Math 310 (10am MWF):  Final Exam  (Solutions)

Prof. S. Smith: Wed 5 May 1999

You must SHOW WORK to receive credit. Problem 1:

(a) (flashback to Hour Exam 1) In physics one sometimes sees the set S of 3 × 3 skew-symmetric

matrices—satisfying A^{T} = −A. Notice such matrices have the general form ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −c & 0 \end{matrix} ).

Show that S forms a subspace of the space R^{3×3} of all 3 × 3 matrices.

(closure,+) ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −c & 0 \end{matrix} ) + ( \begin{matrix}0 & d & e \\ −d & 0 & f \\ −e & −f & 0 \end{matrix} ) = ( \begin{matrix}0 & a + d & b + e \\ −a − d & 0 & c + f \\ −b − e & −c − f & 0 \end{matrix} ).

The sum is also skew, and so still lies in the set S.

(closure,sc.mult.) f ( \begin{matrix}0 & a & b \\ −a & 0 & c \\ −b & −c & 0 \end{matrix} ) = ( \begin{matrix}0 & fa & fb \\ −fa & 0 & fc \\ −fb & −fc & 0 \end{matrix} ), also skew, so in S.

(b) (and to Exam 2:) Let S be the subspace of R^{3} spanned by (1, 1, 1)^{T} and (1, 2, 3)^{T}. Find S^{⊥}. Put in as rows of A = ( \begin{matrix}1 & 1 & 1 \\ 1 & 2 & 3 \end{matrix} ), solve Ax = 0 to get x_{3}(1, −2, 1)^{T}.

Problem 2: Let A = ( \begin{matrix}0 & 2 \\ 2 & 0 \end{matrix} ). (a) Find the eigenvalues of A.

det(A − xI) = x^{2} − 4 = (x − 2)(x + 2) so eigenvalues are 2, −2.

(b) Find the eigenspaces for those eigenvalues. Is A diagonalizable? Why/why not? For 2: Solve (A − 2I)x = 0 with ( \begin{matrix}−2 & 2 \\ 2 & −2 \end{matrix} ) to get a(1, 1)^{T};

For −2: (A + 2I)x = 0 with ( \begin{matrix}2 & 2 \\ 2 & 2 \end{matrix} ) to get b(1, −1)^{T}.

Diagonalizable—distinct eigenvalues (not repeated).

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

GIVEN: the matrix ( \begin{matrix}1 & 3 \\ 3 & 1 \end{matrix} ) has eigenvalues 4, −2; with eigenvectors (1, 1)^{T} and (1, −1)^{T}.

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

( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) ( \begin{matrix}c_{1}e^{4t} \\ c_{2}e^{−2t} \end{matrix} ) so y_{1} = c_{1}e^{4t} + c_{2}e^{−2t} and y_{2} = c_{1}e^{4t} − c_{2}e^{−2t}.

(b) Now determine the values of c_{1}, c_{2} for the initial value problem y_{1}(0) = 8, y_{2}(0) = 4.

Solve ( \begin{matrix}1 & 1 & 8 \\ 1 & −1 & 4 \end{matrix} ) to get c_{1} = 6, c_{2} = 2. So y_{1} = 6e^{4t} + 2e^{−2t} and y_{2} = 6e^{4t} − 2e^{−2t}.

Problem 4: (a) Let A be the (Markov) matrix ( \begin{matrix}\frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{2}{3} \end{matrix} ) .

Indicate the “steady-state” vector; that is, a vector v such that Av = v—and the coordinates of v add up to 1.

We need an eigenvector for eigenvalue 1. Solve (A − 1.I)x = 0

using ( \begin{matrix}−\frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & −\frac{2}{3} \end{matrix} ), to get solutions a(2, 3)^{T}; so use vector \frac{1}{5}(2, 3)^{T}.

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

No: for example, determinants of principal minors are 1,-1,0; not all positive.

Problem 5: (a) Let A be the matrix ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} ). GIVEN: A has eigenvalues 0, 1, 2.

Find eigenvectors, and diagonalize A (that is, give X such that X^{−1}AX is diagonal).

For 0: get a(1, 0, −1)^{T}. For 1: get b(0, 1, 0)^{T} For 2: get c(1, 0, 1)^{T}. So X = ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ −1 & 0 & 1 \end{matrix} ).

(b) The matrix A = ( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} ) has eigenvalues 2, 0 and eigenvectors (1, 1)^{T} and (1, −1)^{T}. Find e^{A}.

We can diagonalize A with X = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ). So from A = XDX^{−1} we have e^{A} = Xe^{D}X^{−1} =

( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) ( \begin{matrix}e^{2} & 0 \\ 0 & e^{0} = 1 \end{matrix} ) \frac{1}{2} ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) = \frac{1}{2} ( \begin{matrix}e^{2} + 1 & e^{2} − 1 \\ e^{2} − 1 & e^{2} + 1 \end{matrix} )

Problem 6: Let A be the symmetric matrix ( \begin{matrix}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} ). GIVEN: A has eigenvalues 1, 1, 4.

(a) Find a basis for each eigenspace.

For 4: find eigenspace spanned by (1, 1, 1)^{T}. For 4: Get eigenvectors (−b − c, b, c)^{T}.

So one basis is (1, −1, 0)^{T} and (1, 0, −1)^{T}.

(b) Use Gram-Schmidt to find an orthonormal basis for each eigenspace. Use that to build an orthogonal matrix X (that is, X^{−1} = X^{T}) with X^{−1}AX diagonal.

For 4, just use \frac{1}{\sqrt{3}}(1, 1, 1)^{T}. For 1, apply Gram-Schmidt to get \frac{1}{\sqrt{2}}(1, −1, 0)^{T} and \frac{1}{\sqrt{6}}(1, 1, −2)^{T}.

So 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} ).