Math 310: Final Exam (Solutions)

Prof. S. Smith: Monday 5 December 1994

Problem 1:

(a) Find the eigenvalues of A = ( \begin{matrix}1 & 0 & 2 \\ 0 & 3 & 0 \\ 5 & 0 & 4 \end{matrix} ).

Using second row, we see

det(A−xI) = −(3−x)[(1−x)(4−x)−5·2] = (x−3)(x^{2} −5x−6) = (x−3)(x+1)(x−6)

So eigenvalues −1, 3, 6.

(b) For each eigenvalue, find a basis for the corresponding eigenspace of A.

Solving (A − λI)x = 0 by Chapter 1 for each λ gives the columns of:

( \begin{matrix}−1 & 0 & \frac{2}{5} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} ).

(c) Is A diagonalizable? If possible, give X such that X^{−1}AX = D is diagonal. Yes. The eigenvectors form a basis, so can use matrix in (b) for X.

Problem 2: I give you that X^{−1}AX = D where

A = ( \begin{matrix}−1 & 2 \\ 2 & −1 \end{matrix} ) , X = ( \begin{matrix}−1 & 1 \\ 1 & 1 \end{matrix} ) , D = ( \begin{matrix}−3 & 0 \\ 0 & 1 \end{matrix} ) .

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

Using eigenvalues −3, 1 and eigenvectors from given diagonalization,

( \begin{matrix}y_{1} \\ y_{2} \end{matrix} ) = c_{1} ( \begin{matrix}−1 \\ 1 \end{matrix} ) e^{−3t} + c_{2} ( \begin{matrix}1 \\ 1 \end{matrix} ) e^{t}

so y_{1} = −c_{1}e^{−3t} + c_{2}e^{t} and y_{2} = c_{1}e^{−3t} + c_{2}e^{t}.

(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 & 1 & 1 \end{matrix} ) ,

Then Chapter 1 methods give solution c_{1} = −1, c_{2} = 2;

so y_{1} = e^{−3t} + 2e^{t}, y_{2} = −e^{−3t} + 2e^{t}.

Problem 3:

(a) Let M be the Markov matrix ( \begin{matrix}.9 & .3 \\ .1 & .7 \end{matrix} ).

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

For eigenvalue 1, M − 1I = \frac{1}{10} ( \begin{matrix}−1 & 3 \\ 1 & −3 \end{matrix} ),

so eigenvector is span of (3, 1)^{T}. Hence (\frac{3}{4}, \frac{1}{4}) is steady-state vector.

(b) Find eigenvectors and eigenvalues for A = ( \begin{matrix}1 & −1 \\ 1 & −1 \end{matrix} ).

Is A diagonalizable (why/why not) ? If possible diagonalize A.

det(A) = (1 − x)(−1 − x) + 1 = x^{2} − 1 + 1 = x^{2}, so eigenvalues are 0, 0. Only solutions of A − 0x = 0 are multiples of (1, 1)^{T}. Not diagonalizable: (geometric multiplicity = 1) < (algebraic multiplicity = 2).

Problem 4: Let A = ( \begin{matrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} ). I give you that the eigenvalues are −1, −1, 2.

(a) Find eigenvectors for the eigenvalue -1. (I give you that a 2-eigenvector is (1, 1, 1)^{T} ).

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

so a basis for 0-eigenvectors is (−1, 1, 0)^{T} and (0, −1, 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 2, just divide by length to get \frac{1}{\sqrt{3}}(1, 1, 1)^{T}. For −1, divide first by length to get \frac{1}{\sqrt{2}}(−1, 1, 0)^{T} and then compute (0, −1, 1)^{T} − [\frac{1}{\sqrt{2}}(−1, 1, 0)(0, −1, 1)^{T}]\frac{1}{\sqrt{2}}(−1, 1, 0)^{T} = (0, −1, 1)^{T} + \frac{1}{2}(−1, 1, 0)^{T}

= \frac{1}{2}(−1, −1, 2)^{T}; divide by length to get \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} ).

Problem 5:

(a) Find the matrix of the quadratic form 6x^{2} + 5y^{2} + 6z^{2} + 8xy − 4xz + 6yz, and use determinants to decide if it is/is not positive definite.

A = ( \begin{matrix}6 & 4 & −2 \\ 4 & 5 & 3 \\ −2 & 3 & 6 \end{matrix} )

Determinants of principal minors are 6, 6.5 − 4.4 = 14 and det(A) = −38. The last is negative, so not positive definite.

(b) Write ( \begin{matrix}4 & 2 \\ 2 & 10 \end{matrix} ) as LDL^{T} for L lower triangular and D diagonal.

Row operation A_{2}^{− \frac{1}{2} ×1} gives ( \begin{matrix}4 & 2 \\ 0 & 9 \end{matrix} )

so factor out diagonal elements, and use inverse of the row operation for L:

A = ( \begin{matrix}1 & 0 \\ \frac{1}{2} & 1 \end{matrix} ) ( \begin{matrix}4 & 0 \\ 0 & 9 \end{matrix} ) ( \begin{matrix}1 & \frac{1}{2} \\ 0 & 1 \end{matrix} ) .