Felix' Math Place » tensor product 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 Homomorphisms, Tensor Products and Certain Canonical Maps. http://math.fontein.de/2010/01/29/homomorphisms-tensor-products-and-certain-canonical-maps/ http://math.fontein.de/2010/01/29/homomorphisms-tensor-products-and-certain-canonical-maps/#comments Fri, 29 Jan 2010 07:20:57 +0000 Felix Fontein http://math.fontein.de/?p=560 Let V and W be vector spaces over a field K and V^* = \Hom_K(V, K), W^* = \Hom_K(W, K) their duals. In case V is finite dimensional, one obtains a non-canonical isomorphism V \cong V^*, a canonical isomorphism V \cong V^{**} and a canonical isomorphism W^* \tensor_K V \cong \Hom_K(W, V).

In case \dim_K V = \infty, V and V^* are not isomorphic: a basis of V^* has a cardinality strictly larger than the one of V. Moreover, the canonical map V \to V^{**} is still a monomorphism, but no longer surjective. In the case of V \tensor_K V^*, one has as well a canonical monomorphism V \tensor_K V^* \to \Hom_K(V, V), but it is no longer surjective as well. We want to study the images of the canonical maps V \to V^{**} and V \tensor_K V^* \to \Hom_K(V, V).

We begin with an auxiliary lemma.

Lemma.
Let v \in V with v \neq 0. Then there exists some \varphi \in V^* with \varphi(v) = 1. Hence, if v \in V satisfies \varphi(v) = 0 for all \varphi \in V^*, then v = 0.
Proof.
Choose a K-basis (v_i)_{i \in I} of V such that there exists some t \in I with v_t = v. Define \pi_t : V \to K by \sum_{i \in I} \lambda_i v_i \mapsto \lambda_t. Then \pi_t(v) = \pi_t(v_t) = 1 and \pi_t \in V^*.
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Proposition.
The map \displaystyle \Psi : V \to V^{**}, \qquad v \mapsto \begin{cases} V^* \to K, \\ \alpha \mapsto \alpha(v) \end{cases} is a monomorphism and its image is \displaystyle \biggl\{ \varphi \in V^{**} \;\biggm|\; \bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0 \biggr\} \cup \{ 0 \}. In particular, if \bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0 for some \varphi \in V^{**}, then \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1.
Proof.
Clearly, for v \in V, \Psi(v) : V^* \to K is K-linear. Moreover, one quickly sees that \Psi is K-linear itself. To see that \Psi is injective, let v \in V with v \neq 0. Now, by the lemma, there exists a \pi \in V^* with \pi(v) = 1; this shows that \Psi(v)(\pi_t) = 1, whence \Psi(v) \neq 0. Therefore, \ker \Psi = 0 an \Psi is injective.
Now, if \alpha \in \ker \Psi(v), \alpha(v) = \Psi(v)(\alpha) = 0, whence v \in \bigcap_{\alpha \in \ker \Psi(v)} \ker \alpha. This shows that the image of \Psi is contained in the given set. Now assume that \varphi \in V^{**} \setminus \{ 0 \} satisfies \bigcap_{\alpha \in \ker\varphi} \ker \alpha \neq 0; say, v \in \bigcap_{\alpha \in \ker\varphi} \ker \alpha \setminus \{ 0 \}. Then \alpha(v) = 0 for all \alpha \in \ker\varphi, whence \ker \varphi \subseteq \ker \Psi(v). By the Homomorphism Theorem, there exists a homomorphism \tilde{\varphi} : V^* / \ker \varphi \to K such that \displaystyle \xymatrix{ V \ar[rr]^{\Psi(v)} \ar[dr]_{\pi} & & K \\ & V / \ker \varphi \ar[ru]_{\tilde{\varphi}} & } commutes. Now V^* / \ker \varphi \cong \varphi(V) = K, whence \dim_K V^* / \ker \varphi = 1. As \tilde{\varphi} \neq 0 (as v \neq 0), \tilde{\varphi} is an isomorphism and we must have \Psi(v) = \lambda \varphi for some \lambda \in K \setminus \{ 0 \}. But then, \varphi = \Psi(\lambda^{-1} v) lies in the image of \Psi.
Finally, if \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha > 0, we saw that we have \varphi = \lambda_v \Phi(v) for any non-zero v \in \bigcap_{\alpha \in \ker \varphi} \ker \alpha, with \lambda_v \in K \setminus \{ 0 \} depending on v. Since \Phi : V \to V^{**} is injective, this shows that we must have \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1.
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This allows us to show that V \to V^{**} is surjective if, and only if, \dim V < \infty.

Corollary.
We have that \Psi : V \to V^{**} is surjective if, and only if, \dim V < \infty.
Proof.
First, if \dim V < \infty, we see that \dim V^{**} = \dim V^* = \dim V. Since \Psi is injective, it follows that \Psi is in fact an isomorphism.
Now assume that \dim V = \infty. It suffices to construct a hyperplane H in V^* with \bigcap_{\alpha \in H} \ker \alpha = 0; this defines an element of V^{**} which is not contained in the image of \Psi by the above proposition. For that, chose a basis (v_i)_{i\in I} of V (using Zorn’s lemma). This defines a family of elements of V^* by \pi_i : V \to K, \sum_{j\in I} \lambda_j v_j \mapsto \lambda_i. Let H' be the subspace of V^* generated by the \pi_i‘s. If we would have H' \subsetneqq V^*, we could emply Zorn’s lemma a second time to obtain a hyperplane H \subseteq V^* with H' \subseteq H; this would prove our claim.
Hence, we have to show that H' \neq V^*. Note that for v \in V, v = \sum_{i\in I} \pi_i(v) v_i; in particular, for every v \in V, only finitely many of the \pi_i(v)‘s are non-zero. Hence, it makes sense to define \pi : V \to K, v \mapsto \sum_{i\in I} \pi_i(v). We claim that \pi \not\in H' in case \abs{I} = \infty: for that, note that \{ \pi_i \}_{i\in I} is a linear independent set in V^*, since for every linear combination \sum \lambda_i \pi_i = 0 \in V^*, we get 0 = \bigl(\sum \lambda_i \pi_i \bigr)(v_j) = \lambda_j for every j \in I.
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Note that in fact, the proof shows that V^* is isomorphic to a \dim V-fold direct product of K, while V is isomorphic to a \dim V-fold direct sum of K. In case \dim V < \infty, these are of the same dimension, but in case \dim V = \infty, they are not.

We continue with the canonical map W^* \tensor_K V \to \Hom_K(W, V).
Proposition.
The map \displaystyle \Phi : W^* \tensor_K V \to \Hom_K(W, V), \qquad \alpha \tensor v \mapsto \begin{cases} W \to V, \\ w \mapsto \alpha(w) v \end{cases} is a monomorphism and its image is \displaystyle \Hom_K^{fin}(W, V) := \{ \varphi \in \Hom_K(W, V) \mid \dim_K \varphi(W) < \infty \}, the K-vector space of finite dimensional K-homomorphisms W \to V.
Proof.
One quickly sees that w \mapsto \varphi(w) v defines an element of \Hom_K^{fin}(W, V), whence \Phi is well-defined and its image is contained in \Hom_K^{fin}(W, V). Moreover, one quickly sees that \Phi is a homomorphism.
Let x = \sum_{i=1}^n \alpha_i \tensor v_i \in W^* \tensor V with \Phi(x) = 0, i.e. with \sum_{i=1}^n \alpha_i(w) v_i = 0 for all w \in W. Without loss of generality, we can assume that our representation of x satisfies that the v_i‘s are linearly independent. In that case, \sum_{i=1}^n \alpha_i(w) v_i = 0 implies \alpha_i(w) = 0 for all i. But since this is true for all w \in W, it follows that \alpha_i = 0 for all i. But then, x = 0. Therefore, \ker \Phi = 0, whence \Phi is injective.
Now let \varphi \in \Hom_K^{fin}(W, V), and let (v_1, \dots, v_n) be a basis of \varphi(W). Let \pi_i : \varphi(W) \to K be the projections with \pi_i(v_i) = 1 an \pi_i(v_j) = 0 for i \neq j. Set \alpha_i := \pi_i \circ \varphi. Then \varphi(w) = \sum_{i=1}^n \alpha_i(w) v_i for all w \in W since v = \sum_{i=1}^n \pi_i(v) v_i for all v \in \varphi(W); therefore, \varphi = \Phi(\sum_{i=1}^n \alpha_i \tensor v_i). This shows that \Hom_K^{fin}(W, V) \subseteq \Phi(W^* \tensor_K V), whence we have equality.
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Now \Hom_K^{fin}(V, V) is a K-algebra, whence for \varphi, \psi \in \Hom_K^{fin}(V, V), it makes sense to define \varphi \circ \psi : V \to V. We are interested on how \Psi^{-1}(\varphi \circ \psi) can be described in terms of \Psi^{-1}(\varphi) and \Psi^{-1}(\psi). This is resolved by the following result:

Proposition.
Let V, W, U be K-vector spaces. The map m :{} & (W^* \tensor_K V) \times (U^* \tensor_K W) \to U^* \tensor_K V, \\ & \biggl(\sum_{i=1}^n \beta_i \tensor v_i, \sum_{j=1}^m \alpha_j \tensor w_j\biggr) \mapsto \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i is K-linear and the following diagram commutes: \xymatrix{ (W^* \tensor_K V) \times (U^* \tensor_K W) \ar[r]^{\qquad\quad m} \ar[d]^{\cong} & U^* \tensor_K V \ar[d]^{\cong} \\ \Hom_K^{fin}(W, V) \times \Hom_K^{fin}(U, W) \ar[r]_{\qquad\quad \circ} & \Hom_K^{fin}(U, V) }
Proof.
Let \Psi_1 : W^* \tensor_K V \to \Hom_K^{fin}(W, V), \Psi_2 : U^* \tensor_K W \to \Hom_K^{fin}(U, W) and \Psi_3 : U^* \tensor_K V \to \Hom_K^{fin}(U, V) be the canonical maps. Since these are isomorphisms, we have to show that for x = \sum_{i=1}^n \beta_i \tensor v_i \in W^* \tensor_K V, y = \sum_{j=1}^m \alpha_j \tensor v_j U^* \tensor_K W and z = \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i \in U^* \tensor_K V, we have \Psi_1(x) \circ \Psi_2(y) = \Psi_3(z). For that, let u \in U. Then (\Psi_1(x) \circ \Psi_2(y))(u) ={} & \Psi_1(x)(\Psi_2(y)(u)) = \Psi_1(x)\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) \\ {}={} & \sum_{i=1}^n \beta_i\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) v_i \\ {}={} & \sum_{i=1}^n \sum_{j=1}^m \alpha_j(u) \beta_i(v_j) v_i = \Psi_3(z)(u), what we had to show.
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In particular, V^* \tensor_K V is a K-algebra isomorphic to \Hom_K^{fin}(V, V); it posseses a 1 if, and only if, \dim_K V < \infty.

Now consider transposition \displaystyle T : \Hom_K(V, W) \to \Hom_K(W^*, V^*), \quad \varphi \mapsto \begin{cases} W^* \to V^*, \\ \psi \mapsto \psi \circ \varphi. \end{cases} Clearly, transposition is injective:

Lemma.
The map T is K-linear and injective.
Proof.
It is clear that T is K-linear. To see that it is injective, let \varphi \in \Hom_K(V, W) with T(\varphi) = 0. Let v \in V; then \psi(\varphi(v)) = 0 for all \psi \in W^*, whence \varphi(v) = 0 by the first lemma. But that means \varphi = 0.
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We show that transposition restricts to the subspaces of the homomorphism spaces of homomorphisms with finite-dimensional image.

Lemma.
Let \varphi \in \Hom_K(V, W). The map \displaystyle \varphi(V)^* \to T(\varphi)(W^*), \qquad \alpha \mapsto \alpha \circ \varphi is an isomorphism. In particular, T^{-1}(\Hom_K^{fin}(W^*, V^*)) = \Hom_K^{fin}(V, W).
Proof.
Let \varphi \in \Hom_K(V, W). The map \psi : \varphi(V)^* \to T(\varphi)(W^*), \alpha \mapsto \alpha \circ \varphi is well-defined and a homomorphism as T(\varphi)(W^*) \subseteq V^* and \varphi(V) \subseteq W. Now let \alpha \in \varphi(V)^* with \psi(\alpha) = 0, i.e. with \alpha \circ \varphi = 0. But since \alpha is defined on \varphi(V), this means that \alpha = 0. Hence, \psi is injective.
Now let \beta \in T(\varphi)(W^*), i.e. there exists some \hat{\psi} \in W^* with \beta = \hat{\psi} \circ \varphi. Set \alpha := \hat{\psi}|_{\varphi(V)}; then \alpha \in \varphi(V)^* and \psi(\alpha) = \hat{\psi}|_{\varphi(V)} \circ \varphi = \hat{\psi} \circ \varphi = \beta. Therefore, \psi is injective.
Finally, in case \dim_K \varphi(V) < \infty, we have \dim_K \varphi(V)^* < \infty and \dim_K T(\varphi)(W^*) < \infty, whence T(\varphi) \in \Hom_K^{fin}(W^*, V^*). On the contrary, if \dim_K \varphi(V) = \infty, we have \infty = \dim_K \varphi(V)^* = \dim_K T(\varphi)(W^*), whence T(\varphi) \not\in \Hom_K^{fin}(W^*, V^*).
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Now we have seen that \Hom_K^{fin}(V, W) \cong V^* \tensor_K W and \Hom_K^{fin}(W^*, V^*) \cong W^{**} \tensor_K V^* in a canonical way, and we have the canonical monomorphism \Psi : W \to W^{**}. We show that these maps behave nicely with transposition:

Proposition.
The map \displaystyle T : V^* \tensor_K W \to W^{**} \tensor_K V^*, \qquad \sum_{i=1}^n v_i^* \tensor w_i \mapsto \sum_{i=1}^n \Psi(w_i) \tensor v_i^* is the unique homomorphism which makes the diagram \displaystyle \xymatrix{ \Hom_K^{fin}(V, W) \ar[r]^T \ar[d]_{\cong} & \Hom_K^{fin}(W^*, V^*) \ar[d]^{\cong} \\ V^* \tensor_K W \ar[r]_T & W^{**} \tensor_K V^* } commuting.
Proof.
Let x = \sum_{i=1}^n v_i^* \tensor w_i \in V^* \tensor_K W and y = \sum_{i=1}^n \Psi(w_i) \tensor v_i^* \in W^{**} \tensor_K V^*. Then \Phi(x)(v) = \sum_{i=1}^n v_i^*(v) w_i \in W for v \in V, and & T(\Phi(x))(w^*)(v) = (w^* \circ \Phi(x))(v) = w^*(\Phi(x)(v)) \\ {}={} & w^*\biggl(\sum_{i=1}^n v_i^*(v) w_i\biggr) = \sum_{i=1}^n v_i^*(v) w^*(w_i) \\ {}={} & \sum_{i=1}^n v_i^*(v) \Psi(w_i)(w^*) = \biggl( \sum_{i=1}^n w^*(w_i) v_i^* \biggr)(v) \\ {}={} & \biggl( \sum_{i=1}^n \Psi(w_i)(w^*) v_i^* \biggr)(v) = \biggl( \sum_{i=1}^n \Phi(\Psi(w_i) \tensor v_i^*)(w^*) \biggr)(v) \\ {}={} & \Phi(y)(w^*)(v) for all w^* \in W^* and v \in V. Hence, T(\Phi(x)) = \Phi(y), what we had to show.
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Now consider double transposition, i.e. \displaystyle T \circ T : \Hom_K(V, W) \to \Hom_K(V^{**}, W^{**}), and its finite-dimensional image restriction \displaystyle T \circ T : \Hom_K^{fin}(V, W) \to \Hom_K^{fin}(V^{**}, W^{**}). The above shows that using the canonical isomorphisms \Hom_K^{fin}(V, W) \cong V^* \tensor_K W and \Hom_K^{fin}(V^{**}, W^{**}) \cong V^{***} \tensor_K W^{**}, we can describe double transpotition by the following commuting diagram: \displaystyle \xymatrix@R-0.85cm{ \Hom_K^{fin}(V, W) \ar[r]^{T \circ T \;\;} \ar[dddd]_{\cong} & \Hom_K^{fin}(V^{**}, W^{**}) \ar[dddd]^{\cong} \\ {\vphantom{x}} \\ {\vphantom{y}} \\ {\vphantom{z}} \\ V^* \tensor_K W \ar[r]^{T \circ T \;\;} & V^{***} \tensor_K W^{**} \\ \sum_{i=1}^n v_i^* \tensor w_i \ar@{|->}[r] & \sum_{i=1}^n \Psi(v_i^*) \tensor \Psi(w_i) }

If \psi \in \Hom_K(W^*, V^*), we obtain a map \displaystyle H(\psi) : V \to W^{**}, \qquad v \mapsto \begin{cases} W^* \to K \\ \alpha \mapsto \psi(\alpha)(v). \end{cases}

Lemma.
The map H : \Hom_K(W^*, V^*) \to \Hom_K(V, W^{**}) is K-linear and injective.
Proof.
First, if \psi \in \Hom_K(W^*, V^*) is fixed, H(\psi)(v + \lambda v') = H(\psi)(v) + \lambda H(\psi)(v') for all v, v' \in V, \lambda \in K; hence, H(V) \subseteq W^{**}. Now, if \psi, \psi' \in \Hom_K(W^*, V^*) and \lambda \in K, v \in V, we have H(\psi + \lambda \psi')(v) ={} & (\psi + \lambda \psi')(\alpha)(v) = \alpha((\psi + \lambda \psi')(v)) \\ {}={} & \alpha(\psi(v) + \lambda \psi'(v)) = H(\psi)(v) + \lambda H(\psi)(v'), whence H is K-linear.
To see that H is injective, let \psi \in \Hom_K(W^*, V^*) be such that H(\psi) = 0. Let \alpha \in W^* and v \in V; since \psi(\alpha)(v) = 0 for all v, we see that \psi(\alpha) = 0, but since this is the case for all \alpha we get \psi = 0.
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Note that we have the following diagram: \displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) & & \Hom_K(V, W^{**}) } Moreover, using the canonical embeddings \Psi : V \to V^{**} and \Psi : W \to W^{**}, we can define a map \Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**}) by \varphi \mapsto \varphi \circ \Phi, and a map \Hom_K(V, W) \to \Hom_K(V, W^{**}) by \varphi \mapsto \Phi \circ \varphi. It turns out that these map make the diagram commute:

Lemma.
The maps \hat{H} : \Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**}), \varphi \mapsto \varphi \circ \Phi and \tilde{H} : \Hom_K(V, W) \to \Hom_K(V, W^{**}), \varphi \mapsto \Phi \circ \varphi are K-linear and make the diagram \displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \ar[dd]^{\tilde{H}} \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) \ar[rr]_{\hat{H}} & & \Hom_K(V, W^{**}) } commute. In particular, \tilde{H} is injective.
Proof.
That \hat{H} and \tilde{H} are K-linear is clear. For the lower triangle, let \varphi \in \Hom_K(W^*, V^*); we have to show that \hat{H}(T(\varphi)) = H(\varphi). For that, let v \in V and \alpha \in W^*; then H(\varphi)(v)(\alpha) ={} & \varphi(\alpha)(v) = \Phi(v)(\varphi(\alpha)) = (\Phi(v) \circ \varphi)(\alpha) \\ {}={} & T(\varphi)(\Phi(v))(\alpha) = \hat{H}(T(\varphi))(v)(\alpha). For the right triangle, let \varphi \in \Hom_K(V, W); we have to show that H(T(\varphi)) = \tilde{H}(\varphi). For that, let v \in V and \alpha \in W^*; then H(T(\varphi))(v)(\alpha) ={} & T(\varphi)(\alpha)(v) = (\alpha \circ \varphi)(v) = \alpha(\varphi(v)) \\ {}={} & \Phi(\varphi(v))(\alpha) = (\Phi \circ \varphi)(v)(\alpha) = \tilde{H}(\varphi)(v)(\alpha).
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Now note that H is injective. We can use this to determine the image of T. For example, for \psi \in \Hom_K(V^*, W^*), & \exists \varphi \in \Hom_K(V, W) : T(\varphi) = \psi \\ {}\Leftrightarrow{} & \forall v \in V : H(\psi)(v) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : (\alpha \mapsto \psi(\alpha)(v)) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : \bigcap_{\alpha \in V^* : \psi(\alpha)(v) = 0} \ker \alpha = 0 \text{ implies } \psi(\bullet)(v) = 0; the last equivalence follows from the first proposition. Unfortunately, this criterion does not really helps in practice.

In case anyone knows a better description of the image of T or \Psi, I’d be happy to know.

]]> http://math.fontein.de/2010/01/29/homomorphisms-tensor-products-and-certain-canonical-maps/feed/ 2 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.

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