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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.


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.

\xymatrix{ V \times W \ar[r]^\phi \ar[rd]_\psi & T \ar@{-->}[d]^{\exists! \varphi} \\ & A }

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'.

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.


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.

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 \R-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.


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).


Set A := W_1 \otimes W_2 and define

\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

\phi(v_1 \otimes v_2) = \psi(v_1, v_2) = \phi_1(v_1) \otimes \phi_2(v_2).

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

\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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