Math 317: Linear Algebra Homework 11 (Extra Credit)
Due: December 11, 2015
The following problems are for additional practice and are not to be turned in: (All
problems come from Linear Algebra: A Geometric Approach, 2nd Edition by ShifrinAdams.)
Turn in the following problems.
1. Section 6.1, Problem 1c
10 −6
We want to find the eigenvalues and eigenvectors for A =
. To begin,
18 −11
we solve the characteristic equation det(A − λI) = 0 =⇒
10 − λ
−6
det
= 0 =⇒ (10 − λ)(−11 − λ) − (18)(−6) = 0 =⇒
18
−11 − λ
(λ + 2)(λ − 1) = 0 =⇒ λ1 = −2, λ2 = 1. To find the corresponding eigenvectors,
we find bases for N (A − (−2)I) and N (A − 1I). To this extent, we obtain
12 −6
2 −1
A + 2I =
→
=⇒ 2x1 − x2 = 0, x2 = x2 =⇒ x1 =
18 −9
0 0 1/2
(1/2)x2 , x2 = x2 =⇒ v1 =
is an eigenvector for λ1 . The second eigenvec1
tor comes from N (A − I) and so
9 −6
3 −2
2/3
A−I =
→
=⇒ x1 = (2/3)x2 , x2 = x2 =⇒ v2 =
is
18 −12
0 0
1
the second eigenvector that corresponds to λ2 .
2. Section 6.1, Problem 2
Proof : Recall that A is singular if and only if det(A) = 0 which is true if and only
det(A − 0I) = 0 which is true if and only 0 is an eigenvalue for A. (Alternatively,
you could have argued that A is singular if and only Ax = 0 has a nontrivial
solution which implies that Ax = 0x for some nonzero x which implies that 0 is
an eigenvalue with the corresponding eigenvector x.)
3. Section 6.1, Problem 4
Recall that if A is the projection matrix that cooresponds to a subspace V ⊂ Rn
then Av = v if v ∈ V and Av = 0 if v ∈ V ⊥ . Since Rn = V + V ⊥ we know that
all vectors in Rn will either be in V or V ⊥ . Thus Av = 1v or Av = 0v. This
tells us that λ1 = 1 and λ2 = 0. The eigenvectors that correspond to λ1 = 1
come from the nonzero vectors in V = C(A). The eigenvectors that correspond
to λ2 = 0 come from the nonzero vectors in V ⊥ = N (AT ). Similarly, we find the
eigenvalues and eigenvectors for a reflection matrix in the same manner. Noting
that R(x) = projV (x) − projV ⊥ (x) for all x ∈ Rn , we have that Ax = x if
x ∈ V and Ax = −x if x ∈ V ⊥ . Thus the eigenvalues of A are 1, −1 and the
eigenvectors that correspond to 1 are given by those nonzero vectors in V and
1
Math 317: Linear Algebra Homework 11 (Extra Credit)
Due: December 11, 2015
the eigenvectors that correspond to −1 are given by those nonzero vectors in V ⊥
4. Section 6.1, Problem 7
(a) Proof : Suppose that x is an eigenvector for A that corresponds to an
eigenvalue λ. Then Ax = λx. Multiply both sides by A to obtain
A2 x = λ(Ax) = λ(λx) = λ2 x. If we continue this process by multiplying A n − 1 times to A, we obtain An x = λn x which implies that λn is an
eigenvalue of An with the corresponding eigenvector x.
(b) Proof : We claim that x is an eigenvector of A + I. This is true since
(A + I)x = Ax + Ix = λx + x = (λ + 1) x which implies that x is an
eigenvector of A + I with the corresponding eigenvalue λ + 1.
(c) Proof : This claim is true. Let x be an eigenvector of B with the corresponding eigenvalue µ. Recalling that x is an eigenvector for A that corresponds
to an eigenvalue λ, we have that (A+B)x = Ax+Bx = λx+µx = (λ+µ)x.
10 −6
3 1
(d) This claim is false. To see why, let A =
and B =
. An
18 −11
−3 7
eigenvalue for A is given by λ =1. An eigenvalue forB is given
by λ = 4.
13 −5
8 −5
We see that A + B =
. (A + B) − 5I =
. Note that
15 −4
15 −9
det((A + B) − 5I)) = 8(−9) − (15)(−5) 6= 0 which means that (A + B) − 5I
is nonsingular. Hence 5 = 1 + 4 is not an eigenvalue for A + B.
5. Section 6.1, Problem 10
(a) Proof : It is sufficient to show that A and AT have the same characteristic
equation. Recalling that det(A) = det(AT ) we have that det(A − λI) =
det((A − λI)T ) = det(AT − (λI)T ) = det(AT − λI). So A and AT have the
same characteristic equation and hence they have the same eigenvalues.
1 1
(b) This claim is false. To see why, let A =
. Then A has eigenvalues
0 0
1
1
0, 1 with associated eigenvectors v1 =
, v2 =
. However, AT has
−1
0
0
1
eigenvectors w1 =
and w2 =
.
1
1
6. Section 6.2, Problem 2
(a) Proof : Suppose that A is an n × n matrix with n distinct eigenvalues. Then
the algebraic multiplicity of each eigenvalue is 1. That is, alg mult(λi ) = 1
for all i = 1, . . . , n. Recalling that alg mult(λi ) ≥ geo mult(λi ) and that
geo mult(λi ) > 0, we have that geo mult(λi ) = 1 for each i = 1, . . . n. Thus,
alg mult(λi ) = geo mult(λi ) for each i and hence A is diagonalizable.
2
Math 317: Linear Algebra Homework 11 (Extra Credit)
Due: December 11, 2015
1 0
(b) This claim is false. To see why let A =
which is a diagonal matrix
0 1
2 1
and hence is diagonalizable. Let B =
. We have shown that B is not
0 2
diagonalizable. Now, AB = BA but B is not diagonalizable, even though
A is diagonalizable.
(c) Proof : To show that A and B have the same eigenvalues, it is sufficient to
show that A and B have the same characteristic equation. Using the fact
that A = P −1 BP we have that
det(A − λI) = det(P −1 BP − λI) = det(P −1 BP − P −1 λIP ) = det(P −1 (B −
λI)P ) = det(P −1 ) det(B − λI) det(P ) = det(B − λI). Thus A and B
have the same characteristic polynomial and hence A and B have the same
eigenvalues.
2 0
2 1
(d) This is not true. To see why, let A =
and B =
. Then A and
0 2
0 2
B have the same eigenvalues but are not similar to each other as we have
shown in a previous homework assignment.
1
1
(e) This statement is false. Let A =
. Then A2 = −I.
−2 −1
(f) Proof : Suppose that A and B are diagonalizable with the same eigenvalues
and algebraic multiplicities. Then A and B are similar to the same diagonal
matrix D. That is, there is an invertible P such that A = P DP −1 and an
invertible matrix Q such that B = QDQ−1 . Thus Q−1 BQ = D and hence
A = P (Q−1 BQ)P −1 . This implies that A = (QP −1 )−1 B(QP −1 ), and hence
A is similar to B.
7. Section 6.2, Problem 11
(a) Proof : Suppose that A is an n × n matrix with the property that A2 =
A. Furthermore, suppose that λ is an eigenvalue of A. Then Ax = λx.
Multiplying both sides by A we obtain A2 x = A(Ax) = A(λx) = λ2 x =
Ax = λx, where we have used the fact that A2 = A. Thus λ2 x = λx =⇒
(λ2 − λ)x = 0 =⇒ λ2 − λ = 0 =⇒ λ(λ − 1) = 0 =⇒ λ = 0 or λ = 1.
(b) By definition, the eigenvectors associated with λ = 0 is given by N (A).
The eigenvectors associated with λ = 1 is given by N (A − I) =⇒ Ax =
1x =⇒ x ∈ C(A) by exercise 3.2.13. Since C(A) ∩ N (A) = {0} and
C(A) + N (A) = Rn , then the eigenvectors from a linearly independent
spanning set for Rn . That is, the eigenvectors of A form a basis for Rn and
hence A is diagonalizable.
8. Problem 6.3, Problem 1
3
Math 317: Linear Algebra Homework 11 (Extra Credit)
Due: December 11, 2015
We wish to calculate Ak for any k ≥ 1. To begin, we find the
eigenvalues
2−λ
5
for A by solving det(A − λI) = 0 =⇒ det
= 0 =⇒
1
−2 − λ
(2 − λ)(−2 − λ) − (5) = 0 =⇒ (λ + 3)(λ − 3) = 0 =⇒ λ1 = −3, λ2 = 3. Next
5 5
we find the corresponding eigenvectors. For λ = −3, A − (−3I) =
=⇒
1 1
1
−1 5
x1 = −x2 , x2 = x2 =⇒ v1 =
. Similarly, A − 3I =
=⇒ x1 =
−1
1
−5
5
5x2 , x2 = x2 =⇒ v2 =
. Recalling that Ak = P Dk P −1 where the columns
1
of P contain the eigenvectors of A, we have that
k
−1
5 1
3
0
5 1
A =
1 −1 0 (−3)k 1 −1
k
4