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

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  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.


    Another 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} spans V. To this end, suppose that v is a vector in V. We must show that there exist scalars k1, k2, and k3 for which

    k1u1 + k2u2 + k3u3=v.

    Substituting from the given formulas, we wish to show that there exist k1, k2, k3 for which

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

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

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

    As a basis for V, the set {v1, v2, v3} spans V. Therefore we know there exist (uniquely determined) scalars c1, c2, c3 such that

    c1v1 + c2v2+c3v3=v.

    So we need to show that there is a solution for the system

     

    k1+k2+k3=c1

    k2+k3=c2

    k3=c3.

    Solving this system by back substitution, we find k3= c3, k2= c2-k3 = c2-c3, k1= c1- k2-k3= c1- c2+c3-c3,= c1- c2. Thus the desired scalars do exist. (This last step in the proof could be established in other ways, as well, for instance by using the Long Theorem.)

  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.

    Another 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 spans P3. To this end, suppose that p is a vector in P3, with, say, p(x) = a+bx+cx2+dx3. We must show that there exist scalars k1, k2, k3, and k4 for which

    k1p1 + k2p2 + k3p3 + k4p4 =p.

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

    k1(1+x)+ k2 (2x2+2x3)+ k3(3+3x3)+ k4(4x+4x2+4x3)=a+bx+cx2+dx3

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

    (k1+3k3)+(k1+4k4)x+(2k2+4k4)x2+(2k2+3k3+4k4)x3=a+bx+cx2+dx3

     

    for every x. A theorem from high-school algebra tells us this happens iff the respective coefficients are equal, so we must have

    k1+3k3 = a
    k1+4k4=b
    2k2+4k4=c
    2k2+3k3+4k4=d

    One possibility at this point is to solve this system. Using Joy of Mathematica to reduce the augmented matrix I found

    k1=a+c-d, k2 = (a-b+2c-d)/2, k3 = (-c+d)/3, k4=(-a+b-c+d)/4.

    Thus the required scalars do exist.

    Alternatively, one could check the determinant for the coefficient matrix for this system. The determinant works out to be nonzero (in fact the determinant is 24), so the coefficient matrix is invertible and the Long Theorem assures us that the given system does have a solution (no matter what a, b, c, d might happen to be). Again, we conclude that the required scalars do exist.

    Since our fixed but arbitrary vector p in P3 has been shown to be a linear combination of p1, p2, p3, p4 we have now shown that {p1, p2, p3, p4}spans P3.

     

     

     

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