Solutions to Assignment 9

Math 217, Fall 2002

5.1.24 Construct an example of a 2× 2 matrix with only one distinct eigenvalue.

We know that if A

[a bc d

]then the eigenvalues of A are the roots of the char-

acteristic equation (a− λ)(d− λ)− bc = 0. This equation will have a repeated

root if a = d and bc = 0. So for example, A

[a 0c a

]will work. To give a specific

instance of this, A =

[1 01 1

]has only one distinct eigenvalue, λ = 1.

5.1.26 Show that if A2 is the zero matrix, then the only eigenvalue of A is 0.

Suppose that λ is an eigenvalue of A. Then there is an eigenvector v such thatAv = λv. Multiplying both sides by A, we get that 0 = A2v = Aλv = λAv =λ(λv) = λ2v. Thus λ2v = 0. Recall that v is non-zero by the definition of aneigenvector. We conclude that λ2 = 0, and hence that λ = 0.

5.1.32 Let A be the matrix of the linear transformation T : R3 → R3 that rotates points

about some line through the origin. Without writing A, find an eigenvalue ofA and describe the eigenspace.

Suppose we use L to indicate the line about which T rotates the points of R3.Then if v ∈ R3 lies on L, we have that T (v) = Av = v (rotating points arounda line does not move points on the line). Thus we can see that λ = 1 is aneigenvalue of A. The corresponding eigenspace is either just L if T does notrotate by 360z degrees for z ∈ N, and all of R3 otherwise (in the first case,points end up in a different place then they begin unless they lie on L; in thesecond, rotation by 360z just takes every point back to its starting position sothat every vector is an eigenvector for λ = 1).

1

5.2.10 Find the characteristic equation of

0 3 13 0 21 2 0

.

This amounts to finding det

−λ 3 13 −λ 21 2 −λ

. I will do this by cofactor ex-

pansion. The answer is −λ det

[−λ 22 −λ

]− 3 det

[3 21 −λ

]+ det

[3 −λ1 2

]=

−λ(λ2 − 4)− 3(−3λ− 2) + (6 + λ) = −λ3 + 14λ+ 12.

5.2.18 Find h in the matrix A =

5 −2 6 −10 3 h 00 0 5 40 0 0 1

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

A− 5I =

0 −2 6 −10 −2 h 00 0 0 40 0 0 −4

.Going as far as we can without knowing h, we get

B =

0 1 −h/2 00 0 6− h 00 0 0 10 0 0 0

.Now rank(A) + dim(Nul(A)) = rank(B) + dim(Nul(B)) = 4, so unless h = 6 wesee that B has 3 pivot columns, and then dim(Nul(B)) = dim(Nul(A−5I)) < 2.If h = 6, then rank(B) = 2 so that dim(Nul(B)) = 2. We conclude that h mustequal 6 in order for the eigenspace corresponding to λ = 5 to be two-dimensional.

5.2.27 Let A =

.5 .2 .3.3 .8 .3.2 0 .4

, v1 =

.3.6.1

, v2 =

1−32

, v3 =

−101

, and w =

111

.

(a) Show that v1, v2, and v3 are eigenvectors of A.

2

To do this we calculate Av1 =

.3.6.1

, Av2 =

.5−1.5

1

, and Av3 =

−.20.2

.

It is not difficult to see that

.361

=

.3.6.1

,

.5−1.5

1

= (1/2)

1−32

, and−.20.2

= (1/5)

−101

. Thus v1, v2, and v3 are eigenvectors of A.

(b) Let x0 be an vector in R3 with nonnegative entries whose sum is 1. Ex-plain why there are constants c1, c2, c3 such that x0 = c1v1 + c2v2 + c3v3.Compute wTx0 and conclude that c1 = 1.

We saw in the part (a) that A has 3 distinct eigenvalues (1, 1/2, and1/5). Thus we know that the three eigenvectors corresponding to theseeigenvalues are linearly independent (theorem 2, pg 307). Because thedimension of R3 is 3, we know by the basis theorem that v1, v2, and v3

give a basis for R3 (in particular, they must span R3). Thus there mustexist constants c1, c2, c3 as required.

If we write x0 =

a1

a2

a3

, then wTx0 = a1 + a2 + a3 = 1. Now multiplying

both sides of the equation x0 = c1v1 + c2v2 + c3v3 by wT, we obtain1 = wTx0 = c1w

Tv1 + c2wTv2 + c3w

Tv3 = c1(1) + c2(0) + c3(0). From thiswe see that c1 = 1.

(c) For k = 1, 2, . . . , define xk = Akx0 with x0 as in part (b). Show thatxk → v1 as k increases.

In part (a) we learned that Av2 = (1/2)v2, so

Akv2 = Ak−1Av2 = Ak−1(1/2)v2 = (1/2)Ak−1v2 = (1/2)Ak−2Av2 =

(1/2)2Ak−2v2 = · · · = (1/2)kv2.

We can similarly show that Akv3 = (1/5)kv3. From part (b) we knowthat x = v1 + c2v2 + c3v3. Thus xk = Akx0 = Ak(v1 + c2v2 + c3v3) =Akv1 + c2A

kv2 + c3Akv3 = v1 + c2(1/2)kv2 + c3(1/5)kv3. For k →∞, this

limits to v1.

5.3.18 Diagonalize A =

−7 −16 46 13 −212 16 1

if possible. One of the eigenvalues is λ = 5,

and one eigenvector is

−212

.

3

We need to calculate the eigenvalues, and then demonstrate that there are 3linearly independent eigenvectors.

First we form the characteristic equation:

det

−7− λ −16 46 13− λ −212 16 1− λ

= (−7− λ) detA1,1 − (−16) detA1,2 + 4 detA1,3

= (−7−λ) det

([13− λ −2

16 1− λ

])+16 det

([6 −212 1− λ

])+4 det

([6 13− λ12 16

])= (−7− λ)((13− λ)(1− λ) + 32) + 16(6− 6λ+ 24) + 4(96− 12(13− λ))

= −λ3 + 7λ2 + 5λ− 75.

We know that λ = 5 is a root of this equation, so we can write −λ3 +7λ2 +5λ−75 = (λ− 5)(−λ2 + 2λ+ 15) (you can either do this by inspection, or use longdivision of polynomials). We can factor this further as (λ−5)(−λ+5)(λ+3)) =−(λ− 5)2(λ+ 3). Thus A has two distinct eigenvalues.

So now we look for the eigenvectors of A. We do this by row reducing [(A−λI) 0]for each of the eigenvalues of A.

For λ = 5, −7− λ −16 4 06 13− λ −2 012 16 1− λ 0

=

−12 −16 4 06 8 −2 012 16 −4 0

,which has row reduced echelon form

1 4/3 1/3 00 0 0 00 0 0 0

. From this matrix we

can read the eigenspace corresponding to the eigenvalue λ = 5. It is the set ofall vectors of the form

=

−43x2 − 1

3x3

x2

x3

= x2

−43

10

+ x3

−13

01

.We choose two linearly independent eigenvectors from this set, v1 =

430

and

v2 =

−103

(I just cleared denominators).

For λ = −3, −7− λ −16 4 06 13− λ −2 012 16 1− λ 0

=

−4 −16 4 06 16 −2 012 16 4 0

,4

which has row reduced echelon form

1 0 1 00 1 −1/2 00 0 0 0

. From this matrix we

can read the eigenspace corresponding to the eigenvalue λ = −3. It is the setof all vectors of the form −x3

12x3

x3

= x3

−112

1

.We choose a representative vector from this set, v3 =

−212

.

Now row reducing the matrix

4 −1 −23 0 10 3 2

. The result is

1 0 00 1 00 0 1

, so by

the invertible matrix theorem the vi are linearly independent.

So let P = [v1 v2 v3] =

4 −1 −23 0 10 3 2

and D =

5 0 00 5 00 0 −3

. By the diagonal-

ization theorem, A = PDP−1, so we have diagonalized A.

5.3.28 Show that if A has n linearly independent eigenvectors, then so does AT.

A matrix A has n linearly independent eigenvectors if and only if A is diag-onalizable, so it is enough to show that AT is diagonalizable. We are giventhat A is diagonalizable, so there is a diagonal matrix D and an invertiblematrix P such that A = PDP−1. Taking the transpose of both side yieldsAT = (PDP−1)T = (P−1)TDTPT = (PT)−1DPT. For the last equality I usedthat fact that (BC)T = CTBT, and (B−1)T = (BT)−1 (see theorems 3, pg 115and 6, pg 121). I also used that the transpose of a diagonal matrix D is simplyD. So let Q = (PT)−1. Then our equation says that AT = QDQ−1, and henceAT is diagonalizable.

5.3.28 Construct a nondiagonal 2× 2 matrix that is diagonalizable but not invertible.

Let A =

[a bc d

]. Then A will fail to be invertible if ad − bc = 0. So con-

sider the matrix A =

[2 61 3

]. Clearly A is not invertible. Note also that

det

[2− λ 6

1 3− λ

]= λ2− 5λ = λ(λ− 5). Thus A has two distinct eigenvectors

and hence by theorem 2 (pg 307), two linearly independent eigenvectors. Thisimplies by the Diagonalization theorem that A is diagonalizable.

5