Felix' Math Place » universal property 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 About Base Changes and Tensor Products. http://math.fontein.de/2009/08/15/about-base-changes-and-tensor-products/ http://math.fontein.de/2009/08/15/about-base-changes-and-tensor-products/#comments Sat, 15 Aug 2009 19:48:25 +0000 Felix Fontein http://math.fontein.de/?p=330 In introductionary Linear Algebra classes, one often has the following problems: let A \in \R^{n \times n} be a real valued matrix, say an orthogonal one, then the eigenvalues are complex numbers of absolute value 1. the only two such values inside \R are \pm 1; hence, most eigenvalues of orthogonal matrices are not elements of \R. Now, let (V, \ggen{\bullet, \bullet}) be a finite-dimensional Euclidean space and \phi : V \to V an orthogonal map. If one fixes an orthogonal basis B of V, one obtains a orthogonal matrix A = M_B(\phi) which represents \phi. One can talk about complex eigenvalues of A, but what about complex eigenvalues of \phi? What should these be? \lambda v does not make sense for a complex number \lambda, if V is a vector space over \R.

The usual solution to this is to complexify V: define V_\C := V \oplus V, and define an action & \C \times V_\C \to V_\C, \\ & (a + i b, (v, w)) \mapsto (a + i b) (v + i w) = (a v - b w, b v + a w); this turns V_\C into a \C-vector space. If one identifies V by its image under V \to V_\C, v \mapsto (v, 0), then \lambda v = (\lambda + 0 i) (v, 0) for all \lambda \in \R, v \in V. Now we are left to extend \phi to V_\C. It turns out that there is exactly one choice to extend \phi to a \C-linear map \phi_\C : V_\C \to V_\C, i.e. that \phi_\C|_V = \phi. Namely, one has to define \phi_\C(v, w) := (\phi(v), \phi(w)); this is obviously \R-linear, whence it suffices to show that \phi_\C(i (v, w)) = i \phi_\C(v, w): \phi_\C(i (v, w)) ={} & \phi_\C(-w, v) = (\phi(-w), \phi(v)) = (-\phi(w), \phi(v)) \\ {}={} & i (\phi(v), \phi(w)) = i \phi_\C(v, w). Now if B is a \R-basis of V, it is as well an \C-basis of V_\C; moreover, M_B(\phi) = M_B(\phi_\C). If now \lambda \in \C is a complex eigenvalue of M_B(\phi), then there exists some \hat{v} \in V_\C \setminus \{ 0 \} such that \phi_\C(\hat{v}) = \lambda \hat{v}. So \lambda is indeed an eigenvalue of \phi_\C. Abusing notation, we say that \lambda is an eigenvalue of \phi; this will always mean that we are talking of \phi_\C. This process is called complexification of V and \phi.

But does this generalize? What if K = \F_2 is the base field and one has an eigenvalue \lambda \in L = \F_8 of the matrix? Can we do the same thing here? And what if K = \Q and we have an eigenvalue in L = \C? The answer is yes. The idea is as follows. A basis of \C over \R is given by 1, i. Hence, we defined V_\C = V \oplus V, where the first V corresponds to 1 and the second to i: i.e. (v, w) \in V_\C should mean v + i w. Now \F_8 / \F_2 has a basis with three elements, so one could define V_L := V \oplus V \oplus V. And for V_L if K = \Q, L = \C, we need an infinite basis and an infinite direct sum.

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

Definition.
Let R be a ring and V, W R-modules. A pair (T, \phi), where T is a R-module and \phi : V \times W \to T is R-bilinear, is said to be a tensor product of V and W over R if the following universal property holds:
If A is any R-module and \psi : V \times W \to A is R-bilinear, there exists exactly one homomorphism \varphi : T \to A such that \psi = \varphi \circ \phi. \displaystyle \xymatrix{ V \times W \ar[r]^\phi \ar[rd]_\psi & T \ar@{->}[d]^{\exists! \varphi} \\ & A }
Theorem.
Tensor products exist and are unique up to unique isomorphism. More precisely, if (T, \phi) and (T', \phi') are tensor products of V and W over R, there exists exactly one R-isomorphism \varphi : T \to T' with \varphi \circ \phi = \phi'.
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From now on, we write V \otimes_R W for T and v \otimes_R w for \phi(v, w), v \in V, w \in W. In case the base R is clear, we will drop the subscript.

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

Theorem.
Let V and W be K-vector spaces. Let (v_i)_{i\in I} be a basis of V and (w_j)_{j\in J} be a basis of W. Then (v_i \otimes w_j)_{(i, j) \in I \times J} is a basis of V \otimes_K W. In particular, \dim_K (V \otimes_K W) = \dim_K V \cdot \dim_K W.
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A different interpretation is that V \otimes_R W is the set of linear combinations of elements of W, where the coefficients are elements of V. Hence, we extend the range of the coefficients of elements of W from K to V. Every element of V \otimes_R W can be written in the form \sum_{i=1}^n v_i \otimes w_i with v_i \in V, w_i \in W, 1 \le i \le n.

Now let L be a field extension of K. Then L is a K-vector space, whence we can consider the tensor product V_L := L \otimes_K V. As expected, this turns out to be a L-vector space with scalar multiplication \C \times V_L \to V_L, (\lambda, \sum_{i=1}^n \lambda_i \otimes v_i) \mapsto \sum_{i=1}^n (\lambda \lambda_i) \otimes v_i. In case \lambda \in K \subseteq L, this definition coincides with the natural K-vector space structure of V_L.

Let us consider the special case K = \R, L = \C. Then (1, i) is a K-basis of L; if (v_j)_{j\in J} is an \R-basis of V, then (v_j, i v_j)_{j \in J} is an Formula does not parse: \IR-basis of V_\C: every element of V_\C can be written in the form v + i w with v, w \in V. Moreover, (v_j)_{j\in J} is a \C-basis of V_\C. Compare this with the ad-hoc definition of V_\C at the beginning of this post.

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

Theorem.
Let V_i, W_i be R-modules, i = 1, 2, and let \phi_i : V_i \to W_i be R-module homomorphisms. Then there exists exactly one R-homomorphism \phi : V_1 \otimes V_2 \to W_1 \otimes W_2 with \phi(v_1 \otimes v_2) = \phi_1(v_1) \otimes \phi_2(v_2).
Proof.
Set A := W_1 \otimes W_2 and define \displaystyle \psi : V_1 \times V_2 \to A, \quad (v_1, v_2) \mapsto \phi_1(v_1) \otimes \phi_2(v_2). One quickly checks that \psi is bilinear. Hence, by the definition of the tensor product V_1 \otimes V_2, there exists exactly one R-homomorphism \phi : V_1 \otimes V_2 \to A with \displaystyle \phi(v_1 \otimes v_2) = \psi(v_1, v_2) = \phi_1(v_1) \otimes \phi_2(v_2).
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Now let us consider K-vector spaces V, W, a K-linear map \varphi : V \to W and the identity map \id_L : L \to L. By the theorem, there exists exactly one K-linear map \displaystyle \varphi_L : V_L = L \otimes_K V \to L \otimes_K W = W_L with \varphi_L(\lambda \otimes v) = \id_L(\lambda) \otimes \varphi(v). But since \lambda \otimes v is \lambda v, using the L-vector space structure of V_L, we obtain \varphi_L(\lambda v) = \lambda \varphi_L(v), i.e. \varphi_L is L-linear.

Finally, let B = (v_i)_{i\in I} be a K-basis of V and B' = (w_j)_{j\in J} be a K-basis of W. Then B is as well a L-basis of V_L and B' is as well a L-basis of W, whence we can consider the matrices M_{B,B'}(\varphi) \in K^{J \times I} and M_{B,B'}(\varphi_L) \in L^{J \times I}. Write \varphi(v_i) = \sum_{j\in J} \lambda_{ij} w_j; then M_{B,B'}(\varphi) = (\lambda_{ij})_{i \in I, \atop j \in J}. Now \varphi_L(v_i) ={} & \varphi_L(1 \otimes_K v_i) = \id_L(1) \otimes_K \varphi(v_i) \\ {}={} & \id_L(1) \otimes_K \sum_{j\in J} \lambda_{ij} w_j = \sum_{j\in J} \lambda_{ij} (\id_L(1) \otimes w_j). Therefore, M_{B,B'}(\varphi_L) = (\lambda_{ij})_{i \in I, \atop j \in J} = M_{B,B'}(\varphi) as well.

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

Finally, recall that every field K has an algebraical closure \overline{K}, which is unique up to K-isomorphism. For K-vector spaces V, W and K-linear maps \phi : V \to W we get \overline{K}-vector spaces V_{\overline{K}}, W_{\overline{K}} and a \overline{K}-linear map \phi_{\overline{K}} : V_{\overline{K}} \to W_{\overline{K}}. We have seen that every K-basis of V resp. W is also an \overline{K}-basis of V_{\overline{K}} resp. W_{\overline{K}}, and that the matrix representation of \phi with respect to the bases equals the one of \phi_{\overline{K}}. Hence, we can not just talk of arbitrary elements of \overline{K} being eigenvalues of matrices M_{B,B'}(\phi) over K, but also of endomorphisms \phi defined over K, by referring to M_{B,B'}(\phi_{\overline{K}}) resp. \phi_{\overline{K}} instead.

]]> http://math.fontein.de/2009/08/15/about-base-changes-and-tensor-products/feed/ 0 A Note on the Jordan Decomposition. http://math.fontein.de/2009/05/05/a-note-on-the-jordan-decomposition/ http://math.fontein.de/2009/05/05/a-note-on-the-jordan-decomposition/#comments Tue, 05 May 2009 01:33:30 +0000 Felix Fontein http://math.fontein.de/?p=77 This time, I want to share an observation on the Jordan decomposition, which is the main tool needed to show the existence of the Jordan normal form. Let me begin by introducing a more general notation, and show that the Jordan decomposition satisfies a kind of universal property.

Let V be a vector space over a field K and \varphi : V \to V a linear map. We say that a subspace W \subseteq V is \varphi-invariant if \varphi(W) \subseteq W. Another way to interpret this is to consider the K-algebra A = K[\varphi]; then V is an A-module and the A-submodules of V are exactly the \varphi-invariant subspaces of V.

Definition.
An A-decomposition of V is a decomposition V = \bigoplus_{i \in I} V_i of A-submodules such that, for every A-submodule W of V, one has W = \bigoplus_{i \in I} (V_i \oplus W).

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

Definition.
Let V = \bigoplus_{i \in I} V_i and V = \bigoplus_{j \in J} W_j be two V decompositions. We say that \bigoplus_{i \in I} V_i \le \bigoplus_{j \in J} W_j if, for every i \in I, there exists an j \in J with V_i \subseteq W_j.

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

Lemma.
Assume that V = \bigoplus_{i \in I} V_i is a minimal A-decomposition. Let W = \bigoplus_{j \in J} W_j be another A-decomposition. Then, for every j \in J, there exists a subset I_j \subseteq I with W_j = \bigoplus_{i \in I_j} V_i.
Proof.

Define I_j := \{ i \in I \mid V_i \subseteq W_j \}. Then I = \bigcup_{j \in J} I_j is a disjoint union. Now, \bigcup_{i \in I_j} U_i \subseteq W_j and \bigoplus_{i \in I_j} U_i form a direct sum, whence \bigoplus_{i \in I_j} U_i \subseteq W_j for all j.

Now assume that \bigoplus_{i \in I_j} U_i \subsetneqq W_j for some j; let w \in W_j \setminus \bigoplus_{i \in I_j} U_i. Now, as V = \bigoplus_{i \in I} V_i, we can write w = \sum_{i \in I} v_i with v_i \in V_i. Moreover, write w = \sum_{t \in J} w_t with w_t \in W_t. Clearly, we must have w_t = \sum_{i \in I_t} v_i for every t \in J. As w \in W_j we have w_t = 0 for all t \neq 0, whence v_i = 0 for all i \not\in I_j. But this implies w = \sum_{i \in I_j} v_i \in \bigoplus_{i \in I_j} V_i, a contradiction.

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

Definition.
Let \lambda \in K. The generalized eigenspace of \varphi with respect to \lambda is \displaystyle \GEig(\varphi, \lambda) := \{ v \in V \mid \exists n \in \N : (\varphi - \lambda \id)^n v = 0 \}.

In case \dim V = n < \infty, one has that \GEig(\varphi, \lambda) = \ker (\varphi - \lambda)^n. Hence, generalized eigenspaces can be efficiently computed. Moreover, we have \Eig(\varphi, \lambda) \subseteq \GEig(\varphi, \lambda), and a simple argument shows that either both are trivial or both are non-trivial. Hence, the \lambda \in K with \GEig(\varphi, \lambda) \neq \{ 0 \} are exactly the zeroes of the characteristic polynomial of \varphi.

Now note that \varphi(\GEig(\varphi, \lambda)) \subseteq \GEig(\varphi, \lambda). Hence, \GEig(\varphi, \lambda) is \varphi-invariant. We now have three lemmas:

Lemma.
Let \lambda_1, \dots, \lambda_t be t different eigenvalues of \varphi. Then \bigoplus_{i=1}^t \GEig(\varphi, \lambda_i) is a direct sum.
Proof.

Let v_i \in \GEig(\varphi, \lambda_i with \sum_{i=1}^t v_i = 0. We have to show that v_i = 0 for all i. Assume that not all v_i are zero, and that the relation is chosen minimal with respect to the number of nonzero v_i.

Let j \in \{ 1, \dots, t \} with v_j \neq 0, and choose n \in \N with (\varphi - \lambda_j \id)^n v_j = 0. If \psi := (\varphi - \lambda_j \id)^n, then \sum_{i=1}^t \psi(v_i) = 0 yields a second relation with \psi(v_i) = 0. By minimality, we must have \psi(v_j) = 0 for all j.

We will show that (\varphi - \lambda_i \id)|_{\GEig(\varphi, \lambda)} is injective for \lambda \neq \lambda_i, which gives v_j = 0 for all j \neq 0 and, therefore, v_i = 0, a contradiction.

Let v \in \GEig(\varphi, \lambda) with (\varphi - \lambda_i \id) v = 0. Assume that v \neq 0 and let n \in \N be maximal with w := (\varphi - \lambda \id)^n v \neq 0; in that case, \varphi(w) = \lambda w and \displaystyle (\varphi - \lambda_i) w = (\varphi - \lambda_i) (\varphi - \lambda)^n v = (\varphi - \lambda)^n (\varphi - \lambda_i) v = (\varphi - \lambda)^n 0 = 0, whence we get \lambda_i w = \varphi(w) = \lambda w, which is only possible for w = 0, a contradiction. Hence, we must have v = 0, i.e. \varphi - \lambda_i \id is injective on \GEig(\varphi, \lambda).

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Lemma.
Assume that \dim V < \infty and let \lambda \in K. Then there exists an \varphi-invariant subspace W \subseteq V such that V = W \oplus \GEig(\varphi, \lambda).
Proof.

Set \psi := \varphi - \lambda \id. ConsiderS the chains \displaystyle \{ 0 \} \subseteq \ker \psi \subseteq \ker \psi^2 \subseteq \ker \psi^3 \subseteq \dots and \displaystyle V \supseteq \image \psi \supseteq \image \psi^2 \supseteq \image \psi^3 \supseteq \dots Clearly, there exists an s \in \N with \image \psi^s = \image \psi^{s+1} as \dim V < \infty. Now one easily shows \image \psi^s = \image \psi^{s+i} for all i \in \N. By the Dimension Formula, we have \dim \ker \psi^{s+i} ={} & \dim V - \dim \image \psi^{s+i} \\ {}={ } & \dim V - \dim \image \psi^s = \dim \ker \psi^s for all i \in \N, whence \ker \psi^{s+i} = \ker \psi^s for all i \in \N. But then \GEig(\varphi, \lambda) = \ker \psi^s and \dim \GEig(\varphi, \lambda) + \dim \image \psi^s = \dim V.

Set W := \image \psi^s and let w \in W, i.e. let v \in V with \psi^s(v) = w. Then \displaystyle \varphi(w) = \varphi (\varphi - \lambda \id)^s v = (\varphi - \lambda \id)^s \varphi(v) \in \image \psi^s, whence W is \varphi-invariant. Now it suffices to show that W \cap \GEig(\varphi, \lambda) = 0. Now \image \psi^{s+1} = \image \psi^s, whence for every w \in W there exists some v \in \image \psi^s with \psi(v) = w. But this means that \psi|_W is surjective, whence \ker(\psi|_W) = \{ 0 \}. But then \ker \psi^s \cap W = \ker (\psi^s|_W) = \ker (\psi|_W)^s = \{ 0 \}.

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Lemma.
Assume that \dim V < \infty and that the characteristic polynomial \chi_\varphi of \varphi splits into linear factors. Let \lambda_1, \dots, \lambda_t be all eigenvalues of \varphi. Then V = \bigoplus_{i=1}^t \GEig(\varphi, \lambda).
Proof.

We proceed by induction on \dim V. For \dim V = 0 this is cleary. Hence, assume \dim V \ge 1 and let \lambda be an eigenvalue of \varphi. Choose an \varphi-invariant subspace W with V = \GEig(\varphi, \lambda) \oplus W. We have \dim \GEig(\varphi, \lambda) \ge \dim \Eig(\varphi, \lambda) \ge 1, whence \dim W < \dim V. Now \displaystyle \chi_\varphi = \chi_{\varphi|_W} \cdot \chi_{\varphi|_{\GEig(\varphi, \lambda)}}, whence the characteristic polynomial of \varphi|_W splits into linear factors as well.

Let \lambda'_1, \dots, \lambda'_s be the eigenvalues of \varphi|_W. Then, by induction, we have W = \bigoplus_{i=1}^s \GEig(\varphi|_W, \lambda_i'). Now \GEig(\varphi|_W, \lambda_i') = W \cap \GEig(\varphi, \lambda_i'), whence W \subseteq \bigoplus_{i=1}^s \GEig(\varphi, \lambda_i').

Finally, note that \lambda \neq \lambda_i' for all i, as this would contradict W \cap \GEig(\varphi, \lambda) = \{ 0 \}. Therefore, V = \GEig(\varphi, \lambda) \oplus \bigoplus_{i=1}^s \GEig(\varphi, \lambda_i'). Moreover, we must have \{ \lambda_1, \dots, \lambda_t \} = \{ \lambda, \lambda'_1, \dots, \lambda_s' \} as the dimensions of the generalized eigenspaces for all \lambda_i must be non-zero, whence “\supseteq” must hold. The converse holds because every non-trival generalized eigenvector yields a non-trivial eigenvector to the same value.

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Therefore, we get:

Corollary (Jordan Decomposition).
Let K be algebraically closed and assume that \dim V < \infty. Then, for every endomorphism \varphi of V, there exist \lambda_1, \dots, \lambda_t \in K such that \displaystyle V = \bigoplus_{i=1}^t \GEig(\varphi, \lambda_i) is an K[\varphi]-decomposition.
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Proof.

We have to show that this yields an K[\varphi]-decomposition. For that, let W be a \varphi-invariant subspace of V. Consider \varphi|_W; this is an endomorphism of W whose set of eigenvalues is a subset of the set of eigenvalues of \varphi. Hence, by the previous lemma applied to \varphi|_W, we have \displaystyle W = \bigoplus_{i=1}^t \GEig(\varphi|_W, \lambda_i) = \bigoplus_{i=1}^t (\GEig(\varphi, \lambda_i) \cap W), what we had to show.

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We can now prove our main result, namely that the generalized eigenspace decomposition is exactly the minimal K[\varphi]-decomposition of V:

Theorem.
Let K be algebraically closed and \dim V < \infty. Then the minimal K[\varphi]-decomposition of V is given by \displaystyle V = \bigoplus_{\lambda \in K} \GEig(\varphi, \lambda).

Note that we do not need that K is algebraically cloesd, but only that \chi_\varphi splits over K.

Proof.

Let V = \bigoplus_{i=1}^n V_i be a K[\varphi]-decomposition. Assume that there exists some \lambda \in K and 1 \le i < j \le n such that V_i \cap W \neq \{ 0 \} \neq V_j \cap W with W := \GEig(\varphi, \lambda); if this would not exist, we would have \bigoplus_{\lambda \in K} \GEig(\varphi, \lambda) \le \bigoplus_{i=1}^n V_i.

Assume that we can find eigenvectors v \in V_i \cap W and w \in V_j \cap W. Then \varphi(v + w) = \lambda (v + w), whence v + w is an eigenvector as well. But then W' := \langle v + w \rangle is an \varphi-invariant subspace of V with W' \subseteq V_i \oplus V_j, but (W' \cap V_i) \oplus (W' \cap V_j) = \{ 0 \} \subsetneqq W', a contradiction that \bigoplus_{i=1}^n V_i is a K[\varphi]-decomposition.

We now show that W \cap V_i contains an eigenvector. As V_i is \varphi-invariant, we can consider \psi := \varphi|_{V_i}. Now \GEig(\psi, \lambda) = W \cap V_i \neq \{ 0 \}, whence we must have \Eig(\psi, \lambda) \neq \{ 0 \}. Hence, there exists some v \in W, v \neq 0 with \varphi(v) = \psi(v) = \lambda v.

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