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Unformatted text preview: Solutions to problems from section 2.8 2. Give a specific reason why the set H ⊂ R 2 displayed below is not a subspace of R 2 . Solution: We can find two vectors u and v in H with u + v not in H . For example: 4. Give a specific reason why the set H ⊂ R 2 displayed below is not a subspace of R 2 . Solution: We can find a vector u ∈ H with ( − 1) u = − u negationslash∈ H . For example: 8. Let v 1 = − 3 6 , v 2 = − 2 2 3 , v 3 = − 6 3 , and p = 1 14 − 9 . Determine if p ∈ Col A where A = [ v 1 v 2 v 3 ]. Solution: We must determine if the augmented matrix [ A p ] corresponds to a consistent system: − 3 − 2 1 2 − 6 14 6 3 3 − 9 NR 3 = R 3 +2 R 1 −→ − 3 − 2 1 2 − 6 14 − 1 3 − 7 NR 2 = R 2 +2 R 3 −→ − 3 − 2 0 1 0 0 − 1 3 − 7 R 2 ↔ R 3 −→ 3 − 2 0 11 3 − 7 0 0 We can see that the corresponding linear system is consistent, thus p is in Col A . 10. With u = − 2 3 1 and A as in exercise 8, determine if u ∈ Nul A . Solution: We just need to see if A u = . Well, A u = − 3 − 2 2 − 6 6 3 3 − 2 3 1 = ....
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 Fall '08
 AXENOVICH
 Vectors, Vector Space, vp ∈ Rn

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