Section 12.2.
Homework #7
Masaya Sato
1. Prove that similar linear transformations of V (or n × n matrices) have the same charac-
teristic and the same minimal polynomials. Proof. Let ϕ and ψ be similar transformations on V with representation matrices A and B ,
resepectiv resepectively ely.. Then observe observe by assumption that there there exists exists some invertible invertible matrix T such that −1 . A = T BT So for an indeterminate x −1 −1 −1 (B )T (B )T xI T − A = xI − T ⇒ xI − A = xT − T −1 (xI ⇒ xI − A = T − B )T
and thus −1 det(xI det(xI det (T (xI ) ⇒ det(xI det(xI (det T )−1 det(xI det(xI − A) = det − B )T − A) = det T − B) det(xI det (xI ⇒ det(xI − A) = det − B ).
This shows that any similar linear transformations on V have the same characteristic polynomial. Now let mϕ (x) be the minimal polynomial for the transformation ϕ, whose dimension is m < n. algorithm for polyno p olynomials mials there exist exist a unique unique pair of polynomials n . By the division algorithm q(x) adn r(x) such that det(xI det(xI − B ) = q(x)mϕ (x) + r (x), where 0 ≤ deg r(x) ≤ m − 1. Sinc Sincee det det (xI det (xI − A) = det − B ), det(xI det(xI − A) = q(x)mϕ (x) + r(x). By minimality of mϕ (x), r (x) is the zero polynomial. Moreover, 0 = det det (BI − B ) = q(B )mϕ (B ) and therefore mϕ (B ) = 0 since F [x] is an integral domain. Hence mϕ (x) is also the minimal polynomial for the transformation ψ. 3. Prove that two 2 × 2 matrices over F which are not scalar matrices are similar if and only
if they have the same characteristic polynomial. similar. r. Then Then by result result Proof. (⇒) Suppose that 2 × 2 non-scalar matrices A and B are simila from Exercise 1, A and B share the common characteristic polynomial. (⇐) Conversely suppose that A and B have the same characteristic polynomial p(x) of degree 2, i.e. p(x) = x2 + ax + b, where a, b ∈ F . Observ Observee that that p(x) is not of the form 2 , since A and B are non-scalar non-scalar matrices. matrices. Thu Thuss p(x) is also the p(x) = (x − λ) , where λ ∈ F minimal polynomial. Therefore both A and B are similar to the companion matrix
0 −b 1 −a
and hence A and B are similar. similar. by Dummit and Foote 1 Abstract Abstract Algebra Algebra
Section 12.2.
Homework #7
Masaya Sato
4. Prove that two 3 × 3 matrices are similar if and only if they have the same characteristic
and same minimal polynomials. Give an explicit counterexample to this assertion for 4 × 4 matrices. similar. So, similar to Exercise Exercise 3, A Proof. (⇒) Suppose that 3 × 3 matrices A and B are similar. and B have the same characteristic and minimal polynomials by result from Exercise 1. (⇐) Conversely suppose that A and B have the same characteristic and minimal polynomials, p(x) and m(x). Observe that deg p(x) = 3 and consider the following 3 cases. . Case 1. Suppose that deg m(x) = 1. Then m(x) = x − a and p(x) = (x − a)3 for some a ∈ F So invariant factors are ai (x) = x − a for i = 1, 2, 3. . Su Suppo ppose se that that deg deg m(x) = 2. 2. Then Then there there exists exists some some linear linear factor factor a1 (x) = x − a, Case 2 where a ∈ F , such that (x ( x − a)m(x) = p(x). Moreover a2(x) = m(x). . Suppose that deg m(x) = 3. Then a1 (x) = p(x) = m(x). Case 3 In any cases discussed above, the characteristic and minimal polynomial uniquely determine invariant invariant factors. factors. Therefore A and B have the same rational canonical form and hence A and B are similar. In terms of a counterexample for 4 × 4 matric matrices, es, consid consider er the follo followin wingg distin distinct ct lists lists of invariant invariant factors (i) a1 (x) = x, a2 (x) = x, a3 (x) = x2 , and (ii) a1 (x) = x2 and a2 (x) = x2 . Observe that both (i) and (ii) have the same characteristic and minimal polynomials x4 and x2 , respectively. For (i), the matrix is given by
0 0 0 0
0 0 0 0
0 0 0 1
and, and on the other hand (ii) gives the matrix
However those matrices are not similar.
0 1 0 0
0 0 0 0
0 0 0 1
0 0 0 0
0 0 . 0 0
all linear transformations of an n 5. Prove directly from the fact that the collection of
dimensional vector space V over F to itself form a vector space over F of dimension n2 that the minimal polynomial of a linear transformation T has degree at most n2 .
by Dummit and Foote 2 Abstract Abstract Algebra Algebra
Section 12.2.
Homework #7
Masaya Sato
arbitrarily taken taken linear linear transformat transformation ion ϕ on V , let A = (aij ) denote the Proof. For an arbitrarily representation matrix for ϕ, where 1 ≤ i ≤ n and 1 ≤ j ≤ n. Moreover let E ij ij be an n × n matrix defined by 1 at the (i, ( i, j ) entry E ij ij = 0 elsewhere .
Then
n
A=
aij E ij ij
i,j =1 n and thus A is generated by the set {E )}i,j the following ij ij ∈ M n×n (F =1 . Moreover for cij ∈ F equation n
cij E ij ij = 0,
i,j =1
where 0 is the zero matrix, has only the trivial solution since n
= 0 [ C = (cij )] cij E ij ij = 0 ⇒ C
i,j =1
⇒ cij = 0
n for all i = 1, . . . , n and j = 1, . . . , n. Therefore ore {E )}i,j n. Theref ij ij ∈ M n×n (F =1 is linearly indepen2 dent and a basis for a vector space of dimension n . This indicates indicates that the character characteristic istic 2 polynomial is of degree n and therefore the degree of its minimal polynomial is at most n2 .
6. Prove that the constant term in the characteristic polynomial of the n × n matrix A is
(−1)n det A and that the coefficient coefficient of xn−1 is the negative of the sum of the diagonal entries of A (the sum of the diagonal entries of A is called the trace of A). Prove that det A is the product of the eigenvalues of A and that the trace of A is the sum of the eigenvalues of A. Proof. For an n × n matrix A its characteri characteristic stic polynomial is given by det ( xI − A). Then
det(xI det(xI − A) = xn + an−1 xn−1 + · · · + a1 x + a0 , which is the monic polynomial of degree n. Evaluati Evaluating ng the abov ab ovee identity identity in terms of the indeterminate x at x = 0 ∈ F , a0 = det(−A) = (−1)n det A. On the other hand, the expansio expansion n of det ( xI − A) gives det(xI det(xI − A) =
)(xδσ(1)1 − aσ(1)1) · · · (xδσ(n)n − aσ(n)n ), (σ)(xδ
σ ∈S n
by Dummit and Foote 3 Abstract Abstract Algebra Algebra
Section 12.2.
Homework #7
Masaya Sato
where δij denotes the Kronecker delta. Moreover, applying nth derivative of the identity and comparing the constant terms,
d dx
1
n
−
n
1
−
to both sides
((n − 1)!)a 1)!)a11 − · · · − ((n ((n − 1)!)a 1)!)ann = ((n ((n − 1)!)a 1)!)an−1 ⇒ −((n ((n − 1)!)(Tr(A 1)!)(Tr(A)) = ((n ((n − 1)!)a 1)!)an−1 −((n r(A). ⇒ an−1 = −Tr(A The argument above implies that for eigenvalues λ1 , . . . , λn of A det(λI det(λI (λ1 + · · · + λn)xn−1 + · · · + a1 x + (λ (λ1 · · · λn ). − A) = xn + (λ Therefore
n
Tr(A r(A) =
λi
i=1
and
n
det A =
λi .
i=1
8. Verify that the characteristic polynomial of the companion matrix
0 1 0 .. .
0 0 1 .. .
0 ··· 0 ··· 0 ··· .. .
0 0 0 ···
is
0 0 0 .. .
−a0 −a1 −a2 .. .
1 −an−1
xn + an−1xn−1 + · · · a1 x + a0 . Proof. For an indeterminate x observe first that xI − A is given by
0 x −1 x 0 −1 .. .. . . 0 0
0 ··· 0 ··· x ··· .. .
0 0 0 .. .
a0 a1 a2 .. .
0 · · · −1 x + an−1
,
where A denotes the companion matrix. By induction on the size n of the matrix, suppose that n = 2. Th Then en det det (xI now suppose suppose that that the det det (xI − A) = x2 + a1 x + a0 . So now − A) has the desired polynomial for n. Then for n + 1
0 x −1 x 0 −1 .. .. . . 0 0 0 0
0 ··· 0 ··· x ··· .. .
0 0 0 .. .
a0 a1 a2 .. .
0 ··· x an−1 0 · · · −1 x + an
by Dummit and Foote 4 Abstract Abstract Algebra Algebra
Section 12.2.
Homework #7
Masaya Sato
and the expansion along the 1st column gives the determinant of minors, i.e. det(xI det(xI det A˜21 , − A) = (−1)1+1 x det A˜11 + (−1)2+1(−1) det where A˜11 and A˜21 denote cofactors. Then det A˜11 = x(xn + an xn−1 + · · · + a1 ) = xn+1 + anxn + · · · + a1 x. To evaluate det A˜21 , keep expanding along the 1st column and then det A˜21 = a0 . Therefore Therefore det det ( xI − A) = xn+1 + an xn + · · · + a1 x + a0 as desired. 11. Find all similarity classes of 6 × 6 matrices matrices over over C with characte characteristi risticc polynomial polynomial
(x4 − 1)(x 1)(x2 − 1). ( x4 −1)(x 1)(x2 − 1) factors into irreducibles Solution: Observe that the characteristic polynomial (x in C[x] as (x (x − i)(x )(x + i)(x )(x − 1)2 (x + 1)2 . To find all similarit similarity y classes, classes, consider their minimal polynomials m(x). Since m(x) and (x (x4 − 1)(x 1)(x2 − 1) have the same roots, the candidates for (x − i)(x )(x + i)(x )(x − 1)(x 1)(x + 1), 1), (x (x − i)(x )(x + i)(x )(x − 1)2 (x + 1), 1), (x (x − i)(x )(x + i)(x )(x − 1)(x 1)(x + 1)2 , m(x) are (x and (x (x − i)(x )(x + i)(x )(x − 1)2 (x + 1)2 . So consider all 4 cases given above. Case 1. Sup Suppose pose that that m(x) = (x − i)(x )(x + i)(x )(x − 1)(x 1)(x + 1). 1). Then the invarian invariantt factors are 1)(x + 1) and a2(x) = m(x). a1 (x) = (x − 1)(x . Suppose that m(x) = (x − i)(x )(x + i)(x )(x − 1)2 (x + 1). Then the invariant factors are Case 2 a1 (x) = x + 1 and a2(x) = m(x). . Suppose that m(x) = (x − i)(x )(x + i)(x )(x − 1)(x 1)(x + 1)2 . Then the invariant factors are Case 3 a1 (x) = x − 1 and a2 (x) = m(x). )(x + i)(x )(x − 1)2 (x + 1)2. Then a1 (x) = m(x) is the only Case 4. Suppose that m(x) = (x − i)(x invariant invariant factor. Therefore, according the classification discussed above, there are 4 similarity classes. 17. Determine representatives for the conjugacy classes for GL3 (F2 ). [Compare your answer
with Theorem 15 and Proposition 14 of Chapter 6.] Solution: For F2 [x], since the only nonzero element in F2 is 1,
(i) F2[x]/(x − 1) ⊕ F2 [x]/((x ((x − 1)2 ), (ii) F2[x]/(x − 1) ⊕ F2 [x]/(x − 1) ⊕ F2 [x]/(x − 1), and (iii) F2[x]/(Q(x)), where Q is a cubic polynomial. For (i) and (ii), the corresponding matrices are given by
1 0 0 0 1 0 0 1 1
and
1 0 0 0 1 0 , respectively. 0 0 1
by Dummit and Foote 5 Abstract Abstract Algebra Algebra
Section 12.2.
Homework #7
Masaya Sato
Then for (iii) let Q(x) = x3 + ax2 + bx + 1. The matrix is given by
0 0 1 1 0 −b . 0 1 −a
Classifying Q(x) as the product of primes factors, Q(x) has the following forms. (1) x3 + x + 1, (2) x3 + x2 + 1, (3) (x (x − 1)(x 1)(x2 + x + 1), and (4) (x (x − 1)3 . Therefore from (1) to (4) the matrices are given by
0 0 1 1 0 1 , 0 1 0
0 0 1 1 0 0 , 0 1 1
1 0 0 0 0 1 , and 0 1 1
1 0 0 1 1 0 . 0 1 1
is nonsingular linear 18. Let V be a finite dimensional vector space over Q and suppose T 2 −1 transformation of = T + T . Prove that the dimension of V such that T V is divisible by 3. If the dimension of are similar. V is precisely 3 prove that all such transformations T 2 −1 be an n dimensional vector space over Q. Observe that T = T + T implies Proof. Let V 3 2 + T = 0. So the minimal polynomial m(x) is given by T − I
m(x) = x3 + x2 − 1, which is irreducible. Therefore
∼ Q[x]/(m(x)) ⊕ · · · ⊕ Q[x]/(m(x)), V = where each quotient space is 3 dimensional. Hence the dimension of is 3k 3 k for some k ∈ Z>0 V as desired. Now suppose that V is 3 dimensional. Then
∼ V = Q[x]/(m(x)). 2 −1 So all nonsingular linear transformations T satisfying T = T + T have the same companion matrix because m(x) is also the characteristic polynomial for T . Th Thus us all all suc such h T ’s are similar.
by Dummit and Foote 6 Abstract Abstract Algebra Algebra
