Math 310: Final Exam  (Solutions)  Prof. S. Smith: Wednesday 13 December 2006

Problem 1: Let A = ( \begin{matrix}6 & −4 \\ 3 & −1 \end{matrix} ) . Work by hand; do not use a calculator on this problem.

(Except to check your work.)

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

det(A − xI) = (6 − x)(−1 − x) − (−4.3) = (x^{2} − 5x − 6) − (−12) = x^{2} − 5x + 6 = (x − 3)(x − 2), so eigenvalues are 2, 3.

(b) Find the eigenspaces for those eigenvalues.

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

For 3: A − 31.I = ( \begin{matrix}3 & −4 & 0 \\ 3 & −4 & 0 \end{matrix} ), has rref ( \begin{matrix}1 & −\frac{4}{3} & 0 \\ 0 & 0 & 0 \end{matrix} ), get solutions β ( 1^{\frac{4}{3}} ).

Problem 2: Given the differential equation system (functions of t): ( \begin{matrix}y′_{1} & = & y_{1} & −2y_{2} \\ y′_{2} & = & 3y_{1} & −4y_{2} \end{matrix} ).

I GIVE you the information that eigenvalues of the coefficient matrix A for this system are −1, −2.

(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 −1, get α ( \begin{matrix}1 \\ 1 \end{matrix} ); for −2, get β ( 1^{\frac{2}{3}} ), so can use ( \begin{matrix}2 \\ 3 \end{matrix} ).

Then soln.vec. c_{1} ( \begin{matrix}1 \\ 1 \end{matrix} ) e^{−t} + c_{2} ( \begin{matrix}2 \\ 3 \end{matrix} ) e^{−2t} , so y_{1} = c_{1}e^{−t} + 2c_{2}e^{−2t} and y_{2} = c_{1}e^{−t} + 3c_{2}e^{−2t}.

(b) Now find the particular solution (values of c_{1}, c_{2}) given initial values y_{1}(0) = 2, y_{2}(0) = 3.

Solve ( \begin{matrix}1 & 2 & 2 \\ 1 & 3 & 3 \end{matrix} ) to get c_{1} = 0, c_{2} = 1. So y_{1} = 2e^{−2t} and y_{2} = 3e^{−2t}.

Problem 3:

(a) GIVEN: the eigenvalues of A = ( \begin{matrix}2 & −8 \\ 1 & −4 \end{matrix} ) are 0, −2. Diagonalize A: that is, give matrices

X, X^{−1}, and D such that X^{−1}AX = D with D a diagonal matrix.

Find eigenvectors for 0, say ( \begin{matrix}4 \\ 1 \end{matrix} ); and for −2, say ( \begin{matrix}2 \\ 1 \end{matrix} ).

We can use X = ( \begin{matrix}4 & 2 \\ 1 & 1 \end{matrix} ), X^{−1} = ( \begin{matrix}\frac{1}{2} & −1 \\ −\frac{1}{2} & 2 \end{matrix} ) with D = X^{−1}AX = ( \begin{matrix}0 & 0 \\ 0 & −2 \end{matrix} ).

(b) Let A = ( \begin{matrix}4 & 6 & −2 \\ −1 & −1 & 1 \\ 0 & 0 & 2 \end{matrix} ). GIVEN: the eigenvalues of A are 2, 2, 1.

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

Check that rref(A − 2.I_{3}) has 2 free variables, so the dimension of the 2-eigenspace is 2. However also rref(A − 1.I_{3}) has 1 free variable, so the dimension of the 1-eigenspace is 1. Then A is diagonalizable—since for each eigenvalue, 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 each eigenvalue).

Problem 4: For the symmetric matrix A = ( \begin{matrix}3 & −2 & 4 \\ −2 & 6 & 2 \\ 4 & 2 & 3 \end{matrix} ),

I GIVE you the eigenvalues −2, 7, 7 of A; and an eigenvector ( \begin{matrix}−2 \\ −1 \\ 2 \end{matrix} ) for eigenvalue −2.

(a) Find a basis of the eigenspace of A for eigenvalue 7.

The row-reduced echelon form of A − (0).I = A has (1, \frac{1}{2}, −1) as its only nonzero row.

So eigenvectors are ( \begin{matrix}−\frac{1}{2}b + c \\ b \\ c \end{matrix} ); and one possible basis is ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ) and ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ).

(b) Now find an orthonormal basis for the 7-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. Show WORK (Gram-Schmidt) in obtaining your orthonormal basis (no calculators!)

For −2: eigenspace is 1-dimensional; divide original ( \begin{matrix}−2 \\ −1 \\ 2 \end{matrix} ) by its length 3: x_{3} = \frac{1}{3} ( \begin{matrix}−2 \\ −1 \\ 2 \end{matrix} ).

For 7: Start with above basis like v_{1} = ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ) and v_{2} = ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ).

Apply Gram-Schmidt: first q_{1} = ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ); and then q_{2} = v_{2} − \frac{v_{2}·q_{1}}{q_{1}·q_{1}} q_{1}

= ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ) − ^{\frac{( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} )·( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} )}{( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} )·( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} )}} ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ) = ( \begin{matrix}1 \\ 0 \\ 1 \end{matrix} ) − \frac{−1}{5} ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ) = ( \frac{4}{5} \frac{2}{5} 1 ), so use = ( \begin{matrix}4 \\ 2 \\ 5 \end{matrix} ).

Now divide each by its length, to get x_{1} = \frac{1}{\sqrt{5}} ( \begin{matrix}−1 \\ 2 \\ 0 \end{matrix} ) and x_{2} = \frac{1}{\sqrt{45}} ( \begin{matrix}4 \\ 2 \\ 5 \end{matrix} ).

So now putting x_{3} first, can use X = ( \begin{matrix}−\frac{2}{3} & −\frac{1}{\sqrt{5}} & \frac{4}{\sqrt{45}} \\ −\frac{1}{3} & \frac{2}{\sqrt{5}} & \frac{2}{\sqrt{45}} \\ \frac{2}{3} & 0 & \frac{5}{\sqrt{45}} \end{matrix} ).

Problem 5: (a) Is the matrix ( \begin{matrix}3 & 4 \\ 4 & 1 \end{matrix} ) positive definite? (Why/why not—any method) No; it has a negative eigenvalue −2.1231 …; Or, the principal minor determinants are 3 and 3 · 1 − 4 · 4 = −13, and the latter is negative.

(b) Give the LU-decomposition of the symmetric matrix A = ( \begin{matrix}4 & 2 \\ 2 & 10 \end{matrix} ). Then give the Cholesky decomposition of A—that is, find a matrix B such that A = BB^{T}.

The row operation A_{2}^{− \frac{1}{2} ×1} converts A to the row echelon form ( \begin{matrix}4 & 2 \\ 0 & 9 \end{matrix} ), which we use for U. And then L is the inverse of the matrix for that operation, namely ( \begin{matrix}1 & 0 \\ \frac{1}{2} & 1 \end{matrix} ). That is, the LU-decomposition is given by A = LU: ( \begin{matrix}4 & 2 \\ 2 & 10 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ \frac{1}{2} & 1 \end{matrix} ) ( \begin{matrix}4 & 2 \\ 0 & 9 \end{matrix} ) .

Next, we factor U = DL^{T} ( \begin{matrix}4 & 2 \\ 0 & 9 \end{matrix} ) = ( \begin{matrix}4 & 0 \\ 0 & 9 \end{matrix} ) ( \begin{matrix}1 & \frac{1}{2} \\ 0 & 1 \end{matrix} )

as the product of the diagonal matrix D of its eigenvalues, with L^{T}.

This gives the decomposition A = LDL^{T}: ( \begin{matrix}4 & 2 \\ 2 & 10 \end{matrix} ) = ( \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} ).

We then take the “square root” of D: namely E = ( \begin{matrix}2 & 0 \\ 0 & 3 \end{matrix} );

and finally we set B = LE = ( \begin{matrix}1 & 0 \\ \frac{1}{2} & 1 \end{matrix} ) ( \begin{matrix}2 & 0 \\ 0 & 3 \end{matrix} ) = ( \begin{matrix}2 & 0 \\ 1 & 3 \end{matrix} ).

We then have the Cholesky decomposition A = BB^{T}: ( \begin{matrix}4 & 2 \\ 2 & 10 \end{matrix} ) = ( \begin{matrix}2 & 0 \\ 1 & 3 \end{matrix} ) ( \begin{matrix}2 & 1 \\ 0 & 3 \end{matrix} ).