About Base Changes and Tensor Products.

Abstract.
In Linear Algebra, one often has the problem that one wants to talk about complex eigenvalues of objects defined over the reals. If the object is a matrix, it is clear what that means. But what if the object is an endomorphism of a non-canonical real vectorspace? This question is strongly related an important use of tensor products, namely base changes.

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.

Tags: base change, complexification, tensor product, universal property

This entry was posted on Saturday, August 15th, 2009 at 19:48 Europe/Berlin and is filed under Algebra, Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

About Base Changes and Tensor Products.

Abstract.

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