# Vectors in C^n

• Dec 31st 2009, 12:22 PM
Bilbo Baggins
Vectors in C^n
Hello:

I have just completed an introductory Linear Algebra course at university, and I have a question. Is it possible to graph a vector in C^n if, perhaps, n is small e.g. n=2 or 3? I can't linearly order the complex numbers, but I can impose a lexicographic order upon them. However, this order precludes placing them on an axis because they become "infinitely dense".

I'm still new to mathematics, and I have not studied complex analysis at all. I worked out of David Lay's "Linear Algebra and Its Applications" which I thought was a decent but not stellar text. Indeed, I had hoped that my text would go into real detail and abstraction. Alas, it did not.
• Dec 31st 2009, 04:55 PM
math2009
if n=2 , \$\displaystyle dim(C^2)=3\$ , you can draw in 3-dimensions space,
if n>2, you only draw projection to 3-dimensions space.

Current coordination couldn't support higher dimensions.
• Dec 31st 2009, 05:05 PM
Bilbo Baggins
Quote:

Originally Posted by math2009
if n=2 , \$\displaystyle dim(C^2)=3\$ , you can draw in 3-dimensions space,
if n>2, you only draw projection to 3-dimensions space.

Current coordination couldn't support higher dimensions.

Hi,

Again, I know nothing of more advanced algebra or complex analysis. How would I graph a complex vector exactly? For example, <1+2i, 3+i> Where would this lie? You said it would have three dimensions, right? Then it's isomorphic to R^3? Thanks.
• Jan 1st 2010, 06:52 AM
HallsofIvy
Quote:

Originally Posted by math2009
if n=2 , \$\displaystyle dim(C^2)=3\$ , you can draw in 3-dimensions space,

How did you get that? As a vector space over the complex numbers, \$\displaystyle dim(C^2)= 2\$, of course, but as a vector space over the real numbers, \$\displaystyle dim(C^2)= 4\$, not 3.

Quote:

if n>2, you only draw projection to 3-dimensions space.

Current coordination couldn't support higher dimensions.
• Jan 2nd 2010, 08:27 AM
Bilbo Baggins
Quote:

Originally Posted by Bilbo Baggins
Hello:

I have just completed an introductory Linear Algebra course at university, and I have a question. Is it possible to graph a vector in C^n if, perhaps, n is small e.g. n=2 or 3? I can't linearly order the complex numbers, but I can impose a lexicographic order upon them. However, this order precludes placing them on an axis because they become "infinitely dense".

I'm still new to mathematics, and I have not studied complex analysis at all. I worked out of David Lay's "Linear Algebra and Its Applications" which I thought was a decent but not stellar text. Indeed, I had hoped that my text would go into real detail and abstraction. Alas, it did not.

Actually, I came up with a bijection between C^n and R^2n that preserves vector addition and scalar multiplication as long as I restrict the scalar factors to the reals. So I suppose that the two are isomorphic. This leads to an interesting result. Given any basis for C^n, I can't span R^2n (and therefore C^n) with the images of that basis. I'm always n vectors short in R^2n.
• Jan 3rd 2010, 04:34 AM
HallsofIvy
Quote:

Originally Posted by Bilbo Baggins
Actually, I came up with a bijection between C^n and R^2n that preserves vector addition and scalar multiplication as long as I restrict the scalar factors to the reals. So I suppose that the two are isomorphic. This leads to an interesting result. Given any basis for C^n, I can't span R^2n (and therefore C^n) with the images of that basis. I'm always n vectors short in R^2n.

Yes, you can. Each a+ bi in C is mapped into (a, b) in \$\displaystyle R^2\$.