Why do we use Gram-Schmidt process?

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In summary, the conversation discusses finding an orthonormal basis using Gram-Schmidt for a set of three vectors. The speaker notes that without Gram-Schmidt, they get two vectors, but with Gram-Schmidt, they get a different third vector. They ask for an explanation and the other person clarifies that Gram-Schmidt adjusts the other vectors to be perpendicular to the first vector. The speaker also asks if there is another method to achieve this and if Gram-Schmidt can be used with a unit circle.
  • #1
soopo
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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?
 
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  • #2
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.
 
  • #3
Dick said:
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:
[tex] v_{1} = [1\ -2\ 0\ 1]^{T}[/tex]
[tex] v_{2} = [-1\ 0\ 0\ -1]^{T}[/tex]
[tex] v_{3} = [1\ 1\ 0\ 0]^{T}[/tex]
in the space of S.

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

--edit here---
I get the same vectors with G-S but the the second and the third ones differ:
[tex] u_{2} = \frac {[-1\ -1\ 0\ -1]^{T}} {\sqrt{3}}[/tex]
[tex] u_{3} = \frac {[1\ 0\ 0\ -1]^{T}} {\sqrt{2}}[/tex]Is the reason that the second and the third ones differ that they are the only vectors which are
normalised already?
 
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  • #4
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.
 
  • #5
Dick said:
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.
 
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  • #6
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.
 
  • #7
Dick said:
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?
 

Related to Why do we use Gram-Schmidt process?

1. Why is the Gram-Schmidt process important in linear algebra?

The Gram-Schmidt process is important because it allows us to find an orthonormal basis for a given vector space. This is useful for solving a variety of problems in linear algebra, such as finding the best fit line for a set of data or solving systems of equations.

2. How does the Gram-Schmidt process work?

The Gram-Schmidt process involves taking a set of linearly independent vectors and transforming them into an orthonormal set of vectors. This is done by subtracting the projections of each vector onto the previously processed vectors, ensuring that the resulting vectors are orthogonal to each other.

3. What is the significance of an orthonormal basis?

An orthonormal basis is significant because it simplifies many calculations in linear algebra. It allows us to easily determine the length and angle of a vector, and also makes it easier to solve systems of equations or find the best fit line for a set of data.

4. Can the Gram-Schmidt process be used for any vector space?

Yes, the Gram-Schmidt process can be used for any vector space. However, it is most commonly used for finite-dimensional vector spaces with an inner product defined, such as Euclidean spaces.

5. Are there any limitations to the Gram-Schmidt process?

One limitation of the Gram-Schmidt process is that it can be computationally expensive for large systems. Additionally, it may not work for vector spaces with non-standard inner products. In these cases, alternative methods may need to be used.

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