Solutions to Assignment 9

Transcription

1 Solutions to Assignment 9 Math 7, Fall 5.. Construct an example of a matrix with only one distinct eigenvalue. [ ] a b We know that if A then the eigenvalues of A are the roots of the characteristic equation (a λ)(d λ) bc =. [ This ] equation will have a repeated c d a root if a = d and bc =. So for example, A will work. To give a specific [ ] c a instance of this, A = has only one distinct eigenvalue, λ = Show that if A is the zero matrix, then the only eigenvalue of A is. Suppose that λ is an eigenvalue of A. Then there is an eigenvector v such that Av = λv. Multiplying both sides by A, we get that = A v = Aλv = λav = λ(λv) = λ v. Thus λ v =. Recall that v is non-zero by the definition of an eigenvector. We conclude that λ =, and hence that λ =. 5.. Let A be the matrix of the linear transformation T : R R that rotates points about some line through the origin. Without writing A, find an eigenvalue of A and describe the eigenspace. Suppose we use L to indicate the line about which T rotates the points of R. Then if v R lies on L, we have that T (v) = Av = v (rotating points around a line does not move points on the line). Thus we can see that λ = is an eigenvalue of A. The corresponding eigenspace is either just L if T does not rotate by 6z degrees for z N, and all of R otherwise (in the first case, points end up in a different place then they begin unless they lie on L; in the second, rotation by 6z just takes every point back to its starting position so that every vector is an eigenvector for λ = ).

2 5.. Find the characteristic equation of. λ This amounts to finding det λ. I will do this by cofactor expansion. The answer is λ det det + det = [ λ ] [ ] [ ] λ λ λ λ λ(λ ) ( λ ) + (6 + λ) = λ + λ Find h in the matrix A = h 5 such that the eigenspace for λ = 5 is two-dimensional. The eigenspace corresponding to λ = 5 is the Null space of the matrix A 5I. We can calculate the dimension of Nul(A 5I) row reducing 6 A 5I = h. Going as far as we can without knowing h, we get h/ B = 6 h. Now rank(a) + dim(nul(a)) = rank(b) + dim(nul(b)) =, so unless h = 6 we see that B has pivot columns, and then dim(nul(b)) = dim(nul(a 5I)) <. If h = 6, then rank(b) = so that dim(nul(b)) =. We conclude that h must equal 6 in order for the eigenspace corresponding to λ = 5 to be two-dimensional Let A =..8., v =.6, v =, v =, and w =.... (a) Show that v, v, and v are eigenvectors of A.

3 ..5. To do this we calculate Av =.6, Av =.5, and Av = It is not difficult to see that 6 =.6,.5 = (/), and... = (/5). Thus v, v, and v are eigenvectors of A. (b) Let x be an vector in R with nonnegative entries whose sum is. Explain why there are constants c, c, c such that x = c v + c v + c v. Compute w T x and conclude that c =. We saw in the part (a) that A has distinct eigenvalues (, /, and /5). Thus we know that the three eigenvectors corresponding to these eigenvalues are linearly independent (theorem, pg 7). Because the dimension of R is, we know by the basis theorem that v, v, and v give a basis for R (in particular, they must span R ). Thus there must exist constants c, c, c as required. If we write x = a a a, then w T x = a + a + a =. Now multiplying both sides of the equation x = c v + c v + c v by w T, we obtain = w T x = c w T v + c w T v + c w T v = c () + c () + c (). From this we see that c =. (c) For k =,,..., define x k = A k x with x as in part (b). Show that x k v as k increases. In part (a) we learned that Av = (/)v, so A k v = A k Av = A k (/)v = (/)A k v = (/)A k Av = (/) A k v = = (/) k v. We can similarly show that A k v = (/5) k v. From part (b) we know that x = v + c v + c v. Thus x k = A k x = A k (v + c v + c v ) = A k v + c A k v + c A k v = v + c (/) k v + c (/5) k v. For k, this limits to v Diagonalize A = 6 if possible. One of the eigenvalues is λ = 5, 6 and one eigenvector is.

4 We need to calculate the eigenvalues, and then demonstrate that there are linearly independent eigenvectors. First we form the characteristic equation: 7 λ 6 det 6 λ 6 λ = ( 7 λ) det = ( 7 λ) det A, ( 6) det A, + det A, ([ ]) ([ ]) ([ ]) λ 6 6 λ +6 det + det 6 λ λ 6 = ( 7 λ)(( λ)( λ) + ) + 6(6 6λ + ) + (96 ( λ)) = λ + 7λ + 5λ 75. We know that λ = 5 is a root of this equation, so we can write λ +7λ +5λ 75 = (λ 5)( λ + λ + 5) (you can either do this by inspection, or use long division of polynomials). We can factor this further as (λ 5)( λ+5)(λ+)) = (λ 5) (λ + ). Thus A has two distinct eigenvalues. So now we look for the eigenvectors of A. We do this by row reducing [(A λi) ] for each of the eigenvalues of A. For λ = 5, 7 λ λ = 6 8, 6 λ 6 / / which has row reduced echelon form. From this matrix we can read the eigenspace corresponding to the eigenvalue λ = 5. It is the set of all vectors of the form x x = x = x + x. x We choose two linearly independent eigenvectors from this set, v = and v = (I just cleared denominators). For λ =, 7 λ λ = 6 6, 6 λ 6

5 which has row reduced echelon form /. From this matrix we can read the eigenspace corresponding to the eigenvalue λ =. It is the set of all vectors of the form x = x. x We choose a representative vector from this set, v =. Now row reducing the matrix. The result is, so by the invertible matrix theorem the v i are linearly independent. 5 So let P = [v v v ] = and D = 5. By the diagonalization theorem, A = P DP, so we have diagonalized A Show that if A has n linearly independent eigenvectors, then so does A T. A matrix A has n linearly independent eigenvectors if and only if A is diagonalizable, so it is enough to show that A T is diagonalizable. We are given that A is diagonalizable, so there is a diagonal matrix D and an invertible matrix P such that A = P DP. Taking the transpose of both side yields A T = (P DP ) T = (P ) T D T P T = (P T ) DP T. For the last equality I used that fact that (BC) T = C T B T, and (B ) T = (B T ) (see theorems, pg 5 and 6, pg ). I also used that the transpose of a diagonal matrix D is simply D. So let Q = (P T ). Then our equation says that A T = QDQ, and hence A T is diagonalizable Construct a nondiagonal matrix that is diagonalizable but not invertible. [ ] a b Let A =. Then A will fail to be invertible if ad bc =. So consider the matrix A =. Clearly A is not invertible. Note also that c d [ ] 6 [ ] λ 6 det = λ λ 5λ = λ(λ 5). Thus A has two distinct eigenvectors and hence by theorem (pg 7), two linearly independent eigenvectors. This implies by the Diagonalization theorem that A is diagonalizable. 5