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Let V be a K-vector space, \varphi : V \to V an K-endomorphism of V and f : K \to K a function. Here, we want to make sense of f(\varphi); this should be another endomorphism of V which is somehow related to both \varphi and f.

Let us make this more precise. For that, let A be a subalgebra of the K-algebra \End_K(V) of endomorphisms of V, containing the identity \id_V, and let F be a K-subalgebra of the K-algebra Fun(K) of functions K \to K, containing the identity \id_K and the constant functions. We say that \Psi : F \times A \to A is a functional calculus if \Psi satisfies the following conditions:

  1. for a fixed \varphi \in A, the map \Psi(\bullet, \varphi) : F \to A, f \mapsto \Psi(f, \varphi) is a K-algebra homomorphism with \Psi(1, \varphi) = \id_V and \Psi(\id_K, \varphi) = \varphi;
  2. for a fixed f \in F, the map \Psi(f, \bullet) : A \to A, \varphi \mapsto \Psi(f, \varphi) is K-algebra with \Psi(f, \id_V) = f(1) \id_V.

We usually write f(\varphi) for \Psi(f, \varphi) if it is clear which \Psi is meant.

Note that F contains all polynomial functions K \to K, i.e. the functions of the type \lambda \mapsto f(\lambda), where f \in K[x] is a polynomial. Note that for polynomial functions f, the value of \Psi(f(\id_K), \varphi) for f \in K[x] is completely determined by the fact that \Psi(\bullet, \varphi) is an K-algebra homomorphism with \Psi(1, \varphi) = \id_V and \Psi(\id_K, \varphi) = \varphi, as \Psi(\lambda, \varphi) = \lambda \Psi(1, \varphi) = \lambda \id_V: if f = \sum_{i=0}^n a_i x^i, then

\Psi(f(\id_K), \varphi) ={} & \Psi\biggl(\sum_{i=0}^n a_i (\id_K)^n, \varphi\biggr) \\ {}={} & \sum_{i=0}^n a_i \Psi(\id_K, \varphi)^n = \sum_{i=0}^n a_i \varphi^i = f(\varphi).

In particular, this gives a canonical functional calculus K[\id_K] \times \End_K(V) \to \End_K(V), where K[\id_K] is the image of the canonical map K[x] \to Fun(K). (In case you are curious, K[x] \cong K[\id_K] \subsetneqq Fun(K) if, and only if, K is infinite; in the other case, Fun(K) = K[\id_K] \cong K[x] / (x^q - x), where q = \abs{K} < \infty.)

What about functions which are not polynomial? In case \varphi is diagonalizable, i.e. V has a basis consisting of eigenvectors of \varphi, one can define f(\varphi) for an arbitrary function f : K \to K by defining f(\varphi) as the linear map which maps an eigenvector v of \varphi with eigenvalue \lambda to f(\lambda) v. If one sets A_\varphi := \{ f(\varphi) \mid f \in Fun(K) \}, one obtains a functional calculus Fun(K) \times A_\varphi \to A_\varphi.

In Functional Analysis, one is interested in such functional calculi with K = \R or \C, and one obtains ones for holomorphic functions f, for continuous functions f and even for certain Borel-measureable functions f. But for today, we want to stick to the situation of an arbitrary K. We will use A = \End_K(V) and F = K[\id_K], i.e. the canonical functional calculus.


Let \varphi \in \End_K(V). In case the canonical map K[x] \to \End_K(V), f \mapsto f(\varphi) is not injective, the unique normed generator of K[x] \to \End_K(V) is called the minimal polynomial of \varphi and denoted by \mu_f.

In case \dim_K V < \infty, every \varphi \in \End_K(V) has a minimal polynomial, as \dim_K \End_K(V) = (\dim_K V)^2 < \infty and \dim_K K[x] = \infty. In case \dim_K V = \infty, certain elements of \End_K(V) do have a minimal polynomial; for example, \varphi = \id_V has the minimal polynomal x - 1; other elements of \End_K(V) do not possess a minimal polynomial, for example any endomorphism with infinitely many different eigenvalues.


Assume that \varphi possesses a minimal polynomial. Then \lambda \in K is an eigenvalue of \varphi if, and only if, \mu_\varphi(\lambda) = 0.


In case \lambda is an eigenvalue, let v be an corresponding eigenvector and let W := \gen{v}. Then \End_K(W) \cong K, and \varphi|_W corresponds to \lambda. Clearly, 0 = \mu_\varphi(\varphi)|_W = \mu_\varphi(\varphi|_W) = \mu_\varphi(\lambda \id_W) = \mu_\varphi(\lambda) \id_W, whence \mu_\varphi(\lambda) = 0.

Conversely, assume that \mu_\varphi(\lambda) = 0. Write \mu_\varphi = (x - \lambda)^n f, where n \in \N and f \in K[x] satisfies f(\lambda) \neq 0. As \mu_\varphi(\lambda) = 0, n > 0. Now 0 = \mu_\varphi(\varphi) = (\varphi - \lambda \id_V)^n \circ f(\varphi). In case \ker (\varphi - \lambda \id_V)^n \neq 0, \lambda is an eigenvalue of \varphi (let v \in \ker (\varphi - \lambda \id_V)^n \setminus 0 and choose i \in \N maximal with w := (\varphi - \lambda \id_V)^i v \neq 0; then \varphi(w) = \lambda w); hence, assume \ker (\varphi - \lambda \id_V)^n = 0. In that case, we must have f(\varphi) = 0. But as f is a proper divisor of \mu_\varphi, this cannot be.

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


Let f be a prime divisor of \mu_\varphi. Then

\GEig(\varphi, f) := \{ v \in V \mid \exists n : f(\varphi)^n(v) = 0 \}

is an \varphi-invariant subspace of V. If g is another prime divisor of \mu_\varphi coprime to f, then g(\varphi) is \GEig(\varphi, f)-invariant and g(\varphi)|_{\GEig(\varphi, f)} is an monomorphism.

If f is an arbitrary prime polynomial, \GEig(\varphi, f) \neq 0 if, and only if, f \mid \mu_\varphi.


Clearly, \GEig(\varphi, f) = \bigcup_{n=0}^\infty \ker f(\varphi)^n. As \ker f(\varphi)^n \subseteq \ker f(\varphi)^{n+1}, this is a subspace of V. As \ker f(\varphi)^n is \varphi-invariant (as f(\varphi) \circ \varphi = (f x)(\varphi) = (x f)(\varphi) = \varphi \circ f(\varphi)), it follows that \GEig(\varphi, f) is \varphi-invariant as well.

As g(\varphi) f(\varphi) = (f g)(\varphi) = f(\varphi) g(\varphi), \GEig(\varphi, f) is g(\varphi)-invariant as well. Let v \in \GEig(\varphi, f) \cap \ker g(\varphi) and let n \in \N be minimal with f(\varphi)^n(v) = 0. As f^n, g are coprime, there exist h, h' \in K[x] with 1 = h f^n + h' g. Then 0 = h(\varphi) f(\varphi)^n(v) + h'(\varphi) g(\varphi)(v) = (h f^n + h' g)(v) = v. Therefore, g(\varphi)|_{\GEig(\varphi, f)} is injective.

Finally, the last statement can be proven in exactly the same way as the previous lemma.


Let \mu_\varphi = \prod_{i=1}^n f_i^{e_i}, where f_1, \dots, f_n is a set of pairwise distinct monic prime polynomials and e_i \in \N_{\ge 1}. Then \bigoplus_{i=1}^n \GEig(\varphi, f_i) is a direct sum.


Assume that this is not a direct sum. Then there exists v_i \in \GEig(\varphi, f_i), not all zero, such that 0 = \sum_{i=1}^n v_i. Assume that the number of non-zero v_i is minimal under this condition. Let i be with v_i \neq 0, and let n \in \N satisfy f_i(\varphi)^n(v_i) = 0. Then 0 = \sum_{j=1}^n f_i(\varphi)^n(v_j), and f_i(\varphi)^n(v_i) = 0. If j \neq i, then f_i(\varphi)^n(v_j) \neq 0 as f_i(\varphi)|_{\GEig(\varphi, f_j)} is injective and so is its n-th power. But this is only possible by the minimality assumption of v_j = 0 for all j \neq i, i.e. 0 = v_i, contradicting the choice of i. Therefore, the sum is a direct sum.


Let f be a prime factor of \mu_\varphi and let e \in \N be the maximal exponent of f appearing in \mu_\varphi, i.e. f^e \mid \mu_\varphi and f^{e+1} \nmid \mu_\varphi. Then \ker f(\varphi)^e = \GEig(\varphi, f).


Let v \in \GEig(\varphi, f), and write \mu_\varphi = f^e \cdot g with g \in K[x]; then g and f are coprime. We have to show f(\varphi)^e(v) = 0. Clearly w := f(\varphi)^e(v) lies in the kernel of g(\varphi), as g f^e = \mu_\varphi. Let n \in \N be such that w \in \ker f(\varphi)^n; as f^n and g are coprime, there exist h, h' \in K[x] with f^n h + g h' = 1. Therefore,

0 ={} & h(\varphi) f(\varphi)^n(w) + h'(\varphi) g(\varphi)(w) \\ {}={} & (h f^n + h' g)(\varphi)(w) = w = f(\varphi)^e(v).

Note that one can in fact show that \image f(\varphi)^{e+1} = \image f(\varphi)^e.


Let f be a prime factor of \mu_\varphi. Then there exists a \varphi-invariant subspace W \subseteq V with V = W \oplus \GEig(\varphi, f).


Write \mu_\varphi = f^e \cdot g with e \in \N, g \in K[x] and f \nmid g. Then \GEig(\varphi, f) = \ker f(\varphi)^e and, in particular, \ker f(\varphi)^e = \ker f(\varphi)^{e+i} for every i \in \N. Let v \in \image f(\varphi)^e \cap \ker f(\varphi)^e = 0; we can write v = f(\varphi)^e(w) for some w \in V. Now f(\varphi)^{2 e}(w) = 0, whence w \in \ker f(\varphi)^{2 e} = \ker f(\varphi)^e, i.e. v = f(\varphi)^e(w) = 0. Therefore, \image f(\varphi)^e \cap \GEig(\varphi, f) = 0.

Let h, h' \in \N such that 1 = h f^e + h' g. Let v \in V; then v = f^e(\varphi) (h(\varphi)(v)) + g(\varphi) (h'(\varphi)(v)) = f(\varphi)^e(w_1) + g(\varphi)(w_2) with w_1, w_2 \in V. Now f^e g = \mu_\varphi, whence 0 = \mu_\varphi(w_2) = f(\varphi)^e g(\varphi)(w_2), i.e. g(\varphi)(w_2) \in \ker f(\varphi)^e. As f(\varphi)^e(w_1) \in \image f(\varphi)^e, we see v \in \image f(\varphi)^e + \ker f(\varphi)^e = \image f(\varphi)^e + \GEig(\varphi, f).

Hence, we get V = \image f(\varphi)^e \oplus \GEig(\varphi, f). Finally, note that W := \image f(\varphi)^e is clearly \varphi-invariant.

Theorem (Generalized Jordan Decomposition).

Let \mu_\varphi = \prod_{i=1}^n f_i^{e_i}, where f_1, \dots, f_n is a set of pairwise distinct monic prime polynomials and e_i \in \N_{\ge 1}. Then V = \bigoplus_{i=1}^n \GEig(\varphi, f_i).


We show this by induction on n. In case n = 0, we have \mu_\varphi = 1, which is only possible if V = 0. In that case, the statement is obvious. Hence, assume n > 0.

Let W_1 := \GEig(\varphi, f_n) and W_2 be an \varphi-invariant direct complement of W_1. Clearly f_n(\varphi)^{e_n} is injective on W_2, whence f_n(\varphi)^{e_n} \circ \prod_{i=1}^{n-1} f_i(\varphi)^{e_i} = \mu_\varphi(\varphi|_{W_2}) = 0 implies \prod_{i=1}^{n-1} f_i(\varphi)^{e_i} = 0. Therefore, \mu_{f|_{W_2}} has strictly less than n distinct prime factors, whence W_2 = \bigoplus_{i=1}^{n-1} \GEig(\varphi|_{W_2}, f_i). In particular,

V = W_1 \oplus W_2 ={} & \GEig(\varphi, f_n) \oplus \bigoplus_{i=1}^{n-1} \GEig(\varphi|_{W_2}, f_i) \\ {}\subseteq{} & \GEig(\varphi, f_n) + \sum_{i=1}^{n-1} \GEig(\varphi, f_i),

whence the claim follows.

Note that this is a generalization of the Jordan decomposition. Note that in fact, \bigoplus_{i=1}^n \GEig(\varphi, f_i) is the minimal K[\varphi]-decomposition of V in case \mu_\varphi = \prod_{i=1}^n f_i^{e_i}. This completes the task started in my post on such decompositions, namely finding minimal K[\varphi]-decompositions in case the characteristic polynomial of \varphi (assuming \dim_K V < \infty) does not splits into linear factors.


Assume that \varphi has a minimal polynomial \mu_\varphi of the form f^e, where f is prime and e \in \N. Let \varphi_n := f(\varphi) and \varphi_d := \varphi - \varphi_n. Then \varphi_d \varphi_n = \varphi_n \varphi_d and \varphi_n is nilpotent of index e. Moreover, \varphi_d is diagonalizable if, and only if \deg f = 1. Finally, \varphi is diagonalizable if, and only if, \deg f = 1 and e = 1.


Clearly, \varphi_d \varphi_n = \varphi_n \varphi_d as both are elements of K[\varphi] \cong K[x]/(\mu_\varphi). Moreover, \varphi_n = f(\varphi), whence \varphi_n is nilpotent of index e.

If \deg f = 1, f = x - \lambda for some \lambda \in K. In that case, \varphi_d = \varphi - \varphi_n = \lambda \id_V. Conversely, if \varphi_d is diagonalizable, any eigenvalue of \varphi_d must be a zero of f^e. This is only possible if \deg f = 1.

Finally, assume that \varphi is diagonalizable. Hence, \varphi_n is diagonalizable as well; but the only diagonalizable and nilpotent endomorphism is 0, whence e = 1 and \varphi_d = \varphi is diagonalizable, i.e. \deg f = 1. Conversely, assume \deg f = 1 and e = 1; then \varphi_n = 0 and \varphi = \varphi_d is diagonalizable.


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


Write \mu_\varphi = \prod_{i=1}^n f_i^{e_i} with e_i \in \N and pairwise distinct, monic prime polynomials f_i. Then V = \bigoplus_{i=1}^n \GEig(\varphi, f_i) by the generalized Jordan decomposition. Hence \varphi is diagonalizable if, and only if, \varphi|_{\GEig(\varphi, f_i)} is diagonalizable for every i. For a fixed i, we have that \mu_{\varphi|_{\GEig(\varphi, f_i)}} = f_i^{e_i}, whence by the previous lemma, \varphi|_{\GEig(\varphi, f_i)} is diagonalizable if, and only if, \deg f_i = 1 and e_i = 1.


Assume that \varphi has a minimal polynomial. Then there exist polynomials f_d, f_n \in K[x] such that \varphi_n = f_n(\varphi) and \varphi_d = f_d(\varphi), where \varphi_n, \varphi_d are the endomorphisms from the previous corollary.


As \varphi_n + \varphi_d = \varphi, it suffices to show the existence of f_n. Write \mu_\varphi = \prod_{i=1}^n f_i^{e_i} with e_i \in \N and pairwise distinct monic primes f_i, and set V_i := \GEig(\varphi, f_i). We want a polynomial f \in K[x] such that f(\varphi)|_{V_i} = f_i(\varphi)|_{V_i}. Now the minimal polynomial of \varphi|_{V_i} is f_i^{e_i}, whence f(\varphi)|_{V_i} = (f \mymod f_i^{e_i})(\varphi)|_{V_i}, i.e. it suffices to solve the congruences f \equiv f_i \pmod{f_i^{e_i}}, i = 1, \dots, n. But since f_i^{e_i}, 1 \le i \le n, are pairwise coprime, such an f exists by the Chinese Remainder Theorem.

Corollary (Generalized Jordan Decomposition).

Assume that \varphi 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 \varphi_d, \varphi_n \in \End_K(V) such that

  1. \varphi = \varphi_n + \varphi_d and \varphi_d \varphi_n = \varphi_n \varphi_d;
  2. \varphi_n is nilpotent;
  3. if L is a splitting field of \mu_\varphi over L, \varphi_n \otimes_K L \in \End_L(V \otimes_K L) is diagonalizable.

By the previous lemma and corollary, there exist polynomials f_n, f_d \in K[x] with \varphi_n = f_n(\varphi), \varphi_d = f_d(\varphi) such that \varphi_n, \varphi_d satisfy the conditions. (Note that \mu_{\varphi_d \otimes_K L} = \mu_{\varphi_d} = \prod_{i=1}^n f_i, and since L/K is separable, \prod_{i=1}^n f_i is squarefree and splits into linear factors over L. Hence, by the second-previous lemma, \varphi_d \otimes_K L is diagonalizable.)

Now let \varphi'_n, \varphi'_d be any two endomorphisms which satisfy the conditions above. As \varphi'_n + \varphi'_d = \varphi and \varphi'_n \varphi'_d = \varphi'_d \varphi'_n, all of \varphi'_n, \varphi'_d, \varphi_n and \varphi_d commute with each other. Hence, we have \varphi'_n - \varphi_n = \varphi_d - \varphi'_d, and \varphi'_n - \varphi_n is nilpotent and (\varphi_d - \varphi'_d) \otimes_K L is diagonalizable. But this is possible if, and only if, \varphi'_n - \varphi_n = \varphi_d - \varphi'_d = 0, i.e. if \varphi_n = \varphi'_n and \varphi_d = \varphi'_d.

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 \varphi has a minimal polynomial which is separable. Let \varphi = \varphi_d + \varphi_n be the generalized Jordan decomposition of \varphi, and let f \in K[x]. Finally, let e be the nilpotence index of \varphi_n, i.e. let e satisfy \varphi_n^e = 0. Then

f(\varphi) = \sum_{i=0}^{e-1} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.

Consider L := K(x), the rational function field. The Taylor expansion of f(t) \in L[t] around \lambda = x \in K(x) is given by f(t) = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(x) (t - x)^i. Here, we have in fact \frac{f^{(i)}}{i!}(x) \in K[x]. As \varphi_n, \varphi commute, we can plug in x = \varphi_d and t = \varphi and obtain

f(\varphi) = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) (\varphi - \varphi_d)^i = \sum_{i=0}^{\deg f} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.

Now \varphi_n^i = 0 for i \ge e gives the formula.

Note that in case K = \R or \C, this formula holds also for arbitrary analytic functions f : K \to K. In fact, the function only needs to be analytic on an open set which contains the complex eigenvalues of \varphi. The most important example is the exponential function \exp : \C \to \C, z \mapsto \sum_{i=0}^\infty \frac{z^i}{i!}. The above shows that every \varphi \in \End_\C(V) possessing a minimal polynomial can be decomposed into a diagonalizable part \varphi_d and a nilpotent part \varphi_n of finite index e, and in that case,

\exp(\varphi) = \sum_{i=0}^{e-1} \frac{\exp(\varphi_d)}{i!} \varphi_n^i = \exp(\varphi_d) \sum_{i=0}^{e-1} \frac{\varphi_n^i}{i!}.

Now let \dim_K V < \infty. Recall the the characteristic polynomial of \varphi \in \End_K(V) is defined as c_\varphi := \det(\varphi - t \id_V) \in K[t]. 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 f is an irreducible prime, then f divides \mu_\varphi if, and only if, f divides c_\varphi.


Let L be a splitting field of f over K, and consider \varphi_L := \varphi \otimes_K L \in \End_L(V \otimes_K L). We have \mu_{\varphi_L} = \mu_\varphi and c_{\varphi_L} = c_\varphi, whence it suffices to show that c_{\varphi_L}(\lambda) = 0 if, and only if, \mu_{\varphi_L}(\lambda) = 0 for every \lambda \in L.

For that, note that \mu_{\varphi_L}(\lambda) = 0 if, and only if, \lambda is an eigenvalue of \varphi_L. But this is equivalent to \varphi_L - \lambda \id_{V \otimes_K L} not being injective, which in turn is equivalent (as \dim_L (V \otimes_K L) = \dim_K V < \infty) to that \varphi_L - \lambda \id_{V \otimes_K L} is not invertible, which is the case if, and only if, \det(\varphi_L - \lambda \id_{V \otimes_K L}) = 0, i.e. c_{\varphi_L}(\lambda) = 0.

In fact, we can show that \mu_\varphi divides c_\varphi, which implies the Cayley-Hamilton theorem as \mu_\varphi(\varphi) = 0. For that, we show that \dim \GEig(\varphi, f) = \nu_f(c_\varphi) \deg f for every prime polynomial f, where \nu_f : K[x] \setminus \{ 0 \} \to \N gives the exponent of f in the prime factor decomposition of a non-zero element of K[x].


Assume \dim_K V < \infty. If \varphi = \varphi_d + \varphi_n with \varphi_n = f(\varphi) being nilpotent, where f \in K[x], then c_\varphi = c_{\varphi_d}.


Let L be a splitting field of c_\varphi over K, and let \varphi_L := \varphi \otimes_K L, \varphi_{d,L} := \varphi_d \otimes_K L and \varphi_{n,L} := \varphi_n \otimes L. It then suffices to show that the statement holds for these L-endomorphisms of V \otimes_K L. Hence, we can assume that c_\varphi splits over K. In that case, there exists a basis B of V such that the representation matrix M_B(\varphi) of \varphi with respect to B is in upper triangular form. Then c_\varphi = \prod (x - \lambda), where \lambda ranges over the diagonal elements of M_B(\varphi). Now \varphi_n = f(\varphi), whence M_B(\varphi_n) is in upper triangular form as well. As \varphi_n is nilpotent, the diagonal elements of M_B(\varphi_n) must all be zero. As M_B(\varphi) = M_B(\varphi_d) + M_B(\varphi_n), we see that c_{\varphi_d} = c_\varphi.


Assume that \varphi has a minimal polynomial. Let f := \prod_{i=1}^n f_i^{e_i}, where f_1, \dots, f_n are pairwise distinct irreducible polynomials. Set

W := \{ v \in V \mid \exists n \in \N : f(\varphi)^n(v) = 0 \}.


W = \bigoplus_{i=1}^n \GEig(\varphi, f_i).

Let v_i \in \GEig(\varphi, f_i) and let t_i \in \N with f_i(\varphi)^{t_i}(v_i) = 0; then, if t := \max\{ t_1, \dots, t_n \}, w = \sum_{i=1}^n v_i satisfies

f(\varphi)^t(v) = \sum_{i=1}^n f(\varphi)^t(v_i) = \sum_{i=1}^n \prod_{j=1 \atop j \neq i}^n f_j(\varphi)^{e_j} \circ f_i(\varphi)^{e_i t}(v_i);

as f_i(\varphi)^{e_i t}(v_i) = 0 since e_i t \ge t \ge t_i, we get \bigoplus_{i=1}^n \GEig(\varphi, f_i) \subseteq W.

For the converse, first note that W is \varphi-invariant. Assume \mu_{\varphi|_W} has a monic prime factor p distinct from p_1, \dots, p_n. Then \dim \GEig(\varphi|_W, p) > 0; let w \in \GEig(\varphi|_W, p) \setminus \{ 0 \}. Let t \in \N be such that p(\varphi)^t(w) = 0 and s \in \N be such that f(\varphi)^s(w) = 0. As p and f are coprime, there exist polynomials h, h' \in K[x] with h p^t + h' f^s = 1. Hence,

0 = h(\varphi) p(\varphi)^t(w) + h'(\varphi) f(\varphi)^s(w) = (h p^t + h' f^s)(\varphi)(w) = w,

a contradiction. Hence, all prime factors \mu_{\varphi|_W} lie in \{ p_1, \dots, p_n \}. Therefore,

W = \bigoplus_{i=1}^n \GEig(\varphi|_W, p_i) \subseteq \bigoplus_{i=1}^n \GEig(\varphi, p_i) \subseteq W,

which shows the claim.


Assume \varphi has a minimal polynomial, and let f be a prime polynomial. Let L be a field extension of K over which f splits; write f = \prod_{i=1}^n (x - \lambda_i)^{e_i} with distinct elements \lambda_1, \dots, \lambda_n \in L and e_i \in \N. Then

\GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^n \GEig(\varphi \otimes_K L, x - \lambda_i).

Notice that \GEig(\varphi, f) \otimes_K L = \{ v \in V \otimes_K L \mid \exists n \in \N : f(\varphi \otimes_K L)^n(v) = 0 \}. Hence, the corollary follows from the previous lemma.


Let \dim_K V < \infty and f be a prime polynomial. Then \dim \GEig(\varphi, f) = \nu_f(c_\varphi) \deg f.


We first show that “\le” holds. Let L be a splitting field of f over K, and write f = \prod_{i=1}^t (x - \lambda_i)^{e_i} with \lambda_1, \dots, \lambda_t \in L pairwise distinct and e_i \in \N. We have \GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^t \GEig(\varphi \otimes_K L, x - \lambda_i); since \nu_{f_i}(c_{\varphi \otimes_K L}) = e_i \nu_f(c_\varphi), it suffices to know that the theorem holds in case \deg f = 1, as then \dim \GEig(\varphi \otimes_K L, x - \lambda_i) = \nu_{x - \lambda_i}(c_{\varphi \otimes_K L}) and, therefore,

\dim_L (\GEig(\varphi, f) \otimes_K L) ={} & \sum_{i=1}^t \dim_L \GEig(\varphi \otimes_K L, x - \lambda_i) \\ {}={} & \sum_{i=1}^t e_i \nu_f(c_\varphi) = \deg f \cdot \nu_f(c_\varphi).

Hence, assume that f = x - \lambda with \lambda \in K. In that case, W := \GEig(\varphi, f) = \GEig(\varphi, \lambda). Let e = \nu_f(c_\varphi) and write c_\varphi = (x - \lambda)^e g with g \in K[x]. Note that c_{\varphi|_W} divides c_\varphi. But \varphi - f(\varphi) is diagonalizable on W with only the eigenvalue \lambda, whence c_{\varphi|_W} = c_{(\varphi - f(\varphi))|_W} = (x - \lambda)^{\dim W}. Therefore, \dim W \le e.

The above argument shows \dim \GEig(\varphi, f) \le \nu_f(c_\varphi) \deg f. If p_1, \dots, p_n are all distinct prime factors of c_\varphi, we get

\dim_K V ={} & \sum_{i=1}^n \dim_K \GEig(\varphi, p_i) \ {}\le{} & \sum_{i=1}^n \nu_{p_i}(c_\varphi) \deg p_i = \deg c_\varphi = \dim_K V;

as all summands are \ge 0, the theorem follows.

Corollary (Cayley-Hamilton over Fields).

If \dim_K V < \infty and \varphi \in \End_K(V), then c_\varphi(\varphi) = 0.


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