MA-305 Homework 6 Solutions


See hw6_sol.mws for a maple worksheet of problem 1 and the two bonus problems.

  1. Consider the general equation for the following surface:
    z = ax2 + by + c
    and the following data:
    i
    1.
    2.
    3.
    4.
    5.     		xi
    0
    1
    -1
    1
    -1     		yi
    0
    2
    2
    -1
    -2     		zi
    3 - 1/10
    15 + 2/10
    15 - 3/5
    0 + 7/8
    -5 - 1/5

    Using a least squares fit, (i) determine the constants a, b, and c, and estimate z for (x,y) = (25,10); (ii) determine the matrix A and the vectors x' and b' for this system such that Ax' = b'; (iii) Calculate the Eucidean norm of the residual, ||Ax'-b||; (iv) determine a basis for the orthogonal complement of A; and (v) write the residual Ax'-b as a linear combination of vectors in the orthogonal complement of A.
    (Note: This problem will be worth 20 points.)


    Solution

    Solve the Normal Equation for x'
         ATAx' = ATb
    using the following matrix and vector
    A = [
    [
    [
    [
    [     		0
    1
    1
    1
    1     		0
    2
    2
    -1
    -2     		1
    1
    1
    1
    1     		]
    ]
    ]
    ]
    ]     		, b = [
    [
    [
    [
    [     		3 - 1/10
    15 + 2/10
    15 - 3/5
    0 + 7/8
    -5 - 1/5     		]
    ]
    ]
    ]
    ]

    1. x' = (a,b,c), say; so x' = (2239/1020, 1997/408, 29/10). for (x,y) = (25,10), z = 242043/170.
    2. The matrix A is stated above; b' = Ax' = (29/10, 7591/510, 7591/510, 409/2040, -399/85)
    3. The residual is b' - b = (0, -161/510, 247/510, -172/255, 43/85); The euclidean norm of this vector is approximately 1.02.
    4. The orthogonal complement to a matrix is the space orthogonal to the space spanned by the columns of the matrix. Thus, the othogonal complement to A is just the nullspace of AT. A basis for this space is:
      {v1,v2} = { [
      [
      [
      [
      [       		0
      -1
      1
      0
      0       		]
      ]
      ]
      ]
      ]       		, [
      [
      [
      [
      [       		0
      1/3
      0
      -4/3
      1       		]
      ]
      ]
      ]
      ]       		}

    5. We look for constants c1,c2 such that
      c1v1 + c2v2 = (Ax' - b)
      This is equivalent to solving a linear system. We find (c1,c2) = (247/510,43/85).

  2. Determine which of the following are inner products. Justify your answer. For those that are inner products, describe the corresponding norm.
    1. For x=(x1, x2) and y=(y1, y2),
      <x,y> = 10x1y1 + 4x2y2.
    2. For x=(x1, x2, x3) and y=(y1, y2, y3),
      <x,y> = x1y1 + x3y3.
    3. Let f and g be two polynomials of degree 2 or less, and <f,g> = f(-1)g(-1) + f(0)g(0) + f(1)g(1).

    Solution

    1. The axioms are easily verified for this inner product. The corresponding norm would be
      ||x||<,> = (10*x12 + 4*x22)1/2
    2. Axiom 1 is violated, for consider x = <0,x2,0), where x2 <> 0. Then <x,x> = 0, but x<>0.
    3. This is an inner product. The difficult axiom to verify is the first; in particular, if f is a non-zero polynomial of degree 2 or less, can we verify that <f,f> > 0?
      Well, suppose not. Clearly <f,f> >= 0.
      So suppose <f,f> = 0.
      Then f(-1)f(-1) + f(0)f(0) + f(1)f(1) = 0,
      and f(-1) = f(0) = f(1) = 0,
      impossible, as this implies that f has 3 distinct roots, and f is of degree 2 or less.
      So the axiom holds. The corresponding norm would be
      ||f||<,> = (f(-1)f(-1) + f(0)f(0) + f(1)f(1))1/2.

  3. For the following, determine if the given function T is a linear transformation. Justify your answer. If T is a linear transformation, the find the matrix A such that T = fA.
    1. T(x,y) = (y,x)
    2. T(x,y,z) = (z-x,z-y)
    3. T:M2x2 -> M2x2 where
      T( [
      [       		w
      y       		x
      z       		]
      ]       		) = [
      [       		w+z
      0       		0
      x+y       		]
      ]

    Solution

    Each of these is a linear transformation; the axioms are easily verified. The associated matrices are:
    1. A= [
      [       		0
      1       		1
      0       		]
      ]
    2. A= [
      [       		-1
      0       		0
      -1       		1
      1       		]
      ]
    3. A= [
      [
      [
      [       		1
      0
      0
      0       		0
      0
      0
      1       		0
      0
      0
      1       		1
      0
      0
      0       		]
      ]
      ]
      ]
      (See the forum for an explanation of pt. c)

(Bonus #1) For the following matrix:
A = [
[
[ 		-3
20
2 		1
3
-2 		-3
10
4 		]
]
]

  1. Find its characteristic polynomial (expressed as a product of linear factors).
  2. Find its eigenvalues.
  3. Find a basis for its eigenspace.

Solution

  1. (lambda+2)*(lambda-3)2
  2. -2, 3, 3
  3. { [
    [
    [     		-1
    2
    1     		]
    ]
    ]     		, [
    [
    [     		1
    0
    -2     		]
    ]
    ]     		}
(Bonus #2) Find (i) the General Solution to the following system of differential equations, and (ii) the particular solution that satisfies the given intial conditions:
     du/dt=-u-3v
     dv/dt=-3u+7v
     u=4, v=7, when t=0.

Solution

  1. The general solution looks like:
          c1e-2tw1 + c2e8tw2
    where
    w1 = [
    [     		3
    1     		]
    ]     		, and w2 = [
    [     		1
    -3     		]
    ]
  2. For the exact solution, with the given initial conditions, we have:
    c1 = (19/10), and c2 = (-17/10).
Problems 2 and 3, and Bonus #2 from Elementary Linear Algebra with Applications by Hill.

Bonus Problem #1 from Matrix Analysis and Applied Linear Algebra by Meyer.