Felix' Math Place » Analysis http://math.fontein.de Focussed on, but not limited to Computational Number Theory Sat, 30 Jul 2011 12:35:49 +0000 en hourly 1 http://wordpress.org/?v=3.1 Inequalities. http://math.fontein.de/2010/02/09/inequalities/ http://math.fontein.de/2010/02/09/inequalities/#comments Tue, 09 Feb 2010 06:29:04 +0000 Felix Fontein http://math.fontein.de/?p=732 Inspired by a tweet, I decided to play a bit with Graphviz to create a graph on which inequalities imply which ones. Here’s the result:

Click here to obtain a larger version. Click any box on the image to obtain the Wikipedia page on the inequality.

]]> http://math.fontein.de/2010/02/09/inequalities/feed/ 3 The Hasse derivative, part II: Multivariate partial Hasse derivatives. http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/ http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/#comments Fri, 02 Oct 2009 19:47:27 +0000 Felix Fontein http://math.fontein.de/?p=474 suggestion by A. Maevskiy, we show how the Hasse derivative can be extended to partial Hasse derivative in arbitrary multivariate polynomial rings. We show multivariate versions of Taylor's Formula, of the Identity Theorem, and of the Generalized Leibnitz Rule.]]> Let R be a commutative unitary ring, and let x_i, i \in I be a family of indeterminates over R. We consider the polynomial ring S = R[\{ x_i \mid i \in I \}], obtained from R by adjoining all indeterminates x_i, i \in I. The two most used examples are I = \{ \bullet \}, where S = R[x], and I = \{ 1, \dots, n \}, where S = R[x_1, \dots, x_n]. (Note that if we use finitely many polynomials f_1, \dots, f_r, there exist a finite subset J \subset I such that f_1, \dots, f_r \in R[\{ x_j \mid j \in J \}], i.e. it suffices to consider finitely many variables.)

In classical analysis, one defines the partial derivative by fixing all variables but one, and then using the classical one-dimensional derivative. This can be done similarly in the case of Hasse derivatives. Given f \in S, i \in I and s \in R^I, we can define f_i(s) = f|_{x_j = s(j) \text{ for } j \neq i}, which is an element of R[x_i]. Then D^{(k)} f_i(s) is another element of R[x_i], whence we can define D^{(k)}_{x_i} f(s) := (D^{(k)} f_i(s))|_{x_i = s(i)} = (D^{(k)} f_i(s))(s(i)). This gives a function \displaystyle D^{(k)}_{x_i} f : R^I \to R, \quad s \mapsto D^{(k)}_{x_i} f(s), which can be considered as the k-th partial Hasse derivative of f with respect to x_i. Now we would like the derivative to be another element of S. In classical analysis, polynomial functions are in bijection to polynomials, and one can show using certain rules that the partial derivative of a polynomial is again a polynomial. In case R is an arbitrary ring, one has in general no longer the bijection between polynomials and polynomial functions.

In the case of derivatives over arbitrary rings, we have the advantage that we can naturally identify S with R_i[x_i], where R_i := R[\{ x_j \mid j \in I, j \neq i \}]; here, R_i is another commutative unitary ring. Now R_i[x_i] is a univariate polynomial ring, whence we have the usual Hasse derivative. Denote the k-th Hasse derivative D^{(k)} : R_i[x_i] \to R_i[x_i] by D^{(k)}_{x_i}; then we obtain an R-linear (in fact, R_i-linear) operator D^{(k)}_{x_i} : S \to S. In fact, if we evaluate D^{(k)} f at a point s \in R^I, we obtain the same value as D^{(k)} f(s) defined above. We will prove this in a minute; before that we will prove a more general result on D^{(k)}_{x_i}.

Proposition.
Let \varphi : R \to R' be a homomorphism of unitary commutative rings R and R' (i.e. it additionally satisfies \varphi(1_R) = 1_{R'}). If x is an indeterminate over R and R', and \varphi^* : R[x] \to R'[x] the natural continuation of \varphi with \varphi^*(x) = x and \varphi^*(r) = \varphi(r) for all r \in R, then for f \in R[x], we have \displaystyle D^{(k)} \varphi^*(f) = \varphi^*(D^{(k)} f).
Proof.
This follows directly from the definition of the Hasse derivative: if f = \sum_{i=0}^n a_i x^i \in R[x] and k \in \N, we have \varphi^*(D^{(k)} f) ={} & \varphi^*\biggl( \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \biggr) \\ {}={} & \sum_{i=k}^n \binom{i}{k} \varphi(a_i) x^{i - k} = D^{(k)} \varphi^*(f).
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Corollary.
For all f \in S and s \in R^I, we have (D^{(k)}_{x_i} f)(s) = D^{(k)} f(s).
Proof.
Let \varphi : R_i \to R be the substitution homomorphism f \mapsto f|_{x_j = s(j) \text{ for } j \neq i}. We interpret f as an element of R_i[x_i]; note that the above defined f_i equals \varphi^*(f). Then, by the proposition, \displaystyle D^{(k) }\varphi^*(f) = \varphi^*(D^{(k)} f) = \varphi^*(D^{(k)}_{x_i} f). Now \varphi^*(D^{(k)}_{x_i} f)|_{x_i = s(i)} = (D^{(k)}_{x_i} f)(s), whence \displaystyle (D^{(k)}_{x_i} f)(s) = (D^{(k)} \varphi^*(f))|_{x_i = s(i)} = (D^{(k)} f_i)|_{x_i = s(i)} = D^{(k)}_{x_i} f(s) by definition of D^{(k)}_{x_i} f(s).
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Hence, the following definition makes sense:

Definition.
The R_i-linear opeator D^{(k)}_{x_i} : S \to S is called the k-th partial Hasse derivative with respect to x_i.

As in the case of usual partial derivatives (in the case of “nice” functions, i.e. the partial derivatives are continuous in a neighborhood of the point we are interested in), the Hasse derivatives commute:

Proposition.
Let i, j \in I, k, \ell \in \N and f \in S. Then D^{(k)}_{x_i} D^{(\ell)}_{x_j} f = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f.

For this, consider the notation R_{i,j} = R[\{ t \in I \mid t \neq i, j \}]; then S = R_{i,j}[x_i, x_j] = R_{i,j}[x_j][x_i] = R_{i,j}[x_i][x_j].

Proof.
By R_{i,j}-linearity of D^{(k)}_{x_i} and D^{(\ell)}_{x_j}, it suffices to show the result for f = x_i^s x_j^t with s, t \in \N. Now D^{(k)}_{x_i} D^{(\ell)}_{x_j} f ={} & D^{(k)}_{x_i} \binom{t}{\ell} x_i^s x_j^{t - \ell} = \binom{s}{k} \binom{t}{\ell} x_i^{s - k} x_j^{t - \ell} \\ {}={} & D^{(\ell)}_{x_j} \binom{s}{k} x_i^{s - k} x_j^t = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f.
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Now, let us define the following notation. We let \N^{(I)} be the set of functions I \to \N which are zero for all but finitely many elements of I. In case I is finite, \N^{(I)} = \N^I in the usual sense. The set \N^{(I)} will be the set of multiindices we use. For s \in \N^{(I)}, define D^{(s)} : S \to S by D^{(s)} := D^{(s(i_1))}_{x_{i_1}} \circ \dots \circ D^{(s(i_t))}_{x_{i_t}}, where \{ i_1, \dots, i_t \} \subseteq I is the support of s. By the above proposition, D^{(s)} is independent of the order of i_1, \dots, i_t, and as D^{(0)}_{x_i} = \id_S, we can also include elements outside the support. We sometimes use the more suggestive notation \displaystyle D^{(s)} = \prod_{i \in I} D^{(s(i))}_{x_i}, which is hence also justified. Moreover, for s \in \N^{(I)} and \lambda \in R^I, define (x - \lambda)^s := \prod_{j=1}^t (x_j - \lambda(j))^{s(j)} \in S. Again, this is well-defined. Using this notation, we obtain Taylor’s formula:

Theorem (Taylor formula).
For f \in S and \lambda \in R^I, we have \displaystyle f = \sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s.
Proof.
By the R-linearity of the equality, it suffices to consider f = x_{i_1}^{e_1} \cdots x_{i_t}^{e_t} with i_1, \dots, i_t \in I pairwise distinct and e_1, \dots, e_t \in \N. Define I' := \{ i_1, \dots, i_t \}. If s \in \N^{(I)} satisfies s(i) \neq 0 for i \not\in I', then D^{(s)} f = 0. Hence, \displaystyle \sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s \in \N^{(I')}} (D^{(s)} f)(\lambda) (x - \lambda)^s; moreover, one can restrict to the subring S' = R[x_{i_1}, \dots, x_{i_t}], as every appearing object lies in that ring. Hence, it suffices to show the result for I = I', i.e. for a finite index set I. We can assume I = \{ 1, \dots, n \}.
But now we can prove this by induction on n. If we write s = (s', s_n) with s' \in \N^{n-1}, s_n \in \N and \lambda = (\lambda', \lambda_n) with \lambda' \in R^{n-1}, \lambda_n \in R for s \in \N^n and \lambda \in R^n, then \displaystyle D^{(s)} f = D^{(s_n)} D^{(s')} f \quad \text{and} \quad (x - \lambda)^s = (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n}. Hence, & \sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s \\ {}={} & \sum_{s \in \N^n} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} \bigl(D^{(s_n)} (D^{(s')} f)(\lambda') \bigr)(\lambda_n) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} D^{(s_n)} \biggl( \sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} \biggr)(\lambda_n) (x_n - \lambda_n)^{s_n}. Now, by induction hypothesis (n - 1 with the base ring R[x_n]), \displaystyle \sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} = f, whence \displaystyle \sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s_n \in \N} D^{(s_n)} f(\lambda_n) (x_n - \lambda_n)^{s_n}. But by the classical univariate case which we already proved, this equals f itself.
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Again, we get the Identity Theorem as a direct corollary:

Corollary (Identity Theorem).
Let f \in S and \lambda \in R^I. If D^{(s)} f(\lambda) = 0 for all s \in \N^{(I)}, then f = 0.
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In a similar manner as Taylor’s formula, we can also adapt other one-dimensional results to the multivariate case.

For example, for s, t \in \N^{(I)}, let us define \binom{s + t}{s} := \prod_{j=1}^u \binom{s(i_j) + t(i_j)}{s(i_j)} if both s and t have no non-zero values outside \{ i_1, \dots, i_u \}. Note that \binom{s + t}{s} = \binom{s + t}{t}. We now obtain the following:

Proposition.
For s, t \in \N^{(I)}, we have D^{(s)} \circ D^{(t)} = \binom{s + t}{s} D^{(s + t)}. In particular, D^{(s)} \circ D^{(t)} = D^{(t)} \circ D^{(s)}.
Proof.
Let R' denote the ring R[x_j \mid j \neq i_1, \dots, i_u\}], where s and t have no non-zero values outside \{ i_1, \dots, i_u \}; then we have reduced to the ring R'[x_{i_1}, \dots, x_{i_u}], i.e. to the case of I being finite, say I = \{ 1, \dots, n \}. Write s = (s_1, \dots, s_n) and t = (t_1, \dots, t_n). As D^{(k)}_{x_i} and D^{(\ell)}_{x_j} commute for i \neq j, we obtain \displaystyle D^{(s)} \circ D^{(t)} = \prod_{i=1}^n (D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i}). Using the corresponding univariate result, we have D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i} = \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i}. Hence, \displaystyle D^{(s)} \circ D^{(t)} = \prod_{i=1}^n \biggl( \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i} \biggr) = \binom{s + t}{s} D^{(s + t)}.
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We also obtain the Leibniz rule:

Proposition (Generalized Leibniz Rule).
For f_1, \dots, f_m \in S and s \in \N^{(I)}, we have \displaystyle D^{(s)} \prod_{i=1}^m f_i = \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i; here, the sum ranges over all (s_1, \dots, s_m) \in (\N^{(I)})^m with s_1 + \dots + s_m = s. As a special case, for f, g \in S, we have \displaystyle D^{(s)}(f g) = \sum_{s' + s'' = s} D^{(s')}(f) D^{(s'')}(g).
Proof.
Using the standard argument, we reduce to the case I = \{ 1, \dots, n \} and we use the standard decomposition \N^n = \N^{n-1} \times \N. Hence, \displaystyle \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i = \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i, where the second sum ranges over all (s'_1, \dots, s_ m') \in (\N^{n-1})^m and the third sum ranges over all (s_{1,n}, \dots, s_{m,n}) \in \N^m. Now & \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i \\ {}={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_{i,n})} f_i, and applying the univariate result to the base ring R[x_1, \dots, x_{n-1}], we obtain \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i ={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} D^{(s_n)}_{x_n} \prod_{i=1}^m f_i \\ {}={} & D^{(s_n)}_{x_n} \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \prod_{i=1}^m f_i. By induction hypothesis, this equals \displaystyle D^{(s_n)}_{x_n} D^{(s')} \prod_{i=1}^m f_i = D^{(s)} \prod_{i=1}^m f_i.
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]]> http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/feed/ 2 Functional Calculus in Linear Algebra, the Jordan Decomposition Reloaded and Cayley-Hamilton’s Theorem. http://math.fontein.de/2009/08/13/functional-calculus-in-linear-algebra-the-jordan-decomposition-reloaded-and-cayley-hamiltons-theorem/ http://math.fontein.de/2009/08/13/functional-calculus-in-linear-algebra-the-jordan-decomposition-reloaded-and-cayley-hamiltons-theorem/#comments Thu, 13 Aug 2009 06:22:41 +0000 Felix Fontein http://math.fontein.de/?p=308 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.

Definition.
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.

Lemma.
Assume that \varphi possesses a minimal polynomial. Then \lambda \in K is an eigenvalue of \varphi if, and only if, \mu_\varphi(\lambda) = 0.
Proof.
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.
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The minimal polynomial is a rather powerful tool. In case it exists, one gets an \varphi-invariant decomposition of V as follows:

Lemma.
Let f be a prime divisor of \mu_\varphi. Then \displaystyle \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.
Proof.
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.
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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.
Proof.
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.
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Lemma.
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).
Proof.
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).
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Note that one can in fact show that \image f(\varphi)^{e+1} = \image f(\varphi)^e.

Lemma.
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).
Proof.
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.
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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).
Proof.
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.
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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.

Lemma.
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.
Proof.
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.
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Corollary.
Assume that \varphi has a minimal polynomial. Then \varphi is diagonalizable if, and only if, \mu_\varphi is squarefree and splits over K.
Proof.
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.
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Lemma.
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.
Proof.
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 Formula does not parse: 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.
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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.
Proof.
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 Formula does not parse: \arphi_n = \varphi'_n and \varphi_d = \varphi'_d.
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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 \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 \displaystyle f(\varphi) = \sum_{i=0}^{e-1} \frac{f^{(i)}}{i!}(\varphi_d) \varphi_n^i.
Proof.
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 \displaystyle 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.
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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, \displaystyle \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.

Lemma.
If f is an irreducible prime, then f divides \mu_\varphi if, and only if, f divides c_\varphi.
Proof.
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.
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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].

Lemma.
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}.
Proof.
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.
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Lemma.
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 \displaystyle W := \{ v \in V \mid \exists n \in \N : f(\varphi)^n(v) = 0 \}. Then \displaystyle W = \bigoplus_{i=1}^n \GEig(\varphi, f_i).
Proof.
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 \displaystyle 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, \displaystyle 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, \displaystyle 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.
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Corollary.
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 \displaystyle \GEig(\varphi, f) \otimes_K L = \bigoplus_{i=1}^n \GEig(\varphi \otimes_K L, x - \lambda_i).
Proof.
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.
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Lemma.
Let \dim_K V < \infty and f be a prime polynomial. Then \dim \GEig(\varphi, f) = \nu_f(c_\varphi) \deg f.
Proof.
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.
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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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]]> http://math.fontein.de/2009/08/13/functional-calculus-in-linear-algebra-the-jordan-decomposition-reloaded-and-cayley-hamiltons-theorem/feed/ 0 The Hasse derivative. http://math.fontein.de/2009/08/12/the-hasse-derivative/ http://math.fontein.de/2009/08/12/the-hasse-derivative/#comments Wed, 12 Aug 2009 05:43:38 +0000 Felix Fontein http://math.fontein.de/?p=277 In real and complex analysis, one has a powerful tool, namely the Taylor expansion, which expands an analytic function into a power series. In algebra, one can define the derivative of a polynomial aswell; for f = \sum_{i=0}^n a_i x^i \in R[x], R being a unitary ring, define f' := \sum_{i=1}^n i a_i x^i \in R[x]. This satisfies the same rules as the usual derivative, for example K-linearity and the product rule (f g)' = f' g + f g'. One can also define f^{(k)} recursively by f^{(0)} = f, f^{(k + 1)} = (f^{(k)})' for k \in \N. Unfortunately, one looses certain properties; for example, if R is of finite characteristic m > 0, the polynomial f = x^m \in R[x] satisfies f' = 0, but is not constant as f(0) = 0 \neq 1 = f(1) (assuming R is not the zero ring). In particular, the Identity Theorem does not work. This example also shows one problem with a possible Taylor expansion: for that, one needs to compute \frac{f^{(i)}(a)}{i!} for i = 0, \dots, \deg f; but m = 0 in R, whence m! has no inverse in R! Hence, a Taylor expansion in the classical sense cannot be defined. A “fix” for this problem is offered by Hasse derivatives: they are defined to make both the Identity Theorem and Taylor expansions work again.

Definition.
Let f = \sum_{i=0}^n a_i x^i \in R[x] and k \in \N. Define \displaystyle D^{(k)} f := \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \in R[x]. The function D^{(k)} : R[x] \to R[x] is called the k-th Hasse derivative.

The Hasse derivative shares several properties with the usual derivative, but not all of them; for example, D^{(k)} D^{(\ell)} \neq D^{(k+\ell)} in general. But we have the following properties:

Theorem.
Let f, g \in R[x] and \lambda \in R, and k, \ell \in \N.
  1. We have that D^{(k)} is R-linear, i.e. D^{(k)} (f + g) = D^{(k)} f + D^{(k)} g and D^{(k)}(\lambda f) = \lambda D^{(k)} f.
  2. We have k! \cdot D^{(k)} f = f^{(k)}; in particular, D^{(1)} f = f'.
  3. We have D^{(k)} D^{(\ell)} f = \binom{k + \ell}{\ell} D^{(k+\ell)} f.
  4. (Leibniz Rule) We have \displaystyle D^{(k)}(f g) = \sum_{i=0}^k D^{(i)} f \cdot D^{(k-i)} g; more generally, for f_1, \dots, f_t \in R[x], we have \displaystyle D^{(k)} \prod_{i=1}^t f_i = \sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i, where the sum goes over all such tuples (m_1, \dots, m_t) \in \N^t.
  5. (Faà di Bruno’s Formula) We have \displaystyle D^{(k)} (f \circ g) = \sum \binom{n}{c_0, c_1, \dots, c_k} (D^{(c_0)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}, where the sum goes over all tuples (c_0, \dots, c_k) \in \N^{k+1} with \sum_{i=0}^k c_i = n and \sum_{i=0}^k i c_i = k; here, \binom{n}{c_0, c_1, \dots, c_k} is a multinomial coefficient having the value \displaystyle \frac{n!}{c_0! \cdot c_1! \cdots c_k!}.
  6. (Taylor Formula) We have \displaystyle f = \sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^i.
  7. (Identity Theorem) If we have (D^{(i)} f)(\lambda) = 0 for all i \ge 0, then f = 0.

For that reason, one can define \frac{f^{(k)}}{k!} := D^{(k)} f, so that we can write Taylor’s formula in a more tempting form as \displaystyle f = \sum_{i=0}^n \frac{f^{(i)}}{i!}(\lambda) (x - \lambda)^i, which almost equals the classical form.

Note that the Leibniz rule, Faà di Bruno’s formula and Taylor’s formula take simpler forms than their classical counterparts; this is due to the fact that the additional factorial terms or binomial coefficients already hide in the definition of the Hasse derivative.

Proof.
  1. This follows from the definition of D^{(k)}.
  2. Write f = \sum_{i=0}^n a_i x^i; then k! \cdot D^{(k)} f ={} & k! \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} = \sum_{i=k}^n k! \binom{i}{k} a_i x^{i-k} \\ {}={} & \sum_{i=k}^n a_i \cdot i (i - 1) (i - 2) \cdots (i - k + 1) x^{i - k} \\ {}={} & \sum_{i=k}^n a_i (x^i)^{(k)} = f^{(k)}.
  3. By 1., it is suffices to show this for f = x^i, i \ge k + \ell (for smaller i, both sides will be zero). We have \displaystyle D^{(k)} D^{(\ell)} f = D^{(k)} \binom{i}{\ell} x^{i - \ell} = \binom{i - \ell}{k} \binom{i}{\ell} x^{i - k - \ell} and \displaystyle D^{(k + \ell)} f = \binom{i}{k + \ell} x^{i - k - \ell}. But since \binom{k + \ell}{\ell} \binom{i}{k + \ell} ={} & \frac{(k + \ell)!}{\ell! k!} \frac{i!}{(k + \ell)! (i - k - \ell)!} \\ {}={} & \frac{i!}{\ell! k! (i - k - \ell!)} \\ {}={} & \frac{(i - \ell)!}{k! (i - k - \ell)!} \frac{i!}{\ell! (i - \ell)!} = \binom{i - \ell}{k} \binom{i}{\ell}, these terms are equal.
  4. Note that if we fix f, we get an R-linear function R[x] \to R[x], g \mapsto D^{(k)} (f g). Hence, it suffices to show this for arbitrary f \in R[x] and g = x^m, m \in \N. By the same argument, for g = x^m, we get an R-linbear function R[x] \to R[x], f \mapsto D^{(k)} (f x^m); therefore, it suffices to consider f = x^n, n \in \N. But now, D^{(k)} (x^n x^m) ={} & D^{(k)} x^{n + m} = \binom{n + m}{k} x^{n + m - k} \\ \text{and} \quad D^{(i)} x^n \cdot D^{(k-i)} x^m ={} & \binom{n}{i} x^{n - i} \binom{m}{k - i} x^{m - (k - i)} \\ {}={} & \binom{n}{i} \binom{m}{k - i} x^{n + m - k}. Hence, it suffices to show \sum_{i=0}^k \binom{n}{i} \binom{m}{k - i} = \binom{n + m}{k}. By reorganizing the binomial coefficients, one transforms this into the equality \displaystyle \sum_{i=0}^k \binom{k}{i} \binom{n + m - k}{n - i} = \binom{n + m}{n}, which is Vandermonde’s Identity and, hence, true.
    The more general equation is shown by induction on t. For t = 1, we have \sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i = D^{(k)} f_1. Now, assume that the equation is true for all k for one t \ge 1. Then, for any k, & D^{(k)} \prod_{i=1}^{t+1} f_i = D^{(k)} \biggl( f_{t+1} \cdot \prod_{i=1}^t f_i \biggr) \\ {}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot D^{(k-m_{t+1})} \biggl( \prod_{i=1}^t f_i \biggr) \\ {}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot \sum_{m_1 + \dots + m_t = k - m_{t+1}} \prod_{i=1}^t D^{(m_i)} f_i by the Leibniz rule and by the induction hypothesis; here, the second sum goes over all such tuples (m_1, \dots, m_t) \in \N^t. But this equals \sum_{m_1+\dots+m_t+m_{t+1}=k} \prod_{i=1}^{t+1} D^{(m_i)} f_i, what we had to show.
  5. Again, by 1., it suffices to show this for f = x^n, n \in \N. Now, by the second part of 4., \displaystyle D^{(k)}(g^n) = D^{(k)}(g \cdots g) = \sum_{m_1 + \dots + m_n = k} \prod_{i=1}^n D^{(m_i)} g, where the sum goes over all such (m_1, \dots, m_n) \in \N^n. The formula we want is now obtained by sorting the summands by the different powers of D^{(i)} g appearing, 0 \le i \le k.
    Consider the map \displaystyle \varphi : \N^n \to \N^{k+1}, \quad m = (m_1, \dots, m_n) \mapsto (c_0(m), \dots, c_k(m)), there c_i(m) := \abs{\{ j \in \{ 1, \dots, n \} \mid m_j = i \}}, 0 \le i \le k. Now, if m = (m_1, \dots, m_n) \in \N^n satisfies \sum_{i=1}^n m_i = k, then \sum_{j=0}^k j c_j(m) = k and \sum_{j=0}^k c_j(m) = n. Now, for a fixed (c_0, \dots, c_k) \in \N^{k+1} with \sum_{i=0}^k c_i = n and \sum_{i=0}^k i c_i = k, the \abs{\varphi^{-1}(c_0, \dots, c_k)} equals the multinomial coefficient \displaystyle \binom{n}{c_0, c_1, \dots, c_k}, whence we get that the above formula for D^{(k)}(g^n) equals \displaystyle \sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}, where the sum goes over all tuples (c_0, c_1, \dots, c_k) \in \N^k with \sum_{i=0}^k i c_i = k and \sum_{i=0}^k c_i = n.
  6. By 1., it suffices to show this for f = x^n, n \in \N. Now \sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^k ={} & \sum_{i=0}^n \biggl(\binom{n}{i} x^{n - i}\biggr)(\lambda) (x - \lambda)^i \\ {}={} & \sum_{i=0}^n \binom{n}{i} \lambda^{n-i} (x - \lambda)^i \\ {}={} & ((x - \lambda) + \lambda)^n = x^n by the Binomial Theorem, what we had to show.
  7. This follows directly from 6.
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]]> http://math.fontein.de/2009/08/12/the-hasse-derivative/feed/ 3 A Topological Proof of the Cayley-Hamilton Theorem over all Commutative Unitary Rings. http://math.fontein.de/2009/05/04/a-topological-proof-of-the-cayley-hamilton-theorem-over-all-commutative-unitary-rings/ http://math.fontein.de/2009/05/04/a-topological-proof-of-the-cayley-hamilton-theorem-over-all-commutative-unitary-rings/#comments Mon, 04 May 2009 06:52:19 +0000 Felix Fontein http://math.fontein.de/?p=27 In this post, I want to present a very elegant proof of the Cayley-Hamilton Theorem which works over all commutative unitary rings by reducing to the case over the complex numbers, where a topological argument is used to reduce to the case of diagonalizable matrices. First of all, let us state the definitions and the theorem itself.

Definition.
Let R be a commutative unitary ring and A \in R^{n \times n} a n \times n-matrix over R. The characteristic polynomial of A is the polynomial \chi_A := \det(x E_n - A) \in R[x].

Then the theorem says:

Theorem (Cayley-Hamilton).
Let R be a commutative unitary ring and A \in R^{n \times n}. Then \chi_A(A) = 0.

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 R = \C. So assume that it holds for \C.

Let R be any commutative unitary ring and A \in R^{n \times n}, A = (a_{ij})_{ij}. Set S := \Z[x_{ij} \mid 1 \le i, j \le n] and consider the ring homomorphism \varphi : S \to R, f \mapsto f(a_{11}, a_{12}, \dots, a_{nn}). Over S, consider the matrix B := (x_{ij})_{ij}. Now \varphi induces S-algebra homomorphisms \varphi^* : S^{n \times n} \to R^{n \times n} and \varphi' : S[x] \to R[x] with \varphi^*(B) = A. Clearly, they satisfy \varphi'(\chi_B) = \chi_A and \varphi^*(\chi_B(B)) = \chi_A(A). Therefore, it suffices to prove \chi_B(B) = 0.

Now \C has infinite transcendence degree over \Q (otherwise, it could be countable), whence there exists an embedding \psi : S \to \C; simply choose n^2 algebraically independent elements in \C and map the x_{ij} to them. Again, we get maps \psi^* : S^{n \times n} \to \C^{n \times n} and \psi' : S[x] \to \C[x] which are injective and satisfy \psi'(\chi_B) = \chi_{\psi^*(B)} and \chi_{\psi^*(B)}(\psi^*(B)) = \psi'(\chi_B)(\psi^*(B)) = \psi^*(\chi_B(B)). But by assumption, Cayley-Hamilton holds over \C, whence \chi_{\psi^*(B)}(\psi^*(B)) = 0. Since \psi^* is injective, \chi_B(B) = 0, which implies \chi_A(A) = 0 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 A \in R^{n \times n} is said to be diagonalizable if there exists an invertible matrix T \in GL_n(R) such that \displaystyle T^{-1} A T = \Matrix{ \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & \lambda_n } =: diag(\lambda_1, \dots, \lambda_n) for \lambda_1, \dots, \lambda_n \in R.

We then have:

Lemma.
The Theorem of Cayley-Hamilton holds for diagonalizable matrices.
Proof.

We first assume that A = diag(\lambda_1, \dots, \lambda_n). Then one gets \chi_A = \prod_{i=1}^n (x - \lambda_i), and since \displaystyle (A - \lambda_i E_n) = diag(\lambda_1 - \lambda_i, \dots, \lambda_{i-1} - \lambda_i, 0, \lambda_{i+1} - \lambda_i, \dots, \lambda_n - \lambda_i) one gets \chi_A(A) = 0.

Now assume that A is diagonalizable, and let T \in GL_n(R) such that T^{-1} A T = diag(\lambda_1, \dots, \lambda_n). Clearly, \det T^{-1} = (\det T)^{-1} and, therefore, \chi_A ={} & \det(x E_n - A) = \det T^{-1} \cdot \det(x E_n - A) \cdot \det T \\ {}={} & \det (T^{-1} (x E_n - A) T) = \det(x E_n - T^{-1} A T) = \chi_{T^{-1} A T}. Now write \chi_A = \sum_{i=0}^n a_i x^i with a_i \in R. Then \displaystyle T^{-1} \chi_A(A) T = \sum_{i=0}^n a_i T^{-1} A^i T = \sum_{i=0}^n a_i (T^{-1} A T)^i = \chi_A(T^{-1} A T), whence T^{-1} \chi_A(A) T = \chi_{T^{-1} A T}(T^{-1} A T). But now T^{-1} A T = diag(\lambda_1, \dots, \lambda_n), whence we get T^{-1} \chi_A(A) T = 0 and, hence, \chi_A(A) = 0.

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

Lemma.
Endow \C^{n \times n} with the Euclidean topology and consider the set \displaystyle D := \{ A \in \C^{n \times n} \mid A \text{ diagonalizable } \}. Then D is dense in \C^{n \times n}.

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

  • Every matrix over \C 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 \C.
  • An n \times n-matrix with n distinct eigenvalues is diagonalizable.
Proof.

Let A \in \C^{n \times n} be an arbitrary matrix. Then there exists a matrix T \in GL_n(\C) such that \displaystyle T^{-1} A T = \Matrix{ \lambda_1 & * & \cdots & * \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \cdots & 0 & \lambda_n } with \lambda_1, \dots, \lambda_n \in \C. As the transcendence degree of \C over \Q is infinite, there exist elements \mu_1, \dots, \mu_n \in \C such that for every j \in \N_{>0}, the set \{ \lambda_i + \frac{1}{j} \mu_i \mid 1 \le i \le n \} has exactly n elements. Define A_j := A + \frac{1}{j} diag(\mu_1, \dots, \mu_n), j \in \N_{>0}. Then A_j \to A for j \to \infty and A_j has n distinct eigenvalues for every j, namely \lambda_1 + \frac{1}{j} \mu_1, \dots, \lambda_n + \frac{1}{j} \mu_n. But this implies that A_j \in D, whence we found a sequence in D converging to A.

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Now, we are able to conclude:

Theorem (Cayley-Hamilton over the complex numbers).
Let A \in \C^{n \times n}. Then \chi_A(A) = 0.
Proof.

Set S := \{ A \in \C^{n \times n} \mid \chi_A(A) = 0 \}. Clearly, D \subseteq S and D is dense in \C^{n \times n} by the previous lemma. Hence, it suffices to show that S is closed.

But note that the map \Phi : \C^{n \times n} \to \C^{n \times n}, A \mapsto \chi_A(A) is defined by polynomials; hence, it is continuous. Now S = \Phi^{-1}(\{ 0 \}) is the preimage of a closed set, whence S 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 \C. But this is accomplished by the previous theorem.

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]]> http://math.fontein.de/2009/05/04/a-topological-proof-of-the-cayley-hamilton-theorem-over-all-commutative-unitary-rings/feed/ 0 Fundamental Theorem of Algebra. http://math.fontein.de/2009/05/04/fundamental-theorem-of-algebra/ http://math.fontein.de/2009/05/04/fundamental-theorem-of-algebra/#comments Mon, 04 May 2009 03:33:18 +0000 Felix Fontein http://math.fontein.de/?p=5 As a warm-up, I want to give probably the most beautiful proof of the Fundamental Theorem of Algebra which I know, using the theory of one complex variable. In case you don’t know the theorem:

Theorem (Fundamental Theorem of Algebra).
Let f \in \C[x] be a polynomial, f \neq 0. If n = \deg f, there exist constants \alpha_1, \dots, \alpha_n, \beta \in \C, \beta \neq 0 such that \displaystyle f = \beta \prod_{i=1}^n (x - \alpha_i).

The main ingredient of the proof is the following statement, which is in fact eqiuvalent to the Fundamental Theorem:

Lemma.
Let f \in \C[x] \setminus \C be a non-constant polynomial. Then there exists an \alpha \in \C with f(\alpha) = 0.
Proof.

Assume that on the contrary, f is zero-free. In that case, g : z \mapsto \frac{1}{f(z)} defines a entire function, i.e. a function defined on \C which is holomorphic everywhere. We will show that g is bounded, whence it follows by Liouville’s Theorem that g is constant. This implies that f is constant, a contradiction.

First, write f = \sum_{i=0}^n a_i x^i with a_n \neq 0; then \displaystyle g(z) = z^{-n} \cdot \frac{1}{a_n + \sum_{i=0}^{n-1} a_i z^{i - n}}. Clearly, for z \to \infty, a_n + \sum_{i=0}^{n-1} a_i z^{i - n} \to a_n \neq 0 uniformly, whence g(z) \to 0 uniformly for z \to \infty. Therefore, g is bounded on B_1 := \{ z \in \C \mid \abs{z} > R \} for some R > 0.

Now consider g on B_2 := \{ z \in \C \mid \abs{z} \le R \}. We have that g is continuous on B_2, and since B_2 is compact, we know that |g| attains its maximum on B_2, whence g is bounded on B_2.

Therefore, g is bounded on B_1 \cup B_2 = \C, and we can conclude.

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Now we are able to prove the Fundamental Theorem:

Proof (Fundamental Theorem).

We proceed by induction on \deg f, the degree of f. If \deg f = 0, then f \in \C, whence we can set n := 0 and \beta := f \in \C \setminus \{ 0 \}.

Now assume that the statement holds for polynomials of degree n. Let f be a polynomial of degree n + 1. By the lemma, there exists some \alpha_{n+1} \in \C with f(\alpha_{n+1}) = 0. Now, using the Division Algorithm, write f = q \cdot (x - \alpha_{n+1}) + r with polynomials q, r \in \C[x] satisfying \deg r < \deg (x - \alpha_{n+1}) = 1, i.e. r \in \C. Now \displaystyle 0 = f(\alpha_{n_1}) = q(\alpha_{n+1}) \cdot (\alpha_{n+1} - \alpha_{n+1}) + r = 0 + r, whence we have r = 0 and, therefore, f = q \cdot (x - \alpha_{n+1}). As \deg f = \deg q + \deg (x - \alpha_{n+1}) = \deg q + 1, we have \deg q = n.

Therefore, by the induction hypothesis, there exist \alpha_1, \dots, \alpha_n, \beta \in \C, \beta \neq 0 with q = \beta \prod_{i=1}^n (x - \alpha_i), whence \displaystyle f = q \cdot (x - \alpha_{n+1}) = \beta \prod_{i=1}^{n+1} (x - \alpha_i), i.e. the induction hypothesis holds for f, too.

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