Math 310: Final Exam (Solutions)

Prof. S. Smith: Wed 6 Dec 1995

Problem 1: (a) Find the eigenvalues of A = ( \begin{matrix}1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3 \end{matrix} ).

We see det(A−xI) = (3−x)[(1−x)^{2} −2·2] = (3−x)(x^{2} −2x−3) = (3−x)(x−3)(x+1). So eigenvalues −1, 3, 3.

(b) Find the eigenspaces for each eigenvalue. Is A diagonalizable? (Why/why not?) For −1 get (−1, 1, 0)^{T}; for 3, span of (1, 1, 0)^{T} and (0, 0, 1)^{T}. So YES diagonalizable (in particular, 2 lin.indep. eigenvectors for 3).

(c) Note A is symmetric—is it positive definite? Give two methods of deciding. No. From above, not all eigenvalues are positive. Alternatively, the three “upper left” determinants are 1, −3, −9, not all positive.

Problem 2: (a) Give the general solution of the differential system \begin{aligned}y′_{1} & = & y_{1} & + & y_{2} \\ y′_{2} & = & −2y_{1} & + & 4y_{2} \end{aligned} .

Find eigenvalues 2 and 3, and corresponding eigenvectors (1, 1)^{T} and (1, 2)^{T}.

( \begin{matrix}y_{1} \\ y_{2} \end{matrix} ) = 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) Give the particular solution when y_{1}(0) = 3 and y_{2}(0) = 1.

Putting in t = 0 gives system with augmented matrix ( \begin{matrix}1 & 1 & 3 \\ 1 & 2 & 1 \end{matrix} ) ,

Then Chapter 1 methods give solution c_{1} = 5, c_{2} = −2; so y_{1} = 5e^{2t} − 2e^{3t}, y_{2} = 5e^{2t} − 4e^{3t}.

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

(a) Find the “steady state” eigenvector for M. (Components of vector should add to 1).

For eigenvalue 1, M − 1I = \frac{1}{3} ( \begin{matrix}−2 & 2 \\ 2 & −2 \end{matrix} ),

so eigenvector is span of (1, 1)^{T}. Hence (\frac{1}{2}, \frac{1}{2}) is steady-state vector.

(b) Diagonalize M and use this to get an expression for the power M^{n}.

Compute det(M − xI) = (x − \frac{1}{3})^{2} − \frac{2}{3}^{2} = x^{2} − \frac{2}{3}x − \frac{1}{3} = (x − 1)(x + \frac{1}{3}).

So other eigenvalue is −\frac{1}{3}; from M + \frac{1}{3}I = \frac{1}{3} ( \begin{matrix}2 & 2 \\ 2 & 2 \end{matrix} ) get eigenvector (1, −1)^{T}.

Using S = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ), get S^{−1}MS = Λ so M = SΛS^{−1} hence M^{n} = S(Λ^{n})S^{−1}.

Thus M^{n} = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) ( \begin{matrix}1^{n} & 0 \\ 0 & (−\frac{1}{3})^{n} \end{matrix} ) \frac{1}{2} ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} )

= ( \begin{matrix}1 & (−\frac{1}{3})^{n} \\ 1 & −(−\frac{1}{3})^{n} \end{matrix} ) \frac{1}{2} ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ) = \frac{1}{2} ( \begin{matrix}1 + (−\frac{1}{3})^{n} & 1 − (−\frac{1}{3})^{n} \\ 1 − (−\frac{1}{3})^{n} & 1 + (−\frac{1}{3})^{n} \end{matrix} ).

Problem 4: Let A = ( \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} ). I give you that the eigenvalues are 3, 0, 0.

(a) Find eigenvectors for these eigenvalues.

A − 3I row-reduces to ( \begin{matrix}1 & 0 & −1 \\ 0 & 1 & −1 \\ 0 & 0 & 0 \end{matrix} ), so a 3-eigenvector is (1, 1, 1)^{T}.

A − 0I = A row-reduces to ( \begin{matrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ),

so independent 0-eigenvectors are (1, −1, 0)^{T} and (1, 0, −1)^{T}.

(b) Note A is symmetric. So find orthogonal S with S^{−1}AS diagonal. (Remember this means the columns of S must be orthonormal). We need to apply Gram-Schmidt to each eigenspace.

For 3, just divide by length to get \frac{1}{\sqrt{3}}(1, 1, 1)^{T}. For 0, first is (1, −1, 0)^{T} and second is (column form of)

(1, 0, −1) − (1, −1, 0)\frac{1}{(1,−1,0)(1,−1,0)^{T}} (1, −1, 0)(1, 0, −1)^{T}

= (1, 0, −1) − \frac{1}{2}(1, −1, 0) = \frac{1}{2}(1, 1, −2).

Divide by lengths to get \frac{1}{\sqrt{2}}(1, −1, 0)^{T} and \frac{1}{\sqrt{6}}(1, 1, −2)^{T}.

So use S = ( \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} ).