# What is C^2?

1. Nov 29, 2005

### loli12

I was asked to find a basis beta for V such that [T]_beta is a diagonal matrix.
V=C^2 and T is defined by T(z,w) = (z+iw, iz+w)
for C = complex.
I don't quite understand the expression C^2, I know C can be represented by a plane with the real and imaginery axis, but how abt C^2?
and also, how do i figure out the basis for this? I used (1,0) and (0,1) as my basis and I got the right answer.. but the basis doesn't seem right to me intuitively.

2. Nov 30, 2005

### matt grime

C^2 is two dimensional complex vector space.

Presumably you're happy with R^2

T is not diagonal with respect to that standard basis.

Last edited: Nov 30, 2005
3. Nov 30, 2005

### dextercioby

The cartesian product between $\mathbb{C}$ and $\mathbb{C}$ is written $\mathbb{C}\times \mathbb{C}$ and commonly written as $\mathbb{C}^{2}$.

Daniel.

4. Nov 30, 2005

### loli12

"C^2 is two dimensional complex vector space."
does this means that geometrically it has 4 axis in total?

so, what is the basis for C^2?
intuitively I thought the basis is {(1,0), (0,1), (i,0),(0,i)} , but i think there're way too many vectors..
and (1+i , 0), (0, 1+i) seems like can't span the whole space..

anyone can let me know what the basis for C^2 is?

5. Dec 1, 2005

### shmoe

You're considering C^2 as a vector space over what field?

6. Dec 1, 2005

### loli12

the question said for the linear operators T on a vector space V, if T is diagonalizable, find a basis beta for V suh that [T]_beta is a diagonal matrix.
the answer for this quesiton is beta = {(1,1) , (1,-1)}
but I still have no clue on the basis... please give me some hint!

7. Dec 1, 2005

### dextercioby

Well, we usually say that $\mathbb{C}^2$ is a 2-dimensional vector space over the field of complex numbers, i.e. $\mathbb{C}$. A basis in this vector space is made up of
$$\left(\begin{array}{c} 1\\0 \end{array} \right) \ \mbox{and} \ \left(\begin{array}{c} 0\\1 \end{array} \right)$$

Daniel.

Last edited: Dec 1, 2005
8. Dec 1, 2005

### matt grime

C^2 is two dimensional over C, you are already using a basis implicitly when you define T(z,w)=(z+iw,w+iz). It is the 'standard basis' and it is given by (0,1) and (1,0) which is a basis for C^2 because you're allowed to multiply these by any complex number ie (3+i,1+2i) = (3+i)*(1,0)+(1+2i)*(0,1)

Suppose I asked you to diagonalize the REAL operator

T(x,y)=(x+y,x-y)

given with respect to the standard basis on R^2. Then you'd have no trouble doing it. This is exactly the same except that you're allowed to use complex numbers not just real ones.

9. Dec 1, 2005

### JasonRox

If you know how to find a basis for R^2, well then that's exactly how you do it. The same principle applies to C^2.

Note: I try not to worry about the visual representation of it. That might be bad, but I'm doing fine without the visual representation. I have some for things like linear independence/depence, but it's still not necessary because it all comes from the definition.

10. Dec 2, 2005

### loli12

Thanks all for the reply! I got it now!

11. Dec 2, 2005

### matt grime

It is not bad to do without visualization, it is very necessary, and it is bad to rely on visualization; the amount of maths that it visualizable is tiny and you should not rely on visualizing it (which I disinguish from having geometric intuition)