Math 310: Final Exam (Solutions) Prof. S. Smith: Wed 10 Dec 1997

You must SHOW WORK to receive credit.

Problem 1: Let A = ( \begin{matrix}3 & 1 & −1 \\ 2 & 2 & −1 \\ 2 & 2 & 0 \end{matrix} ).

(a) Find the eigenvalues of A. (Hint: they are small, positive integers, including 1). det(A − xI) = −x(x^{2} − 5x + 6 − 2) + 1(6 − 2x − 2) − 1(4 − (4 − 2x)) = −x^{3} + 5x^{2} − 8x + 4 =

−(x − 1)(x^{2} − 4x + 4) = −(x − 1)(x − 2)^{2}, so values 1, 2, 2.

(b) Find the eigenspaces for those eigenvalues.

For 1: A − I = ( \begin{matrix}2 & 1 & −1 \\ 2 & 1 & −1 \\ 2 & 2 & −1 \end{matrix} ) via A_{2,3}^{−1×1} ( \begin{matrix}2 & 1 & −1 \\ 0 & 0 & 0 \\ 0 & −1 & 0 \end{matrix} ) so solutions a(\frac{1}{2}, 0, 1)^{T}.

For 2: A − 2I = ( \begin{matrix}1 & 1 & −1 \\ 2 & 0 & −1 \\ 2 & 2 & −2 \end{matrix} ) via A_{2,3}^{−2×1} ( \begin{matrix}1 & 1 & −1 \\ 0 & −2 & 1 \\ 0 & 0 & 0 \end{matrix} ) so solutions b(\frac{1}{2}, \frac{1}{2}, 1)^{T}.

(c) Is A diagonalizable? Why/why not? If so, find X with X^{−1}AX diagonal.

No: 2 is an eigenvalue twice, but only one dimension of 2-eigenvectors.

Problem 2: Consider the differential equation system (functions of t): \binom{y′_{1} = 2y_{2}}{y′_{2} = 3y_{1} −y_{2}}.

I GIVE you that X^{−1}AX = D where A = ( \begin{matrix}0 & 2 \\ 3 & −1 \end{matrix} ), X = ( \begin{matrix}1 & 2 \\ 1 & −3 \end{matrix} ), D = ( \begin{matrix}2 & 0 \\ 0 & −3 \end{matrix} ).

(a) Give the general solution of the system (with undetermined constants c_{1}, c_{2}).

c_{1} ( \begin{matrix}1 \\ 1 \end{matrix} ) e^{2t} + c_{2} ( \begin{matrix}2 \\ −3 \end{matrix} ) e^{−3t} so y_{1} = c_{1}e^{2t} + 2c_{2}e^{−3t} and y_{2} = c_{1}e^{2t} − 3c_{2}e^{−3t}.

(b) Now determine the particular solution (values of c_{1}, c_{2}) for the initial value problem y_{1}(0) = 5, y_{2}(0) = −5.

Solve ( \begin{matrix}1 & 2 & 5 \\ 1 & −3 & −5 \end{matrix} ) to get c_{1} = 1, c_{2} = 2.

Problem 3: Let A be the Markov matrix ( \begin{matrix}\frac{1}{2} & \frac{2}{3} \\ \frac{1}{2} & \frac{1}{3} \end{matrix} ) .

(a) Find the eigenvalues and eigenvectors of A. Indicate the “steady-state” vector. det(A − xI) = x^{2} − \frac{5}{6}x − \frac{1}{6} + \frac{2}{6} = x^{2} − \frac{5}{6}x + \frac{1}{6} = (x − 1)(x + \frac{1}{6}) so values 1, −\frac{1}{6}. Corresponding eigenvectors a(4, 3)^{T} and b(−1, 1)^{T}. Steady-state is (\frac{4}{7}, \frac{3}{7})^{T}.

(b) Write A = XDX^{−1} with D diagonal—and then get a formula for the power A^{m}.

We can use X = ( \begin{matrix}4 & −1 \\ 3 & 1 \end{matrix} ) so X^{−1} = \frac{1}{7} ( \begin{matrix}1 & 1 \\ −3 & 4 \end{matrix} )

so A^{m} = XD^{m}X^{−1} = \frac{1}{7} ( \begin{matrix}4 + 3(−\frac{1}{6})^{m} & 4 − 4(−\frac{1}{6})^{m} \\ 3 − 3(−\frac{1}{6})^{m} & 3 + 4(−\frac{1}{6})^{m} \end{matrix} )

Problem 4: Let A be the symmetric matrix ( \begin{matrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} ).

I GIVE you that the eigenvalues of A are 2, −1, −1.

(a) Find an orthonormal basis for each eigenspace.

For 2: eigenvectors are a(1, 1, 1)^{T}, so use \frac{1}{\sqrt{3}}(1, 1, 1)^{T}.

For -1: Get eigenvectors (−b − c, b, c)^{T}. Start with basis like (−1, 1, 0)^{T} and (−1, 0, 1)^{T}. Apply Gram-Schmidt to get \frac{1}{\sqrt{2}}(−1, 1, 0)^{T} and \frac{1}{\sqrt{6}}(1, 1, −2)^{T}.

(b) Now find an orthogonal matrix X (that is, X^{−1} = X^{T}) with X^{−1}AX diagonal.

From (a) can use X = ( \begin{matrix}\frac{1}{\sqrt{3}} & −\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{−2}{\sqrt{6}} \end{matrix} ).

Problem 5:

(a) Give the (symmetric) matrix for the quadratic form x_{1}^{2} − 2x_{1}x_{3} + 2x_{2}^{2} + 4x_{2}x_{3} + 6x_{3}^{2}. Indicate new variables in which the form is diagonal. Is it positive definite?

( \begin{matrix}1 & 0 & −1 \\ 0 & 2 & 2 \\ −1 & 2 & 6 \end{matrix} ).

Completing squares, we see (x_{1} − x_{3})^{2} + 2(x_{2} + x_{3})^{2} + 3x_{3}^{2}.

These give new diagonalizing variables; all with positive coefficient, so positive definite.

(b) Write the matrix of (a) in the form LDL^{T} with L lower triangular, and D diagonal.

Operations A_{3}^{1×1} and A_{3}^{−1×2} give U = ( \begin{matrix}1 & 0 & −1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix} )

so D = ( \begin{matrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix} ) and L^{T} = ( \begin{matrix}1 & 0 & −1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) and get L as (L^{T})^{T}.