Gram-schmidts orthonormalization

1. Nov 11, 2009

jbear12

Apply Gram-Schmidts process to the sebust S of the inner product space V to obtain an orthogonal basis for span(S). Then normalize the vectors in this basis to obtain an orthognormal basis for span(S)

V=span(S) where S={(1,i,0), (1-i,2,4i)} and x=(3+i,4i,-4).

Isn't the length of (1,i,0) zero? I'm confused about finding orthonormal basis for a complex set. Can anyone solve this problem step by step for me?
Thank you very much!

2. Nov 11, 2009

lanedance

I won;t solve the problem for you, but am happy to have a look at your work

the length of a= (1,i,0) is not zero, it is usually given by the complex innner product, which in matrix notation, if a is a complex column vector this becomes
$$||a||^2 = <a,a> = (a^*)^Ta[/$$
where * denotes the complex conjugate

so just like a dot product, but with the complex conjugate of itself

3. Nov 11, 2009

jbear12

Thank you lanedance. I realized it after some searching on the internet. :)
I have another dumb question...:P
What's the length of (1-5i/2, 7-i/2,4i)? Is it the square root of (1-5i/2)squared+(7-i/2)squared+(4i)squared, where i2=-1?

4. Nov 11, 2009

lanedance

the length squred will be the following dot product
$$(1-5i/2, 7-i/2,4i)* \bullet (1-5i/2, 7-i/2,4i)$$

so taking the congujate
$$= (1+5i/2, 7+i/2,-4i) \bullet (1-5i/2, 7-i/2,4i)$$

then the multiply out as a normal dot product

if you're still confused, start with: what is the magnitude of the 1+5i/2 in the 1D complex space?

Last edited: Nov 11, 2009
5. Nov 11, 2009

jbear12

I get it.
Thank you very much, lanedance. :)