# Complexification of a vector space

1. Sep 4, 2007

### quasar_4

Hello all,

I've just learned a bit about the complexification of a real vector space V to include scalar multiplication by complex numbers. A bit of confusion has ensued, which I am hoping someone can help me with conceptually: 1) how does one generate a basis for the new space Vc? It seems that one obtains the basis by somehow extending the basis for V, but I am very confused about this. In fact, I'm not exactly sure how vectors in the new space should be defined at all. :yuck:

2) does anyone know how one would prove that the dim(Vc)=dim(V)? I'm not asking for homework; I've just heard that this is the case, but haven't seen anything proved.

3) Under what circumstances would one want to complexify V? anyone have some good examples?

Thanks.

2. Sep 4, 2007

### dextercioby

1. If the vectors in the real space are C-linear independent, then they provide a basis for the complexfication.
2. This part is valid in the case the condition in 1. is satisfied.
3. Complexification of Lie algebras is a tool useful for representation theory.

3. Sep 4, 2007

### matt grime

I don't like the previous answer.

1. the bases are the same.
2. you mean dim_C of one and dim_R of the other - since, by 1, the basis sets are the same this is a vacuous question.
3. extension of scalars is in general an important concept, far beyond dexter's note. Please google for it.

4. Sep 5, 2007

### mathwonk

complexification makes all characteristic polynomials split.

5. Sep 18, 2007

### quasar_4

the basis thing is still a bit odd. According to my class if the basis for V is given by {e1,e2,...,en} then the basis for Vc is given by {(e1,0),(e2,0),...,(en,0)}. But then, I guess that is basically the same except that we've sneakily made the vectors into ordered pairs. Somehow this is now representative of VxV?

6. Sep 18, 2007

### mathwonk

the complexification of the space with real basis (1,0), (0,1), is the complex space with absis (1,0),(0,1).

i.e. the complexification of the space of ordered pairs of reals, is the space of ordered pairs of complexes.

or try this one: if V is a real vector space, then since C is also a real vector space, you can consider all real linear maps from V to C. this is a complex vector space since you can multiply the values by complex numbers.

or you could let V* be the real dual of V and consider all real linear maps V*-->C.

then given an element v of V, you could define a map v**:V*-->C by sending the real linear map t in V*, to the complex number t(v) = v**(t). this shows how to embed V into the new space HomR(V*,C). so we could regard HomR(V*,C) as a complex vector space which contains V, namely as the subspace of maps with images in R, i.e,. then V is the subspace HomR(V*,R) = V** = V (in finite dimensions only of course.)

then any real basis v1,...vn of V corresponds to a real basis of HomR(V*,R) and also a complex basis v1**,...,vn** of HomR(V*,C).

how do you like them apples? i made that up just for you.

Last edited: Sep 18, 2007
7. Sep 18, 2007

### mathwonk

so i guess i want my complexification of V to be a complex vector space W containing V as a real subspace and such that W equals V + iV, which is a direct sum decomposition of the real space underlying W, as a sum of two real subspaces.

it should follow that a subset of V is real independent if and only if it is complex independent as a subset of W, and spans V over R iff it spans W over C.

In particular, any real basis of V is a complex basis of W.

8. Sep 18, 2007

### mathwonk

i am getting no feedback on my post 6, of which i am somewhat proud. i.e. i have given you a construction that does not mention tensor products, hence is elementary.

9. Sep 18, 2007

### mathwonk

as stated above, after complexifying, there should be more eigenvectors.

i.e. the main point is not just complexifying the spaces but also the maps.

i.e. every real linear map of V-->V should become a complex linear map from the complexification to itself. and presumably with the same characteristic polynomial?

so a 90 degree rotation, becomes multiplication by i or -i.

presumably then one can transform the proof of diagionalizability of the complex operator into a classification of the real operators. i.e. we should be able to use the complex spectral theorem to deduce a real classification for orthogonal matrices.

i.e. just as in ode, the diff eq f''+f = 0 has compklex basis of solutions either e^(ix) or e^(-ix), but realsolutiions cosc and sinx.

i guess the complexification of the real solution space should be the complex solutiion space. and just as letting d act on sin, cos, rotates them into each other, letting it act on thiose exponentials, gives eigenfunctiions.

you are beginnin to make this seem interesting, once you combine it with ode, as all linear algebra should be!!!! but seldom is.

Last edited: Sep 18, 2007
10. Sep 18, 2007

### mathwonk

here is alittle easier sounding version of that construction. take any n objects and call them e1,...,en.

then define a vector space V as the set of all functions from the set {e1,...,en}-->R.

Then define the complexification of V to be the set of all functiuions from

{e1,...,en}-->C. notice that in both cases a basis consist of the n functions ei*, each taking exactly one of the ei to 1, and the others to 0.

thus the same set of n things is an R basis for V and a C basis for its compl.....tion.

but i regress, as it gets late. i will retire before i become rude again.

11. Sep 20, 2007

### mathwonk

notice that the complexification of a real space has more structure than a general complex space, i.e. it also has a real conjugation operator on it, defined by composing the functioins above by complex conjugation.

or if you like by writing each vector as an element of V + iV, and using the identity on the first summand and the minus map on the second one.

12. Sep 20, 2007

### mathwonk

13. Sep 27, 2007

### quasar_4

wow - it's taken a few days for my brain to absorb all of this. It is making more sense now though. So, should one care to take a coordinate vector [x] on the basis B for V and rewrite [x] on the basis B' for the complexified space, why would the components of [x] change at all? wouldn't our change of basis matrix just be the identity?

14. Sep 27, 2007

### quasar_4

in fact, that seems like a trivial question. There isn't effectively a "change" of basis at all, the only thing happening is that you'd have to rewrite your vectors as an ordered pair (right?) so [x]={x1, x2,...,xn} rewritten for the basis for W becomes [x]={(x1,0),(x2,0),...,(xn,0)}. Is that correct?

15. Sep 27, 2007

### quasar_4

oh, and, I liked the ode stuff. To be honest, no one's even mentioned odes to me since about 3 years ago. I wish they did tie them more into linear algebra. It would make for a lovely depth of understanding on my diff eqs.