# Solutions - Solutions to homework 1 35 First note ~ u 1 ~ u 2 = ~ u 3 i.e the vectors ~ u 1 ~ u 2 ~ u 3 are linearly dependent i.e they lie in the

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Unformatted text preview: Solutions to homework # 1 35. First note ~ u 1 + ~ u 2 = ~ u 3 , i.e., the vectors ~ u 1 , ~ u 2 , ~ u 3 are linearly dependent, i.e., they lie in the same plane. The vectors ~ u 1 and ~ u 2 , on the other hand, are not not proportional, hence linearly independent. So, it is enough to determine whether ~ a and ~ b can be expressed as linear combinations of ~ u 1 and ~ u 2 only. Since the second component of ~ u 2 is zero, the coefficient of ~ u 1 must match the second component of a vector we want to write as a linear combination of ~ u 1 and ~ u 2 . So, for ~ a we must take 2 ~ u 1 , and ~ a- 2 ~ u 1 =- ~ u 2 , so ~ a = 2 ~ u 1- ~ u 2 . Geometrically, this means that ~ a lies in the same plane as ~ u 1 , ~ u 2 , and ~ u 3 . On the other hand, to express ~ b , we must take 3 ~ u 1 , but ~ a- 3 ~ u 1 = (- 2 , ,- 1) is not a multiple of ~ u 2 , so ~ b cannot be expressed as a combination of ~ u 1 , ~ u 2 , and ~ u 3 . Geometrically, this means that ~ b does not lie in the plane defined by these vectors. 40. We know that tr( AB ) = tr( BA ) (see p. 44 of the textbook). Hence tr( CAC- 1 ) = tr(( CA ) C- 1 ) = tr( C- 1 ( CA )) = tr(( C- 1 C ) A ) = tr( IA ) = tr( A ) . Here I denotes the identity matrix and the associativity of matrix products is used. Note that the product of matrices is generally not commutative, so we cannot rearrange the factors C , A , and C- 1 at will....
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## This note was uploaded on 12/12/2009 for the course MATHEWS SOC 2000 taught by Professor M during the Spring '09 term at AIB College of Business.

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Solutions - Solutions to homework 1 35 First note ~ u 1 ~ u 2 = ~ u 3 i.e the vectors ~ u 1 ~ u 2 ~ u 3 are linearly dependent i.e they lie in the

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