In introductionary Linear Algebra classes, one often has the following problems: let 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
are
; hence, most eigenvalues of orthogonal matrices are not elements of
. Now, let
be a finite-dimensional Euclidean space and
an orthogonal map. If one fixes an orthogonal basis
of
, one obtains a orthogonal matrix
which represents
. One can talk about complex eigenvalues of
, but what about complex eigenvalues of
? What should these be?
does not make sense for a complex number
, if
is a vector space over
.
The usual solution to this is to complexify : define
, and define an action
this turns into a
-vector space. If one identifies
by its image under
,
, then
for all
,
. Now we are left to extend
to
. It turns out that there is exactly one choice to extend
to a
-linear map
, i.e. that
. Namely, one has to define
; this is obviously
-linear, whence it suffices to show that
:
Now if is a
-basis of
, it is as well an
-basis of
; moreover,
. If now
is a complex eigenvalue of
, then there exists some
such that
. So
is indeed an eigenvalue of
. Abusing notation, we say that
is an eigenvalue of
; this will always mean that we are talking of
. This process is called complexification of
and
.
But does this generalize? What if is the base field and one has an eigenvalue
of the matrix? Can we do the same thing here? And what if
and we have an eigenvalue in
? The answer is yes. The idea is as follows. A basis of
over
is given by
,
. Hence, we defined
, where the first
corresponds to 1 and the second to
: i.e.
should mean
. Now
has a basis with three elements, so one could define
. And for
if
,
, we need an infinite basis and an infinite direct sum.
It would be nice if we could avoid working with bases, both of and of the field extension
. This can indeed be done, using the tensor product. We begin with a very abstract defintion.
Let be a ring and
-modules. A pair
, where
is a
-module and
is
-bilinear, is said to be a tensor product of
and
over
if the following universal property holds:
If is any
-module and
is
-bilinear, there exists exactly one homomorphism
such that
.
Tensor products exist and are unique up to unique isomorphism. More precisely, if and
are tensor products of
and
over
, there exists exactly one
-isomorphism
with
.
From now on, we write for
and
for
,
,
. In case the base
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 and
be
-vector spaces. Let
be a basis of
and
be a basis of
. Then
is a basis of
. In particular,
.
A different interpretation is that is the set of linear combinations of elements of
, where the coefficients are elements of
. Hence, we extend the range of the coefficients of elements of
from
to
. Every element of
can be written in the form
with
,
,
.
Now let be a field extension of
. Then
is a
-vector space, whence we can consider the tensor product
. As expected, this turns out to be a
-vector space with scalar multiplication
,
. In case
, this definition coincides with the natural
-vector space structure of
.
Let us consider the special case ,
. Then
is a
-basis of
; if
is an
-basis of
, then
is an
-basis of
: every element of
can be written in the form
with
. Moreover,
is a
-basis of
. Compare this with the ad-hoc definition of
at the beginning of this post.
Now, let us consider what to do with -linear maps
, where
and
are
-vector spaces. We begin with a general result on tensor products.
Let be
-modules,
, and let
be
-module homomorphisms. Then there exists exactly one
-homomorphism
with
.
Set and define
One quickly checks that is bilinear. Hence, by the definition of the tensor product
, there exists exactly one
-homomorphism
with
Now let us consider -vector spaces
,
, a
-linear map
and the identity map
. By the theorem, there exists exactly one
-linear map
with . But since
is
, using the
-vector space structure of
, we obtain
, i.e.
is
-linear.
Finally, let be a
-basis of
and
be a
-basis of
. Then
is as well a
-basis of
and
is as well a
-basis of
, whence we can consider the matrices
and
. Write
; then
. Now
Therefore, as well.
Hence, the tensor product allows us to describe , as a generalization of the complexification of real vector spaces, in a very clean and abstract manner.
Finally, recall that every field has an algebraical closure
, which is unique up to
-isomorphism. For
-vector spaces
,
and
-linear maps
we get
-vector spaces
,
and a
-linear map
. We have seen that every
-basis of
resp.
is also an
-basis of
resp.
, and that the matrix representation of
with respect to the bases equals the one of
. Hence, we can not just talk of arbitrary elements of
being eigenvalues of matrices
over
, but also of endomorphisms
defined over
, by referring to
resp.
instead.
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