# Planes in C^n

1. May 6, 2012

### WWGD

Hi, All:

Just curious:

What is the general equation of a linear n-subspace S of C^n (n>1)? I can figure out a line

in C^n, but can't think of strict higher-d subspaces S, i.e., S is not the whole space.

Thanks for any refs, hints.

2. May 6, 2012

### DonAntonio

If you know the real spaces then you're done, as $\,\,\mathbb{C}^n_\mathbb{R}\cong \mathbb{R}^{2n}_\mathbb{R}\,\,$ ...

Now, if you're talking of $\,\,\mathbb{C}^n_\mathbb{C}\,\,$ then this is "basically the same" as $\,\,\mathbb{R}^n_\mathbb{R}\,\,$: a k-dimensional

subspace can be given as the solution to some system of linear equations.

DonAntonio

3. May 10, 2012

### WWGD

Thanks, Don Antonio, I thought something very similar:

We use the fact that linear maps take subspaces to subspaces, and linear isomorphisms

actually preserve the dimension of these subspaces. I believe then, this isomorphism

would transform the equation of an n-subspace in R^2n into one in C^n.

4. May 10, 2012

### WWGD

I guess the/a method for defining subspaces S<V ; V,W vector space, where DimW>DimV-DimS, as kernel of a linear map T:V-->W is:

1) Choose a basis B_S for S , and extend to a basis B_V for all of V. Choose a basis B_W

for W

2)Define a linear map by defining:

T(b_V)==0 , for all b_S in B_S

3)Set up a bijection between basis vectors in B_V\B_S and basis vectors in B_W.

4)Extend by linearity.

5. May 31, 2012

### WWGD

Sorry, I forgot the original question I had:

Can we represent a line in C^n as a function of z ( a single complex variable)?

If f: R-->R is linear, we can represent it as f(x)=ax+b, i.e., as a function of a single

real variable. Can the same be done for lines in C^n ?

6. Jun 1, 2012

### HallsofIvy

Staff Emeritus
No, you cannot write a line, a one dimensional subspace, of Cn as a single equation in z. Saying that a subspace "has dimension i" means that you need to know i of its coordinates of a given point but can calculate the other n- i. And that means you will need to have n- i equations to calculate those.

This has nothing to do with "C" versus "R". Back in Calculus you saw that a single equation can determine a line in two dimension, but a single equation in R3 determines a plane, of dimension 2. In order to specify a line, of dimension 1, you need two equations. Geometrically, that can interpreted as the line of intersection of the two planes given by the two equations. Of course, you can specify a line in R3 by 3 parametric equations, giving x, y, and z as functions of the single variable, t. But that is now three equations in four variables and so still has 4- 3= 1 degree of freedom. In general, to specify an i dimensional subspace of an n dimensional space, you need n- i equations.

7. Jun 1, 2012

### WWGD

But , given ℂ1 has complex dimension 1, and a line is 1-dimensional, so that would lead to n-i=1-1=0 .

And ℂ2, has complex dimension 2, and so we would need 2-1=1 , so what am I missing?

8. Jun 2, 2012

### chiro

The dimension is usually representative of the number of independent parameters you need to describe something.

You could embed a line or a two-dimensional plane (proper dimension, not dimension of embedded space) instead any higher dimensional space and the intrinsic dimension is still unchanged.

9. Jun 3, 2012

### WWGD

Right, but the issue was how to write a line or plane as a function of complex variables.

According to Ivy, to write an i-dim space in an ambient n-space, n-i variables would be

enough. But , if I understood well, this would lead to a line in ℂ1 requiring

n-i =1-1=0 equations and for a line embedded in ℂ2 needing n-i=2-1

equations, which does not work, from previous posts.

10. Jun 4, 2012

### chiro

Complex numbers have a dimension of two when it comes to their linear structure. The real and the imaginary component for the standard definitions are orthogonal which implies independence which implies that they are completely distinct parameters and therefore you need to specify both to describe the space in its completion. This means it is two-dimensional for a single complex number.

If you want to describe a plane for n-dimensional complex numbers, then simply define a linear object with your embedded space, and your number of intrinsic dimensions for your object (exactly as HallsOfIvy has pointed out) and then check that they obey the plane equation and also linearity.

If you want to deal with projections and so on, define the appropriate projection that maintains the properties of a projection (P^2 = P) and make sure you get the other expected properties that you get for cartesian-type projections (assuming you are dealing with a flat plane).

If are only dealing with the situation where are only dealing with addition and scalar multiplication of complex numbers, then you are done.

If you wish to however deal with multiplication of complex numbers, then this will change things a lot, but multiplication is not really in a linear context: linear usually refers to having axioms f(X+Y) = f(X) + f(Y) and f(aX) = af(X): anything involving multiplication is different.

11. Jun 4, 2012

### WWGD

Thanks, Chiro, I was thinking along those lines. And, yes, my problem

is multiplication, because I am also considering the complexes as a vector space

over C, which implies that for f(aX)=af(X) , both aX and af(X) are multiplications of

complexes.

And, I guess , re your 1st paragraph, that a line from C to C is not a function of

a complex variable. My idea was to use the isomorphism between C as a V.S over

R and R^2 also as a V.S over R, to map lines to lines, since linear isomorphisms not

only send subspaces to subspaces, but also preserve the dimension of the subspaces.

Ultimately, I am trying to get a better understanding of complex projective spaces,

and for that I need to set up the lines before understanding what their projections

look like. There seems to be a big difference between a line in C as a V.S over R,

and a line in C as a V.S over itself.

12. Jun 4, 2012

### WWGD

I am trying to replicate the idea/method to describe how any subspace of a real vector

space is the kernel of a linear map*, to complex vector spaces. My confusion had to see

with the difference between the identification z~λz (with z seen as a vector (x,y)), when λ is

real, vs. when it is complex ( i.e., has non-zero imaginary part ). In the first case z~λz gives

us a plane thru the origin ; we get all rotations and scalings of z, since multiplying by a

complex number has the effect of rotating it and stretching it. In the last case, we get a

line, since real multiples of z only scale but do not rotate z.

*The method I know for having a subspace S of V described as the kernel of a map

T:V-->W (where W has to be large-enough ), is that of defining T(v_s)==0 for any v_s in

the basis of S , extending that basis of S into a basis for V, and then setting up a bijection

between vectors b_v in the extended basis into any basis for W (so we need W to be large-

enough dimension wise to be able to define that bijection. )

13. Jun 4, 2012

### chiro

Have you looked at the theory for complex vector spaces?

14. Jun 5, 2012

### WWGD

Not really; up to know I thought all the results from vector spaces over ℝ would extend

in a standard, clear way. Can you suggest a good source?

15. Jun 5, 2012

### chiro

I don't have a deep knowledge in this area: I only know the relevance of it from my readings dealing with things like Quantum Mechanics and Hilbert-Space Theory.

Doing a quick google search (for complex vector spaces) gives this:

http://math.stackexchange.com/questions/103193/complex-vector-spaces

There are plenty of PDF's for the geometry of complex vector spaces which indirectly answers all your questions (look at the inner and outer products and related projections involved in complex vector spaces).