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Math 60 – Linear Algebra
Solutions to Homework 5
8.1 #7a We compute

6−λ
−24
−10 − λ
det  2
1
4
the determinant of A − λI by expanding along the first row:

−4
−2  = (6−λ) [(−10 − λ)(1 − λ) + 8]+24 [2 − 2λ + 2]−4 [8 + 10 + λ]
1−λ
= (6 − λ)(λ2 + 9λ − 2) + (−48λ + 96) + (−4λ − 72) = −λ3 − 3λ2 + 4λ + 12 = 0
Factoring, we get −(λ − 2)(λ + 2)(λ + 3) = 0. So the eigenvalues of this matrix are 2, −2
and −3. Now we find the eigenvectors.
λ = 2: We must find the solutions to the homogeneous system of linear equations (A−2I)~x =
~0.








4 −24 −4 0
1 −6 −1 0
1 −6 −1 0
1 −6 −1 0
2 −12 −2 0 −→ 1 −6 −1 0 −→ 0 0
0 0 −→ 0 1
0 0
1 4 −1 0
1 4 −1 0
0 10 0 0
0 0
0 0


1 0 −1 0
−→ 0 1 0 0
0 0 0 0
 
 
r
1



So the eigenvectors of A with eigenvalue 2 are all vectors of the form 0 = r 0 with
r
1
 
1

r 6= 0. In other words, all nonzero multiples of 0.
1
λ = −2: Again, we find solutions to the homogeneous system of equations (A + 2I)~x = ~0.




1 0 1 0
8 −24 −4 0
2 −8 −2 0 −→ · · · −→ 0 1 1/2 0
0 0 0 0
1 4
3 0


−1
So the eigenvectors of A with eigenvalue −2 are all nonzero multiples of −1/2.
1
λ = −3: a similar
  computation will show that the associated eigenvalues are all nonzero
−4
multiples of −2.
3
8.1 #12 Suppose ~v ∈ V is a nonzero vector (nonzero so that it fits the first eigenvector
criterion). Then idV (~v ) = ~v = 1 · ~v . Therefore, ~v is an eigenvector of idV with eigenvalue 1.
This is the same as saying that all nonzero vectors are eigenvectors of the identity matrix,
with eigenvalue 1. (Note that we used axiom 8 in our calculations).
8.1 #13 We wish to show that the vector P −1~v is an eigenvector of the matrix P −1 AP ,
with the same eigenvalue λ. (P −1 AP )(P −1~v ) = (P −1 A)(P P −1 )(~v ) (by associativity of matrix
multiplication) = (P −1 A)(In )(~v ) (by the defining property of the inverse) = (P −1 A)(~v ) (since
the identity matrix is a multiplicative identity element among matrices) = (P −1 )(A~v ) (again
by associativity) = P −1 (λ~v ) (since we’re given that ~v is an eigenvector of A with eigenvalue
λ) = λP −1~v (since λ is a scalar). So we have proved that
(P −1 AP )(P −1~v ) = λ(P −1~v )
8.1 # 15a A2~v = (AA)~v = A(A~v ) (by associativity) = A(λ~v ) (since we’re given that ~v is an
eigenvector of A with eigenvalue λ) = λ(A~v ) (since λ is a scalar) = λ(λ~v ) (again using the
given fact that ~v is an eigenvector of A with eigenvalue λ) = λ2~v (using the associativity of
scalar multiplication axiom), so that ~v is an eigenvector of A2 with associated eigenvalue λ2 .
8.1 # 15b Using a similar technique, but iterated k times, we can show that Ak~v =
λ(Ak−1~v ) = λ2 (Ak−2~v ) = · · · = λk~v , so that ~v is an eigenvector of Ak with associated
eigenvalue λk .
8.1 #15c We proved this in class! Briefly, since A~v = λ~v , we multiply both sides by A−1 to
get ~v = λA−1~v , and now we multiply by 1/λ (we know λ 6= 0) to get A−1~v = (1/λ)~v , so that
~v is an eigenvector of A−1 with eigenvalue 1/λ. Applying part b above, we can show that ~v
is an eigenvector of (A−1 )k with eigenvalue 1/λk .
8.3 #2a We found three linearly independent eigenvectors in #7a above (since, in part,
there were three distinct eigenvalues) so we know that this matrix is diagonalizable, and
−1


 

1 −1 −4
2 0
0
1 −1 −4
6 −24 −4
2 −10 −2 = 0 −1/2 −2 0 −2 0  0 −1/2 −2
1
1
3
0 0 −3
1
1
3
1 4
1
8.3 #7a Since A = A2 , then any eigenvalue of A will also be an eigenvalue of A2 . But, from
problem 15a above, we know that if λ is an eigenvalue of A, then λ2 is an eigenvalue of A2 .
Combining these two observations, we see that any eigenvalue λ of A must satisfy λ2 = λ,
or λ(λ − 1) = 0, so that λ must equal either 0 or 1.
8.3 #7b Suppose ~v is a vector in Rn . If A~v = ~0, then ~v is an eigenvector of A with eigenvalue
0. If not, observe that A(A~v ) = A2~v (by associativity of matrix multiplication) = A~v (by
the given fact that A2 = A) = 1 · (A~v ) (by axiom 8) so that A~v is an eigenvector of A
with eigenvalue 1. Now observe that A(~v − A~v ) = A~v − A2~v (by distributivity of matrix
multiplication over matrix addition) = A~v −A~v (since we’re given A2 = A) = ~0 = 0·(~v −A~v )
(since 0 times any vector is the zero vector). So (~v −A~v ) is an eigenvector of A with eigenvalue
0. That proves the asserion, since A~v + (~v − A~v ) = ~v .
8.3 #7c For A to be diagonalizable, there must be a set of eigenvectors which form a basis
for Rn . Now suppose that {~v1 , · · · , ~vk } is a basis for EA (1) and {w
~ 1, · · · , w
~ l } is a basis
for EA (0). In particular, the ~vi ’s and w
~ j ’s are eigenvectors of A with eigenvalues 1 and 0,
respectively. Now if ~v ∈ Rn is any vector, we’ve shown that
A~v = a1~v1 + · · · + ak~vk
(since A~v ∈ EA (1)), and
(~v − A~v ) = b1 w
~ 1 + · · · bl w
~l
(since (~v − A~v ) ∈ EA (0)), and so
~v = a1~v1 + · · · + ak~vk + b1 w
~ 1 + · · · bl w
~l
so that the set {~v1 , · · · , ~vk , w
~ 1, · · · , w
~ l } spans Rn , and by the contraction theorem, we can
find a subset which forms a basis for Rn . That basis would consist solely of eigenvectors of
A, proving that A is indeed diagonalizable.