Linear Algebra: Orthonormal Basis

In summary, the conversation discusses finding an orthonormal basis for the subspace of R^4 spanned by the given vectors. The attempt at using the Gram-Schmidt process resulted in an incorrect basis, but after clarification, it was understood that the basis only needs to span the subspace and not R^4. The vector (0,0,0,0) is not included in the basis as it is in every subspace and is not a part of the orthonormal basis.
  • #1
tylerc1991
166
0

Homework Statement



Find an orthonormal basis for the subspace of R^4 that is spanned by the vectors: (1,0,1,0), (1,1,1,0), (1,-1,0,1), (3,4,4,-1)

The Attempt at a Solution



When I try to use the Gram-Schmidt process, I am getting (before normalization): (1,0,1,0), (0,1,0,0), (1,0,-1,2), (0,0,0,0). So obviously there is some mistake that I am making but I have checked this at least 3 times. Can someone help me and let me know if it is something on my end or the problem. Thank you.
 
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  • #2
Without have actually worked it out fully, why do you think you made a mistake? The dimension of the subspace is less than 4. You are only going to get a number of orthonormal vectors equal to the dimension of the subspace.
 
  • #3
Dick said:
Without have actually worked it out fully, why do you think you made a mistake? The dimension of the subspace is less than 4. You are only going to get a number of orthonormal vectors equal to the dimension of the subspace.

I thought that the basis had to span R^4? And since of of the elements was (0,0,0,0) the basis can't span R^4.
 
  • #4
tylerc1991 said:
I thought that the basis had to span R^4? And since of of the elements was (0,0,0,0) the basis can't span R^4.

It doesn't span R^4. It spans the subspace of R^4 spanned by the given vectors.
 
  • #5
Dick said:
It doesn't span R^4. It spans the subspace of R^4 spanned by the given vectors.

I see. So will (0,0,0,0) be included in the orthonormal basis?
 
  • #6
tylerc1991 said:
I see. So will (0,0,0,0) be included in the orthonormal basis?

(0,0,0,0) is in EVERY subspace. You throw that away. It's never part of a basis. A basis is a set of linearly independent vectors. And it certainly isn't normal.
 

1. What is an orthonormal basis in linear algebra?

An orthonormal basis is a set of vectors in a vector space that are both orthogonal (perpendicular) and normalized (unit length). This means that the vectors are all independent from each other and form a complete basis for the vector space.

2. Why is an orthonormal basis important in linear algebra?

An orthonormal basis is important because it simplifies calculations and makes it easier to understand and manipulate vectors in a vector space. It also allows for the representation of any vector in the space as a unique linear combination of the basis vectors.

3. How do you determine if a set of vectors form an orthonormal basis?

To determine if a set of vectors form an orthonormal basis, you need to check if the vectors are orthogonal and normalized. To check for orthogonality, calculate the dot product of each pair of vectors and make sure it equals 0. To check for normalization, calculate the magnitude (length) of each vector and make sure it equals 1.

4. Can you have more than one orthonormal basis for a vector space?

Yes, a vector space can have multiple orthonormal bases. This is because there are infinitely many ways to choose a set of orthogonal vectors, and then each set can be normalized in different ways.

5. How does an orthonormal basis relate to the Gram-Schmidt process?

The Gram-Schmidt process is a method for finding an orthonormal basis from a set of linearly independent vectors in a vector space. It involves orthogonalizing the vectors by subtracting their projections onto previously determined basis vectors, and then normalizing them to have a unit length. This process ensures that the resulting basis is orthonormal.

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