Math 310: Final Exam  (Solutions)  Prof. S. Smith: Monday 8 Dec 2003

Problem 0: (review: linear transformations) Let L denote the function from R^{2} to R^{2} defined by L(( \begin{matrix}x_{1} \\ x_{2} \end{matrix} )) = ( \begin{matrix}x_{1} − 2x_{2} \\ 3x_{1} \end{matrix} ). Show that L is a LINEAR transformation, via the following steps:

(1) Write down a general vector (that is, name its coordinates). Answer: ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} )

(2) Write down another general vector (name its coordinates): Answer: ( \begin{matrix}y_{1} \\ y_{2} \end{matrix} )

(3) Now add your two vectors from (1) and (2). Answer: ( \begin{matrix}x_{1} + y_{1} \\ x_{2} + y_{2} \end{matrix} ) (3+) Now apply L to the sum vector in (3): Answer: ( \begin{matrix}(x_{1} + y_{1}) − 2(x_{2} + y_{2}) \\ 3(x_{1} + y_{1}) \end{matrix} )

(4) Now apply L separately to the two vectors in (1) and (2).

Answers: ( \begin{matrix}x_{1} − 2x_{2} \\ 3x_{1} \end{matrix} ) and ( \begin{matrix}y_{1} − 2y_{2} \\ 3y_{1} \end{matrix} )

(4+) Now add the two vectors in (4). Answer: ( \begin{matrix}(x_{1} − 2x_{2}) + (y_{1} − 2y_{2}) \\ 3x_{1} + 3y_{1} \end{matrix} ) FINALLY: Are the answers in (3+) and (4+) the same? Answer: Yes, using distributive laws (5) Write down a general vector (you can use (1) again). Answer: ( \begin{matrix}x_{1} \\ x_{2} \end{matrix} )

(6) Write down a general scalar (name it): Answer: c

(7) Now form the product of the vector in (5) and the scalar in (6). Answer: ( \begin{matrix}c x_{1} \\ c x_{2} \end{matrix} ) (7+) Now apply L to the product vector in (7): Answer: ( \begin{matrix}(cx_{1}) − 2(cx_{2}) \\ 3(cx_{1}) \end{matrix} )

(8) Now apply L to the vector in (5). Answer: ( \begin{matrix}x_{1} − 2x_{2} \\ 3x_{1} \end{matrix} ). (8+) Now multiply the vector in (8) by the scalar in (6). Answer: ( \begin{matrix}c (x_{1} − 2x_{2}) \\ c (3x_{1}) \end{matrix} ) FINALLY: Are the answers in (7+) and (8+) the same? Answer: Yes, using distributive laws Problem 1: Let A = ( \begin{matrix}2 & 3 \\ 2 & 1 \end{matrix} ) .

(a) Find the characteristic polynomial, and the eigenvalues, of A.

det(A − xI) = (2 − x)(1 − x) − 6 = (x^{2} − 3x + 2) − 6 = x^{2} − 3x − 4 = (x − 4)(x + 1), so eigenvalues are 4, −1.

(b) Find the eigenspaces for those eigenvalues.

For 4: A − 4.I = ( \begin{matrix}−2 & 3 \\ 2 & −3 \end{matrix} ) has rref ( \begin{matrix}1 & −\frac{3}{2} \\ 0 & 0 \end{matrix} ), get solutions a(\frac{3}{2}, 1)^{T}.

For -1: A + 1.I = ( \begin{matrix}3 & 3 \\ 2 & 2 \end{matrix} ), has rref ( \begin{matrix}1 & 1 \\ 0 & 0 \end{matrix} ), get solutions b(−1, 1)^{T}.

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

I GIVE you the information that eigenvalues of the coefficient matrix A for this system are 0, −5,

(a) Find eigenvectors for these eigenvalues of A; then use them to give the general solution of the system (with undetermined constants c_{1}, c_{2}).

For 0, get a(\frac{1}{2}, 1)^{T}; for −5, get b(−2, 1)^{T}.

Then solution vector c_{1} ( 1^{\frac{1}{2}} ) e^{0t} + c_{2} ( \begin{matrix}−2 \\ 1 \end{matrix} ) e^{−5t} so y_{1} = \frac{1}{2}c_{1} − 2c_{2}e^{−5t} and y_{2} = c_{1} + c_{2}e^{−5t}.

(b) Now find the particular solution (values of c_{1}, c_{2}) given initial values y_{1}(0) = 0, y_{2}(0) = 5.

Solve ( \begin{matrix}\frac{1}{2} & −2 & 0 \\ 1 & 1 & 5 \end{matrix} ) to get c_{1} = 4, c_{2} = 1. So y_{1} = 2 − 2e^{−5t} and y_{2} = 4 + e^{−5t}.

Problem 3:

(a) GIVEN: the eigenvalues of A = ( \begin{matrix}1 & −1 \\ 2 & 4 \end{matrix} ) are 2, 3, with eigenvectors (−1, 1)^{T} and (−1, 2)^{T}.

Find matrices X, X^{−1}, and D such that X^{−1}AX = D with D a diagonal matrix. Use this to information find a matrix B such that B^{2} = A (use square-root symbols, not decimals).

We can use X = ( \begin{matrix}−1 & −1 \\ 1 & 2 \end{matrix} ), X^{−1} = ( \begin{matrix}−2 & −1 \\ 1 & 1 \end{matrix} ) with D = X^{−1}AX = ( \begin{matrix}2 & 0 \\ 0 & 3 \end{matrix} ).

Then E = ( \begin{matrix}\sqrt{2} & 0 \\ 0 & \sqrt{3} \end{matrix} ) satisfies E^{2} = D (indeed we could multiply either row of E by ±1

so B = XEX^{−1} = ( \begin{matrix}−1 & −1 \\ 1 & 2 \end{matrix} ) ( \begin{matrix}\sqrt{2} & 0 \\ 0 & \sqrt{3} \end{matrix} ) ( \begin{matrix}−2 & −1 \\ 1 & 1 \end{matrix} ) = ( \begin{matrix}2\sqrt{2} − \sqrt{3} & \sqrt{2} − \sqrt{3} \\ −2\sqrt{2} + 2\sqrt{3} & −\sqrt{2} + 2\sqrt{3} \end{matrix} )

satisfies B^{2} = A.

(b) Let A = ( \begin{matrix}5 & −6 & −6 \\ −1 & 4 & 2 \\ 3 & −6 & −4 \end{matrix} ). GIVEN: the eigenvalues of A are 1, 2, 2.

Find the DIMENSIONS of the eigenspaces for these eigenvalues. Is A diagonalizable? Say why/why not.

Check that rref(A − 1.I_{3}) has 1 free variable, so the dimension of the 1-eigenspace is 1.

However rref(A − 2.I_{3}) has 2 free variables, so the dimension of the 2-eigenspace is 2.

Then A is diagonalizable—since for the eigenvalue 2, the dimension of the eigenspace is equal to the number of times the eigenvalue appears as a root of the characteristic polynomial. (That is, geometric multiplicity = algebraic multiplicity for 2, as well as for 1).

Problem 4: For the symmetric matrix A = ( \begin{matrix}1 & −2 & 2 \\ −2 & 1 & −2 \\ 2 & −2 & 1 \end{matrix} ),

I GIVE you the eigenvalues 5, −1, −1 of A; and an eigenvector (1, −1, 1)^{T} for eigenvalue 5.

(a) Find a basis of the eigenspace of A for eigenvalue −1.

The row-reduced echelon form of A − (−1).I = A + I has (1, −1, 1) as its only nonzero row. So eigenvectors are (b − c, b, c)^{T}; and one possible basis is (1, 1, 0)^{T} and (−1, 0, 1)^{T}.

(b) Now find an orthonormal basis for the eigenspace in (a). Use it to give an orthogonal diagonalization of A; that is, find an orthogonal matrix X (satisfying X^{−1} = X^{T}) with X^{−1}AX diagonal.

For 5: eigenspace is 1-dimensional; divide original (1, −1, 1) by its length \sqrt{3}: x_{3} = \frac{1}{\sqrt{3}}(1, −1, 1)^{T}. For -1: Start with above basis like v_{1} = (1, 1, 0)^{T} and v_{2} = (−1, 0, 1)^{T}. Apply Gram-Schmidt: first q_{1} = (1, 1, 0), and then q_{2} = v_{2} − \frac{v_{2}·q_{1}}{q_{1}·q_{1}} q_{1} = (−1, 0, 1)^{T} − \frac{−1}{2}(1, 1, 0)^{T} = (−\frac{1}{2}, \frac{1}{2}, 1)^{T}

so may as well use the more convenient multiple q_{2} = (−1, 1, 2)^{T}. Now divide each by its length, to get x_{1} = \frac{1}{\sqrt{2}}(1, 1, 0)^{T} and x_{2} = \frac{1}{\sqrt{6}}(−1, 1, 2)^{T}.

So now putting x_{3} first, 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) Find a steady-state vector (with coordinates adding to 1) for the Markov matrix A = ( \begin{matrix}.6 & .3 \\ .4 & .7 \end{matrix} ).

Need eigenvector for eigenvalue 1, so A − 1 · I_{2} has rref ( \begin{matrix}1 & −\frac{3}{4} \\ 0 & 0 \end{matrix} ),

so divide eigenvector (3, 4)^{T} by coordinate sum 7 to get proportions (\frac{3}{7}, \frac{4}{7})^{T}.

(b) Is the symmetric matrix A = ( \begin{matrix}1 & 2 \\ 2 & 1 \end{matrix} ) positive definite?

Give the decomposition A = LDL^{t} with L lower triangular and D diagonal.

NOT positive definite: eigenvalues -1,3 not all positive (or, upper-left determinants 1, -3 not all positive) For decomposition, we start with the LU-decomposition of chapter 1:

we use A_{2}^{−2×1} to get row-reduced form U = ( \begin{matrix}1 & 2 \\ 0 & −3 \end{matrix} ), with L = ( \begin{matrix}1 & 0 \\ 2 & 1 \end{matrix} ); from U,

we take the diagonal entries to get D = ( \begin{matrix}1 & 0 \\ 0 & −3 \end{matrix} ); that is, U = DL^{t} where L^{t} = ( \begin{matrix}1 & 2 \\ 0 & 1 \end{matrix} ).