18.06 Problem Set #6 Solutions

1. a) If we denote the number of subscribers to Star at k-th year by s_{k}, and the number of subscribers to Times by t_{k}, then the conditions in the problems tell us s_{k+1} = .6s_{k} +.1t_{k}, t_{k+1} = .4s_{k} + .9t_{k}. So the Markov matrix describing transition of subscribers is M =

[ \begin{matrix}.6 & .1 \\ .4 & .9 \end{matrix} ].

b) The characteristic equation of M is λ^{2} − 1.5λ + .5 = 0. So the eigenvalues of M are 1 and 0.5. The steady state vector, which is a eigenvector corresponding to eigenvalue 1, is (1, 4).

c) The formula for (s_{k}, t_{k}) is, (s_{k}, t_{k}) = c_{1}1^{k}(1, 4) + c_{2}(.5)^{k}(1, −1), where c_{1}, c_{2} are some constants. When k goes to infinity, (.5)^{k} goes to zero. So (s_{k}, t_{k}) goes to c_{1}(1, 4). This means after many years, approximately 80% of the subscribers will subscribe to the Times.

2. a) \overline{(UV)}^{T} (UV) = \overline{V}^{T} \overline{U}^{T} UV = \overline{V}^{T} V = I, so UV is unitary if both U and V are. In the first step, we basically use the associativity of matrix multiplication;in the second step we use the fact that U is unitary;in the third step we use the fact that V is unitary. \overline{U^{−1}}^{T} = (\overline{U}^{T})^{−1} = (U^{−1})^{−1}, so U^{−1} is unitary if U is. In the first step we use the fact that we can commute the operation of taking complex conjugation, taking transpose or taking inverse of a matrix; in the second step we use the fact U is unitary, so \overline{U}^{T} = U^{−1}.

b) Suppose A is a hermitian matrix, i.e. \overline{A}^{T} = A. Then det(\overline{A}^{T}) = det(\overline{A}) = \overline{det(A)}. But det(\overline{A}^{T}) = det(A) because A is hermitian. So \overline{det(A)} = det(A), det(A) has to be a real number.

3. a) x_{k} = A^{k}x_{0} = A^{k}(2v_{1} +5v_{2}) = 2A^{k}v_{1} +5A^{k}v_{2} = 2v_{1} +5(\frac{1}{3})^{2}v_{2} , The last step is because v_{1}, v_{2} are eigenvectors, use Av_{i} = λ_{i}v_{i} iteratively, one can have A^{k}v_{i} = λ_{i}^{k}v_{i}. Here λ_{1} = 1, λ_{2} = 1/3.

b) In above formula for x_{k}, when k tends to infinity, (\frac{1}{3})^{k} tends to zero. So x_{k} tends to 2v_{1}.

4. The coefficient matrix for the ODE is A = [ \begin{matrix}10 & −3 \\ 5 & 2 \end{matrix} ]. First we find the characteristic

equation det(λI − A) = 0, i.e. λ^{2} − 12λ + 35 = 0. So the two eigenvalues are 5 and 7.

Substitute λ by 5 or 7 respectively into the matrix λI − A and solve for (λI − A)v = 0 to find eigenvectors. The eigenvector associated to 5 is v_{1} = (3, 5), and eigenvector associated to 7 is v_{2} = (1, 1).

The general solution to the differential equation is (r(t), w(t)) = c_{1}e^{5t}v_{1} + c_{2}e^{7t}v_{2}. Let t = 0 and use the condition r(0) = 30 = w(0), we find c_{1} = 0, c_{2} = 30. So the final solution is r(t) = w(t) = 30e^{7t}

5. a) If Av = λv then ABv = BAv = Bλv = λBv. The first step we use the fact that AB = BA, the second step we use the fact that v is an eigenvector of A corresponding to λ, the third step is because λ is a scalar, so it commute with matrix multiplications. So we have A(Bv) = λ · Bv. On the other hand, Av = λv ≠ 0, so v ∉ N(A). But by condition of the problem, N(A) = N(B), we know v ∉ N(B). So Bv ≠ 0, combined with the equality A(Bv) = λ · Bv, this means Bv is also a λ -eigenvector for A. b) First when A has distinct n eigenvalues, the corresponding eigenvectors form a basis of

ℝ^{n}. Secondly when λ_{i} is non-zero, from part a), we know x_{i}, Bx_{i} are both eigenvectors of A corresponding to λ_{i}. But the eigenspace corresponding to λ_{i} is 1-dimensional, namely spanned by x_{i}. So Bx_{i} is a nonzero multiple of x_{i}, say Bx_{i} = μ_{i} · x_{i}. So x_{i} is also an eigenvector of B. When λ_{i} = 0, by the condition N(A) = N(B), we know Bx_{i} = 0, so x_{i} is still an eigenvector of B. B has n linearly independent eigenvectors, so is diagonalizable.