# Thread: Orthogonal Basis Using Gram-Schmidt

1. ## Orthogonal Basis Using Gram-Schmidt

Consider the inner product space $\displaystyle \mathbb{C}^3$ with the usual inner product. Starting with the basis

{$\displaystyle u_1, u_2, u_3$} = {$\displaystyle (1,1,1),(4i,-i,0),(2,0,-1)$},

find an orthogonal basis using the Gram-Schmidt process.

2. Originally Posted by funnyinga
Consider the inner product space $\displaystyle \mathbb{C}^3$ with the usual inner product. Starting with the basis

{$\displaystyle u_1, u_2, u_3$} = {$\displaystyle (1,1,1),(4i,-i,0),(2,0,-1)$},

find an orthogonal basis using the Gram-Schmidt process.
Hello funnyinga,

Can you show us where you are getting stuck? If you are not understanding the whole process, then I suggest working out a solved example...

4. Im really stuck here. The complex numbers are confusing me. Here is what I have done (although its probably all wrong).

Let $\displaystyle v_1 = u_1 = (1,1,1), u_2 = (4i,-i,0), u_3 = (2,0,-1)$

Then $\displaystyle v_2 = u_2 - \frac{<u_2,v_1>}{||v_1||^2} v_1$

So $\displaystyle v_2 = (4i, -i, 0) - \frac{(4i, -i, 0) * (1,1,1)}{3} (1,1,1) \ = ??$

and $\displaystyle v_3 = (2,0,1) - \frac{(2,0,-1) * (v_2)}{3} (1,1,1) - \frac{(2,0,-1)*(v_2)}{||v_2||^2} v_2 \ = ???$

5. [QUOTE=funnyinga;324406]Im really stuck here. The complex numbers are confusing me. Here is what I have done (although its probably all wrong).

Let $\displaystyle v_1 = u_1 = (1,1,1), u_2 = (4i,-i,0), u_3 = (2,0,-1)$

Then $\displaystyle v_2 = u_2 - \frac{<u_2,v_1>}{||v_1||^2} v_1$

So $\displaystyle v_2 = (4i, -i, 0) - \frac{(4i, -i, 0) * (1,1,1)}{3} (1,1,1) \ = ??$
Just treat "i" as if it were a variable.
That is $\displaystyle (4i,-i,0)- \frac{4i- i}{3}(1,1,1)= (4i, -i, 0)- \frac{3i}{3}(1,1,1) = (4i, -i, 0)- (i, i, i)= (3i, 0, -i)$.

and $\displaystyle v_3 = (2,0,1) - \frac{(2,0,-1) * (v_2)}{3} (1,1,1) - \frac{(2,0,-1)*(v_2)}{||v_2||^2} v_2 \ = ???$