Download Exam 3 Solutions

Math 220
Exam 3 Solutions
Spring ’99
3 1
1. Let A =
.
0 3
a) What is the characteristic polynomial chA (λ)?
λ − 3 −1 = (λ − 3)2 .
Solution: chA (λ) = 0
λ − 3
b) For each eigenvalue of A, find a basis for the associated eigenspace.
Solution: The only eigenvalue
is 3. The eigenspace for 3 is
0 1
the nullspace of A − 3I =
, which is already reduced.
0 0
There is one free variable: x1 , so we set x1 = t. The first row
of A − 3I gives x2 = 0. Thus, a vector x is in the eigenspace
of 3 if
t
1
x=
=t
,
0
0
1
so the eigenspace is one-dimensional with basis
.
0
c) Is A diagonalizable? If so, find a matrix P such that P −1 AP
is diagonal, and display the diagonal matrix P −1 AP .
Solution: The geometric multiplicity of 3 is 1 and the algebraic multiplicity is 2. Since these are not equal, A is not
diagonalizable.
2
Exam 3 Solutions


7 0 −10
2. Let A = 5 2 −10. Then the characteristic polynomial of A
5 0 −8
2
is (λ − 2) (λ + 3).
a) For each eigenvalue of A, find a basis for the associated eigenspace.
Solution: The eigenspace of 2 is the nullspace of


5 0 −10
A − 2I = 5 0 −10 .
5 0 −10


1 0 −2
This reduces to 0 0 0 . There are two free variables: x2
0 0 0
and x3 . We set x2 = s and x3 = t. The first row of the
reduction gives x1 = 2x3 = 2t, so x is in the eigenspace if
 
 
 
2t
2
0





x = s = t 0 + s 1 .
t
1
0
   
2
0



Thus, a basis for the eigenspace of 2 is given by 0 , 1.
1
0


10 0 −10
The eigenspace for −3 is the nullspace of A+3I =  5 5 −10.
5 0 −5


1 0 −1

This reduces to 0 1 −1, which has one free variable: x3 .
0 0 0
We set x3 = t. The first row then gives x1 = x3 = t, and the
second row gives x2 = x3 = t, so x is in the eigenspace if

 
t
1
x = t = t 1 .
t
1
 
1
Thus, the eigenspace is one-dimensional with basis 1.
1
Exam 3 Solutions
3
b) Is A diagonalizable? If so, find a matrix P such that P −1 AP
is diagonal, and display the diagonal matrix P −1 AP .
Solution: The algebraic and geometric multiplicities are equal
for each of the two eigenvalues, so A is diagonalizable.
basis
   A 

2
0
1
of R3 consisting of eigenvectors is given by 0, 1, 1.
1
0
1
We make these the columns of the matrix P :


2 0 1
P = 0 1 1 .
1 0 1


2 0 0
Then P −1 AP = 0 2 0 , the diagonal matrix whose diag0 0 −3
onal entries are the eigenvalues corresponding to the eigenvectors in the same-numbered columns of P .
3. Let T : R2 → R2 be the linear transformation
x1
2x1 − x2
T
=
.
x2
x 1 + x2
a) What is the matrix for T with respect to the standard basis
B = {e1 , e2 } of R2 ?
Solution: The matrix of T with respect to the
basis
standard
2 −1
.
is the standard matrix of T : [T (e1 )|T (e2 )] =
1 1
3
2
b) Let B = {v2 , v2 } be the basis given by v1 =
and v2 =
.
2
1
What is the matrix [T ]B0 of T with respect to B0 ? (Use the
transition matrix P = [I]B,B0 from B 0 to B to find it.)
0
Solution: We have P = [I]B,B0 = [ [v1 ]B | [v2 ]B ] = [v1 | v2 ],
3 2
because B is the standard basis. Thus P =
. Since
2 1
4
Exam 3 Solutions
1 −2
−1 2
1
−1
det P = −1, P = −1
=
. We get
−2 3
2 −3
[T ]B0 = P −1 [T ]B P
−1 2
2 −1 3 2
=
2 −3 1 1
2 1
6
3
=
−7 −3
4. Let T : P2 → P2 be given by T (p) = p(3x + 2).
a) What is the matrix of T with respect to the standard basis
B = {1, x, x2 }?
Solution:
[T ]B = [[T (1)]B | [T (x)]B | [T (x2 )]B ]
= [[1]B | [3x + 2]B | [(3x + 2)2 ]B ]


1 2 4
= 0 3 12
0 0 9
b) What is det(T )?
Solution: det T = det[T ]B . Since [T ]B is upper triangular,
its determinant is the product of its diagonal entries. Thus,
det T = 27.
c) What are the rank and nullity of T ?
Solution: The rank and nullity of T are the rank and nullity
of [T ]B . Because det[T ]B 6= 0, [T ]B is invertible, so its rank is
3 and its nullity is 0.
d) What is the characteristic polynomial
chT (λ)?
λ − 1 −2
−4
λ − 3 −12 . Again,
Solution: chT (λ) = ch[T ]B (λ) = 0
0
0
λ − 9
the matrix is upper triangular, so chT (λ) = (λ−1)(λ−3)(λ−9).
Exam 3 Solutions
5
e) What are the eigenvalues of T ?
Solution: The eigenvalues are 1, 3, 9.
f) Is T diagonalizable?
Solution: Yes: Each eigenvalue has algebraic multiplicity 1.
Eigenvalues always have geometric multiplicity at least 1, so
the two multiplicities must be equal for each eigenvalue.