Math 310 (2 pm MWF):  Final Exam  (Solutions)

Prof. S. Smith (originally Prof. H. Alexander): Tues 4 May 1999 There are 6 problems, worth 20 % each (so you can get more than 100 %). You must SHOW WORK to get credit; if by calculator, show WHERE and HOW you used it.

Problem 1: (a) (as promised, a flashback to the midterm):

Let S be the set of all lower triangular 2 × 2 matrices. Remember that these matrices have the

general form ( \begin{matrix}a & 0 \\ b & c \end{matrix} ). Show that S is a subspace of the vector space R^{2×2} of all 2 × 2 matrices.

(closure, +:) ( \begin{matrix}a & 0 \\ b & c \end{matrix} ) + ( \begin{matrix}d & 0 \\ e & f \end{matrix} ) = ( \begin{matrix}a + d & 0 \\ b + e & c + f \end{matrix} ), also lower-triangular, so in S.

(closure, scalar mult:) f ( \begin{matrix}a & 0 \\ b & c \end{matrix} ) = ( \begin{matrix}fa & 0 \\ fb & fc \end{matrix} ), also lower-triangular, so in S.

(b) Give the standard matrix representation for linear transformation from R^{3} to R^{2} defined by L[(x_{1}, x_{2}, x_{3})^{T}] = (x_{1} −3x_{2} +2x_{3}, 7x_{2} +3x_{3})^{T}. (That is, just use the standard basis for each space).

Apply L to each basis vector, and write as columns: ( \begin{matrix}1 & −3 & 2 \\ 0 & 7 & 3 \end{matrix} )

Problem 2: (a) Find the orthogonal complement S^{⊥} to the space spanned by the two vectors (1, 1, 1)^{T} and (1, 0, 2)^{T}.

Write as rows of A: ( \begin{matrix}1 & 1 & 1 \\ 1 & 0 & 2 \end{matrix} ). Then Ax = 0 has solutions x_{3}(−2, 1, 1)^{T}.

(b) Find the least-squares solution x̂ of the inconsistent system Ax = b given by: ( \begin{matrix}0 & 1 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 4 \end{matrix} ).

Give the projection p of b in the column space of A. What is the error?

Multiply by A^{T} = ( \begin{matrix}0 & 1 & 2 \\ 1 & 1 & 1 \end{matrix} ) to get normal equations ( \begin{matrix}5 & 3 & 10 \\ 3 & 3 & 7 \end{matrix} ).

Solve to get x̂ = \frac{1}{6}(9, 5)^{T}. Then p = Ax̂ = \frac{1}{6}(5, 14, 23)^{T}. And error = b − p = \frac{1}{6}(1, −2, 1)^{T}.

Problem 3: (a) Use the Gram-Schmidt process to find an orthonomal basis for the space spanned by the two vectors b_{1} = (0, 1, 2)^{T} and b_{2} = (1, 1, 1)^{T}.

(Working with rows) x_{1} = (0, 1, 2) and x_{2} = b_{2} − \frac{x_{2}·x_{1}}{x_{1}·x_{1}} x_{1} = (1, 1, 1)− = \frac{3}{5}(1, 1, 1) = (1, \frac{2}{5}, −\frac{1}{5}) so u_{1} = u_{1}/|u_{1}| = \frac{1}{\sqrt{5}}(0, 1, 2) and u_{2} = \frac{1}{\sqrt{30}}(5, 2, −1).

(b) Find the projection p of the vector b = (1, 2, 3)^{T} in the 2-dimensional subspace spanned by the vectors b_{1} = (1, −1, 0)^{T} and b_{2} = (1, 1, 1)^{T}.

Many ways of solving. Easiest: note b_{1} and b_{2} are orthogonal.

So p has coordinates (b · b_{1})/(b_{1} · b_{1}) = −\frac{1}{2} and similarly for b_{2} to get 2.

so p = −\frac{1}{2}(1, −1, 0)^{T} + 2(1, 1, 1)^{T} = \frac{1}{2}(3, 5, 4)^{T}.

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

(a) Find the eigenvalues of A.

det(A − xI) = x^{2} − 5x + 4 − 4 = x^{2} − 5x = x(x − 5), so eigenvalues are 0, 5.

(b) For each eigenvalue, determine the corresponding eigenvectors.

(λ = 0) For A − 0I = A solve Ax = 0: solutions are multiples of (−2, 1)^{T}.

(λ = 5) solve (A − 5I)x = 0 where A − 5I = ( \begin{matrix}−4 & 2 \\ 2 & −1 \end{matrix} ) solutions are multiples of (1, 2)^{T}.

Problem 5: Given the differential equation system (y as 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 are 3, −1; with eigenvectors are (1, 1)^{T} and (1, −1)^{T}.

(a) Give the general solution of the system (with undetermined constants c_{1}, c_{2}).

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

(b) Now determine the values of c_{1}, c_{2} for the initial value problem y_{1}(0) = 4, y_{2}(0) = 2.

Solve ( \begin{matrix}1 & 1 & 4 \\ 1 & −1 & 2 \end{matrix} ) to get c_{1} = 3, c_{2} = 1. So y_{1} = 3e^{3t} + e^{−t} and y_{2} = 3e^{3t} − e^{−t}.

Problem 6: Let A be the matrix ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ). I GIVE you that the eigenvalues of A are −1, 1.

(a) Diagonalize A: that is, find the eigenvalues and eigenvectors, and give a matrix X such that X^{−1}AX is a diagonal matrix D.

For 1: eigenvectors are a(1, 1)^{T}, For -1: Get eigenvectors b(1, −1)^{T}. So X = ( \begin{matrix}1 & 1 \\ 1 & −1 \end{matrix} ).

(b) Now let A be the symmetric matrix ( \begin{matrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} ).

I GIVE you that the eigenvalues of A are −1, −1, 2; and that corresponding eigenvectors are (1, −1, 0)^{T}, (1, 0, −1)^{T}, and (1, 1, 1)^{T}. Give an orthogonal diagonalization of A:

That is, use Gram-Schmidt to find an orthonormal basis for each eigenspace. Then use that to build an orthogonal matrix X (that is, X^{−1} = X^{T}) with X^{−1}AX diagonal.

Apply Gram-Schmidt to the −1-eigenspace to get \frac{1}{\sqrt{2}}(1, −1, 0)^{T} and \frac{1}{\sqrt{6}}(1, 1, −2)^{T}.

For the 2-eigenspace, use \frac{1}{\sqrt{3}}(1, 1, 1)^{T}.

So can use X = ( \begin{matrix}\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}} & \frac{1}{\sqrt{3}} \end{matrix} ).