# Complex number sum that should be easy

1. Dec 5, 2012

### trap101

Hey,

So I have a sum of complex numbers that really should be easy, but I'm not getting the right solution. It is with respct to using the Gram Schmidt process

U1 = (i, -1, i) U2 = (1,1,0)

So I perform the Gram Schmidt with U1 being my initial vector selection and I get:

V2 = (1,1,0) - [(-1 + i)/3 ] (i, -1 , i)

and here is where the issue arises. Now I did regular subtraction w.r.t complex numbers and got:

[ (4 + i)/ 3, (2+i)/3 , (1+i)/3]

but the solution is:

[(2+i)/3 , (2-i)/3, (-1+i)/3 ]

my vector is not orthogonal, so my question is what am I missing with respect to this?

thanks

2. Dec 5, 2012

### Dick

You are probably missing that dot product of complex vectors involves taking a complex conjugate of one of the vectors. And <u1,u2> is not necessarily equal to <u2,u1>. Do it again and be careful.

3. Dec 5, 2012

### Staff: Mentor

I checked the solution vector by dotting u1 with (their) v2, and got a value of zero. Are you forgetting that if u and v are complex vectors, then $u \cdot v = \sum u_i * \overline{v_i}$? IOW, you need to use the complex conjugate?

Show us how you calculated v2.

4. Dec 6, 2012

### HallsofIvy

Staff Emeritus
Also, you are dividing by "3" when you should be dividing by $\sqrt{3}$.

5. Dec 6, 2012

### trap101

This is how I got my V2

V2 = U2 - {[<(i, -1 , i) , (1,1,0)>]/ [<(i, -1 , i), (i, -1 , i)>]} (i, -1, i)

ok, well I figured it out, and well I'm having a big conundrum here. Essentially it came down to the order in which I performed the hermitian inner product. But I had dubbed
U1 = (i, -1 , i) and when I used the GS formula I did my inner product as:

<V1, U2 >. Doing it that way I got the wrong result, but doing it <U2, V1> I got the right result. Which brings up the question, of which is the way it is supposed to be and is that way fixed? Because when doing the inner product w.r.t just the reals this problem doesn't arise.

Thanks for the help.