore, as well.

Hence, the tensor product allows us to describe , as a generalization of the complexification of real vector spaces, in a very clean and abstract manner.

Finally, recall that every field has an algebraical closure , which is unique up to -isomorphism. For -vector spaces , and -linear maps we get -vector spaces , and a -linear map . We have seen that every -basis of resp. is also an -basis of resp. , and that the matrix representation of with respect to the bases equals the one of . Hence, we can not just talk of arbitrary elements of being eigenvalues of matrices over , but also of endomorphisms defined over , by referring to resp. instead.

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:

1. for a fixed , the map , is a -algebra homomorphism with and ;
2. 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.

Definition.

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.

Lemma.

Assume that possesses a minimal polynomial. Then is an eigenvalue of if, and only if, .

Proof.

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.

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The minimal polynomial is a rather powerful tool. In case it exists, one gets an -invariant decomposition of as follows:

Lemma.

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, .

Proof.

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.

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Lemma.

Let , where is a set of pairwise distinct monic prime polynomials and . Then is a direct sum.

Proof.

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.

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Lemma.

Let be a prime factor of and let be the maximal exponent of appearing in , i.e. and . Then .

Proof.

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,

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Note that one can in fact show that .

Lemma.

Let be a prime factor of . Then there exists a -invariant subspace with .

Proof.

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.

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Theorem (Generalized Jordan Decomposition).

Let , where is a set of pairwise distinct monic prime polynomials and . Then .

Proof.

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.

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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.

Lemma.

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 .

Proof.

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.

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Corollary.

Assume that has a minimal polynomial. Then is diagonalizable if, and only if, is squarefree and splits over .

Proof.

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 .

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Lemma.

Assume that has a minimal polynomial. Then there exist polynomials such that and , where are the endomorphisms from the previous corollary.

Proof.

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.

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Corollary (Generalized Jordan Decomposition).

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

1. and ;
2. is nilpotent;
3. if is a splitting field of over , is diagonalizable.

Proof.

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 .

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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:

Theorem (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

Proof.

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.

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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.

Lemma.

If is an irreducible prime, then divides if, and only if, divides .

Proof.

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. .

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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 .

Lemma.

Assume . If with being nilpotent, where , then .

Proof.

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 .

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Lemma.

Assume that has a minimal polynomial. Let , where are pairwise distinct irreducible polynomials. Set Then

Proof.

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.

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Corollary.

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

Proof.

Notice that . Hence, the corollary follows from the previous lemma.

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Lemma.

Let and be a prime polynomial. Then .

Proof.

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.

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Corollary (Cayley-Hamilton over Fields).

If and , then .

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Let be a vector space over a field and a linear map. We say that a subspace is -invariant if . Another way to interpret this is to consider the -algebra ; then is an -module and the -submodules of are exactly the -invariant subspaces of .

Definition.

An -decomposition of is a decomposition of -submodules such that, for every -submodule of , one has .

Clearly, there always exists a trivial -decomposition of , namely itself. One can define a partial order on the set of -decompositions:

Definition.

Let and be two decompositions. We say that if, for every , there exists an with .

Clearly, the trivial -decomposition is the maximum with respect to this order. One can ask whether a minimal -decomposition exists. In case it exists, it has a nice property:

Lemma.

Assume that is a minimal -decomposition. Let be another -decomposition. Then, for every , there exists a subset with .

Proof.

Define . Then is a disjoint union. Now, and form a direct sum, whence for all .

Now assume that for some ; let . Now, as , we can write with . Moreover, write with . Clearly, we must have for every . As we have for all , whence for all . But this implies , a contradiction.

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Now one can ask when such a decomposition exists, and if it can be computed. An important case in which this is true is the one where is a finitely dimensional vector space over an algebraically closed field ; for example, and .

Definition.

Let . The generalized eigenspace of with respect to is

In case , one has that . Hence, generalized eigenspaces can be efficiently computed. Moreover, we have , and a simple argument shows that either both are trivial or both are non-trivial. Hence, the with are exactly the zeroes of the characteristic polynomial of .

Now note that . Hence, is -invariant. We now have three lemmas:

Lemma.

Let be different eigenvalues of . Then is a direct sum.

Proof.

Let with . We have to show that for all . Assume that not all are zero, and that the relation is chosen minimal with respect to the number of nonzero .

Let with , and choose with . If , then yields a second relation with . By minimality, we must have for all .

We will show that is injective for , which gives for all and, therefore, , a contradiction.

Let with . Assume that and let be maximal with ; in that case, and whence we get , which is only possible for , a contradiction. Hence, we must have , i.e. is injective on .

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Lemma.

Assume that and let . Then there exists an -invariant subspace such that .

Proof.

Set . ConsiderS the chains and Clearly, there exists an with as . Now one easily shows for all . By the Dimension Formula, we have for all , whence for all . But then and .

Set and let , i.e. let with . Then whence is -invariant. Now it suffices to show that . Now , whence for every there exists some with . But this means that is surjective, whence . But then .

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Lemma.

Assume that and that the characteristic polynomial of splits into linear factors. Let be all eigenvalues of . Then .

Proof.

We proceed by induction on . For this is cleary. Hence, assume and let be an eigenvalue of . Choose an -invariant subspace with . We have , whence . Now whence the characteristic polynomial of splits into linear factors as well.

Let be the eigenvalues of . Then, by induction, we have . Now , whence .

Finally, note that for all , as this would contradict . Therefore, . Moreover, we must have as the dimensions of the generalized eigenspaces for all must be non-zero, whence “” must hold. The converse holds because every non-trival generalized eigenvector yields a non-trivial eigenvector to the same value.

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Therefore, we get:

Corollary (Jordan Decomposition).

Let be algebraically closed and assume that . Then, for every endomorphism of , there exist such that is an -decomposition.

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Proof.

We have to show that this yields an -decomposition. For that, let be a -invariant subspace of . Consider ; this is an endomorphism of whose set of eigenvalues is a subset of the set of eigenvalues of . Hence, by the previous lemma applied to , we have what we had to show.

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We can now prove our main result, namely that the generalized eigenspace decomposition is exactly the minimal -decomposition of :

Theorem.

Let be algebraically closed and . Then the minimal -decomposition of is given by

Note that we do not need that is algebraically cloesd, but only that splits over .

Proof.

Let be a -decomposition. Assume that there exists some and such that with ; if this would not exist, we would have .

Assume that we can find eigenvectors and . Then , whence is an eigenvector as well. But then is an -invariant subspace of with , but , a contradiction that is a -decomposition.

We now show that contains an eigenvector. As is -invariant, we can consider . Now , whence we must have . Hence, there exists some , with .

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Definition.

Let be a commutative unitary ring and a -matrix over . The characteristic polynomial of is the polynomial .

Then the theorem says:

Theorem (Cayley-Hamilton).

Let be a commutative unitary ring and . Then .

We first begin with a fascinating reduction argument, which I first saw in a lecture of Paul Balmer at the ethz:

Lemma.

The Theorem of Cayley-Hamilton holds over any commutative unitary ring if, and only if, it holds over the complex numbers.

Proof.

Clearly, if the theorem holds for all rings, so it does for the special case . So assume that it holds for .

Let be any commutative unitary ring and , . Set and consider the ring homomorphism , . Over , consider the matrix . Now induces -algebra homomorphisms and with . Clearly, they satisfy and . Therefore, it suffices to prove .

Now has infinite transcendence degree over (otherwise, it could be countable), whence there exists an embedding ; simply choose algebraically independent elements in and map the to them. Again, we get maps and which are injective and satisfy and . But by assumption, Cayley-Hamilton holds over , whence . Since is injective, , which implies as mentioned above.

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Now we can concentrate on showing the Theorem of Cayley-Hamilton for the complex numbers. We begin with a special case, namely the diagonalizable matrices.

Definition.

A matrix is said to be diagonalizable if there exists an invertible matrix such that for .

We then have:

Lemma.

The Theorem of Cayley-Hamilton holds for diagonalizable matrices.

Proof.

We first assume that . Then one gets , and since one gets .

Now assume that is diagonalizable, and let such that . Clearly, and, therefore, Now write with . Then whence . But now , whence we get and, hence, .

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We now get to the main piece of proving Cayley-Hamilton over :

Lemma.

Endow with the Euclidean topology and consider the set Then is dense in .

For this proof, we need two facts from linear algebra:

• Every matrix over is equivalent to a triagonal matrix; this can be done if, and only if, the characteristic polynomial of the matrix splits into linear factors. But, by the Fundamental Theorem of Algebra, this is always the case over .
• An -matrix with distinct eigenvalues is diagonalizable.

Proof.

Let be an arbitrary matrix. Then there exists a matrix such that with . As the transcendence degree of over is infinite, there exist elements such that for every , the set has exactly elements. Define , . Then for and has distinct eigenvalues for every , namely . But this implies that , whence we found a sequence in converging to .

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Now, we are able to conclude:

Theorem (Cayley-Hamilton over the complex numbers).

Let . Then .

Proof.

Set . Clearly, and is dense in by the previous lemma. Hence, it suffices to show that is closed.

But note that the map , is defined by polynomials; hence, it is continuous. Now is the preimage of a closed set, whence is closed itself.

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This completes the proof of the theorem:

Proof (Cayley-Hamilton over commutative unitary rings).

By the first lemma, it suffices to show the theorem over . But this is accomplished by the previous theorem.

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