Gram-Schmidt Finding Orthonormal Basis The famous Gram-Schmidt - - PowerPoint PPT Presentation

Gram-Schmidt

Finding Orthonormal Basis

The famous Gram-Schmidt process is used to produce an orthogonal basis from a given ba-

• sis. It provides a constructive proof of the earlier

claim that every vector space has an orthonor- mal basis.

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Gram-Schmidt

ALGOR Input: collection x1, . . . , xk of linearly independent vectors. Output: collection y1, . . . , yk of orthogonal vec- tors that span the same space. Process: Generate vectors y1, y2, y3, . . . by yi = xi −

i−1

• j=1

projyj(xi) These vectors can then be normalized, if de- sired.

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Example

Find orthonormal basis of span

• f the vectors

x1 =       1 1       x2 =       2 −1 3       x3 =       3 1 1 −7      

Answer is:

y1 =       1 1       y2 = x2 − 2 2y1 =       1 −1 −1 3       y3 = x3 − 4 2y1 − −20 12 y2 =       8/3 −8/3 −2/3 −2      

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Example Continued

After normalization we have the vectors

1 √ 2(1, 1, 0, 0), 1 √ 12(1, −1, −1, 3), and 1 √ 42(4, −4, −1, −3)

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Another Example

Use Gram-Schmidt to find an orthogonal basis for the space spanned by (1, 1, 0), (1, 0, 1), and (0, 1, 2). We proceed. y1 =    1 1    then y2 =    1 1    − 1 2    1 1    =   

1 2

−1

2

1   

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Another Example Continued

And y3 =    1 2    − 1 2    1 1    − 3/2 3/2   

1 2

−1

2

1    =    −1 1 1    Normalized, we have 1 √ 2    1 1    , 1 √ 6    1 −1 2    , 1 √ 3    −1 1 1    (But note span of the vectors is all of R3)

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Summary

The Gram-Schmidt process produces an orthog-

• nal basis. It considers the basis vectors in turn

and for each, subtracts its projection onto the previous basis vectors. The resultant basis can be made orthonormal by normalization.

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