Math 310: Final Exam  (Solutions)  Prof. S. Smith: Tuesday 3 May 2005

Problem 1: Let A = ( \begin{matrix}3 & 2 \\ 4 & 1 \end{matrix} ) . Work by hand; do not use a calculator on this problem.

(Except possibly to check your work.)

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

det(A − xI) = (3 − x)(1 − x) − 4.2 = (x^{2} − 4x + 3) − 8 = x^{2} − 4x − 5 = (x − 5)(x + 1), so eigenvalues are 5, −1.

(b) Find the eigenspaces for those eigenvalues.

For 5: A − 5.I = ( \begin{matrix}−2 & 2 \\ 4 & −4 \end{matrix} ) has rref ( \begin{matrix}1 & −1 \\ 0 & 0 \end{matrix} ), get solutions a(1, 1)^{T}.

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

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

(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 −5, get a(−1, 1)^{T}; for 1, get b(1, 1)^{T}.

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

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

Problem 3:

(a) GIVEN: the eigenvalues of A = ( \begin{matrix}5 & 6 \\ −2 & −2 \end{matrix} ) are 2, 1. 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 2, say (−2, 1)^{T}; and for 1, say (-3,2).

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

(b) Let A = ( \begin{matrix}−4 & −10 & 2 \\ 3 & 7 & −1 \\ 0 & 0 & 1 \end{matrix} ). GIVEN: the eigenvalues of A are 2, 1, 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 1 free variable, so the dimension of the 2-eigenspace is 1.

However also rref(A − 1.I_{3}) has 1 free variable, so the dimension of the 1-eigenspace is only 1. Then A is not diagonalizable—since for the eigenvalue 1, the dimension of the eigenspace is less than the number of times the eigenvalue appears as a root of the characteristic polynomial. (That is, geometric multiplicity < algebraic multiplicity for 1).

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

I GIVE you the eigenvalues 6, 0, 0 of A; and an eigenvector (−2, −1, 1)^{T} for eigenvalue 6.

(a) Find a basis of the eigenspace of A for eigenvalue 0.

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

So eigenvectors are (−\frac{1}{2}b + \frac{1}{2}c, b, c)^{T}; and one possible basis is (−1, 2, 0)^{T} and (1, 0, 2)^{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. Show WORK in obtaining your orthonormal basis (no calculators!)

For 6: eigenspace is 1-dimensional; divide original (−2, −1, 1) by its length \sqrt{6}: x_{3} = \frac{1}{\sqrt{6}}(−2, −1, 1)^{T}. For 0: Start with above basis like v_{1} = (−1, 2, 0)^{T} and v_{2} = (1, 0, 2)^{T}. Apply Gram-Schmidt: first q_{1} = (−1, 2, 0), and then q_{2} = v_{2} − \frac{v_{2}·q_{1}}{q_{1}·q_{1}} q_{1} = (1, 0, 2)^{T} − \frac{−1}{5}(−1, 2, 0)^{T} = (\frac{4}{5}, \frac{2}{5}, 2)^{T}

so may as well use the more convenient multiple q_{2} = (2, 1, 5)^{T}. Now divide each by its length, to get x_{1} = \frac{1}{\sqrt{5}}(−1, 2, 0)^{T} and x_{2} = \frac{1}{\sqrt{30}}(2, 1, 5)^{T}.

So now putting x_{3} first, can use X = ( \begin{matrix}−\frac{2}{\sqrt{6}} & −\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{30}} \\ −\frac{1}{\sqrt{6}} & \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{30}} \\ \frac{1}{\sqrt{6}} & 0 & \frac{5}{\sqrt{30}} \end{matrix} ).

(c) Give the projection matrices into the two eigenspaces for A. (Calculators OK on this part.) Just compute UU^{T}, where the columns of U are the orthonormal basis from (b):

for 6, \frac{1}{6} ( \begin{matrix}4 & 2 & −2 \\ 2 & 1 & −1 \\ −2 & −1 & 1 \end{matrix} ) and for 0, \frac{1}{6} ( \begin{matrix}2 & −2 & 2 \\ −2 & 5 & 1 \\ 2 & 1 & 5 \end{matrix} )

Problem 5: (a) Can you find an orthonormal basis of eigenvectors for A = ( \begin{matrix}1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix} ) ?

(Why/why not?)

No; the matrix is not normal (does not commute with its transpose).

So the separate eigenspaces are not orthogonal.

(b) For the Markov matrix A = ( \begin{matrix}.8 & .1 \\ .2 & .9 \end{matrix} ), I GIVE you that the eigenvalues are 1 and .7.

Give a formula for the nth power A^{n}.

Compute eigenvectors for 1, say (1, 2)^{T}; and for .7, say (−1, 1)^{T}.

So for X = ( \begin{matrix}1 & −1 \\ 2 & 1 \end{matrix} ) we have X^{−1} = \frac{1}{3} ( \begin{matrix}1 & 1 \\ −2 & 1 \end{matrix} ) and X^{−1}AX = D where D = ( \begin{matrix}1 & 0 \\ 0 & .7 \end{matrix} ).

So A = XDX^{−1} and hence A^{n} = XD^{n}X^{−1} = \frac{1}{3} ( \begin{matrix}1 + 2(.7)^{n} & 1 − (.7)^{n} \\ 2 − 2(.7)^{n} & 2 + (.7)^{n} \end{matrix} ).