Math 593 Assignment # 8 Solutions

Bryden R. Cais

November 20, 2002

2. (a) Any additive (finite) abelian group G may be realized as a ℤ module. The structure theorem tells us that we may decompose G in primary form, that is, we have

G ≃ ⨁ ℤ/(p_{i}^{a_{i}})

, where the (not necessarily distinct) p_{j} satisfy

∏_{i} p_{i}^{a_{i}} = #G.

Since 500 = 2^{2} · 5^{3}, we have the following 6 possibilities for the primary decomposition of G, and hence for G up to isomorphism:

ℤ/(2) ⊕ ℤ/(2) ⊕ ℤ/(5) ⊕ ℤ/(5) ⊕ ℤ/(5)
ℤ/(2) ⊕ ℤ/(2) ⊕ ℤ/(5) ⊕ ℤ/(5^{2})
ℤ/(2) ⊕ ℤ/(2) ⊕ ℤ/(5^{3})
ℤ/(2^{2}) ⊕ ℤ/(5) ⊕ ℤ/(5) ⊕ ℤ/(5)
ℤ/(2^{2}) ⊕ ℤ/(5) ⊕ ℤ/(5^{2})
ℤ/(2^{2}) ⊕ ℤ/(5^{3}).

(b) Let T be a matrix over a field k with characteristic polynomial (x^{2} − x)^{5}. Let us determine all the possible rational canonical forms for T. Let M_{T} be the k[x] module determined by the action of T on k^{10}. Then the structure theorem enables us to write

M_{T} ≃ k[x]/(f_{1}(x)) ⊕ … ⊕ k[X]/(f_{r}(x)),

where f_{r} is the minimal polynomial of T, f_{i}|f_{i+1}, and x^{5}(x − 1)^{5} = ∏_{i} f_{i}. It follows that we have

f_{i} = x^{a_{i}}(x − 1)^{b_{i}}

with ∑_{i} a_{i} = ∑_{i} b_{i} = 5. It follows that for any two partitions of 5 we have a (unique) list of invariant factors f_{i}. The partitions of 5 are precisely

5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1.

Therefore, there are 7 · 7 = 49 similarity classes of matricies T, enumerated by any two partitions p_{1}, p_{2} of 5. If r_{i} denotes the largest part of the partition p_{i}, then the minimal polynomial of a class given by p_{1}, p_{2} is

x^{r_{1}}(x − 1)^{r_{2}}.

(c) Certainly the minimal polynomial of T must divide x(x − 1)^{2}. Moreover, since the characteristic polynomial of T has the same roots as the minimal polynomial and has degree 4 (since T is a 4 × 4 matrix), we can easily list the possible invariant factors for the decomposition of the k[x] module specified by T into cyclic factors. The possibilities for the minimal polynomial are:

x, x − 1, (x − 1)^{2}, x(x − 1), x(x − 1)^{2}.

It follows that the possible invariant factors are:

x, x, x, x
x − 1, x − 1, x − 1, x − 1
(x − 1), (x − 1), (x − 1)^{2}
(x − 1)^{2}, (x − 1)^{2}
x, x, x(x − 1)
x − 1, x − 1, x(x − 1)
x(x − 1), x(x − 1)
x, x(x − 1)^{2}
x − 1, x(x − 1)^{2}

for a total of 9 similarity classes. Each list of invariant factors above gives the matrix T in rational canonical form. For example, the list x(x − 1), x(x − 1) tells us that we have two blocks, each with companion matrix x^{2} − x, corresponding to

(\begin{matrix}0 & 0 \\ 1 & 1 \\ & & 0 & 0 \\ & & 1 & 1 \end{matrix}) .

3. Let A = (a_{ij}) be an n × n matrix with integer entries presenting the ℤ module M. Recall that we can bring A to diagonal form via elementary row and column operations. That is, we have

B = QAP^{−1}

where Q, P are products of elementary matrices, so that det B = ± det A and

B = (\begin{matrix}d_{1} \\ & d_{2} \\ & & . & . & . \\ & & & & & d_{n} \end{matrix}) ,

where

d_{i}|d_{i+1}

and d_{i} > 0 for 1 ≤ i ≤ l ≤ n. Moreover, we have

M ≃ ℤ/(d_{1}) ⊕ ⋯ ⊕ ℤ/(d_{l}) ⊕ ℤ^{n−l}

and

det B = | det A| = |d_{1} ⋯ d_{n}|.

Clearly, | det A| ≠ 0 if and only if d_{i} > 0 for all i, that is, if and only if n = l, if and only if M is finite. If M is finite, we have

#M = #ℤ/(d_{1}) ⊕ ⋯ ⊕ ℤ/(d_{n}) = ∏_{i=1}^{n} #ℤ/(d_{i}) = ∏_{i=1}^{n} d_{i} = det B = | det A|.

4. i Let T be an n×n invertible matrix over a field k, and let P_{T} (x) denote the characteristic polynomial of T. We have

P_{T} (x) = det(xI − T)

= det(T) det(xT^{−1} − I)

= (−x)^{n} det(T) det(\frac{1}{x}I − T^{−1})

= (−x)^{n} det(T)P_{T^{−1}} (\frac{1}{x}) ,

so that

P_{T^{−1}}(x) = \frac{(−x)^{n}}{det(T)}P_{T} (\frac{1}{x}) .

ii Suppose that V_{T} = M_{1}⊕M_{2}⊕⋯⊕M_{r} where M_{i} ⊂ V is a T submodule of V_{T}, isomorphic (with the structure of multiplication by x given as multiplication by T) to k[x]/(d_{i}(x)) with d_{i}|d_{i+1}. Clearly, d_{i} is the characteristic polynomial of multiplication by T on the T -submodule M_{i} of V_{T}. It is not difficult to see that M_{i} also has the structure of a T^{−1} module. Indeed, T M_{i} ⊂ M_{i} since M_{i} is a T -submodule; equality follows from the fact that M_{i} is finite dimensional, so that M_{i} = T^{−1}M_{i}. Moreover, by part (i), the characteristic polynomial of multiplication by T^{−1} on M_{i} is

f_{i}(x) = \frac{(−x)^{deg d_{i}}}{det(T |_{M_{i}})}d_{i} (\frac{1}{x}) = x^{r_{i}} + \frac{a_{1i}}{a_{0i}} x^{r_{i}−1} + ⋯ + \frac{a_{(r_{i}−1)i}}{a_{0i}} ,

where

d_{i}(x) = x^{r_{i}} + a_{(r_{i}−1)i}x^{r_{i}−1} + ⋯ + a_{0i}.

It follows that M_{i} with the structure of multiplication by x as multiplication by T^{−1} is isomorphic to k[x]/(f_{i}(x)). It follows that

V_{T^{−1}} ≃ ⨁_{i=1}^{r} k[x]/(f_{i}(x)).

Moreover, we have f_{i}|f_{i+1} since we can write

d_{i+1}(x) = e_{i}(x)d_{i}(x)

so that

f_{i+1}(x) = \frac{(−x)^{deg d_{i+1}}}{det(T |_{M_{i+1}})}d_{i+1} (\frac{1}{x})

= \frac{(−x)^{deg e_{i}}(−x)^{deg d_{i}}}{det(T |_{M_{i+1}})} d_{i} (\frac{1}{x}) e_{i} (\frac{1}{x})

= \frac{det(T_{M_{i}})}{det(T |_{M_{i+1}})}f_{i}(x)(−x)^{deg e_{i}}e_{i} (\frac{1}{x}) .

There is another way to see this. The decomposition

V_{T} ≃ ⨁ k[x]/d_{i}

is giving us the rational canonical form of T. Let d_{i} be as above. Then the companion matrix to d_{i} is

C_{i} = (\begin{matrix}0 & & & & & −a_{0i} \\ 1 & 0 & & & & −a_{1i} \\ & 1 & 0 & & & ⋮ \\ & & ... & ... & & ⋮ \\ & & & ... & 0 & ⋮ \\ & & & & 1 & −a_{(r_{i}−1)i} \end{matrix}) .

By a direct calculation, it is easy to verify that

C_{i}^{−1} = (\begin{matrix}−\frac{a_{1i}}{a_{0i}} & 1 & 0 & & & 0 \\ −\frac{a_{2i}}{a_{0i}} & & 1 & 0 & & 0 \\ ⋮ & & & ... & ... & ⋮ \\ −\frac{1}{a_{0i}} & & & & 0 & 1 \end{matrix}) .

By appropriately flipping columns and rows, it can be seen that the rational canonical form for C_{i}^{−1} is the companion matrix D_{i} to f_{i}(x), where f_{i} is above. Since the rational canonical form of T is

(\begin{matrix}C_{1} \\ & C_{2} \\ & & . & . & . \\ & & & & & C_{r} \end{matrix})

it follows that the rational canonical form of T^{−1} is

(\begin{matrix}D_{1} \\ & D_{2} \\ & & . & . & . \\ & & & & & D_{r} \end{matrix}) .

This then gives us the decomposition

V_{T^{−1}} ≃ ⨁_{i=1}^{r} k[x]/f_{i}