MHB Understanding Extension of Scalars in a Vector Space

[caffeinemachine]

$\newcommand{\R}{\mathbf R}\newcommand{\C}{\mathbf C}$

Low-Tech Complexification: Let $V$ be a finite dimensional vector space over $\R$. We can forcefully make $W:=V\times V$ into a complex vector space by defining addition component-wise and product $\C\times W\to W$ as
$$
(a+ib)(u, v)=(au-bv, av+bu)
$$
for all $a, b\in \R$ and $u, v\in V$.
The vector space axioms can be readily checked.

Also, a linear transformation $T:V\to V$ induces a linear map $\bar T:W\to W$ as $\bar T(u, v)= (Tu,Tv)$ for all $u, v\in V$.
From here, interesting information about $T$ can obtained. For example, $\bar T$ has an eigenvalue. So there is $(0, 0)\neq (u, v)\in W$ such that
$$
\bar T(u, v)=(\lambda +i\mu)(u, v)=(Tu, Tv)
$$
which gives, $Tu=\lambda u-\mu v$ and $Tv=\lambda v+\mu u$.
So there are vectors $u$ and $v$ in $V$, not both zero, such that the above equations hold. This wasn't obvious (to me) without making the above construction.
___
I want to see how the above relates with the standard notion of extension of scalars.

The wikipedia article on complexification(Complexification - Wikipedia, the free encyclopedia) talks about defining the tensor product $V\otimes_{\R} \C$ as giving it a complex vector space structure by defining $\alpha(v\otimes \beta)=v\otimes(\alpha\beta)$ for $\alpha, \beta\in \C$ and $v\in V$.

I can feebly see how the "low-tech" process described earlier is same as the tensorial construction. The vector $(u, v)\in W$ corresponds to $u\otimes 1+v\otimes i$, and for $T:V\to V$ we get $\bar T(v\otimes \alpha)=(Tv)\alpha$.

Admittedly, this transition is not obvious to me.

Can somebody please throw some light on how the two approaches are really the same? (Assuming they actually are same).

Thanks.

 

[Opalg]

I can feebly see how the "low-tech" process described earlier is same as the tensorial construction. The vector $(u, v)\in W$ corresponds to $u\otimes 1+v\otimes i$, and for $T:V\to V$ we get $\bar T(v\otimes \alpha)=(Tv)\alpha$.

I don't see anything "feeble" about your mapping $(u,v) \mapsto u\otimes 1+v\otimes i$. In fact, that seems to give a very simple and precise correspondence between the low-tech and high-tech (tensorial) constructions of the complexification procedure. If you like, you can think of both constructions as being ways to make sense of the expression $u+vi$ for an element of the complexified space, just as a complex number can be considered in either of the forms $(a,b)$ or $a+bi$ (where $a,b\in\mathbb{R}$).

 

[caffeinemachine]

I don't see anything "feeble" about your mapping $(u,v) \mapsto u\otimes 1+v\otimes i$. In fact, that seems to give a very simple and precise correspondence between the low-tech and high-tech (tensorial) constructions of the complexification procedure. If you like, you can think of both constructions as being ways to make sense of the expression $u+vi$ for an element of the complexified space, just as a complex number can be considered in either of the forms $(a,b)$ or $a+bi$ (where $a,b\in\mathbb{R}$).

Hello opalg,

I did a poor job explaining my question.

The isomorphism between the low-tech and the hi-tech construction is clear. What I wanted to see was if the idea of tensoring with $\mathbf C$ is somehow a natural one. To me it is very unnatural as of now.

 

[Opalg]

What I wanted to see was if the idea of tensoring with $\mathbf C$ is somehow a natural one. To me it is very unnatural as of now.

The big advantage of having complex scalars is that $\mathbb{C}$ is algebraically closed. For example, in the finite-dimensional case, if you have a real $n\times n$ matrix $A$, you might think that it would "naturally" act on the real space $\mathbb{R}^n$. But then it need not have any eigenvectors. If you complexify $\mathbb{R}^n$ so that it becomes $\mathbb{C}^n$ then $A$ acts on the complexified space, where it is guaranteed to have eigenvectors.

At a more advnced level, a Google search for "complexification" will bring up explanations of how the theory of real Lie groups and algebras depends on their complexifications.

 

[caffeinemachine]

The big advantage of having complex scalars is that $\mathbb{C}$ is algebraically closed. For example, in the finite-dimensional case, if you have a real $n\times n$ matrix $A$, you might think that it would "naturally" act on the real space $\mathbb{R}^n$. But then it need not have any eigenvectors. If you complexify $\mathbb{R}^n$ so that it becomes $\mathbb{C}^n$ then $A$ acts on the complexified space, where it is guaranteed to have eigenvectors.

At a more advnced level, a Google search for "complexification" will bring up explanations of how the theory of real Lie groups and algebras depends on their complexifications.

Thanks Opalg.

I guess I just need more time.

One unrelated thing. Should the word "thereon" in your signature be really "thereof"?

 

[Opalg]

One unrelated thing. Should the word "thereon" in your signature be really "thereof"?

The german original has wovon [whereof] ... darüber [thereon]. If it had been wovon ... davon (using the same preposition von (meaning of) then I would agree that it should be translated whereof ... thereof. But über means on (or upon), so I thought it would be better to render it as thereon (or thereupon if you prefer).

Good question though – it makes a nice change from mathematics!

 

[caffeinemachine]

The german original has wovon [whereof] ... darüber [thereon]. If it had been wovon ... davon (using the same preposition von (meaning of) then I would agree that it should be translated whereof ... thereof. But über means on (or upon), so I thought it would be better to render it as thereon (or thereupon if you prefer).

Good question though – it makes a nice change from mathematics!

I don't know any German. Haha. But Danke for this!