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Monthly Archives: February 2014
Linear Algebra and Its Applications, Exercise 3.1.8
Exercise 3.1.8. Suppose that and are orthogonal subspaces. Show that their intersection consists only of the zero vector. Answer: If and are orthogonal then we have for any vectors in and in . Suppose that is an element of both … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.7
Exercise 3.1.7. For the matrix find vectors and such that is orthogonal to the row space of and is orthogonal to the column space of > Answer: The nullspace of is orthogonal to the row space of . We can … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.6
Exercise 3.1.6. What vectors are orthogonal to and in ? From these vectors create a set of three orthonormal vectors (mutually orthogonal with unit length). Answer: If is a vector orthogonal to both and then the inner product of with … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.5
Exercise 3.1.5. Of the following vectors which are orthogonal to one another? Answer: We have the following inner products among the vectors: So and are orthogonal to but not to each other. NOTE: This continues a series of posts containing … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.4
Exercise 3.1.4. If is an invertible matrix, describe why row of and column of are orthogonal in the case . Answer: We have . The identity matrix has ones on the diagonal (i.e., when ) and zeros otherwise (when ). … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.3
Exercise 3.1.3. In the – plane two lines are perpendicular if the product of their slopes is -1. Use this fact to derive the condition for two vectors and being orthogonal. Answer: The line through the origin and has slope … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.2
Exercise 3.1.2. For give an example of linearly independent vectors that are not mutually orthogonal, as well as mutually orthogonal vectors that are not linearly independent. Answer: The vectors and are linearly independent, since the second vector cannot be expressed … Continue reading
Linear Algebra and Its Applications, Exercise 3.1.1
Exercise 3.1.1. For and what is the length of each vector and their inner product? Answer: We have and so that and . The inner product of and is then Note that and are thus orthogonal. NOTE: This continues a … Continue reading
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