# The necessity of defining complexified vector space from an existing real vector space of finite dimension

Let $$(V,\mathbb{R})$$ be a real vector space of finite dimension given to us. As it is finite there always exists a basis of $$V$$, denoted $$\mathcal{B}=\{e_i|i=1,…,n\}$$ such that $$V=\text{Span}(\mathcal{B})$$. Why cannot $$V_\mathbb{C}$$ be simply defined as $$V_\mathbb{C}=\text{Span}_\mathbb{C}(\mathcal{B})$$ ignoring the fact that scalar multiplication only makes sense when the field is $$\mathbb{R}$$?

The way presented to me from a more reliable source is that instead of the above process, we construct from $$(V,\mathbb{R})$$ the vector space $$(\mathbb{C}\otimes V,\mathbb{R})$$, then define a new scalar multiplicaiton when the scalar is a complex number by $$\alpha(\beta\otimes v):=\alpha\beta\otimes v$$. And we call the set $$\mathbb{C}\otimes V$$ equipped with this newly defined multiplication with complex field the complexified vector space of $$V$$, denoted $$V_\mathbb{C}$$.

It is as if this entire construction is to avoid the operation of multiplying a complex number directly with any vector in the original vector space $$V$$, and instead make the multiplication become one that we could understand, i.e. by scalar multiplication in the field of complex numbers $$\mathbb{C}$$ as in the first component of $$\mathbb{C}\otimes V$$. However in doing so we still need to define a new multiplication rule which essentially violates the bilinearity requirement of tensor product. So why cannot it be done just by $$\text{Span}_\mathbb{C}(\mathcal{B})$$ if we are to define some new multiplication rule anyways, we could define a new operation for complex numbers scalar multiplied with vectors in $$V$$ just as we did in the $$\mathbb{C}\otimes V$$ case.

## 1 Answer

But what would $${\rm Span}_\mathbb{C}({\cal B})$$ mean if a multiplication by scalars in $$\mathbb{C}$$ is not defined in $$V$$?

If you do not want to enlarge $$V$$ (as with the $$V\otimes\mathbb C$$ construction) in order to define a complex structure on $$V$$ you need to give a linear transformation $$T:V\longrightarrow V$$ such that $$T^2=-{\rm Id}$$ and let $$(x+yi)v=x+yT(v)\qquad\forall v\in V, x, y\in\mathbb{R}.$$ Equivalently, this amounts to fixing a group homomorphism $$\mathbb{C}^\times\longrightarrow{\rm GL}(V).$$ This cannot always be done: $$V$$ needs to have an even dimension over $$\mathbb{R}$$.