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1 Linear Algebra Professor M. Zuker
Quiz 2 - September 21, 2015. - Solutions
- True or False.
(a) (v 1 , v 2 , v 3 , v 4 , v 5 , v 6 ) is a linearly independent list of vectors in a vector space V. Let U 1 = span(v 1 , v 2 ), let U 2 = span(v 3 , v 4 ) and let U 3 = span(v 5 , v 6 ). Then U 1 + U 2 + U 3 = U 1 โ U 2 โ U 3. That is, U 1 + U 2 + U 3 is a direct sum.
Solution: True. Suppose that ui โ Ui for 1 โค i โค 3. Then u 1 = av 1 + bv 2 , u 2 = cv 3 + dv 4 and u 3 = ev 5 + f v 6 for scalars a, b, c, d, e and f. If u 1 + u 2 + u 3 = 0, then 0 = u 1 + u 2 + u 3 = (av 1 + bv 2 ) + (cv 3 + dv 4 ) + (ev 5 + f v 6 ) = av 1 + bv 2 + cv 3 + dv 4 + ev 5 + f v 6. Since (v 1 , v 2 , v 3 , v 4 , v 5 , v 6 ) is linearly independent, a = b = c = d = e = f = 0. This implies that u 1 = u 2 = u 3 = 0, so the sum is direct. This result generalizes.
(b) (v 1 , v 2 , v 3 , v 4 ) is a linearly independent list of vectors in a complex vector space V. Then ((1 + i)v 1 , (1 โ i)v 2 , (โ 1 โ i)v 3 , (โ1 + i)v 4 ) is linearly independent.
Solution: True. If v = a(1 + i)v 1 + b(1 โ i)v 2 + c(โ 1 โ i)v 3 + d(โ 1 โ i)v 4 = 0, then all scalars in the sum must be zero because the list is linearly independent. That is, a(1 + i) = b(1 โ i) = c(โ 1 โ i) = d(โ 1 โ i) = 0. If the product of two scalars is zero, then at least one must be zero. Since 1 + i 6 == 0, a = 0. Similarly for b, c and d.
(c) In R^3 , ((1, โ 5 , 7), (2, 8 , โ5), (โ 13 , 29 , 51), (23, 31 , โ61), (45, 99 , โ77)) is linearly independent.
2 Linear Algebra Professor M. Zuker
Solution: False. We know that R^3 is spanned by e 1 = (1, 0 , 0), e 2 = (0, 1 , 0) and e 3 = (0, 0 , 1). Furthermore, (e 1 , e 2 , e 3 ) is linearly independent. Suppose a list of m vectors were linearly independent in R^3. Then m โค 3 because (e 1 , e 2 , e 3 ) spans R^3 and we know that the length of a linearly independent list of vectors must be โค the length of a spanning list. Thus, no list of 4 or more vectors in R^3 can be linearly independent.
- In R^2 , consider the list of vectors:
((2, โ3), (โ 1 , 3 /2), (โ 1 , 2), (7, 11), (
Use the linear dependence lemma as many times as necessary to remove vectors that are in the span of the previous vectors. What is the final list that results from this procedure? What is the span of the beginning list? What is the span of the final list?
Solution: (2, โ3) 6 = (0, 0), so it is kept. (โ 1 , 3 /2) = โ^12 (2, โ3), so it is discarded because it is in the span of the previous vector(s). (โ 1 , 2) is not a scalar multiple of (2, โ3), so it is kept. The list ((2, โ3), (โ 1 , 2)) is linearly independent in R^2 , so it must span R^2 since R^2 is spanned by ((1, 0), (0, 1)), a list of two linearly independent vectors. Thus, the remaining vectors in the list are discarded since they must be in the span of ((2, โ3), (โ 1 , 2)). Summary: Final list is ((2, โ3), (โ 1 , 2)). The span does not change as vectors that are in the span of previous vectors are discarded. The span is R^2.
