We prove a slightly stronger statement, namely:
Theorem: \({\mathbf{Z}}\) is initial in the category of unital rings and ring homomorphisms.
This means that if we are given any such ring \(R\), there is exactly one map \({\mathbf{Z}}\to R\).
Then, given an abelian group \(A\), we can take \(R = \hom_{\text{Ab}}(A, A)\), the hom set of abelian group endomorphisms, which is itself a unital ring. This will imply that there is a unique map \({\mathbf{Z}}\to \hom_{\text{Ab}}(A, A)\), and since all such maps induce \({\mathbf{Z}}{\hbox{-}}\)module structures on \(A\), the result will follow.
Proof: Let \(R\) be arbitrary and \(1_R\) be its multiplicative identity. We first show that there exists a ring homomorphism \({\mathbf{Z}}\to R\), namely
\begin{align*}
\phi: {\mathbf{Z}}&\to R \\
n &\mapsto \sum_{i=1}^n 1_R
.\end{align*}
Note that \(\phi(1) = 1_R\) and \(\phi(-1) = -1_R\), and it is routine to check that \(\phi\) is a ring homomorphism.
Now toward a contradiction, suppose there were another such ring homomorphism \(\psi: {\mathbf{Z}}\to R\). From the definition of a ring homomorphism, \(\psi\) must satisfy,
\begin{align*}
\psi(1) &= 1_R \\
\psi(-1) &= -1_R
,\end{align*}
and by \({\mathbf{Z}}{\hbox{-}}\)linearity, we must have \begin{align*} \psi(n) = \psi(\sum_{i=1}^n 1) = \sum_{i=1}^n \psi(1) = \sum_{i=1}^n 1_R = \phi(n), \end{align*}
and so \(\psi(x) = \phi(x)\) for every \(x\in {\mathbf{Z}}\). But this precisely means that \(\psi = \phi\) as ring homomorphisms. \(\hfill\blacksquare\)
2
a
Let \(\phi: {\mathbf{Z}}^4 \to {\mathbf{Z}}^3\) be a linear map which in the standard basis \(\mathcal B\) is represented by
\begin{align*}
T &\coloneqq[\phi]_{\mathcal B} =
[f_1^t, f_2^t, f_3^t, f_4^t] =
\left[\begin{array}{cccc}
1 & 2 & 0 & 3 \\
0 & -3 & 3 & 1 \\
-1 & 1 & 1 & 5
\end{array}\right]
.\end{align*}
Then \(\operatorname{im}T = \mathop{\mathrm{span}}_{\mathbf{Z}}\left\{{f_1 ,f_2, f_3, f_4}\right\} \coloneqq N\) by construction.
We can then compute the echelon form
\begin{align*}
\left(\begin{array}{cccc}
1 & 1 & 1 & 5 \\
0 & 3 & 1 & 8 \\
0 & 0 & 4 & 9
\end{array}\right)
,\end{align*}
which has pivots in columns \(1,2,\) and \(3\), and thus
\begin{align*} N = \mathop{\mathrm{span}}_{\mathbf{Z}}\left\{{f_1, f_2, f_3}\right\} \end{align*}
b
Without loss of generality, we can consider the image of the reduced matrix
\begin{align*}
A' =
\left(\begin{array}{ccc}
-1 & 2 & 0 \\
0 & -3 & 3 \\
1 & 1 & 1
\end{array}\right)
,\end{align*}
since \(N = \operatorname{im}A = \operatorname{im}A'\).
When computing the characteristic polynomial, we find that \(\chi_{A'}(x) = (x+3)(x+2)(x-2)\), which means that \(A'\) has distinct eigenvalues. We can thus immediately write
\begin{align*}
JCF(A) =
\left[\begin{array}{c|c|c}
2 & 0 & 0 \\
\hline
0 & -2 & 0 \\
\hline
0 & 0 & -3
\end{array}\right]
.\end{align*}
From this, we can obtain the Smith normal form,
\begin{align*}
SNF(A') =
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 12 \\
\end{array}\right]
,\end{align*}
which allows us to read off
\begin{align*}
\operatorname{im}A' \cong {\mathbf{Z}}\oplus {\mathbf{Z}}\oplus 12{\mathbf{Z}}
,\end{align*}
and thus
\begin{align*}
{\mathbf{Z}}^3/N \cong \frac{{\mathbf{Z}}\oplus {\mathbf{Z}}\oplus {\mathbf{Z}}}{{\mathbf{Z}}\oplus {\mathbf{Z}}\oplus 12{\mathbf{Z}}} \cong {\mathbf{Z}}/12{\mathbf{Z}}.
.\end{align*}
3
The elementary divisors are given by:
\begin{align*}
(x-1)^3 && (x^2+1)^4 && (x+2) \\
(x-1) && (x^2 + 1)^2 && \\
&& (x^2+1)^2 &&
.\end{align*}
The invariant factors are:
\begin{align*}
d_3 &= (x-1)^3(x^2+1)^4(x+2) \\
d_2 &= (x-1)(x^2+1)^2 \\
d_1 &= (x^2+1)^2
.\end{align*}
4
Lemma: \((2, x) {~\trianglelefteq~}{\mathbf{Z}}[x]\) is not a principal ideal.
Proof: If this ideal were generated by a single element \(p(x)\), then \(p \divides 2\) would force \(p \in {\mathbf{Z}}\). But this means that the element \(x\not\in (p)\), a contradiction. \(\hfill\blacksquare\)
Suppose toward a contradiction that \(J = (2, x) {~\trianglelefteq~}{\mathbf{Z}}[x]\) is a direct sum of cyclic submodules of \(R \coloneqq{\mathbf{Z}}[x]\).
Then write \begin{align*} J = M_1 \oplus M_2 \oplus \cdots \oplus M_n \end{align*}
where each \(M_i = \alpha_i {\mathbf{Z}}[x]\) is a cyclic \({\mathbf{Z}}[x]{\hbox{-}}\)module.
Note that by the lemma, we can not have \(n=1\), since this would mean \(J = \alpha_1 {\mathbf{Z}}[x] = (\alpha_1)\) where we can identify cyclic submodules with principal ideals.
On the other hand, we also can’t have \(n \geq 2\). Since the sum is direct, this forces (for example) \(M_1 \cap M_2 = \emptyset\).
However, take the two generating elements \(\alpha_1, \alpha_2 \in {\mathbf{Z}}[x]\) and consider their product. Noting that \({\mathbf{Z}}[x]\) is a commutative ring, we have
\begin{align*}
\alpha_1 \alpha_2 \in \alpha_1 {\mathbf{Z}}[x] = M_1 \text{ since } \alpha_2 \in {\mathbf{Z}}[x]
\alpha_1 \alpha_2 = \alpha_2 \alpha_1 \in \alpha_2 {\mathbf{Z}}[x] = M_2 \text{ since } \alpha_1 \in {\mathbf{Z}}[x]
,\end{align*}
and so \(\alpha_1\alpha_2 \in M_1 \cap M_2\), a contradiction. So no such direct sum decomposition is possible. \(\hfill\blacksquare\)
5
Irreducible: Let \(a\in M\) be arbitrary; we can then consider the cyclic submodule \(aR {~\trianglelefteq~}M\). Since \(M\) is irreducible, we must have \(aR = 0\) or \(aR = M\). If \(aR = 0\) then \(a\) must be \(0\).
Otherwise, \(aR = M\) implies that \(M\) itself is a cyclic module with generator \(a\). Since \(R\) is a PID, we can find an element \(p\) such that \(\operatorname{Ann}_R(M) = (p) {~\trianglelefteq~}R\), in which case \(M \cong R/(p)\).
It is also necessarily the case that \((p)\) is maximal, for if there were another ideal \((p) \subseteq J {~\trianglelefteq~}R\), then \(J/(p) {~\trianglelefteq~}R/(p) \cong M\) is a submodule by the correspondence theorem for ideals. But this necessarily forces \(J/(p) = 0\) or \(M\) by irreducibility of \(M\), so \(J = (p)\) or \(R\).
Thus irreducible modules are exactly the cyclic modules, or equivalently those of the form \(R/(p)\) where \((p)\) is a maximal ideal.
Indecomposable: We first note that by the structure theorem for modules over a PID, any module \(M\) has a primary decomposition \(M \cong \bigoplus_i R/(p_i^{k_i})\).
This means that if \(M\) is indecomposable, we must have \(M \cong R/(p^n)\) (with a single summand) for some prime \(p \in R\); otherwise the primary decomposition would yield additional summands. Moreover, by the Chinese Remainder Theorem, \(M\) can not be decomposed further.
Thus all indecomposable module are of the form \(R/(p^n)\) for some \(n\geq 1\).
6
Suppose \(T: V \to V\) is not invertible, then \(\dim \operatorname{im}T < n\) and \(\dim \ker T > 0\) by the Rank-Nullity theorem. This means that there is a nontrivial \(\mathbf{v} \in \ker T\), and a nontrivial vector \(\mathbf{w} \in \operatorname{im}(T)\), so let \(S\) be the matrix formed by the outer product \(\mathbf{v} \mathbf{w}^t\).
We then consider how \(ST\) acts on vectors \(\mathbf{x}\):
\begin{align*}
TS\mathbf{x}
&= T\mathbf{v} \mathbf{w}^t \mathbf{x} \\
&= (T\mathbf{v} )\mathbf{w}^t \mathbf{x} \\
&= \mathbf{0} \mathbf{w}^t \mathbf{x} \\
&= \mathbf{0_n} \mathbf{x} \\
&= \mathbf{0}
,\end{align*}
where \(\mathbf{0_n}\) is the \(n\times n\) matrix of all zeros.
Similarly,
\begin{align*}
ST\mathbf{x}
&\coloneqq S \mathbf{y} \\
&= \mathbf{v} \mathbf{w}^t \mathbf{y} \\
&= {\left\langle {\mathbf{w}},~{\mathbf{y}} \right\rangle} \mathbf{v} \\
&= c_i \mathbf{v} \\
&\neq \mathbf{0},
\end{align*}
where \({\left\langle {\mathbf{w}},~{\mathbf{y}} \right\rangle} \coloneqq c_i \neq 0\) because \(\mathbf{y} \in \operatorname{im}(T) = (\operatorname{im}(T)\perp)\perp\), so \(\mathbf{y}\) and \(\mathbf{w}\) can not be orthogonal.
\(\hfill\blacksquare\)
7
a
Note that if \(A = 0\) or \(I\) then \(A\) is patently diagonal, so suppose otherwise. Since \(A^2 = A\), we have \(A^2 - A = 0\) and thus \(A\) satisfies the polynomial \(p(x) = x^2 - 1 = x(x-1)\). Moreover, since \(A\neq 0, I\), the minimal polynomial is at least degree – since \(p\) is monic, it must in fact be the minimal polynomial.
We can immediately deduce that the size of the largest Jordan block corresponding to \(\lambda = 0\) is exactly 1, as is the size of the largest Jordan block corresponding to \(\lambda = 1\). But this says that all Jordan blocks must be size 1, so \(JNF(A)\) has no off-diagonal entries and is thus diagonal.
b
If \(k\) is the multiplicity of \(\lambda = 0\) as an eigenvalue, we have
\begin{align*}
A \sim
\left[\begin{array}{ccc|cccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \ddots & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \ddots & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right]
,\end{align*}
which has a \(k\times k\) block of zeros and an \((n-k)\times(n-k)\) block of 1s.
8
In both cases, we will need the characteristic polynomials \(\chi_A(x)\), since \(RCF(A)\) will depend on the invariant factors of \(A\). We will also use the fact that over the algebraic closure \(\overline {\mathbf{Q}}\), the minimal and characteristic polnyomials must have the same roots.
a
Suppose \(m_A(x) = (x-1)(x^2+1)^2\), which is a degree 5 polynomial. Since \(\deg \chi_A\) must be 6 and \(m_A\) must divide \(\chi_A\) in \({\mathbf{Q}}[x]\), the only possibility in this case is that \begin{align*} \chi_A(x) = (x-1)^2 (x^2+2)^2. \end{align*}
To determine the possible invariant factors \(\left\{{d_i}\right\}\), we can just note that \(\prod d_i = \chi_A(x)\) and \(d_n = m_A(x)\). With these constraints, the only possibility is
\begin{align*}
d_1 &= (x-1) \\
d_2 &= (x-1)(x^2+1)^2.
,\end{align*}
from which we can immediately obtain the elementary divisors:
\begin{align*}
(x-1), (x-1), (x^2+1)^2
.\end{align*}
Then noting that \begin{align*} d_2 =d_2 = (x-1)(x^2+1)^2 = x^5 -x^4 + 4x^3 -4x^2 + 4x - 4, \end{align*}
there is thus only one possible Rational Canonical form:
\begin{align*}
RCF(A) &=
\left[\begin{array}{c|ccccc}
1 & 0 & 0 & 0 & 0 & 0\\
\hline
0 & 0 & 0 & 0 & 0 & 4 \\
0 & 1 & 0 & 0 & 0 & -4 \\
0 & 0 & 1 & 0 & 0 & 4 \\
0 & 0 & 0 & 1 & 0 & -4 \\
0 & 0 & 0 & 0 & 1 & 1 \\
\end{array}\right]
.\end{align*}
b
The constraints \(m_A(x) = (x^2+1)^2(x^3+1)\) with \(\deg m_A(x) = 7\) and \(\deg \chi_A(x) = 10\) forces \begin{align*} \chi_A(x) = (x^2+1)^2 (x^3+1)^2. \end{align*}
Furthermore, the invariant factors are similarly constrained, and so the only possibility is
\begin{align*}
d_1 &= (x_3 + 1) \\
d_2 &= (x^2+1)^2 (x^3+1)
\end{align*}
with corresponding elementary divisors
\begin{align*}
(x^3 + 1), (x^3 + 1), (x^2 + 1)^2
.\end{align*}
Noting that \begin{align*} d_2 = (x^2+1)^2 (x^3+1) = x^5 + x^3 + x^2 + 1, \end{align*}
we have
\begin{align*}
RCF(A) &=
\left[\begin{array}{cc|ccccc}
0 & -1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
0 & 0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
\end{array}\right]
.\end{align*}
9
The standard computation of \(\operatorname{det}(xI - A) = 0\) shows that \(\chi_A(x) = \operatorname{det}(xI - A) = (x-1)^2 (x+1)^2\), and so the eigenvalues of \(A\) are \(1, -1\). We want the minimal polynomial of \(A\), which is given by \(\prod(x-\lambda_i)^{\alpha_i}\) where \(\alpha_i = \dim E_{\lambda_i}\) is the geometric multiplicity of \(\lambda_i\).
Another standard computation shows that \begin{align*} \lambda = 1 \implies \operatorname{rank}(A - 1I) = 2 \implies \dim \ker (A-1I) = 4-2 = 2 \end{align*} and similarly \begin{align*} \lambda = -1 \implies \operatorname{rank}(A + I) = 3 \implies \dim \ker(A + I) = 4 - 3 = 1. \end{align*}
We thus have
\begin{align*}
p_A(x) &= (x-1) (x+1)^2\\
\chi_A(x) &= (x-1)^2 (x+1)^2
.\end{align*}
To compute \(JCF(A)\), we use the following facts:
-
For \(\lambda = 1\),
- Since \((x-1)^1\) occurs in \(p_A(x)\), the largest Jordan block for \(\lambda = 1\) is size 1.
- Since \((x-1)^2\) occurs in \(\chi_A(x)\), the sum of sizes of all such Jordan blocks is 2.
- Since \(\dim E_1 = 2\), there are 2 such Jordan blocks.
-
For \(\lambda = -1\),
- Since \((x+1)^2\) occurs in \(p_A(x)\), the largest Jordan block for \(\lambda = -1\) is size 2.
- Since \((x+1)^2\) occurs in \(\chi_A(x)\), the sum of sizes of all such Jordan blocks is 2.
- Since \(\dim E_{-1} = 1\), there is 1 such Jordan block.
We can thus immediately write
`\begin{align*} JCF(A) = J_{-1}^2 \oplus 2 J_{1}^1
\left[\begin{array}{cccc} -1 & 1 & 0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ \end{array}\right] .\end{align*}`{=tex}
By arguments similar to the previous two problems, the only possible invariant factor decomposition is given by
\begin{align*}
d_1 &= (x+1) \\
d_2 &= (x-1)^2 (x+1)
\end{align*}
and thus
\begin{align*}
RCF(A) &= C(d_1) \oplus C(d_2) =
\left[\begin{array}{c|ccc}
-1 & 0 & 0 & 0 \\ \hline
0 & 0 & 0 & -1 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
\end{array}\right]
.\end{align*}
10
Suppose \(A^* = A\). It is then a fact that \(A\) is self-adjoint, and so for every \(\mathbf{v}\in V\) we have \begin{align*} {\left\langle {A\mathbf{v}},~{\mathbf{v}} \right\rangle} = {\left\langle {\mathbf{v}},~{A^*\mathbf{v}} \right\rangle} = {\left\langle {\mathbf{v}},~{A\mathbf{v}} \right\rangle}. \end{align*}
a
Let \((\lambda, \mathbf{v})\) be an eigenvalue of \(A\) with one of its corresponding eigenvectors, so \(A\mathbf{v} = \lambda{\mathbf{v}}\).
On one hand, `\begin{align*} {\left\langle {A\mathbf{v}},~{\mathbf{v}} \right\rangle}
{\left\langle {\lambda\mathbf{v}},~{\mathbf{v}} \right\rangle}
\lambda {\left\langle {\mathbf{v}},~{\mathbf{v}} \right\rangle}
\lambda {\left\lVert {\mathbf{v}} \right\rVert}^2 ,\end{align*}`{=tex}
while on the other hand,
`\begin{align*} {\left\langle {A\mathbf{v}},~{\mathbf{v}} \right\rangle}
{\left\langle {\mathbf{v}},~{A^*\mathbf{v}} \right\rangle}
{\left\langle {\mathbf{v}},~{A\mathbf{v}} \right\rangle}
{\left\langle {\mathbf{v}},~{\lambda\mathbf{v}} \right\rangle}
\overline{\lambda} {\left\langle {\mathbf{v}},~{\mathbf{v}} \right\rangle}
\overline{\lambda} {\left\lVert {\mathbf{v}} \right\rVert}^2 .\end{align*}`{=tex}
Equating these expressions, we find that \begin{align*} \lambda = \overline{\lambda} \implies \lambda \in {\mathbf{R}}. \hfill\blacksquare \end{align*}
b
We can make use of the following fact:
Theorem (Schur): Every square matrix \(A \in M_n({\mathbf{C}})\) is unitarily similar to an upper triangular matrix, i.e. there exists a unitary matrix \(U\) such that \(A = UTU^{-1}\) where \(T\) is upper-triangular.
Applying this theorem yields \(A = UTU^{-1}\) and thus \(T = U^{-1}A U\). In particular, \(A \sim T\).
Noting that if \(U\) is unitary then \(U^{-1}= U^*\), we have
\begin{align*}
T^*
&= (U^{-1}A U)^* \\
&= U^* A^* (U^{-1})^* \\
&= U^* A^* U^{**} \\
&= U^{-1}A^* U \\
&= T
,\end{align*}
and so \(T^* = T\).
Since \(T\) is upper triangular, this forces \(T_{ij} = 0\) whenever \(i\neq j\) But this makes \(T\) diagonal, so \(A\) is similar to a diagonal matrix. \(\hfill\blacksquare\)
Proof of Schur’s Theorem: We’ll proceed by induction on \(n = \dim_{\mathbf{C}}(V)\), and showing that there is an orthonormal basis of \(V\) such that the matrix of \(A\) is upper triangular.
Lemma: If \(V\) is finite dimensional and \(\lambda\) is an eigenvalue of \(A\), then \(\overline{\lambda}\) is an eigenvalue of \(A^*\).
Proof: \begin{align*} \operatorname{det}(A-\lambda I) = 0 = \overline{\operatorname{det}\left(A^{*} - \overline{\lambda} I\right)}. \hfill\blacksquare \end{align*}
Since \({\mathbf{C}}\) is algebraically closed, every matrix \(A \in M_n({\mathbf{C}})\) will have an eigenvalue, since its characteristic polynomial will have a root by the Fundamental Theorem of Algebra.
So let \(\lambda_1, \mathbf{v}_1\) be an eigenvalue/eigenvector pair of the adjoint \(A^*\).
Consider the space \(S = \mathop{\mathrm{span}}_{\mathbf{C}}\left\{{\mathbf{v}_1}\right\}\); then \(V = S \oplus S^\perp\). The claim is that the original \(A\) will restrict to an operator on \(S^\perp\), which has dimension \(n-1\). The inductive hypothesis will then apply to \({\left.{{A}} \right|_{{S^\perp}} }\).
Note that if this holds, there will be an orthonormal basis \(\mathcal{B}\) of \(S^\perp\) such that the matrix \begin{align*} \mathbf{A}' \coloneqq[ {\left.{{A}} \right|_{{S^\perp}} }]_{\mathcal{B}} \end{align*} will be upper triangular. We would then be able to obtain an orthonormal basis \(\mathcal{C} \coloneqq\mathcal{B} \cup\left\{{\mathbf{v_1}}\right\}\) of \(S \oplus S^\perp = V\).
Since we have a direct sum decomposition, the matrix of \(A\) with respect to \(\mathcal{C}\) can be written in block form as
`\begin{align*} [A]{\mathcal{C}} &= \left[\begin{array}{cc} [ {\left.{{A}} \right|{{S}} }]{\mathcal{C}} & 0 \ 0 & [ {\left.{{A}} \right|{{S^\perp}} }]_{\mathcal{C}} \end{array}\right]
\left[\begin{array}{cc} [ {\left.{{A}} \right|{{S}} }]{\left{{\mathbf{v}1}\right}} & 0 \ 0 & [ {\left.{{A}} \right|{{S^\perp}} }]_{\mathcal{B}} \end{array}\right]
\left[\begin{array}{cc} \lambda_1 & 0 \ 0 & \mathbf{A}’ \end{array}\right] ,\end{align*}`{=tex}
which is upper-triangular since \(\mathbf{A}'\) is upper-triangular.
To see that \(A\) does indeed restrict to an operator on \(S^\perp\), we need to show that \(A(S^\perp) \subseteq S^\perp\). So let \(\mathbf{s} \in S^\perp\); then \({\left\langle {\mathbf{v}_1},~{\mathbf{s}} \right\rangle} = 0\) by definition. Then \(A\mathbf{s} \in S^\perp\) since
\begin{align*}
{\left\langle {\mathbf{v}_1},~{A\mathbf{s}} \right\rangle}
&= {\left\langle {A^* \mathbf{v}_1},~{\mathbf{s}} \right\rangle} \\
&= {\left\langle {\lambda_1 \mathbf{v}_1},~{\mathbf{s}} \right\rangle} \\
&= \lambda_1 {\left\langle {\mathbf{v}_1},~{\mathbf{s}} \right\rangle} \\
&= 0
.\end{align*}
\(\hfill\blacksquare\)