Solution11 - M341 Homework Solution 11 Section 4.5 3 Using the Simpli ed Span Method 1 1 − 1 1 − 1 1 1 − 1 1 − 1 1 − 1 is reduced to 1 1

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Unformatted text preview: M341 Homework Solution 11 Section 4.5: 3: Using the Simpli ed Span Method, 1 1 − 1 1 − 1 1 1 − 1 1 − 1 1 − 1 is reduced to 1 1 1 1 1 So the set spans P 4 . Using the Test For Linear Independence, 1 1 1 1 − 1 − 1 − 1 1 1 1 − 1 1 also reduces to 1 1 1 1 1 So the set is linearly independent and is thus a basis for P 4 . We could have also used just one of these tests (either one) and Thm. 4.13 and the fact that dim ( P 4 ) = 5 . 4(b,d): Since dim ( R 4 ) = 4 by Theorem 4.13(1) we know set (b) cannot span R 4 and it is not a basis. For part (d) 1 3 2 − 2 6 7 6 10 7 2 10 − 3 1 is reduced to 1 3 2 6 7 1 11 / 41 which will not further reduce to the identity matrix. Therefore the set (d) does not span R 4 and it is not a basis. 5(a): We need to check both properties for a maximal linearly independent subset. Since neither vector in B is a scalar multiple of the other they are linearly independent. For the other property [1 , 4 , 1 , − 2] = [2 , 3 , , − 1] + [ − 1 , 1 , 1 , − 1] and [3 , 2 , − 1 , 0] = [2 , 3 , , − 1] − [ − 1 , 1 , 1 , − 1] so adding either of these vectors would make the set linearly dependent. So it is a maximal linearly independent subset.of these vectors would make the set linearly dependent....
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This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas at Austin.

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Solution11 - M341 Homework Solution 11 Section 4.5 3 Using the Simpli ed Span Method 1 1 − 1 1 − 1 1 1 − 1 1 − 1 1 − 1 is reduced to 1 1

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