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homework4_solutions

# homework4_solutions - MATH 211(Siefken Homework#4 Due...

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MATH 211 (Siefken): Homework #4 Due June 27, 2011 Question #1 (a) Consider the set of vectors S = { v 1 , v 2 , v 3 , v 4 , v 5 , v 6 } where v 1 = 1 2 3 4 v 2 = - 2 4 4 6 v 3 = - 3 2 1 2 v 4 = 0 10 8 10 v 5 = 6 6 6 6 v 6 = - 7 0 1 4 . (a) Find two distinct, maximal, linearly independent subsets of S . If A = [ v 1 | v 2 | v 3 | v 4 | v 5 | v 6 ] we compute that rref( A ) = 1 0 - 1 0 2 - 1 0 1 1 0 - 2 3 0 0 0 1 1 - 1 0 0 0 0 0 0 . We see that the rank of A is three and so the largest set of linearly independent vectors among the columns of A has size 3. The pivots of the rref form of A indicate a maximal linearly independent set, and so we may quickly list one such set S 1 = { v 1 , v 2 , v 4 } . The rref form also shows all dependence relations among the columns of A . Thus we see that the sets { v 1 , v 2 } , { v 1 , v 3 } , { v 2 , v 3 } are sets of linearly independent vectors. Further, each of these sets is linearly independent from v 4 and so we can find a different maximal set of linearly independent vectors, S 2 = { v 1 , v 3 , v 4 } . There are many, many other maximal sets of linearly independent vectors, all of which can be obtained by analyzing the dependence relations. Alternatively, one could scramble the columns of A and row reduce again. For example if B = [ v 6 | v 3 | v 4 | v 1 | v 5 | v 2 ], rref( B ) indicates that S 3 = { v 3 , v 4 , v 6 } is a maximal linearly independent set. (b) What can you say about the span of the sets in part (a)? What is dim(span( S ))? We know that a maximal linearly independent subset of S is a basis for span( S ). Thus span( S 1 ) = span( S 2 ) = span( S ). Since there are three vectors in this basis, dim(span( S )) = 3. (c) If A = [ v 1 | v 2 | v 3 | v 4 | v 5 | v 6 ], what is dim(null( A )) and dim(null( A T ))? Computed in part (a), we know rank( A ) = 3. Since dim(row( A )) + dim(null( A )) = 6 and dim(row( A )) = 3, we know dim(null( A )) = 3. Similarly, we know dim(col( A )) + dim(null( A T )) = 4 and dim(col( A )) = 3 and so dim(null( A T )) = 1.

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