Math 310: Final Exam (Solutions) Prof. S. Smith: Tues 2 May 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}3 & 2 \\ 3 & 8 \end{matrix} ) .

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

det(A − xI) = (x − 3)(x − 8) − 6 = x^{2} − 11x + 18 = (x − 9)(x − 2), so eigenvalues are 2, 9.

(b) Find the eigenspaces for those eigenvalues.

For 2: A − 2.I = ( \begin{matrix}1 & 2 \\ 3 & 6 \end{matrix} ), via A_{2}^{−2×1} to ( \begin{matrix}1 & 2 \\ 0 & 0 \end{matrix} ), get solutions a(−2, 1)^{T}.

For 9: A − 9.I = ( \begin{matrix}−6 & 2 \\ 3 & −1 \end{matrix} ), via A_{2}^{\frac{1}{2} ×1} to ( \begin{matrix}−6 & 2 \\ 0 & 0 \end{matrix} ), get solutions b(\frac{1}{3}, 1)^{T}.

Problem 2: Given the differential equation system (functions of t): ( \begin{matrix}y′_{1} & = & 2y_{1} & +3y_{2} \\ y′_{2} & = & −y_{1} & −2y_{2} \end{matrix} ).

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

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

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

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

Solve ( \begin{matrix}−3 & −1 & 3 \\ 1 & 1 & 2 \end{matrix} ) to get c_{1} = −2.5, c_{2} = 4.5.

So y_{1} = 7.5e^{t} − 4.5e^{−t} and y_{2} = −2.5e^{t} + 4.5e^{−t}.

Problem 3: (a) Let A = ( \begin{matrix}4 & 0 & 0 \\ 0 & 5 & 0 \\ 1 & 0 & 4 \end{matrix} ).

Is A diagonalizable? Indicate why/why not.

No: As A is triangular, we see the eigenvalues are 4,4,5. Solving (A − 4I)x = 0, we see the eigenspace for 4 consists of a(0, 0, 1)^{T}: dimension only 1, whereas the eigenvalue is repeated twice. (That is, the geometric multiplicity is less than the algebraic multiplicity). So we cannot get a basis of eigenvectors, and A is not diagonalizable.

(b) For A = ( \begin{matrix}0 & 2 \\ 2 & 0 \end{matrix} ), I GIVE you that eigenvalues are 2, −2; with corresponding eigenvectors

(1, 1)^{T} and (1, −1)^{T}. Use the diagonalization of A (namely use the relevant matrix X with X^{−1}AX diagonal) to determine the 5-th power A^{5}.

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

So A = XDX^{−1} and then A^{5} = X(D^{5})X^{−1} where D^{5} = ( \begin{matrix}2^{5} & 0 \\ 0 & (−2)^{5} \end{matrix} ).

Multiplying out we get A^{5} = ( \begin{matrix}0 & 32 \\ 32 & 0 \end{matrix} ).

Problem 4: Let A be the symmetric matrix ( \begin{matrix}5 & −4 & −2 \\ −4 & 5 & 2 \\ −2 & 2 & 2 \end{matrix} ).

I GIVE the eigenvalues 1, 1, 10 of A; and an eigenvector (−2, 2, 1)^{T} for eigenvalue 10.

(a) Find a basis for the eigenspace of A for eigenvalue 1.

The row-reduced echelon form of A − 1.I has (1, −1, −.5) as its only nonzero row. So eigenvectors are (b + \frac{1}{2}c, b, c)^{T}; and one possible basis is (1, 1, 0)^{T} and (\frac{1}{2}, 0, 1)^{T}.

(b) 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 10: space is 1-dimensional; divide original (−2, 2, 1) by its length 3: x_{3} = \frac{1}{3}(−2, 2, 1)^{T}. For 1: Start with above basis like v_{1} = (1, 1, 0)^{T} and v_{2} = (1, 0, 2)^{T}. Apply Gram-Schmidt: first q_{1} = (1, 1, 0) and then q_{2} = v_{2} − \frac{v_{2}·q_{1}}{q_{1}·q_{1}} q_{1} = (1, 0, 2)^{T} − \frac{1}{2}(1, 1, 0)^{T} = (\frac{1}{2}, −\frac{1}{2}, 2)^{T} so may as well use the more convenient multiple q_{2} = (1, −1, 4)^{T}. Now divide each by length, to get x_{1} = \frac{1}{\sqrt{2}}(1, 1, 0)^{T} and x_{2} = \frac{1}{\sqrt{18}}(1, −1, 4)^{T}.

So can use X = ( \begin{matrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{18}} & −\frac{2}{3} \\ \frac{1}{\sqrt{2}} & −\frac{1}{\sqrt{18}} & \frac{2}{3} \\ 0 & \frac{4}{\sqrt{18}} & \frac{1}{3} \end{matrix} ).

Problem 5: (a) For the Markov matrix A = ( \begin{matrix}.1 & .6 \\ .9 & .4 \end{matrix} ), find the eigenvalues; also find the

“steady-state” vector v. (That is, Av = v, and the coordinates of v add up to 1.)

det(A − xI) = (.1 − x)(.4 − x) − (.9)(.6) = x^{2} − .5x + .04 − .54 = x^{2} − .5x − .5 = (x − 1)(x + .5) so the eigenvalues are 1, −.5. The eigenspace for 1 consists of vectors a(2, 3)^{T}. So the steady-state vector is (.4, .6)^{T}.

(b) Write the matrix ( \begin{matrix}9 & −3 \\ −3 & 2 \end{matrix} ) as a product LDL^{T}, with L lower triangular, and D diagonal.

First get LU-decomposition as in Section 1.4:

A_{2}^{\frac{1}{3} ×1} takes A to U = ( \begin{matrix}9 & −3 \\ 0 & 1 \end{matrix} )

so that L = ( \begin{matrix}1 & 0 \\ −\frac{1}{3} & 1 \end{matrix} )

then factor U = DU^{∗} by using diagonal values of U in D:

D = ( \begin{matrix}9 & 0 \\ 0 & 1 \end{matrix} ) so that U^{∗} = ( \begin{matrix}1 & −\frac{1}{3} \\ 0 & 1 \end{matrix} ), and then observe that U^{∗} is indeed L^{T}.