18.06 Exam 3 Solutions

1. a) M = [ \begin{matrix}0.7 & 0.5 \\ 0.3 & 0.5 \end{matrix} ] .

b) It is the eigenvector for M corresponding to eigenvalue 1, (0.5, 0.3).

c) After many years, the percentage of people drinking two kinds of coffee will converge to one that is proportional to the steady state vector. So people drinking regular coffee will be about 5/8 = 62.5%.

2. a) Column vectors of Q are normalized eigenvectors of A, denote column vectors of Q by v_{1}, v_{2}, v_{3}. Then

(A − (−2)I)v_{1} = 0 ⇒ v_{1} = (0, 1, 0)
(A − 4I)v_{2} = 0 ⇒ v_{2} = (−1, 0, 2)/\sqrt{5}
(A − (−1)I)v_{3} = 0 ⇒ v_{3} = (2, 0, 1)/\sqrt{5}

Q = [ \begin{matrix}0 & −1/\sqrt{5} & 2/\sqrt{5} \\ 1 & 0 & 0 \\ 0 & 2/\sqrt{5} & 1/\sqrt{5} \end{matrix} ] , Λ = [ \begin{matrix}−2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & −1 \end{matrix} ]

b) Λ is the matrix for L under the basis \{[ \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ] , [ \begin{matrix}−1/\sqrt{5} \\ 0 \\ 2/\sqrt{5} \end{matrix} ] , [ \begin{matrix}2/\sqrt{5} \\ 0 \\ 1/\sqrt{5} \end{matrix} ]

3. First we find out the eigenvalues of A^{T} A corresponding to w_{i} respectively.

A^{T} Aw_{1} = [ \begin{matrix}1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{matrix} ] [ \begin{matrix}1 \\ 2 \\ 1 \end{matrix} ] = [ \begin{matrix}3 \\ 6 \\ 3 \end{matrix} ] = 3w_{1}

A^{T} Aw_{2} = [ \begin{matrix}1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{matrix} ] [ \begin{matrix}1 \\ 0 \\ −1 \end{matrix} ] = [ \begin{matrix}1 \\ 0 \\ −1 \end{matrix} ] = w_{2}

A^{T} Aw_{3} = [ \begin{matrix}1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{matrix} ] [ \begin{matrix}1 \\ −1 \\ 1 \end{matrix} ] = [ \begin{matrix}0 \\ 0 \\ 0 \end{matrix} ] = 0

So λ_{1} = 3, σ_{1} = \sqrt{3}, λ_{2} = 1, σ_{2} = 1, and Σ = [ \begin{matrix}\sqrt{3} & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ]

Column vectors of V are normalized eigenvectors, v_{1} = w_{1}/|w_{1}|, v_{2} = w_{2}/|w_{2}|, v_{3} = w_{3}/|w_{3}|,

V = [ \begin{matrix}1/\sqrt{6} & 1/\sqrt{2} & 1/\sqrt{3} \\ 2/\sqrt{6} & 0 & −1/\sqrt{3} \\ 1/\sqrt{6} & −1/\sqrt{2} & 1/\sqrt{3} \end{matrix} ]

Column vectors of U satisfies, u_{1} = Av_{1}/σ_{1} = [ \begin{matrix}1/\sqrt{2} \\ 1/\sqrt{2} \end{matrix} ], u_{2} = Av_{2}/σ_{2} = [ \begin{matrix}1/\sqrt{2} \\ −1/\sqrt{2} \end{matrix} ].

Finally the answer is

SV D(A) = UΣV^{T} = [ \begin{matrix}1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & −1/\sqrt{2} \end{matrix} ] [ \begin{matrix}\sqrt{3} & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ] [ \begin{matrix}1/\sqrt{6} & 2/\sqrt{6} & 1/\sqrt{6} \\ 1/\sqrt{2} & 0 & −1/\sqrt{2} \\ 1/\sqrt{3} & −1/\sqrt{3} & 1/\sqrt{3} \end{matrix} ]

4. a)

L([ \begin{matrix}1 \\ 0 \end{matrix} ]) = [ \begin{matrix}4 \\ 5 \end{matrix} ] = 4 [ \begin{matrix}1 \\ 0 \end{matrix} ] + 5 [ \begin{matrix}0 \\ 1 \end{matrix} ]

L([ \begin{matrix}0 \\ 1 \end{matrix} ]) = L(\frac{1}{2} [ \begin{matrix}2 \\ 2 \end{matrix} ] − [ \begin{matrix}1 \\ 0 \end{matrix} ]) = \frac{1}{2} [ \begin{matrix}8 \\ −6 \end{matrix} ] − [ \begin{matrix}4 \\ 5 \end{matrix} ] = [ \begin{matrix}0 \\ −8 \end{matrix} ] = −8 [ \begin{matrix}0 \\ 1 \end{matrix} ]

So the matrix for L with respect to the standard basis for R^{2} is A = [ \begin{matrix}4 & 0 \\ 5 & −8 \end{matrix} ]

b) The change of basis is equal to the inverse of basis matrix, i.e. P = [ \begin{matrix}1 & 2 \\ 0 & 2 \end{matrix} ]^{−1} =

[ \begin{matrix}1 & −1 \\ 0 & 1/2 \end{matrix} ].

c)

B = P^{−1}AP = [ \begin{matrix}1 & 2 \\ 0 & 2 \end{matrix} ] [ \begin{matrix}4 & 0 \\ 5 & −8 \end{matrix} ] [ \begin{matrix}1 & −1 \\ 0 & 1/2 \end{matrix} ] = [ \begin{matrix}−1 & 14 \\ \frac{5}{2} & −3 \end{matrix} ]