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 \(\mathcal{V}\) is a real vector space, and let \(\mathcal{V}^{+}\) be the set of all ordered pairs \(\langle x, y\rangle\) with both \(x\) and \(y\) in \(\mathcal{V}\) . Define the sum of two elements of \(\mathcal{V}^{+}\) by \[\langle x_{1}, y_{1}\rangle+\langle x_{2}, y_{2}\rangle=\langle x_{1}+x_{2}, y_{1}+y_{2}\rangle,\] and define the product of an element of \(\mathcal{V}^{+}\) by a complex number \(\alpha+i \beta\) ( \(\alpha\) and \(\beta\) real, \(i=\sqrt{-1}\) ) by \[(\alpha+i \beta)\langle x, y\rangle=\langle\alpha x-\beta y, \beta x+\alpha y\rangle.\] (To remember these formulas, pretend that \(\langle x, y\rangle\) means \(x+i y\) .) A straightforward and only slightly laborious computation shows that the set \(\mathcal{V}^{+}\) becomes a complex vector space with respect to these definitions of the linear operations.

The set of those elements \(\langle x, y\rangle\) of \(\mathcal{V}^{+}\) for which \(y=0\) is in a natural one-to-one correspondence with the space \(\mathcal{V}\) . Being a complex vector space, the space \(\mathcal{V}^{+}\) may also be regarded as a real vector space; if we identify each element \(x\) of \(\mathcal{V}\) with its replica \(\langle x, 0\rangle\) in \(\mathcal{V}^{+}\) (it is exceedingly convenient to do this), we may say that \(v^{+}\) (as a real vector space) includes \(\mathcal{V}\) . Since \(\langle 0, y\rangle=i\langle y, 0\rangle\) , so that \(\langle x, y\rangle=\langle x, 0\rangle+i\langle y, 0\rangle\) , our identification convention enables us to say that every vector in \(\mathcal{V}^{+}\) has the form \(x+i y\) , with \(x\) and \(y\) in \(\mathcal{V}\) . Since \(\mathcal{V}\) and \(i\mathcal{V}\) (where \(i \mathcal{V}\) denotes the set of all elements \(\langle x, y\rangle\) in \(\mathcal{V}^{+}\) with \(x=0\) ) are subsets of \(\mathcal{V}^{+}\) with only \(0\) (that is, \(\langle 0,0\rangle\) ) in common, it follows that the representation of a vector of \(\mathcal{V}^{+}\) in the form \(x+i y\) (with \(x\) and \(y\) in \(\mathcal{V}\) ) is unique. We have thus constructed a complex vector space \(\mathcal{V}^{+}\) with the property that \(\mathcal{V}^{+}\) considered as a real space includes \(\mathcal{V}\) as a subspace, and such that \(\mathcal{V}^{+}\) is the direct sum of \(\mathcal{V}\) and \(i \mathcal{V}\) . (Here \(i\mathcal{V}\) denotes the set of all those elements of \(\mathcal{V}^{+}\) that have the form \(i y\) for some \(y\) in \(\mathcal{V}\) .) We shall call \(\mathcal{V}^{+}\) the complexification of \(\mathcal{V}\) .

If \(\{x_{1}, \ldots, x_{n}\}\) is a linearly independent set in \(\mathcal{V}\) (real coefficients), then it is also a linearly independent set in \(\mathcal{V}^{+}\) (complex coefficients). Indeed, if \(\alpha_{1}, \ldots, \alpha_{n}, \beta_{1}, \ldots, \beta_{n}\) are real numbers such that \(\sum_{j}(\alpha_{j}+i \beta_{j}) x_{j}\) \(=0\) , then \(\big(\sum_{j} \alpha_{j} x_{j}\big)+i\big(\sum_{j} \beta_{j} x_{j}\big)=0\) , and consequently, by the uniqueness of the representation of vectors in \(\mathcal{V}^{+}\) by means of vectors in \(\mathcal{V}\) , it follows that \(\sum_{j} \alpha_{j} x_{j}=\sum_{j} \beta_{j} x_{j}=0\) ; the desired result is now implied by the assumed (real) linear independence of \(\{x_{1}, \ldots, x_{n}\}\) in \(\mathcal{V}\) . If, moreover, \(\{x_{1}, \ldots, x_{n}\}\) is a basis in \(\mathcal{V}\) (real coefficients), then it is also a basis in \(\mathcal{V}^{+}\) (complex coefficients). Indeed, if \(x\) and \(y\) are in \(\mathcal{V}\) , then there exist real numbers \(\alpha_{1}, \ldots, \alpha_{n}, \beta_{1}, \ldots, \beta_{n}\) such that \(x=\sum_{j} \alpha_{j} x_{j}\) and \(y=\) \(\sum_{j} \beta_{j} x_{j}\) ; it follows that \(x+i y=\sum_{j}(\alpha_{j}+i \beta_{j}) x_{j}\) , and hence that \(\{x_{1}, \ldots, x_{n}\}\) spans \(\mathcal{V}^{+}\) . These results imply that the complex vector space \(\mathcal{V}^{+}\) has the same dimension as the real vector space \(\mathcal{V}\) .

There is a natural way to extend every linear transformation \(A\) on \(\mathcal{V}\) to a linear transformation \(A^{+}\) on \(\mathcal{V}^{+}\) ; we write \[A^{+}(x+i y)=A x+i A y\] whenever \(x\) and \(y\) are in \(\mathcal{V}\) . (The verification that \(A^{+}\) is indeed a linear transformation on \(\mathcal{V}^{+}\) is routine.) A similar extension works for linear and even multilinear functionals. If, for instance, \(w\) is a (real) bilinear functional on \(\mathcal{V}\) , its extension to \(\mathcal{V}^{+}\) is the (complex) bilinear functional defined. by \begin{align} w^{+}(x_{1}+i y_{1}, x_{2}+i y_{2}) = w(x_{1}, x_{2})-w(y_{1}, y_{2})+i\big(w(x_{1}, y_{2})+w(y_{1}, x_{2})\big). \end{align} 

If, on the other hand, \(w\) is alternating, then the same is true of \(w^{+}\) . Indeed, the real and imaginary parts of \(w^{+}(x+i y, x+i y)\) are \(w(x, x)-w(y, y)\) and \(w(x, y)+w(y, x)\) respectively; if \(w\) is alternating, then \(w\) is skew symmetric ( Section: Alternating forms , Theorem 1), and therefore \(w^{+}\) is alternating. The same proof establishes the corresponding result for \(k\) -linear functionals also, for all values of \(k\) . From this and from the definition of determinants it follows that \(\operatorname{det} A=\operatorname{det} A^{+}\) for every linear transformation \(A\) on \(\mathcal{V}\) .

The method of extending bilinear functionals works for conjugate bilinear functionals also. If, that is, \(\mathcal{V}\) is a (real) inner product space, then there is a natural way of introducing a (complex) inner product into \(\mathcal{V}^{+}\) ; we write, by definition, \[(x_{1}+i y_{1}, x_{2}+i y_{2})=(x_{1}, x_{2})+(y_{1}, y_{2})-i\big((x_{1}, y_{2})-(y_{1}, x_{2})\big).\] Observe that if \(x\) and \(y\) are orthogonal vectors in \(\mathcal{V}\) , then \[\|x+i y\|^{2}=\|x\|^{2}+\|y\|^{2}.\] 

The correspondence from \(A\) to \(A^{+}\) preserves all algebraic properties of transformations. Thus if \(B=\alpha A\) (with \(\alpha\) real), then \(B^{+}=\alpha A^{+}\) ; if \(C=A+B\) , then \(C^{+}=A^{+}+B^{+}\) ; and if \(C=A B\) , then \(C^{+}=A^{+} B^{+}\) . If, moreover, \(\mathcal{V}\) is an inner product space, and if \(B=A^{*}\) , then \(B^{+}=(A^{+})^{*}\) . (Proof: evaluate \(\big(A^{+}(x_{1}+i y_{1}),(x_{2}+i y_{2})\big)\) and \(\big(x_{1}+i y_{1}, B^{+}(x_{2}+i y_{2})\big)\) .)

If \(A\) is a linear transformation on \(\mathcal{V}\) and if \(A^{+}\) has a proper vector \(x+i y\) , with proper value \(\alpha+i \beta\) (where \(x\) and \(y\) are in \(\mathcal{V}\) and \(\alpha\) and \(\beta\) are real), so that \begin{align} & A x=\alpha x-\beta y, \\ & A y=\beta x+\alpha y, \end{align} then the subspace of \(\mathcal{V}\) spanned by \(x\) and \(y\) is invariant under \(A\) . (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 \(1\) or \(2\) .) If, in particular, \(A^{+}\) happens to have a real proper value (that is, if \(\beta=0\) ), then \(A\) has the same proper value (since \(A x=\alpha x\) , \(A y=\alpha y\) , and not both \(x\) and \(y\) can vanish).

We have already seen that every (real) basis in \(\mathcal{V}\) is at the same time a (complex) basis in \(\mathcal{V}^{+}\) . It follows that the matrix of a linear transformation \(A\) on \(\mathcal{V}\) , with respect to some basis \(\mathcal{X}\) in \(\mathcal{V}\) , is the same as the matrix of \(A^{+}\) on \(\mathcal{V}^{+}\) , with respect to the basis \(\mathcal{X}\) in \(\mathcal{V}^{+}\) . 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