Math 206
Graded Problems/Proofs #19
Due Thursday April 18 

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Note: Both of these problems can be done in more than one way. Some ways are easier than others. If you take advantage of theorems, not just definitions, these problems will be easier.

 

  1. Let {v1, v2, v3} be a basis for a vector space V.
    Let u1 = v1, u2= v1+v2, u3= v1+v2+v3.
    Prove that {u1, u2, u3} is a basis for V.


    Proof. The stated hypotheses tell us that V has dimension 3. Since there are 3 vectors in {u1, u2, u3}, one of the theorems in Section 5.4 tells us that it suffices to prove that {u1, u2, u3} is independent. To this end, suppose that k1, k2, and k3 are scalars for which

    k1u1 + k2u2 + k3u3=0.

    Substituting from the given formulas, we have

    k1v1 + k2 (v1+v2 )+ k3 (v1+v2+v3)=0.

    Distributing the scalars and then combining terms, we can rewrite this last equation in the form

    (k1+k2+k3)v1 + (k2+k3)v2+k3v3=0.
     

    As a basis for V, the set {v1, v2, v3} is necessarily independent. Therefore we must have

    k1+k2+k3=0

    k2+k3=0

    k3=0.

    Solving this system by back substitution, we find k3=0, k2=0, k1=0, which establishes the desired independence and completes the proof.

      

  2. In P3, let p1, p2, p3, p4 be the functions defined by
    • p1(x) = 1+x
    • p2(x) = 2x2+2x3
    • p3(x) = 3+3x3
    • p4(x) = 4x+4x2+4x3.

    Prove that {p1, p2, p3, p4} is a basis for P3.

    Proof. We know that P3 has dimension 4, and there are 4 vectors in the given set. According to one of the theorems in 5.4, we need only show that the given set is independent. To this end, suppose that k1, k2, k3, and k4 are scalars for which

    k1p1 + k2p2 + k3p3 + k4p4 =0.

    Using the stipulated definitions for the four vectors we're working with, we can reformulate this condition as

    k1(1+x)+ k2 (2x2+2x3)+ k3(3+3x3)+ k4(4x+4x2+4x3)=0

    for every x. Distributing the scalars and combining terms, we find that

    (k1+3k3)+(k1+4k4)x+(2k2+4k4)x2+(2k2+3k3+4k4)x3=0

     

    for every x. A theorem from high-school algebra tells us this happens iff each of the coefficients is zero, so we must have

    k1+3k3 = 0
    k1+4k4=0
    2k2+4k4=0
    2k2+3k3+4k4=0

    I'll leave it to you to solve this system and verify that, in fact, k1 = k2=k3=k4=0, as required. (Your proof should include details to support this claim.) This completes the proof.

     

     

     

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