Math 416 Midterm 2. Solutions.
1. (a) Let A be a matrix and set B = αA. Show that B has the same eigenvectors as A. Show
that each eigenvalue of B is an eigenvalue of A times α.
(b) Now let C = A2 . Show that C has the same eigenvectors as A. Show that the eigenvalues
of C are the squares of the eigenvalues of A.
Solution: This is homework problem 8.5a plus homework problem 9.2.
(a) Assume that Ax = λx. Then
Bx = αAx = αλx,
so x is an eigenvector of B with eigenvalue αλ.
(b) We also have
Cx = A2 x = A(Ax) = A(λx) = λAx = λ · λx = λ2 x.
So x is an eigenvector of C with eigenvalue λ2 .
2. If a, b, c ∈ R, with c 6= a, and
A=
a b
0 c
,
then compute A10 , A100 , and A1000 .
Solution: This is a modified version of homework problem 9.4.
We first diagonalize!
The eigenvalues are a, c since the matrix is upper triangular. We now obtain the eigenvectors:
λ = a. We have
0
b
0 c−a
a−c b
0
0
A − aI =
and we have (1, 0) as an eigenvector.
λ = c. We have
A − cI =
,
and thus the first equation is (a − c)x + by = 0, and we can choose (b, c − a) as an eigenvector.
Therefore, we have
A = QDQ−1 ,
with
Q=
1
b
0 c−a
,
D=
a 0
0 c
.
We can compute

b
1
 1 −c − a 
·
=
.
1
c−a
0
c−a

Q−1 =
c − a −b
0
1
Since Ak = QDk Q−1 , we have

b
k
 1 −c − a 
b
a 0 



c−a
0 ck 

1
0
c
−a

b
 1 −
bck

c−a 






1
(c − a)ck
0
 c−a
b
(ak − ck ) 
.
c−a
k
c

k
1
0

ak
A =
=
0

=
ak
0
3. For each of the following matrices, either find Q, D so that A = QDQ−1 , or show that this
cannot be done.
3 2
4 −1
1
1
(a) A =
(b) A =
(c) A =
2 3
4 0
−2 −1
Solution: This is homework problem 9.1 with different numbers.
(a) The characteristic polynomial is t2 − 6t + 5 = (t − 5)(t − 1), so the eigenvalues are 5, 1.
Since
−2 2
A − 5I =
,
2 −2
we have v1 = (1, 1) and since
A−I =
2 2
2 2
,
we have v2 = (1, −1), so
Q=
1 1
1 −1
,
D=
5 0
0 1
.
(b) The characteristic polynomial is t2 − 4t + 4 = (t − 2)2 , so we have a repeated eigenvalue
of 2.
We see that
2 −1
A − 2I =
4 −2
has only a one-dimensional nullspace, so the matrix is nondiagonalizable, and we cannot
find Q, D.
(c) The characteristic polynomial is t2 + 1, which has roots λ± = ±i.
λ+ = i. We have
A − iI =
1−i
1
−2 −1 − i
and looking at the first row, we see that we need to solve
(1 − i)x + y = 0,
so the vector v+ = (1, −1 + i) will work.
We showed in class that since A is real, both the eigenvalues and eigenvectors come in
complex conjugate pairs, so with no more work we can write v− = (1, −1 − i). Then we
have
1
1
i 0
Q=
, D=
.
−1 − i −1 + i
0 −i
4. Let A be the 2 × 2 matrix
A=
a b
c d
.
Show that
det(A − tI) = t2 − (TrA)t + det A.
Determine the conditions under which are there two distinct real numbers t1 , t2 such that
det(A − t1 I) = det(A − t2 I) = 0.
Solution: This is homework problem 7.11 exactly, see solutions there.
5. For each of the following matrices, determine whether or not the matrix is invertible. If it is,
compute its inverse.
1 0
1 2
4 1
(a) A =
(b) A =
(c) A =
0 2
3 6
2 1
Solution: We will first use the determinant to determine whether it is invertible, and if it is
we will then use row reductions to put it in RREF, taking care to apply the same operations
to the identity matrix. As we showed in class, this is guaranteed to give us A−1 when A is
invertible.
(a) The determinant is 2, so it is invertible. To turn this into the identity, we need to divide
the second row by 2. Applying that operation to the identity gives us
1 0
−1
A =
.
0 1/2
(b) The determinant is zero, so not invertible.
(c) The determinant is 2, so it is invertible. We perform row operations:
0 1
0 1
0 1
2
1
2
1
2
1
4 1 1
0
,
,
,
,
2 1 0 1
4 1 1 0
0 −1 1 −2
0 1 −1 2
1/2 −1/2
2 0 1
−1
1
0
,
,
0 1 −1 2
0 1 −1
2
so
−1
A
=
1/2 −1/2
−1
2
0 1
6. Let A =
. Use the Cayley–Hamilton Theorem to show that A2 is a linear combination
1 3
of A and I, and determine the coefficients in this linear combination. Use this relation to
compute A4 .
Solution: The characteristic polynomial of this matrix is t2 − 3t − 1. The Cayley–Hamilton
Theorem tells us that
A2 − 3A − I = 0.
From this we can deduce that A2 = 3A + I.
We can also compute
A4 = (A2 )2 = (3A + I)2 = 9A2 + 6A + I = 9(3A + I) + 6A + I = 33A + 10I.
Thus
4
A = 33
0 1
1 3
+ 10
1 0
0 1
=
10 33
33 109
7. Consider the linear recursion relation
xn+2 = 3xn+1 − 2xn ,
x0 = 1,
x1 = 2.
Compute x100 .
Solution: This is Problem 9.6 with different numbers.
Writing yn = xn+1 , we have
yn+1 = xn+2 = 3xn+1 − 2xn = 3yn − 2xn ,
.
so we have the recursive formula
xn+1
yn+1
=
0 1
−2 3
From this it follows that
xn
yn
=
0 1
−2 3
n xn
yn
1
2
.
0 1
Let A =
. The characteristic polynomial of A is t2 − 3t + 2, so the eigenvalues are
−2 3
λ1 = 1, λ2 = 2. We see that
−1 1
A−I =
,
−2 2
so the corresponding eigenvector is (1, 1). We also have that
−2 1
A − 2I =
,
−2 1
so the corresponding eigenvector is (1, 2).
Now, we could use this information to diagonalize A and compute An , but notice that we only
care about An applied to the vector (1, 2), and this vector is an eigenvector! So we have
n n 0 1
1
xn
1
2
n
=
=2
.
=
yn
−2 3
2
2n+1
2
Therefore xn = 2n .
8. Let T : V → V be linear.
(a) Show that T 2 : V → V is also a linear transformation.
(b) Show that T 2 is the zero transformation if and only if R(T ) ⊆ N (T ).
(c) Give an example of a T such that T is not the zero transformation, but T 2 is.
Solution: This is homework problem 6.4, so see solutions there.
9. Show that if A and B are similar, then they have the same eigenvalues.
Solution: If A and B are similar, then there is a Q such that A = QBQ−1 . Then we have
det(A − tI) = det(QBQ−1 − tI) = det(QBQ−1 − tQIQ−1 ) = det(Q(B − tI)Q−1 ).
Since determinant is multiplicative,
det(Q(B −tI)Q−1 ) = det(Q) det(B −tI) det(Q−1 ) = det(Q) det(B −tI)
1
= det(B −tI),
det(Q)
so det(A − tI) = det(B − tI). Since A and B have the same characteristic polynomial, then
they have the same eigenvalues.
10. In each case, you will be given two numbers, n and r, and in each case you should write down
a linear map T : R3 → R3 such that the nullity of T is n and the rank of T is r, or explain
why such a map does not exist.
(a) n = 3, r = 0
(b) n = 2, r = 2
(c) n = 1, r = 2
(d) n = 0, r = 3
Solution: First recall that the Rank–Nullity theorem tells us that r+n = 3. So (b) is impossible
from the get-go. Moreover, it tells us in the three other cases, we only need to get n correct,
and r comes along for the ride.
(a) If n = 3, then dim(N (T )) = dim(V ), so N (T ) = V . Thus we choose T to be the zero
transformation.
(b) Unpossible
(c) We need a transformation with a one-dimensional nullspace. Multiplying by the matrix


1 2 3
 4 5 6 
0 0 0
will certainly work.
(d) Since n = 0, this transformation is invertible. We could choose T to be the identity.