Exam3 Key

Math 203 Practice Midterm 3

Please work out each of the given problems.  Credit will be based on the steps towards the final answer.  Show your work.  Do your work on your own paper.

Problem 1

A physicist has plotted the position of a projectile over time.  Based on Newton’s laws, the projectile should ideally travel in a parabolic path.  Use matrices to find the most likely equation of this parabola.  (You may use a calculator, but show the matrices that are being manipulated).

Solution

We consider the matrix and vector

         \(  A \begin{pmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1\\ 16 & 4 & 1 \end{pmatrix}\)     \(  b =  \begin{pmatrix} 2 \\ 30 \\ 20  \\ 2 \end{pmatrix}   \)

Then we solve the system

        A^TAx  =  A^Tb 

which becomes

              \(  \begin{pmatrix} 354 & 100 & 30 \\ 100 & 30 & 10 \\ 30 & 10 & 4 \end{pmatrix} \begin{pmatrix} a_2 \\ a_2 \\ a_1 \end{pmatrix} = \begin{pmatrix} 334 \\ 130 \\ 54 \end{pmatrix} \)

We now rref the corresponding augmented matrix

   \( \left(\begin{array}{ccc|c} 354 & 100 & 30 & a_2 \\ 100 & 30 & 10 & a_2  \\ 30 & 10 & 4 & a_1  \end{array}\right) = \begin{pmatrix} 334 \\ 130 \\ 54 \end{pmatrix} \)

to get

   \( \left(\begin{array}{ccc|c} 1 & 0 & 0 & -11.5 \\ 0 & 1 & 0 & 56.5  \\ 0 & 0 & 1 & -41.5  \end{array}\right)  \)

We can conclude that the least squares regression parabola is given by

        y  =  -11.5x² + 56.5x - 41.5

Problem 2

A new species of fish is introduced into the Truckee River.  Initially 2 fish were stocked.  It takes one year for this species to spawn, when each fish averages 3 successful children each year.  (So there are 2 at the beginning, 2 at the end of the first year, 8 at the end of the second year, 14 at the end of the third year, etc.)

A.     Assuming no fish die, set up a recursion relationship that gives the number of fish w_n at the end of year n.
 

        We have that the number of fish should be the number previous plus the offspring of those two years ago.  We get

                w_n  =  w_n-1 + 3w_n-2

                w_n-1  =  w_n-1

B.     Find a matrix A such that w_n-1  =  A^n-1 (w₀, w₁)^T .
 

        We can write the recursive equation as            

         \(  \begin{pmatrix} w_n \\ w_{n-1} \end{pmatrix}\) \(   =  \begin{pmatrix} 1 & 3 \\ 1 & 0 \end{pmatrix}   \) \( \begin{pmatrix} w_{n-1} \\ w_{n-2} \end{pmatrix}\) 

        The above 2 x 2 matrix is what is required.

C.     Find a diagonal matrix D such that A is similar to D.

    Solution

        We need to find the eigenvalues and eigenvectors for A.  First for the eigenvalues.  We have

        det(A - lI)  =  (1 - l)(-l) - 3  =  l² - l - 3  =  0

        has roots 

                 \(  \lambda = \frac{1 \pm \sqrt{13}}{2} \)

        The diagonalized matrix is 

                \( D =  \begin{pmatrix} \frac{1 + \sqrt{13}}{2} & 0 \\ 0 & \frac{1 - \sqrt{13}}{2} \end{pmatrix} \) 

Problem 3

Let V be the subspace of differentiable functions spanned by {e^x, e^2x, e^3x} and let

        L:  V --->  V

be the linear transformation with 

        L(f(x))  =  f ''(x) - 3f '(x) + 2f(x)

A.     Write down the matrix A_L with respect to the given basis.
 

Solution

        We have

                L((1,0,0)  =  L(e^x)  =  e^x - 3e^x + 2e^x  =  0  =  (0,0,0)

                L((0,1,0)  =  L(e^2x)  =  4e^2x - 6e^2x + 2e^2x  =  0  =  (0,0,0)

                L((0,0,1)  =  L(e^3x)  =  9e^3x - 9e^3x + 2e^3x  =  2e^3x  =  (0,0,2)

        The matrix of L is the matrix whose columns are the vectors above.

         \(  A_L \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2\\ \end{pmatrix}\)

B.     Find the a basis for the kernel and range of L.

                    A basis for the kernel corresponds to the null space of A_L which has basis 

                            {(1,0,0), (0,1,0)}

                    in other words the basis is given by

                        {e^x, e^2x}

Problem 4 

Let W = Span{(1,1,0,1), (0,1,2,3)}.  Find a basis for the orthogonal complement of W.

Solution

        The orthogonal complement is equal to the null space of the matrix whose rows are the given vectors.  If 

         \(  rref(A) \begin{pmatrix} 1 & 1 & 0 & 1 \\ 0 & 1 & 2 & 3  \end{pmatrix} \)  

        then the rref(A) is given by

         \(  rref(A) \begin{pmatrix} 1 & 0 & -2 & -2 \\ 0 & 1 & 2 & 3  \end{pmatrix} \)  

        so that a basis for the null space is given by

                {(2,-2,1,0), (2,-3,0,1)}

Problem 5

Let 

         \( A = \begin{pmatrix} 0 & 2  \\ -2 & 0  \end{pmatrix} \) and \(  b = \begin{pmatrix} -2 \\ 1 \end{pmatrix}   \)

and T be the affine transformation T(x)  =  Ax + b .  Sketch the image under T of the figure below.

We just compute T(x) for each of the vertices and use the theorem that under an affine transformation line segments go to line segments.  We have

        T(0,0)  =  (0,0) + (-2,1)  =  (-2,1)        T(2,0)  =  (0,-4) + (-2,1)  =  (-2,-3)

        T(2,1)  =  (2,-4) + (-2,1)  =  (0,-3)       T(1,2)  =  (4,-2) + (-2,1)  =  (2,-1)

        T(0,1)  =  (2,0) + (-2,1)  =  (0,1)

The image is shown below in red.  Notice that this is a rotation by -p/2 then a expansion by a factor of 2 then a translation by (-2,1)

Problem 6

Let  L:  V ---> V  be a linear transformation.  Use the fact that 

        dim(Ker L) + dim(Range L)  =  dim(V)

to show that if L is one to one then L is onto. 

Solution

If L is one to one then the kernel of L is just the zero vector space since only zero gets sent to zero.  Hence 

        dim(Ker L)  =  0

The equation tells us that

        dim(Range L)  =  dim(V)

Since the range is a subspace of V and has the same dimension as V, the range of L equal V.  We can conclude that L is onto.

Problem 7 

Let A and B be matrices and let v be an eigenvector of both A and B.  Prove that v is an eigenvector of the product AB.

Proof

If v is an eigenvector of A and B, then there are numbers a and b with

        Av  =  av          and            Bv  =  bv  

We then have

        (AB)v  =  A(Bv)  =  A(bv)  =  b(Av)  =  bav  =  abv 

Hence v is an eigenvector of AB with eigenvalue ab.

Problem 8

Answer True of False and explain your reasoning.

A.    Let A be a 3x3 matrix such that the columns of A form an orthonormal set of vectors.  Then

        \( A^T A \begin{pmatrix} 4 \\ -1  \\ 2 \end{pmatrix}  = \begin{pmatrix} 4 \\ -1  \\ 2 \end{pmatrix}   \)

Solution

True, since A is an orthogonal matrix, we have 

        A^T  =  A^-1 

so that

        A^TA  =  A^-1A  =  I

B.     Let V be the vector space of continuous functions then the expression

        <f, g>  =  f(1) + g(1)

defines an inner product on V.

False, for example if 

        f(x)  =  -x 

then

        <f, f>  =  f(-1) + f(-1)  =  -1 + -1  =  -2  <  0

violating the first property of inner products.