M314 Practice Exam 3 Solutions

Note: These are problems I've used on old exams. The actual exam will probably be longer, so be sure also to study the problems on the syllabus, the quizzes, and what was covered in class. To save space, column vectors are sometimes given as the transpose of a row vector.

Instructions: Show all of your work; your work must justify your answers.

[1] (9 points) (a) Explain why you know, without doing any calculation at all, that the matrix A =
[ 4  -4]
[-4  -2]
is diagonalizable.
Answer: A real square symmetric matrix is not only diagonalizable, but orthogonally diagonalizable, by the Spectral Theorem.
(b) Find an orthogonal matrix P and a diagonal matrix D such that P-1AP = D. Also find P-1.
Answer: First find the eigenvalues; they turn out to be 6 and -4. Then find a basis for each eigenspace. Given an eigenvalue l of a matrix A, recall that its eigenspace is the nullspace of A-lI. Here we have two different eigenvalues for a 2x2 matrix A, so each eigenspace is 1-dimensional. Thus we have to find one eigenvector for each eigenvalue. We find [2, -1] for 6 and [1, 2] for -4. By Theorem 5.19, p. 399, these eigenvectors ought to be orthogonal, and they are, which is a good check on our answer. We normalize the eigenvectors by dividing by their length 51/2 to get the columns of P; D is the diagonal matrix whose entries are the eignevalues (taken in the same order as we took their eigenvectors as columns in P). Since P is an orthogonal matrix, P-1 is just the transpose, PT.

[2] (8 points) (a) Find an orthogonal basis for the subspace V of R3 spanned by the vectors w1 = (1,2,1) and w2 = (0,1,2).
Answer: Use Gram-Schmidt to get an orthogonal basis of V. We take v1 = w1 = (1,2,1) as our first basis vector, and v2 = w2 - (v1.w2/v1.v1)v1 = (1/3)(-2, -1, 4) is our second basis vector. As a check, make sure that v1.v2 = 0. Since I'll be using this basis to find projV(1,3,1) in part (b), I'll replace v2 = (1/3)(-2, -1, 4) by v2 = (-2, -1, 4) to simplify my arithmetic. This is OK since it still gives an orthogonal basis of V.
(b) Find the projection of the vector (1,3,1) into the subspace V of part (a).
Answer: Here we just use the formula projV(1,3,1) = (v1.(1,3,1)/v1.v1)v1 + (v2.(1,3,1)/v2.v2)v2. All that we need to apply this formula is that v1 and v2 give an orthogonal basis of V. We get projV(1,3,1) = (4/3)v1 - (1/21)v2.

[3] (8 points) Consider the matrix A =
[0 1 1]
[1 0 1]
[1 1 0]
(a) Find the eigenvalues for A. Show your work!
Answer: First find det(A-tI) = -(t3 - 3t - 2). Now solve t3 - 3t - 2 = 0. Using a graphing calculator, we see that t = -1 seems to be a double root and t = 2 is a root. As a check, we multiply out (t + 1)2(t - 2) to see if we get t3 - 3t - 2, which we do. Thus the eigenvalues are -1, -1 and 2.
(b) What are the algebraic multiplicities and the geometric multiplicities of each of the eigenvalues of A?
Answer: If l is an eigenvalue of a matrix A, its geometric multiplicity is the nullity of A-lI. Its algebraic multiplicity is just the number of times it occurs as a root in the characteristic polynomial. From (a) we see that the algebraic multiplicities are 2 for -1 and 1 for 2. Since an eigenvalue of algebraic multiplicity 1 must also have geometric multiplicity 1, the geometric multiplicity of 2 is 1. The geometric multiplicity of -1 is 2 since A+I has rank 2 (and hence nullity 3-2=1).

[4] (8 points) Let A be the matrix
 [ 3  2  6]
a[-6  b  2]
 [ 2  6  c].
(a) Choose values for a, b and c so that A is an orthogonal matrix.
Answer: Choose b so that column 1 is orthogonal to column 2; thus b = 3. Choose c so that columns 1 and 2 are orthogonal to column 3; thus c = -3. Now choose a to make each column a unit vector; thus a = 1/7.
(b) Using the values of a, b and c that you chose to make A orthogonal, find A-1.
Answer: Since A is orthogonal, A-1 is just AT.