HW08-solutions - huynh(lth436 HW08 gilbert(57245 This...

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huynh (lth436) – HW08 – gilbert – (57245) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Four linearly independent vectors span R 5 . True or False? 1. TRUE 2. FALSE correct Explanation: The space R 5 is 5-dimensional, so five lin- early independent vectors are needed to form a basis, and hence span R 5 . Consequently, the statement is FALSE . 002 10.0points The pivot columns of rref( A ) form a basis for Col A . True or False? Let H be the set of all vectors a 2 b ab + 3 a b where a and b are real. Determine if H is a subspace of R 3 , and then check the correct answer below. 1. H isnotasubspaceof R 3 becauseitdoes notcontain 0 . 1. TRUE 2. FALSE correct Explanation: The columns of A have the same linear de- pendence relation as the columns of rref( A ). Also, the pivot columns of A are linearly in- dependent and span Col A , hence form a basis for Col A . But the columns of rref( A ) may not span Col A , so the pivot columns of rref( A ) need not form a basis for Col A . For example, when A = 1 2 3 2 5 8 2 4 6 , then rref( A ) = 1 0 1 0 1 2 0 0 0 . 2. H is a subspace of R 3 because it can be writtenasNul ( A ) forsomematrix A . 3. H is a subspace of R 3 because it can be writtenasSpan { v 1 , v 2 } with v 1 , v 2 in R 3 . 4. H is not a subspace of R 3 because it is notclosedundervectoraddition . correct Explanation: To check if the set H of all vectors a 2 b ab + 3 a b is a subspace of R 3 we check the properties defining a subspace: Thus the first two columns of A form a ba- sis for Col A because the first two columns of rref( A ) are pivot columns. But the last en- try is 0 in each pivot column of rref( A ), so the columns of A can never be a linear combina- tion of the pivot columns of rref( A ). Consequently, the statement is FALSE . 003 10.0points
huynh (lth436) – HW08 – gilbert – (57245) 2 1. the zerovector 0 isin H : set a = b = 0. Then 0 0 0 + 0 0 = 0 0 0 , so H contains 0 .
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