# Orthogonal basis

• Feb 4th 2013, 01:08 PM
sirellwood
Orthogonal basis
Could anyone tell me if any of the following sets of vectors is an orthogonal basis for R3

{(1,0,0), (1,1,0), (0,0,1)}
{(1,0,0), (1,-1,0), (0,0,1)}
{(1,0,1), (1,0,-1), (0,1,0)}
{(1,0,0), (1,1,0), (1,1,1)}
• Feb 4th 2013, 02:18 PM
jakncoke
Re: Orthogonal basis
First of all the set of vectors have to be linearly independent, second the dot product of each pair of vectors must be equal to 0.

Can you first of all list all the set of vectors which are lin. independent ?
• Feb 4th 2013, 08:12 PM
hollywood
Re: Orthogonal basis
It's easy to eliminate three of the options by checking if they're orthogonal:

{(1,0,0), (1,1,0), (0,0,1)}:
$\displaystyle (1,0,0) \cdot (1,1,0) = 1$

{(1,0,0), (1,-1,0), (0,0,1)}:
$\displaystyle (1,0,0) \cdot (1,-1,0) = 1$

{(1,0,1), (1,0,-1), (0,1,0)}:
This one is orthogonal:
$\displaystyle (1,0,1) \cdot (1,0,-1) = 0$
$\displaystyle (1,0,1) \cdot (0,1,0) = 0$
$\displaystyle (1,0,-1) \cdot (0,1,0) = 0$
To tell if the vectors are independent, set $\displaystyle r_1(1,0,1)+r_2(1,0,-1)+r_3(0,1,0)=(0,0,0)$. Then:
$\displaystyle r_1+r_2=0$
$\displaystyle r_3=0$
$\displaystyle r_1-r_2=0$
and you can see the only solution is $\displaystyle r_1=r_2=r_3=0$, which means the vectors are linearly independent.

{(1,0,0), (1,1,0), (1,1,1)}:
$\displaystyle (1,0,0) \cdot (1,1,0) = 1$

So the third set is an orthogonal basis, and the others aren't.

- Hollywood