Complexification

In the past few sections we have been treating real and complex vector spaces simultaneously. Sometimes this is not possible; the complex number system is richer than the real. There are theorems that are true for both real and complex spaces, but for which the proof is much easier in the complex case, and there are theorems that are true for complex spaces but not for real ones. (An example of the latter kind is the assertion that if the space is finite-dimensional, then every linear transformation has a proper value.) For these reasons, it is frequently handy to be able to "complexify" a real vector space, that is, to associate with it a complex vector space with essentially the same properties. The purpose of this section is to describe such a process of complexification.

Suppose that is a real vector space, and let be the set of all ordered pairs with both and in . Define the sum of two elements of by and define the product of an element of by a complex number ( and real, ) by (To remember these formulas, pretend that means .) A straightforward and only slightly laborious computation shows that the set becomes a complex vector space with respect to these definitions of the linear operations.

The set of those elements of for which is in a natural one-to-one correspondence with the space . Being a complex vector space, the space may also be regarded as a real vector space; if we identify each element of with its replica in (it is exceedingly convenient to do this), we may say that (as a real vector space) includes . Since , so that , our identification convention enables us to say that every vector in has the form , with and in . Since and (where denotes the set of all elements in with ) are subsets of with only (that is, ) in common, it follows that the representation of a vector of in the form (with and in ) is unique. We have thus constructed a complex vector space with the property that considered as a real space includes as a subspace, and such that is the direct sum of and . (Here denotes the set of all those elements of that have the form for some in .) We shall call the complexification of .

If is a linearly independent set in (real coefficients), then it is also a linearly independent set in (complex coefficients). Indeed, if are real numbers such that , then , and consequently, by the uniqueness of the representation of vectors in by means of vectors in , it follows that ; the desired result is now implied by the assumed (real) linear independence of in . If, moreover, is a basis in (real coefficients), then it is also a basis in (complex coefficients). Indeed, if and are in , then there exist real numbers such that and ; it follows that , and hence that spans . These results imply that the complex vector space has the same dimension as the real vector space .

There is a natural way to extend every linear transformation on to a linear transformation on ; we write whenever and are in . (The verification that is indeed a linear transformation on is routine.) A similar extension works for linear and even multilinear functionals. If, for instance, is a (real) bilinear functional on , its extension to is the (complex) bilinear functional defined. by

If, on the other hand, is alternating, then the same is true of . Indeed, the real and imaginary parts of are and respectively; if is alternating, then is skew symmetric ( Section: Alternating forms , Theorem 1), and therefore is alternating. The same proof establishes the corresponding result for -linear functionals also, for all values of . From this and from the definition of determinants it follows that for every linear transformation on .

The method of extending bilinear functionals works for conjugate bilinear functionals also. If, that is, is a (real) inner product space, then there is a natural way of introducing a (complex) inner product into ; we write, by definition, Observe that if and are orthogonal vectors in , then

The correspondence from to preserves all algebraic properties of transformations. Thus if (with real), then ; if , then ; and if , then . If, moreover, is an inner product space, and if , then . (Proof: evaluate and .)

If is a linear transformation on and if has a proper vector , with proper value (where and are in and and are real), so that then the subspace of spanned by and is invariant under . (Since every linear transformation on a complex vector space has a proper vector, we conclude that every linear transformation on a real vector space leaves invariant a subspace of dimension equal to or .) If, in particular, happens to have a real proper value (that is, if ), then has the same proper value (since , , and not both and can vanish).

We have already seen that every (real) basis in is at the same time a (complex) basis in . It follows that the matrix of a linear transformation on , with respect to some basis in , is the same as the matrix of on , with respect to the basis in . This comment is at the root of the whole theory of complexification; the naive point of view on the matter is that real matrices constitute a special case of complex matrices.

EXERCISES