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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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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