Tag Archives: orthonormal vectors

Linear Algebra and Its Applications, Exercise 3.4.28

Exercise 3.4.28. Given the plane and the following vectors in the plane, find an orthonormal basis for the subspace represented by the plane. Report the dimension of the subspace and the number of nonzero vectors produced by Gram-Schmidt orthogonalization. Answer: … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.27

Exercise 3.4.27. Given the subspace spanned by the three vectors find vectors , , and that form an orthonormal basis for the subspace. Answer: We can save some time by noting that and are already orthogonal. We can normalize these … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.18

Exercise 3.4.18. If is the projection matrix onto the column space of the matrix and , what is a simple formula for ? Answer: The projection matrix onto the column space of can be calculated as . Since the columns … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.16

Exercise 3.4.16. Given the matrix whose columns are the following two vectors and [sic]: factor as . If there are vectors with elements each, what are the dimensions of , , and ? Answer: With and as the two columns … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.15

Exercise 3.4.15. Given the matrix find the orthonormal vectors and that span the column space of . Next find the vector that completes the orthonormal set, and describe the subspace of of which is an element. Finally, for find the … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.14

Exercise 3.4.14. Given the vectors find the corresponding orthonormal vectors , , and . Answer: We first choose . We then have We then have Now that we have calculated the orthogonal vectors , , and , we can normalize … Continue reading

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Linear Algebra and Its Applications, Exercise 3.4.13

Exercise 3.4.13. Given the vectors and the matrix whose columns are , , and , use Gram-Schmidt orthogonalization to factor . Answer: We first choose . We then have We then have We have , so , , and . … Continue reading

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