Download Math 252 — Solutions to Homework Set #9 Number 1. — 5 points

Math 252 — Solutions to Homework Set #9
Number 1. — 5 points
Find the rational canonical forms of

0
 0
−1
−1
0
0

−1
0,
0

c
 0
−1

−1
1
c
0
c
1
and
422
 −420

840
−140

465
−463
930
−155
15
−15
32
−5

−30
30 
.
−60
12
Solution: The respective rational canonical forms are:

0 0 0
1 0 1,
0 1 0


0 0
1 0
0 1

c3 − 2 c
−3 c2 + 2 
3c
and

0
6 0 0
 1 −1 0 0 


0
0 2 0
0
0 0 2

Number 2. — 5 points
Find all similarity classes of 6 × 6 matrices over Q with minimal polynomial (x + 2)2 (x − 1) (it suffices to
give all lists of invariant factors and write out some of their corresponding matrices).
Solution: Permissible lists of invariant factors are:
(x + 2)2 (x − 1), (x + 2)2 (x − 1)
(x + 2)2 (x − 1), (x + 2)2 , (x + 2)
(x + 2)2 (x − 1), (x + 2), (x + 2), (x + 2)
(x + 2)2 (x − 1), (x + 2)(x − 1), (x + 2)
(x + 2)2 (x − 1), (x + 2)(x − 1), (x − 1)
(x + 2)2 (x − 1), (x − 1), (x − 1), (x − 1).
As an example, the rational canonical form for the second list is:
0
1

0

0

0
0

0
4 0
0
0
0 0
0
1 −3 0
0
0
0 0 −4
0
0 1 −4
0
0 0
0

0
0

0
.
0

0
2
Number 3. — 5 points
Find all similarity classes of 6 × 6 matrices over C with characteristic polynomial
(x4 − 1)(x2 − 1).
Solution: First note that in C[x] we have the factorization
x4 − 1 = (x − 1)(x + 1)(x − i)(x + i)
√
polynomial must have the same roots as x4 − 1,
where i = −1. Since x2 − 1 divides x4 − 1, the minimal
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and so must be divisible by x − 1. Since mA (x) cA (x), the possibilities for mA (x) are
(i) mA (x) = x4 − 1,
(ii) mA (x) = (x4 − 1)(x − 1) or (x4 − 1)(x + 1), or
(iii) mA (x) = (x4 − 1)(x2 − 1).
Since the product of all invariant factors divides cA (x), in case (i) there can only be one other invariant
factor: x2 − 1. Likewise in each of the cases in (ii) there is one remaining invariant factor: x + 1 or x − 1
respectively. In case (iii) we have mA (x) = cA (x) is the only invariant factor.
In summary, there are 4 similarity classes.
Number 4. — 5 points
Determine all possible rational canonical forms for a linear transformation with characteristic polynomial
x2 (x2 + 1)2 .
Solution: Note that since the characteristic polynomial has degree 6, we must be considering linear
transformations on a 6-dimensional space (or 6 × 6 matrices). The minimal polynomial must have the
same roots as the characteristic polynomial, so mA (x) must be divisible by x(x2 + 1). [Note: here I’m
assuming we’re over a field where x2 + 1 has distinct roots; this is not the case for all fields—for example,
over F2 we have x2 + 1 = (x + 1)2 .] The possibilities for mA (x) and then the resulting possible list of
invariant factors are: (there is only one possible list for each mA (x))
(i) mA (x) = x(x2 + 1), and other invariant factor x(x2 + 1),
(ii) mA (x) = x2 (x2 + 1), and other invariant factor x2 + 1,
(iii) mA (x) = x(x2 + 1)2 , and other invariant factor x, or
(iv) mA (x) = x2 (x2 + 1)2 = cA (x) (no other invariant factors).
In summary, there are 4 similarity classes for such linear transformations.
Number 5. — 5 points
Determine which of the following matrices are similar:

−1
 2
0
4
−1
−4

−4
3
3

−3
 2
8
−4
3
8

0
0
1

−3
 2
3
2
1
−1

−4
0
3

−1
 0
0
4
−3
−4

−4
2 .
3
Solution: The respective Jordan canonical forms are:

−1 0 0
 0 1 1
0 0 1


−1
 0
0

0 0
1 0
0 1

−1 0 0
 0 1 1
0 0 1


−1
 0
0
so the only similarities are the first and third.
Number 6. — 5 points
Determine the Jordan canonical forms for the following matrices:

5
 −1
−3
4
0
−4

1
0
1

3
 −2
−4
4
−3
−4

2
−1 .
−3
Solution: The Jordan canonical forms are respectively:

2 1
0 2
0 0

0
1
2


−1
1
0
 0 −1
0.
0
0 −1
2

0 0
−1 0 
0 1
Number 7. — 5 points
Verify for yourself that the matrices

−8
A= 7
3

−1
1
0
−10
9
2

−3
B= 4
4
2
−1
−2

−4
4
5
both have (x − 1)2 (x + 1) as characteristic polynomial. Determine the Jordan canonical form for both
matrices. explain why one can be diagonalized and the other cannot.
Solution: The Jordan canonical forms are respectively:




−1 0 0
−1 0 0
 0 1 1
 0 1 0
0 0 1
0 0 1
so only B can be diagonalized. The minimal polynomial of B is (x−1)(x+1), which has no repeated roots;
the minimal polynomial for A is (x + 1)(x − 1)2 which has a repeated root (so A cannot be diagonalized).
Number 8. — 5 points
Show that the characteristic polynomial of
1
 0
A=
−2
−2

0
1
−2
0
0
0
0
−1

0
0

1
−2
is a product of linear factors over Q. Determine the rational and Jordan canonical forms for A over Q.
Solution: The characteristic polynomial of A is cA (x) = (x − 1)2 (x + 1)2 ; and the minimal polynomial is
mA (x) = (x − 1)(x + 1)2 = x3 + x2 − x − 1. The rational and Jordan canonical forms for A are respectively:
0
1

0
0

0
1
0
1
1 −1
0
0

0
0

0
1
1
0

0
0


0
0
0
1
0
0
.
0 −1
1
0
0 −1
Number 9. — 5 points
Determine the Jordan canonical form for the matrix
3
4

2
0

0
−8
−4
2
−2
14
7
−4

−3
−15 
.
−7
3
Solution: The characteristic polynomial of the given matrix is (x− 1)3 (x− 2); and its minimal polynomial
is (x − 1)2 (x − 2). The Jordan canonical form is:
2
0

0
0

0
1
0
0
0
1
1
0
3

0
0
.
0
1
Number 10. — 5 points
Verify for yourself that the matrices
0
1
A=
1
1

1
0
1
1
1
1
0
1

1
1

1
0
5
 −6
B=
−3
3

2
−3
−1
1
−8
8
3
−4

−8
8

4
−5
both have characteristic polynomial (x − 3)(x + 1)3 . Determine the Jordan canonical form for each matrix
and explain whether or not they are similar.
Solution: Both matrices have the specified characteristic polynomial, and both have minimal polynomial
(x − 3)(x + 1) (which has no repeated roots). Thus both are diagonalizable with the same Jordan canonical
form, as follows (and hence they are similar):

3
0
0
0
0
0
 0 −1

.
0
0 −1
0
0
0
0 −1

Number 11. — 10 points
(a) Find all similarity classes of 3 × 3 matrices A over F2 satisfying A6 = I (compare with the answer we
computed over Q).
(b) Find all similarity classes of 4 × 4 matrices B satisfying B 20 = I.
Solution: (a) Since A6 = I, the minimal polynomial of A divides x6 − 1. Since A is a 3 × 3 matrix, cA (x)
has degree 3. In F2 [x] = Z/2Z[x] we have the (unique) factorization into irreducible polynomials:
x6 − 1 = (x3 − 1)2 = (x + 1)2 (x2 + x + 1)2 .
Since since the characteristic and minimal polynomials have the same roots, cA (x) is a product of some of
the above factors. Thus the only possibilities for cA (x) are:
(i) cA (x) = (x + 1)3 , or
(ii) cA (x) = (x + 1)(x2 + x + 1).
In case (i) the permissible lists of invariant factors are:
(ia) x + 1, x + 1, x + 1, or
(ib) (x + 1)2 , x + 1.
(Note that we cannot have mA (x) = (x + 1)3 , since this does not divide x6 − 1.)
In case (ii), mA (x) = cA (x) is the only invariant factor.
In summary, there are 3 similarity classes for such A.
(b) Likewise, the minimal polynomial of B divides x20 − 1, which factors over F2 into irreducible
polynomials as
x20 − 1 = (x5 + 1)4 = (x + 1)4 (x4 + x3 + x2 + x + 1)4 .
Since B is 4 × 4, its characteristic polynomial has degree 4 and is a product of some of the above factors.
It follows that one of the following must hold:
(i) cA (x) = (x + 1)4 , or
(ii) cA (x) = (x4 + x3 + x2 + x + 1).
In case (i) the permissible lists of invariant factors are:
(ia) x + 1, x + 1, x + 1, x + 1
(ib) (x + 1)2 , (x + 1)2 ,
(ic) (x + 1)2 , x + 1, x + 1,
(id) (x + 1)3 , x + 1, or
(ie) (x + 1)4 .
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In case (ii), mA (x) = cA (x) is the only invariant factor.
In summary, there are 6 similarity classes for such B.
Number 12. — 5 points
Show that if A2 = A then A is similar to a diagonal matrix which has only 0’s and 1’s along the diagonal.
Solution: Since A2 = A, the minimal polynomial of A divides x2 − x = x(x − 1). Since therefore the
minimal polynomial has distinct roots, A is similar to a diagonal matrix whose diagonal entries are the
roots of the minimal polynomial, namely zeros or ones—and these roots may be repeated many times
along the diagonal—as desired.
Number 13. — 5 points
Prove there are no 3 × 3 matrices A over Q with A8 = I but A4 6= I.
Solution: Since A8 = I, the minimal polynomial of A divides x8 − 1, which (one checks) factors into
irreducible polynomials in Q[x] as
x8 − 1 = (x − 1)(x + 1)(x2 + 1)(x4 + 1).
Since the characteristic polynomial of A has degree 3, it can have no factor of x4 + 1. Thus the minimal
polynomial of A must divide (x − 1)(x + 1)(x2 + 1) = x4 − 1. This, in turn, shows that A4 − I = 0, as
needed to show that there is no such A.
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