A TOPOLOGICAL PROOF OF THE CAYLEY–HAMILTON THEOREM

Jeffrey A. Rosoff

Gustavus Adolphus College

The Cayley–Hamilton theorem in linear algebra is generally proven by solely algebraic means, e.g. the use of cyclic subspaces, companion matrices, etc. [1,2]. In this article we give a short and basically topological proof of this very algebraic theorem. First the theorem:

\underline{Cayley–Hamilton}. Let V be a finite–dimensional vector space over a field, and let T: V → V be a linear transformation with characteristic polynomial p_{T} (x). Then p_{T} (T) is the zero transformation.

The correspondence between linear transformations and matrices allows us to prove the equivalent matrix form of the theorem.

\underline{Cayley–Hamilton (matrix form)}. Let A be an n × n matrix over a field with characteristic polynomial p_{A}(x) = det(xI − A), where I is the n × n identity matrix. Then p_{A}(A) is the zero matrix.

For simplicity, we will prove the result over the field ℂ of complex numbers and, as promised, use topological techniques. However, essentially the same proof works over any field (see the remark at the end of the proof).

\underline{Notation}. Let

P_{n} = { monic polynomials of degree n over ℂ},

let

M_{n} = {n × n matrices over ℂ},

and let

D_{n} = {A ∈ M_{n} | A is diagonalizable }.

Note that we may identify P_{n} with ℂ^{n} and M_{n} with ℂ^{n^{2}}, making each into a topological (in fact metric) space.

\underline{Lemma 1}. Let F: ℂ^{n} → ℂ be a non–constant polynomial mapping. Then if U is open in ℂ^{n}, F (U) is open in ℂ.

\underline{Proof}. (Induction on n).

1) Say F: ℂ → ℂ is a non–constant polynomial mapping, and suppose that we have an open set U in ℂ. If F (U) is \underline{not} open then there is a point p in U and a sequence \{q_{i}\} in ℂ converging to F (p) such that F^{−1}(q_{i}) ∩ U is empty. Hence, there is a number h > 0 such that whenever t ∈ F^{−1}(q_{i}) for any i, then |t − p| > h.

For each q_{i} as above, we have the factorization

F (x) − q_{i} = a(x − r_{1,i})(x − r_{2,i}) ⋯ (x − r_{k,i}),

where a is the leading coefficient of F and {r_{j,i}}_{j=1}^{k} = F^{−1}(q_{i}); k = degree of F.

Thus,

|F (p) − q_{i}| = |a||p − r_{1,i}| ⋯ |p − r_{k,i}| > |a|h^{k};

letting i get large yields a contradiction, so F (U) is open.

2) Now suppose that F: ℂ^{n} → ℂ is a non–constant polynomial mapping, with U open in ℂ^{n}, and n ≥ 2. Since F (x_{1}, x_{2}, …, x_{n}) is non–constant, we may assume that it is non–constant in one of the n coordinates, say x_{n−1}. Let

H_{t} = {(x_{1}, …, x_{n}) ∈ ℂ^{n} | x_{n} = t}.

Then H_{t} is naturally identified with ℂ^{n−1}, and U ∩ H_{t} is open in H_{t}, i.e. open in ℂ^{n−1}.

Now,

F (U) = ⋃_{t} F (U ∩ H_{t}),

and F |U ∩ H_{t} can be regarded as a non–constant polynomial in the n − 1 variables x_{1}, …, x_{n−1}; by the inductive hypothesis F (U ∩ H_{t}) is open in ℂ, so F (U) is open.

\underline{Corollary}. Let F: ℂ^{n} → ℂ be as in Lemma 1. Then if S is a dense subset of ℂ, F^{−1}(S) is dense in ℂ^{n}; in particular F^{−1}(ℂ − {0}) is dense in ℂ^{n}.

\underline{Proof}. If F^{−1}(S) is \underline{not} dense in ℂ^{n}, there would be a non–empty open set U in the complement of F^{−1}(S). By Lemma 1, F (U) would be open in the complement of S, which contradicts the density of S.

We recall a couple of well–known facts about polynomials, the details of which may be found in [3]. Let x_{1}, x_{2}, …, x_{n} be variables. The kth elementary symmetric function of the {x_{i}} is defined by

μ_{k}(x_{1}, …, x_{n}) = ∑(x_{i_{1}} ⋯ x_{i_{k}}),

where the sum is taken over all subsets {i_{1}, …, i_{k}} of {1, 2, …, n}. Then 1) If G is a symmetric polynomial in the {x_{i}}, i.e. if

G(x_{1}, …, x_{n}) = G(x_{β(1)}, …, x_{β(n)})

for all permutations β on n letters, then G can be written as a polynomial in the elementary symmetric functions μ_{k}(x_{1}, …, x_{n}).

2) If f ∈ P_{n} is given by

f(x) = x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{0},

then the coefficients of f are given by elementary symmetric functions of its roots {r_{i}}, i.e.

a_{n−k} = (−1)^{k}μ_{k}(r_{1}, …, r_{n}),

for k = 1, 2, …, n.

\underline{Lemma 2}. D_{n} is dense in M_{n}.

\underline{Proof}. Let △: P_{n} → ℂ be given by

△(f) = ∏_{i<j}(r_{i} − r_{j})^{2}

where the {r_{i}} are the roots of f; note that △(f) vanishes if and only if f has a multiple root. Clearly △(f) is invariant under any permutation of the roots {r_{i}}, so in light of our observations 1) and 2) above, △(f) is a polynomial in the coefficients of f. Hence, △ can be regarded as a polynomial mapping from ℂ^{n} to ℂ.

Let ∂: M_{n} → P_{n} be given by

∂(A) = det(xI − A).

Then ∂ can be regarded as a polynomial mapping from ℂ^{n^{2}} to ℂ^{n}, and the composition △ · ∂: M_{n} → ℂ can be viewed as a (non–constant) polynomial mapping from ℂ^{n^{2}} to ℂ. By the Corollary to Lemma 1, (△·∂)^{−1}(ℂ−{0}) is dense in M_{n} (i.e. in ℂ^{n^{2}} ); since a sufficient criterion for diagonalizability of an n × n matrix is the existence of n distinct eigenvalues, (△ · ∂)^{−1}(ℂ − {0}) ⊂ D_{n}, proving the lemma.

\underline{Lemma 3}. The Cayley–Hamilton theorem holds for diagonalizable matrices.

\underline{Proof}. Suppose that A is diagonalizable, so A = QDQ^{−1} for some invertible matrix Q and some diagonal matrix

D = ( \begin{matrix}r_{1} & 0 & … & 0 \\ 0 & r_{2} & … & 0 \\ . & . & … & . \\ 0 & . & … & r_{n} \end{matrix} ) .

If

p_{A}(x) = x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{1}x + a_{0}

then we have

\begin{aligned}p_{A}(A) & = (QDQ^{−1})^{n} + a_{n−1}(QDQ^{−1})^{n−1} + ⋯ + a_{1}(QDQ^{−1}) + a_{0}I \\ & = Q(p_{A}(D))Q^{−1}. \end{aligned}

But

p_{A}(D) = ( \begin{matrix}p_{A}(r_{1}) & … & 0 \\ . & & . \\ . & & . \\ 0 & … & p_{A}(r_{n}) \end{matrix} ) ,

which is the zero matrix since the r_{i} are merely the eigenvalues of A.

\underline{Proof of the Cayley–Hamilton Theorem}. The mapping Ω: M_{n} × P_{n} → M_{n} given by Ω(A, f(x)) = f(A) is continuous (it is really a polynomial map from ℂ^{n^{2}} × ℂ^{n} to ℂ^{n^{2}}), and the mapping Φ: M_{n} → M_{n} × P_{n} given by Φ(A) = (A, p_{A}(x)) is similarly continuous. Hence, the composition

Ω · Φ: M_{n} → M_{n},

which is given by Ω · Φ(A) = p_{A}(A), is continuous. By Lemmas 2 and 3, this mapping is identically zero on a dense subset of M_{n}, so by continuity vanishes everywhere.

\underline{Remark}. For an analogous proof of the Cayley–Hamilton theorem over an arbitrary field F, first replace F by its algebraic closure \overline{F}. Then, as in the above proof, identify P_{n} and M_{n} with \overline{F}^{n} and \overline{F}^{n^{2}} respectively, and give each of these spaces the Zariski topology (in which the closed sets are the zero–loci of finite sets of polynomials [4]). As in our proof, Lemma 2 is obtained by noting that D_{n} contains those matrices A for which the characteristic polynomial p_{A}(x) has distinct roots; this set is open and dense in the Zariski topology. The remainder of the proof is identical to that given above.

\underline{References}

More standard (i.e. purely algebraic) proofs of the Cayley–Hamilton theorem can be found in most linear algebra tests, e.g.

1. I. Friedberg and S. Friedberg, Linear Algebra, 2nd edition, Prentice Hall, 1989.

2. Walker, Introduction to Abstract Algebra, Random House, 1987.

3. Van der Waerden, Algebra, Vol. 1, Ungar Publishing Co., 1970.

For a good description of the Zariski topology, one can consult

4. R. Hartshorne, Algebraic Geometry, GTM, Springer–Verlag, 1977.

A complex–analytic proof of the Cayley–Hamilton theorem (valid only over ℂ) based on the Cauchy Integral Theorem can be found in

5. C. A. McCarthy, “The Cayley–Hamilton Theorem,” American Mathematical Monthly, 82 (April 1975). æ