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March 28, 1996, Key for Exam 2, Math 311

  1. (25) Let tex2html_wrap_inline118 , and tex2html_wrap_inline120 . Let A be a tex2html_wrap_inline124 matrix. Suppose that tex2html_wrap_inline126 and that tex2html_wrap_inline128 .
    1. Show that the two vectors tex2html_wrap_inline130 and tex2html_wrap_inline132 form a basis of tex2html_wrap_inline134 .
      Since the set tex2html_wrap_inline136 has two vectors and the dimension of tex2html_wrap_inline134 is two it will suffice to show that the set is linearly independent. So suppose that tex2html_wrap_inline140 . This leads to the system of equations:

      eqnarray25

      The coefficient matrix of this system, tex2html_wrap_inline142 , is row equivalent to the identity matrix. This tells us that the only solution to the above system of equations is tex2html_wrap_inline144 . Thus, tex2html_wrap_inline130 and tex2html_wrap_inline132 are linearly independent and form a basis of tex2html_wrap_inline134 .

    2. Let tex2html_wrap_inline152 . Determine what tex2html_wrap_inline154 must equal. Hint: write tex2html_wrap_inline156 as a linear combination of tex2html_wrap_inline130 and tex2html_wrap_inline132 .
      Since tex2html_wrap_inline162 , we have tex2html_wrap_inline164 . We could of course write tex2html_wrap_inline166 , where tex2html_wrap_inline168 , and tex2html_wrap_inline170 . Then tex2html_wrap_inline172 .
    3. What is the null space of the matrix A?
      The null space of A must consist of just the zero vector. For if not, then 0 must be an eigenvalue. We already know the two eigenvalues of A and neither of them is zero.
  2. (25) Let L be a linear transformation from tex2html_wrap_inline182 to tex2html_wrap_inline184 , where tex2html_wrap_inline186 . Find the matrix representation of this linear transformation with respect to the standard bases in tex2html_wrap_inline182 and tex2html_wrap_inline184 .
    tex2html_wrap_inline192 , tex2html_wrap_inline194 , tex2html_wrap_inline196 . Thus, the matrix representation of L with respect to the standard bases is:

    displaymath200

  3. (25) Let L be a linear transformation which rotates tex2html_wrap_inline182 about the line through the origin parallel to the vector (-1,1,3). Suppose that it is a counterclockwise rotation of 30 degrees when the origin is viewed from the point (-1,1,3). Find the matrix representation of L with respect to the standard basis. Calculate to where the point (1,-6,4) is mapped by L.
    A basis for the plane of rotation is tex2html_wrap_inline216 . An orthonormal basis for this plane is found using Gram-Schmidt. An examination of the direction of rotation shows us that we want to turn the vector (3,0,1) towards the vector (1,1,0). Thus, when we write the ``nice'' basis for tex2html_wrap_inline182 we will need to reverse the order in which we list the vectors from our orthonormal basis of the plane of rotation. Doing this we get the following change of basis matrix P. tex2html_wrap_inline224 . Denote the tex2html_wrap_inline226 column of P by tex2html_wrap_inline230 , we have:

    eqnarray69

    The matrix representation, tex2html_wrap_inline232 of L with respect to the basis consisting of the columns of P is: tex2html_wrap_inline238 . Thus, the matrix representation of L with respect to the standard basis is tex2html_wrap_inline242

    displaymath244

    To compute L(1,-6,4), just compute tex2html_wrap_inline248 .

  4. (25) You are given the following set of data: tex2html_wrap_inline250 . What choice of constants, a, b, and c will cause the graph of the function tex2html_wrap_inline252 to get as close as possible to the data points?
    The system of equations we would like to solve is tex2html_wrap_inline254 , for i running from 1 to 4. The coefficient matrix of this system is:

    displaymath256

    and the equation we want to solve is tex2html_wrap_inline258 , where tex2html_wrap_inline260 . It soon becomes clear that we cannot solve this system. (The last row of the reduced row echelon form of the augmented matrix is tex2html_wrap_inline262 . This of course implies that the associated system of equations does not have a solution.) So, we need to find a least squares solution of this system. Let Q be a matrix whose columns form an orthonormal basis for the column space of A. Then we want to solve the system of equations tex2html_wrap_inline268 , where A=QR or tex2html_wrap_inline272 . These matrices are approximately:

    displaymath274

    Solving the equations tex2html_wrap_inline268 , we get tex2html_wrap_inline278 . Thus,

    displaymath280


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Next: About this document

Mike Stecher
Thu Apr 4 11:02:16 CST 1996