Math 551, Assignment 7

Due November 21, in class

Do 8 of these problems. V is always a f.d.v.s. over a field F.

1. Let M be a torsion module (M = T (M)) over the PID R. For each prime p ∈ R define M[p^{∞}] = {m ∈ M | p^{i}m = 0 for some integer i > 0}. Show that M[p^{∞}] is a submodule of M and

M = ⨁_{p} M[p^{∞}],

the sum over a set of representatives for association classes of primes. (Note: there is no assumption that M is finitely generated.) M[p^{∞}] is called the p-primary part of M.)

2. A linear transformation T : V → V (or square matrix A) is nilpotent if and only if T^{n} = 0 (or A^{n} = 0) for some n. Show that if A is m × m and nilpotent, then A^{m} = 0.

3. Suppose that the square matrix A is in block diagonal form:

A = [ \begin{matrix}A_{1} & 0 & ⋯ & 0 \\ 0 & A_{2} & ⋯ & 0 \\ & & ⋯ \\ 0 & 0 & ⋯ & A_{r} \end{matrix} ] .

Show that ν(A) = ∑_{i=1}^{r} ν(A_{i}), μ_{A} = lcm_{i=1}^{r} μ_{A_{i}} and χ_{A} = ∏_{i=1}^{r} χ_{A_{i}}. Which of these statements remain true if nonzero entries are allowed anywhere above the diagonal?

4. Suppose that A is an n × n matrix over a field F, and μ_{A} ∈ F [X] is irreducible of degree d. Show that d divides n.

5. How many pairwise nonsimilar 5×5 nilpotent complex matrices are there? How many pairwise nonsimilar 5 × 5 nilpotent real matrices are there?

6. A is a complex 5 × 5 matrix such that Tr A = 9, det A = 16 and μ_{T} (x) = (x − 2)^{2}(x − 1). What are the possible values for χ_{T}, and what are the possible Jordan canonical forms of A?

7. Compute the powers J^{2}, J^{3}, …, of J = J(0, n). Then use J(α, n) = αI + J to compute the powers of J(α, n). Finally, if A is an m × m complex matrix and det A ≠ 0, show that for each natural number n there is an m × m complex matrix B such that B^{n} = A.

8. Show that any complex square matrix is similar to its transpose.

9. Suppose that T and U are linear transformations from V to V such that T U = UT. Show that for every λ ∈ F, ker(T − λ) is invariant under U. If T and U are both diagonalizable, show that they are simultaneously diagonalizable, that is, there is a basis of V consisting of vectors which are eigenvectors of both T and U. Write the equivalent version of this statement for square matrices.

10. Suppose that A is a 2 × 2 integer matrix and A^{n} = 1 for some integer n. Show that A^{m} = 1 for some m ≤ 6. Reason with eigenvalues, without doing any matrix multiplications.

11. Let A and B be real square matrices. If there exists a nonsingular complex matrix D such that DAD^{−1} = B, show that a nonsingular real matrix E can be found such that EAE^{−1} = B.

12. Suppose that T : V → V is a linear transformation. Show that μ_{T} = χ_{T} if and only if there is a vector v ∈ V such that V is spanned (as vector space) by the set {v, T v, T^{2}v, …, T^{n}v}. Furthermore, let W ⊆ V be a T -invariant subspace, and set U = T |W. Show that if μ_{T} = χ_{T}, then μ_{U} = χ_{U}. (Hint. Don’t use the first part. This is close to the theorem that subgroups of cyclic groups are cyclic.)

Math 551, Assignment 7

1. M[p^{∞}] is a submodule: Let m, m′ ∈ M[p^{∞}] and r ∈ R. Then p^{i}m = p^{j}m′ = 0 for some positive integers i, j, so p^{k}(m + m′) = 0 where k = max(i, j); and p^{i}(rm) = rp^{i}m = r0 = 0 by the commutativity of R.

M is generated by the M[p^{∞}]: Let m ∈ M. Then Rm is a submodule of M, and the mapping _{R}R → Rm, r ↦ rm, is a surjective homomorphism of R-modules. The kernel is of the form Rr for some r ∈ R as R is a PID. Therefore Rm ≅ R/Rr. Factorizing r = up_{1}^{a_{1}} ⋯ p_{n}^{a_{n}} with u a unit we have Rr = Rur so

Rm ≅ _{R}R/Rp_{1}^{a_{1}} ⋯ p_{n}^{a_{n}} ≅ _{R}R/Rp_{1}^{a_{1}} ⊕ ⋯ ⊕ _{R}R/Rp_{n}^{a_{n}}

by the Chinese Remainder Theorem. Therefore Rm is the direct sum Rm = M_{1} ⊕ ⋯ M_{n} of submodules M_{i} ≅ _{R}R/Rp_{i}^{a_{i}}. Clearly M_{i} ⊆ M[p_{i}^{∞}] so Rm ⊆ ∑_{p} M[p^{∞}]. As m was arbitrary, M = ∑_{p} M[p^{∞}].

The sum of the M[p^{∞}] is direct: From the definition of direct sum, it must be verified that for any p, M[p^{∞}] ∩ ∑_{q≠p} M[q^{∞}] = 0. Suppose that m ∈ M[p^{∞}] ∩ ∑_{q≠p} M[q^{∞}]. Then m = ∑_{i=1}^{n} m_{i} where for each i, m_{i} ∈ M[q_{i}^{∞}], and the primes p, q_{1}, …, q_{n} are pairwise non-associated. Moreover p^{k}m = 0 and q_{i}^{k_{i}}m_{i} = 0 for suitable positive integers k, k_{i}. Let Q = q_{1}^{k_{1}} ⋯ q_{n}^{k_{n}}. Then Qm_{i} = 0 for all i so Qm = 0. Also p^{k}m = 0. But gcd(p^{k}, Q) = 1 so we may write 1 = ap^{k} + bQ for some a, b ∈ R, and then m = ap^{k}m + bQm = 0. QED

2. By nilpotence, μ_{A}(X) divides a power of X so μ_{A}(X) = X^{r} for some r. But μ_{A} divides χ_{A}, which has degree n, so r ≤ n. Therefore A^{n} = 0. QED

4. Let f_{1}|f_{2}| ⋯ |f_{r} be the elementary divisors of A. We know that μ_{A} = f_{r} and χ_{A} = f_{1} ⋯ f_{r}. Since f_{r} = μ_{A} is irreducible, f_{i} = f_{r} for all i and so χ_{A} = μ_{A}^{r}. The degree of χ_{A} is n, so n = dr. QED

5. We can use Jordan canonical form to determine similarity classes over C. Over R, which is not algebraically closed, we can’t always do this, but we can always use rational canonical form. Let A be a nilpotent real or complex 5 × 5 matrix. The minimal polynomial of A is then X^{k} for some i. The elementary divisors of A are f_{1}|f_{2}| ⋯ |f_{r} = X^{k} so f_{i} = X^{k_{i}} for each i, where k_{1} ≤ k_{2} ≤ ⋯ ≤ k_{r} = k. Since χ_{A} = f_{1} ⋯ f_{r} has degree 5, ∑ k_{i} = 5. The rational canonical form of A therefore is a block-diagonal matrix whose blocks are the companion matrices of X^{k_{1}}, …, X^{k_{r}}. (These companion matrices are actually the Jordan blocks J(0, k_{i}).) So the number of similarity classes is the number of partitions of 5, which is 7.

6. Too much information is given. From the given, we know that the characteristic polynomial of A is (X − 2)^{a}(X − 1)^{b} (since μ_{A} and χ_{A} have the same irreducible factors), where a + b = 5, and a ≥ 2, b ≥ 1 (since μ_{A}|χ_{A}). Thus the five eigenvalues are two, three or four 2’s, and the rest 1’s. We also know that det A is the product of the eigenvalues and Tr A is the sum of the eigenvalues. Therefore the eigenvalues are four 2’s and one 1. The total size of the Jordan blocks for each eigenvalue is the multiplicity of that eigenvalue, and μ_{A} is obtained from the Jordan blocks of maximal size for each eigenvalue. Therefore A has Jordan canonical form (J(1, 1), J(2, 2), J(2, 2)) or (J(1, 1), J(2, 2), J(2, 1), J(2, 1)).

8. We have A = DJD^{−1} for some invertible D and some J in Jordan canonical form. Therefore A^{T} = EJ^{T} E^{−1} where E = (D^{T})^{−1}. So it is enough to show that J is similar to its transpose. This easily reduces to the case of a single Jordan block. So it is enough to show that the Jordan canonical form of J(α, n)^{T} is J(α, n). One easily calculates that (J(α, n)^{T} − αI)^{n} = 0 but (J(α, n)^{T} − αI)^{n} = 0, so J(α, n)^{T} has minimal polynomial (X − α)^{n}. As J(α, n)^{T} is n × n, its Jordan canonical form is therefore J(α, n). QED

9. Lemma: If U : V → V is a diagonalizable linear transformation and W is a U-invariant subspace, then U|W is diagonalizable. Proof: U is diagonalizable so μ_{U} is square-free. But μ_{U}(U|W) = 0 so μ_{U|W} divides μ_{U}, hence is square-free too, so U|W is diagonalizable.

Given that T U = UT, if v ∈ ker(T −λ) for some λ, then T v = λv, so T (Uv) = U(T v) = U(λv) = λ(Uv) and thus Uv ∈ ker(T − λ). So ker(T − λ) is U-invariant. Since T is diagonalizable, V is the direct sum V = ⨁_{λ} ker(T − λ), the sum over all eigenvalues. For each λ, U| ker(T − λ) is diagonalizable so the eigenspace ker(T − λ) has a basis B_{λ} consisting of eigenvectors for U. Then ∪_{λ}B_{λ} is a basis of V consisting of simultaneous eigenvectors for T and U.

The matrix-theoretic statement is: If A and B are square matrices such that A and B are diagonalizable and AB = BA, then there exists invertible D such that DAD^{−1} and DBD^{−1} are both diagonal.

11. Write D = D_{1} +iD_{2} where D_{1} and D_{2} are both real matrices. DAD^{−1} = B implies DA = BD. Equate real and imaginary parts of both sides to get D_{1}A = BD_{1} and D_{2}A = BD_{2} (since A and B are real). If D_{i} is invertible for either i = 1 or 2, we are done. In general set D(t) = D_{1} + tD_{2} (t an indeterminate). Then D(t)A = BD(t). This equation remains true when we specialize t to be any particular complex number. It suffices to find t ∈ R such that det D(t) ≠ 0, for then D(t)AD(t)^{−1} = B. But f(t) = det D(t) ∈ R[t], and f(t) is not the zero polynomial because f(i) = det D(i) = det D ≠ 0. Therefore f(t) ≠ 0 for some (infinitely many!) t ∈ R.