Math 310: Final Exam (Solutions) Prof. S. Smith: Tues 5 Dec 2000

All 5 problems are worth 20 points each. You must SHOW WORK to receive credit. (If you use a calculator, WRITE “I used calculator” at those places).

Problem 1: Let A = ( \begin{matrix}6 & −4 \\ 3 & −1 \end{matrix} ) .

(a) Find the characteristic polynomial, and the eigenvalues, of A.

det(A − xI) = (x − 6)(x + 1) + 12 = x^{2} − 5x + 6 = (x − 3)(x − 2), so eigenvalues are 2, 3.

(b) Find the eigenspaces for those eigenvalues.

For 2: A − 2.I = ( \begin{matrix}4 & −4 \\ 3 & −3 \end{matrix} ) has rref ( \begin{matrix}1 & −1 \\ 0 & 0 \end{matrix} ), get solutions a(1, 1)^{T}.

For 9: A − 3.I = ( \begin{matrix}3 & −4 \\ 3 & −4 \end{matrix} ), has rref ( \begin{matrix}1 & \frac{4}{3} \\ 0 & 0 \end{matrix} ), get solutions b(\frac{4}{3}, 1)^{T}.

Problem 2: Given the differential equation system (functions of t): ( \begin{matrix}y′_{1} & = & −y_{1} & +2y_{2} \\ y′_{2} & = & 2y_{1} & −y_{2} \end{matrix} ).

I GIVE you the information that eigenvalues of the coefficient matrix A for this system are 1, −3,

(a) Find eigenvectors for A; then use them to give the general solution of the system (with undetermined constants c_{1}, c_{2}).

For 1, get a(1, 1)^{T}; for −1, get b(−1, 1)^{T}.

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

(b) Now find the particular solution (values of c_{1}, c_{2}) given initial values y_{1}(0) = 3, y_{2}(0) = 1.

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

So y_{1} = 2e^{t} + e^{−3t} and y_{2} = 2e^{t} − e^{−3t}.

Problem 3: (a) For A = ( \begin{matrix}5 & 6 \\ −2 & −2 \end{matrix} ), I GIVE you that eigenvalues are 1, 2. Find eigenvectors

for these eigenvalues; and give a matrix X such that X^{−1}AX is a diagonal matrix D. (Show X^{−1} and D also).

For 1, get a(−3, 2)^{T}; for −1, get b(−2, 1)^{T}.

We can use X = ( \begin{matrix}−3 & −2 \\ 2 & 1 \end{matrix} ), X^{−1} = ( \begin{matrix}1 & 2 \\ −2 & −3 \end{matrix} ) with D = X^{−1}AX = ( \begin{matrix}1 & 0 \\ 0 & 2 \end{matrix} ).

(b) Let A = ( \begin{matrix}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} ).

GIVEN: the eigenvalues of A are 1, 1, 4. Find the DIMENSIONS of the eigenspaces for these eigenvalues. (It is not necessary to give eigenvectors). Is A diagonalizable? Say why/why not.

Check that rref(A − 1.I_{3}) has 2 free variables, so the dimension of the 1-eigenspace is 2. Similarly rref(A − 4.I_{3}) has 1 free variable, so the dimension of the 4-eigenspace is 1. Then A is diagonalizable—since for each eigenvalue, the dimension of the eigenspace is equal to (not less than) the number of times the eigenvalue appears as a root of the characteristic polynomial. (That is, geometric multiplicity = algebraic multiplicity for each).

Problem 4: Let A be the symmetric matrix ( \begin{matrix}4 & 2 & −2 \\ 2 & 1 & −1 \\ −2 & −1 & 1 \end{matrix} ).

I GIVE the eigenvalues 0, 0, 6 of A; and an eigenvector (−2, −1, 1)^{T} for eigenvalue 6.

(a) Find a basis of the eigenspace of A for eigenvalue 0.

The row-reduced echelon form of A − 0.I = has (1, .5, −.5) as its only nonzero row.

So eigenvectors are (−\frac{1}{2}b + \frac{1}{2}c, b, c)^{T}; and one possible basis is (−1, 2, 0)^{T} and (1, 0, 2)^{T}.

(b) Now find an orthonormal basis for the eigenspace in (a). Use it to give an orthogonal diagonalization of A; that is, find an orthogonal matrix X (satisfying X^{−1} = X^{T}) with X^{−1}AX is diagonal.

For 6: eigenspace is 1-dimensional; divide original (−2, −1, 1) by its length \sqrt{6}: x_{3} = \frac{1}{\sqrt{6}}(−2, −1, 1)^{T}. For 1: Start with above basis like v_{1} = (−1, 2, 0)^{T} and v_{2} = (1, 0, 2)^{T}. Apply Gram-Schmidt: first q_{1} = (−1, 2, 0) and then q_{2} = v_{2} − \frac{v_{2}·q_{1}}{q_{1}·q_{1}} q_{1} = (1, 0, 2)^{T} − \frac{−1}{5}(−1, 2, 0)^{T} = (\frac{4}{5}, \frac{2}{5}, 2)^{T} so may as well use the more convenient multiple q_{2} = (2, 1, 5)^{T}. Now divide each by length, to get x_{1} = \frac{1}{\sqrt{5}}(−1, 2, 0)^{T} and x_{2} = \frac{1}{\sqrt{30}}(2, 1, 5)^{T}.

So can use X = ( \begin{matrix}−\frac{2}{\sqrt{6}} & −\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} \\ −\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{30}} \\ \frac{1}{\sqrt{6}} & 0 & \frac{5}{\sqrt{30}} \end{matrix} ).

Problem 5: (a) The Markov matrix A = ( \begin{matrix}.2 & .6 \\ .8 & .4 \end{matrix} ), has eigenvalues 1, −.4 and corresponding

eigenvectors (3, 4)^{T} and (−1, 1)^{T}. Use the diagonalization of A to find a formula for the m-th power A^{m}, in terms of m. What does A^{m} converge to, for large m ?

For X = ( \begin{matrix}3 & −1 \\ 4 & 1 \end{matrix} ) and D = ( \begin{matrix}1 & 0 \\ 0 & .4 \end{matrix} ) we have X^{−1}AX = D, so A = XDX^{−1}.

Thus A^{m} = XD^{m}X^{−1} = ( \begin{matrix}3 & −1 \\ 4 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 \\ 0 & (−.4)^{m} \end{matrix} ) \frac{1}{7} ( \begin{matrix}1 & 1 \\ −4 & 3 \end{matrix} ) = \frac{1}{7} ( \begin{matrix}3 + 4(−.4)^{m} & 3 − 3(−.4)^{m} \\ 4 − 4(−.4)^{m} & 4 + 3(−.4)^{m} \end{matrix} )

Thus A^{m} for large m converges to \frac{1}{7} ( \begin{matrix}3 & 3 \\ 4 & 4 \end{matrix} ).

(b) Is the symmetric matrix A = ( \begin{matrix}4 & 2 & 1 \\ 2 & 3 & −2 \\ 1 & −2 & 5 \end{matrix} ) positive definite? Explain why/why not.

Yes. Hard way: find eigenvalues—messy, but all positive. Easy way: The upper left determinants are 4, 4.3 − 2.2 = 8, det(A) = 13, all positive.