Math 310: Final Exam (Solutions)

Prof. S. Smith: Tues 7 Dec 1999

You must SHOW WORK to receive credit. (If you use a calculator, INDICATE those places where you use it).

Problem 0 (review) is worth 10 points, and Problems 1–5 are worth 20 points. So the maximum score possible is 110.

Problem 0: (As promised, a flashback to Hour Exam 2:)

Let V be the space F^{2×2} of 2 × 2 matrices. Let I be the usual 2 × 2 identity matrix ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

Determine the orthogonal complement, that is, the subspace I^{⊥} of all 2 × 2 matrices orthogonal

to I. (Recall that the inner product of two matrices is defined by ⟨A, B⟩ = ∑_{i,j=1}^{2} A_{ij}B_{ij}).

A is in I^{⊥} if 0 = ⟨A, I⟩ = A_{11}.1 + A_{12}.0 + A_{21}.0 + A_{22}.1 = A_{11} + A_{22}.

Thus I^{⊥} consists of the matrices of form ( \begin{matrix}a & b \\ c & −a \end{matrix} ) .

Problem 1: Let A = ( \begin{matrix}1 & 2 \\ 2 & 1 \end{matrix} ) . (a) Find the eigenvalues of A.

det(A − xI) = (x − 1)^{2} − 4 = x^{2} − 2x − 3 = (x − 3)(x + 1), so eigenvalues are 3, −1.

(b) Find the eigenspaces for those eigenvalues.

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

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

Problem 2: Given the differential equation system (functions of t): ( \begin{matrix}y′_{1} & = & y_{1} & +y_{2} \\ y′_{2} & = & −2y_{1} & +4y_{2} \end{matrix} ).

I GIVE you the information that eigenvalues of the coefficient matrix A for this system are 2, 3, with corresponding eigenvectors (1, 1)^{T} and (1, 2)^{T}.

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

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

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

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

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

I GIVE you that the eigenvalues of A are 0, 1, 1. Is A diagonalizable? Indicate why/why not.

Yes: we compute that the eigenspace of 0 is a(−1, 0, 1)^{T}, and the eigenspace for 1 is (0, b, c) of dimension 2. Thus we CAN find a basis of eigenvectors; alternatively, the geometric multiplicity equals the algebraic multiplicity for each eigenvalue.

(b) For A = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ), I GIVE you that eigenvalues are 3, 1. Diagonalize A: this is, find X with

X^{−1}AX diagonal. Use this to determine the exponential e^{A}.

We compute that eigenvectors are (1, 1)^{T} and (1, −1)^{T}.

So 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 = ( \begin{matrix}3 & 0 \\ 0 & 1 \end{matrix} ).

So A = XDX^{−1} and then e^{A} = X(e^{D})X^{−1} where e^{D} = ( \begin{matrix}e^{3} & 0 \\ 0 & e \end{matrix} ).

Multiplying out we get e^{A} = \frac{1}{2} ( \begin{matrix}e^{3} + e & e^{3} − e \\ e^{3} − e & e^{3} + e \end{matrix} ).

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

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

(a) Find a basis for each eigenspace of A.

For 1: eigenvectors are a(1, 1, 1)^{T}.

For -2: Get eigenvectors (−b − c, b, c)^{T}. So one possible basis is (−1, 1, 0)^{T} and (−1, 0, 1)^{T}.

(b) Give an orthogonal diagonalization of A; that find an orthogonal matrix X (satisfying X^{−1} = X^{T}) with X^{−1}AX is diagonal.

For 1: eigenspace is 1-dimensional; divide previous vector by its length: \frac{1}{\sqrt{3}}(1, 1, 1)^{T}.

For -2: Start with above 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}. 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} ).

Problem 5: (a) For the Markov matrix A = ( \begin{matrix}.7 & .2 \\ .3 & .8 \end{matrix} ), find the “steady-state” vector v. (That

is, Av = v, and the coordinates of v add up to 1.)

The eigenspace for 1 consists of vectors a(2, 3)^{T}. So the steady-state vector is (.4, .6)^{T}.

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

Yes: e.g., the determinants of the principal minors are positive: 1, 1, 1. (Alternatively, the eigenvalues are positive; but the computer is needed to find them...)