Why do we use Gram-Schmidt process?

[soopo]

Homework Statement

I have 3 vectors. I need to find the ortonormal basis by Gram-Schmidt.

The Attempt at a Solution

I observed that, without Gram-Schmidt, I get two vectors which I get too by Gram-Schmidt. I have just divided the vectors by their lengths.

The third vector got without Gram-Schmidt differs however from the vector got by Gram-Schmidt.

Do you know why?

 

[Dick]

How to you 'get' vectors without Gram-Schmidt? If you are dividing the vectors by their length, they will be unchanged if they are unit length. If you apply Gram-Schmidt to list of normalized vectors they will unchanged if they are already orthogonal.

 

[soopo]

How to you 'get' vectors without Gram-Schmidt? If you are dividing the vectors by their length, they will be unchanged if they are unit length. If you apply Gram-Schmidt to list of normalized vectors they will unchanged if they are already orthogonal.

My vectors are:
v_{1} = [1\ -2\ 0\ 1]^{T}
v_{2} = [-1\ 0\ 0\ -1]^{T}
v_{3} = [1\ 1\ 0\ 0]^{T}
in the space of S.

I get these without G-S:
u_{1} = \frac {[1\ -2\ 0\ 1]^{T}} {\sqrt{6}}
u_{2} = \frac {[-1\ 0\ 0\ -1]^{T}} {\sqrt{2}}
u_{3} = \frac {[1\ 1\ 0\ 0]^{T}} {\sqrt{2}}

--edit here---
I get the same vectors with G-S but the the second and the third ones differ:
u_{2} = \frac {[-1\ -1\ 0\ -1]^{T}} {\sqrt{3}}
u_{3} = \frac {[1\ 0\ 0\ -1]^{T}} {\sqrt{2}}Is the reason that the second and the third ones differ that they are the only vectors which are
normalised already?

 

[Dick]

The u2 in the second group isn't orthogonal to the u1 in the second group. The final u2 you gave is. That's what's supposed to happen. u3 had better change as well. It's not orthogonal to anything.

 

[soopo]

The u2 in the second group isn't orthogonal to the u1 in the second group. The final u2 you gave is. That's what's supposed to happen. u3 had better change as well. It's not orthogonal to anything.

I agree with you: only the first vector got without G-S is the same as the one got with G-S.

Is the reason that it is already ortogonal to space?
I am not sure what the 1st vector is perpendicular to. To space?

PS: I corrected my previous post: the two last vectors.

 

[Dick]

I'm not sure I get what your problem is. Gram-Schmidt just says to normalize the first vector. You did that. After that it doesn't change. You adjust the other ones to be perpendicular to the first one. You can pick anyone of the vectors to remain unchanged just by changing the order of the vectors.

 

[soopo]

I'm not sure I get what your problem is. Gram-Schmidt just says to normalize the first vector. You did that. After that it doesn't change. You adjust the other ones to be perpendicular to the first one. You can pick anyone of the vectors to remain unchanged just by changing the order of the vectors.

Thank you! You cleared my confusion.

So we need first to normalize the 1st vector. Then, we change the other ones to be perpendicular to the 1st one. To achieve this we use Gram-Schmidt.

Is there any other method to make vectors perpendicular to the selected vector?

I am interested in unit circle. Does Gram-Schmidt process allow us to use unit circle?