Gramm-Schmidt and Complex Numbers

[hank]

Ok, here's the question.

Apply Gramm-Schmidt procedure with standard inner product to {(i,i,i), (0,i,i),(0,0,i)}.

Ok, here we go.

u1 = (i,i,i)/||(i,i,i)|| = (1/sqrt(-3))(i,i,i) //book says answer is (1/sqrt(3))(i,i,i)


u2 = (0,i,i) - (1/sqrt(3))(0-1-1)(1/sqrt(3))(i,i,i) = (0,i,i) + (2/3)(i,i,i) = (1/3)(2i,5i,5i)
|| of above || = (1/3)(2i,5i,51)/sqrt(-6) //book says this answer is wrong also.

Before going to the next step, which is a nightmare, I was hoping someone could point out where I'm making my mistakes so far.

 

[DrMike]

Remember that over C, the euclidean inner product goes :

\(\displaystyle \langle(x_1,x_2,x_3),(y_1,y_2,y_3)\rangle=\sum x_i\bar{y_i}\mathrm{,\ not\ }\sum x_i y_i\).

Therefore, \(\displaystyle ||(x_1,x_2,x_3)||^2=\sum x_i\bar{x_i}\mathrm{,\ not\ }\sum x_i^2\)

Things like this :

\(\displaystyle ||v_{2}||^{2}=-6\)

should have told you something was wrong - norms are non-negative, strictly positive for nonzero vectors - and certainly not complex!

 

[hank]

Ok, then I should assume the book is wrong when it says the answer to u1 is (i,i,i)/sqrt(3)?

 

[galactus]

I apologize. I am familiar with the Gram-Schmidt technique, but I had some mistakes in my post. As Dr. Mike pointed out, a norm is a distance and can not be negative. I was half asleep when I done this, so I am sorry for the booboos. I will look at it again when I have time. I have to leave in a few minutes.

 

[DrMike]

Ok, then I should assume the book is wrong when it says the answer to u1 is (i,i,i)/sqrt(3)?

No.