Ma-305 Homework 4 Solutions


  1. Define the set P3 by:
    P3 = {a + bx + cx2 + dx3 | a, b, c, d are Rational Numbers}.
    1. Verify that P3 is a vector space with normal polynomial addition and scalar multiplication over polynomials.

    2. Verify that (-1)V = - V, for all V in P3.

    Note: We introduced this problem in Lecture 14, Slide 1.

    Solution

      Let a0 + a1x + a2x2 + a3x3, b0 + b1x + b2x2 + b3x3, and c0 + c1x + c2x2 + c3x3 be elements of P3,
      and p,q be rational numbers.

    1. P3 with normal vector addition and scalar multiplication forms a vector space:

      1. (P3; +, 0, -) forms an abelian group, for

        (a0 + a1x + a2x2 + a3x3) + (b0 + b1x + b2x2 + b3x3)
        = (a0+b0) + (a1+b1)x + (a2+b2)x2 + (a3+b3)x3
        = (b0+a0) + (b1+a1)x + (b2+a2)x2 + (b3+a3)x3
        = (b0 + b1x + b2x2 + b3x3) + (a0 + a1x + a2x2 + a3x3),

        ((a0 + a1x + a2x2 + a3x3) + (b0 + b1x + b2x2 + b3x3)) + c0 + c1x + c2x2 + c3x3
        = ((a0+b0) + (a1+b1)x + (a2+b2)x2 + (a3+b3)x3) + (c0 + c1x + c2x2 + c3x3)
        = (a0+b0+c0) + (a1+b1+c1)x + (a2+b2+c2)x2 + (a3+b3+c3)x3
        = (a0 + a1x + a2x2 + a3x3) + ((b0+c0) + (b1+c1)x + (b2+c2)x2 + (b3+c3)x3)
        = (a0 + a1x + a2x2 + a3x3) + ((b0 + b1x + b2x2 + b3x3) + (c0 + c1x + c2x2 + c3x3)),

        (a0 + a1x + a2x2 + a3x3) + 0 = 0 + (a0 + a1x + a2x2 + a3x3) = a0 + a1x + a2x2 + a3x3, and

        (a0 + a1x + a2x2 + a3x3) + (-a0 + -a1x + -a2x2 + -a3x3) = 0 .

      2. Scalar multiplication distributes over the sum of two vectors, and the sum of two scalars distribute over a vector, for

        (p)((a0 + a1x + a2x2 + a3x3) + (b0 + b1x + b2x2 + b3x3))
        = (p)((a0+b0) + (a1+b1)x + (a2+b2)x2 + (a3+b3)x3)
        = (p)(a0+b0) + (p)(a1+b1)x + (p)(a2+b2)x2 + (p)(a3+b3)x3
        = ((p)a0 + (p)a1x + (p)a2x2 + (p)a3x3) + ((p)b0 + (p)b1x + (p)b2x2 + (p)b3x3))
        = (p)(a0 + a1x + a2x2 + a3x3) + (p)(b0 + b1x + b2x2 + b3x3)

      3. Scalar multiplication is associative, for

        (pq)(a0 + a1x + a2x2 + a3x3)
        = (pq)a0 + (pq)a1x + (pq)a2x2 + (pq)a3x3)
        = p(qa0) + p(qa1x) + p(qa2x2) + p(qa3x3)
        = p(q(a0 + a1x + a2x2 + a3x3))

      4. There exists, with resp. to the scalar mult., an identity scalar (i.e. 1R), for
        1(a0 + a1x + a2x2 + a3x3)
        = (a0 + a1x + a2x2 + a3x3)
    2. -1(a0 + a1x + a2x2 + a3x3)
      = (-1)a0 + (-1)a1x + (-1)a2x2 + (-1)a3x3
      = -a0 + -a1x + -a2x2 + -a3x3
      =-(a0 + a1x + a2x2 + a3x3)

    Note: We introduced this problem in Lecture 14, Slide 1.

  2. *Determine if the given set V is a vector space. Justify your answer.

    1. The set V of all pairs of real numbers, i.e.
    3. V = { (x1,x2) | x1,x2 are Real Numbers}.
    with usual scalar multiplication, and with addition defined by
    (x1,x2) + (y1,y2) = (x1 + y1 + 1, x2 + y2 + 1)

  3. The set V of all ordered triples {(x1,x2,x3)} with usual addition, and with scalar multiplication defined by
    r(x1,x2,x3) = (2rx1,2rx2,2rx3)
    for all scalars r.

Solution

  1. Let v1, v2 be elements of V, a a scalar. The definition of addition violates the following axiom:

    a(v1+v2) = av1+av2
    and hence V is not a vector space.

  2. Let v be an element of the vector space. The definition of multiplication violates the following axiom:

    1*v = v
    and hence V is not a vector space.

  • *A Vector Space V and a subset S are given in the exercises below. For each, determine whether S is a subspace of V. Justify your answer.

    1. Let V = P3 from Ex. 1 above. Define S by S = { ax + bx3 | a and b are Rational Numbers}.

    2. Let V = M22 be the vector space of all 2x2 matrices with real number entries, under the usual operations of matrix addition and scalar multiplication. Define S by
      S = { [
      [       		0
      a       		b
      0       		]
      ]       		| a and b are real numbers }

    Solution

    1. Let ax + bx3 and cx + dx3 be elements of S, and r be a scalar. Then

      ax + bx3 + cx + dx3 = (a+c)x + (b+d)x3
      is an element of S, and

      r(ax + bx3) = rax + rbx3
      is an element of S. Therefor S is a subspace.

    2. As in (a) above, the space is closed under the defined vector addition and scalar multiplication. Therefore s is a subspace.

  • *Find a set of vectors spanning the solution space of Ax=0, where
    A = [
    [
    [     		2
    4
    14     		7
    -3
    15     		-1
    4
    7     		]
    ]
    ]

    Solution

    The solution space consists of a single vector:

    {(0,0,0)T}

  • *Determine which of the following sets of vectors are linearly independent. Justify your answer. For those sets that are linearly dependent, express "extra" vectors in terms of the others.

    1. In R3, {u, v, w} = {(2, -1, 3), (3, 4, 1), (2, -3, 4)}

    2. In R4, {u, v, w} = {(2, 3, -1, 2), (-6, 1, 4, 5), (12, 8, -7, 1)} < LI>In R4, {v1, v2, v3, v4} = {(1, 1, 0, 0), (0, 1, 1, 0), (0, 0, 1, 1), (1, 0, 0, 1)}

    Solution:

    For a set of n vectors {x1,x2, ..., xn}, we only need to check that for scalars a1,a2, ..., an,
    a1x1+a2x1+...+anx1 = 0 => a1=a2=...=an=0.
    1. Using our methods for solving linear systems,
      (i.e. we set A = [u|v|w] and x=(a,b,c)T, and solve Ax=0)
      we see that the only scalars a, b, and c for which au + bv + cw = 0 are a=b=c=0,
      i.e. the vectors are linearly independent.

    2. We use the same method from above, and we find that there is a free variable. If you find the reduced row-echelon form of the associated matrix (with the vectors as columns), it is clear that the third vector is 3 times the first, minus the second, i.e.
      w = 3*u - v.

    3. As in (b) above, the reduced row echelon form reveals
      v4 = v1-v2+v3

  • *Determine which of the following sets of vectors forms a basis for the given vector space. Justify your answer.

    1. For R3, v1 = (3, -2, 1), v2 = (2, 3, 1), v3 = (2, 1, -3)
    2. For R2, v1 = (2, 0), v2 = (3,3)
    3. For P3 (defined in Exercise 1 above),   p1 = 1 + x,   p2 = 1 - x2,   p3 = x3,      p4 = 1 - x

    Solution

    1. These vectors are linearly independent (the row-echelon form of the related matrix has a pivot in every column); there are 3 vectors, and the dimension of the vectors space is 3; therefore they span the space. Hence they form a basis.
    2. These vectors are linearly independent; there are 2 vectors, and the dimension of the vectors space is 2; therefore they span the space. Hence they form a basis.
    3. Note that
      ap1 + bp2 + cp3 + dp4 = 0
      => a(1 + x) + b(1 - x2) + c(x3) + d(1 - x) = 0
      => cx3 - bx2 + (a - d)x + (a + b + d) = 0
      => a=b=c=d=0
      So the vectors are linearly independent.

      Now, let a + bx + cx2 + dx3 be an element of P3. Can we find coefficents a', b', c', and d', such that a'p1 + b'p2 + c'p3 + d'p4 = a + bx + cx2 + dx3?
      Yes; choose a' = (1/2)(a-b+c), b'=-c, c'=d, and d'=(1/2)(a+b+c).
      So our vectors span the space and hence form a basis.

  • **Find a basis for the Null Space associated with the following matrix a:
    a = [
    [
    [     		1
    3
    2     		2
    6
    4     		0
    1
    1     		2
    9
    7     		1
    6
    5     		]
    ]
    ]

    Solution:

    The space = span { [
    [
    [
    [
    [     		-2
    0
    -3
    1
    0     		]
    ]
    ]
    ]
    ]     		, [
    [
    [
    [
    [     		-2
    1
    0
    0
    0     		]
    ]
    ]
    ]
    ]     		, [
    [
    [
    [
    [     		-1
    0
    -3
    0
    1     		]
    ]
    ]
    ]
    ]     		}

    Bonus: How many calls to the recursive determinant function defined in Slide 3, Lecture 13 will be made if we use memoization?

    Solution:


    2n-2

    *: problem from Introductory Linear algebra with applications, by Kolman and Hill

    **: problem from Matrix analysis and Linear algebra, by Carl Meyer