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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 8 Solutions}
\begin{document}

\maketitle
\begin{enumerate}
    \stepcounter{enumi}
    \item\begin{enumerate}
        \item 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
        $$G\simeq \bigoplus \Z/(p_i^{a_i})$$,
        where the (not necessarily distinct) $p_j$ satisfy
        $$\prod_i p_i^{a_i}=\# G.$$  Since $500=2^2\cdot 5^3$, we have the following $6$ possibilities for the primary decomposition
        of $G$, and hence for $G$ up to isomorphism:
        \begin{align*}
            &\Z/(2)\oplus \Z/(2)\oplus \Z/(5)\oplus \Z/(5)\oplus \Z/(5)\\
            &\Z/(2)\oplus \Z/(2)\oplus \Z/(5)\oplus \Z/(5^2)\\
            &\Z/(2)\oplus \Z/(2)\oplus \Z/(5^3)\\
            &\Z/(2^2)\oplus \Z/(5)\oplus \Z/(5)\oplus \Z/(5)\\
            &\Z/(2^2)\oplus \Z/(5)\oplus \Z/(5^2)\\
            &\Z/(2^2)\oplus \Z/(5^3).
        \end{align*}
        \item 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
        $$M_T\simeq k[x]/(f_1(x))\oplus\ldots\oplus k[X]/(f_r(x)),$$
        where $f_r$ is the minimal polynomial of $T$, $f_i|f_{i+1}$, and $x^5(x-1)^5=\prod_i f_i$.
        It follows that we have
        $$f_i=x^{a_i}(x-1)^{b_i}$$
        with $\sum_i a_i=\sum_i b_i=5$.  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
        \begin{align*}
            &5\\
            &4+1\\
            &3+2\\
            &3+1+1\\
            &2+2+1\\
            &2+1+1+1\\
            &1+1+1+1+1.
        \end{align*}
        Therefore, there are $7\cdot 7=49$ 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
        $$x^{r_1}(x-1)^{r_2}.$$

        \item 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 $4\times 4$ 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:
        \begin{align*}
            &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
        \end{align*}
        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
        $$\begin{pmatrix}0 & 0 & & \\ 1 & 1 &  & \\  &  & 0 & 0\\  &  & 1 & 1  \end{pmatrix}.$$
    \end{enumerate}
    \item Let $A=(a_{ij})$ be an $n\times n$ 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 $\det B=\pm \det A$ and
    $$B=\begin{pmatrix}d_1 & & & \\ & d_2 & &  \\ & & \ddots &  \\ & & & d_n  \end{pmatrix},$$
    where
    $$d_i|d_{i+1}$$ and $d_i>0$ for $1\leq i\leq l\leq n$.
    Moreover, we have
    $$M\simeq \Z/(d_1)\oplus\cdots\oplus \Z/(d_l)\oplus \Z^{n-l}$$
    and
    $$\det B=|\det A|=|d_1\cdots d_n|.$$
    Clearly, $|\det A|\neq 0$ 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
    $$\# M=\# \Z/(d_1)\oplus\cdots\oplus \Z/(d_n)=\prod_{i=1}^n \# \Z/(d_i)=\prod_{i=1}^n d_i=\det B=|\det A| .$$

    \item
    \begin{enumerate}[i]
        \item Let $T$ be an $n\times n$ invertible matrix over a field $k$, and let $P_T(x)$ denote the characteristic
        polynomial of $T$.  We have
        \begin{align*}
        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}}\left(\frac{1}{x}\right),
        \end{align*}
        so that
        $$P_{T^{-1}}(x)=\frac{(-x)^n}{\det(T)}P_{T}\left(\frac{1}{x}\right).$$

        \item Suppose that $V_T=M_1\oplus M_2\oplus\cdots\oplus M_r$ where $M_i\subset V$ 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, $TM_i\subset M_i$ 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
        $$f_i(x)=\frac{(-x)^{\deg d_i}}{\det( T|_{M_i})}d_i\left(\frac{1}{x}\right)=x^{r_i}+\frac{a_{1i}}{a_{0i}}x^{r_i-1}+\cdots+\frac{a_{(r_i-1)i}}{a_{0i}},$$
        where
        $$d_i(x)=x^{r_i}+a_{(r_i-1)i}x^{r_i-1}+\cdots +a_{0i}.$$
        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
        $$V_{T^{-1}}\simeq \bigoplus_{i=1}^{r} k[x]/(f_i(x)).$$
        Moreover, we have $f_i|f_{i+1}$ since we can write
        $$d_{i+1}(x)=e_i(x)d_i(x)$$ so that
        \begin{align*}
            f_{i+1}(x)&=\frac{(-x)^{\deg d_{i+1}}}{\det( T|_{M_{i+1}})}d_{i+1}\left(\frac{1}{x}\right)\\
            &=\frac{(-x)^{\deg e_i}(-x)^{\deg d_{i}}}{\det( T|_{M_{i+1}})}d_{i}\left(\frac{1}{x}\right)e_{i}\left(\frac{1}{x}\right)\\
            &=\frac{\det(T_{M_i})}{\det( T|_{M_{i+1}})}f_{i}(x)(-x)^{\deg e_i}e_{i}\left(\frac{1}{x}\right).
        \end{align*}

        There is another way to see this.  The decomposition
        $$V_T\simeq \bigoplus k[x]/d_i$$
        is giving us the rational canonical form of $T$.  Let $d_i$ be as above.  Then the companion matrix to $d_i$ is
        $$C_i=\begin{pmatrix}0 & & & & &-a_{0i}\\ 1&0&&& &-a_{1i}\\ &1&0 & && \vdots\\ &  & \ddots & \ddots   && \vdots\\
         & & & \ddots & 0 & \vdots\\ & & & & 1 & -a_{(r_i-1)i} \end{pmatrix}.$$
         By a direct calculation, it is easy to verify that
         $$C_i^{-1}=\begin{pmatrix}-\frac{a_{1i}}{a_{0i}} & 1&0 & & & 0\\ -\frac{a_{2i}}{a_{0i}}&&1& 0& &0\\ \vdots && & \ddots &\ddots& \vdots\\
         -\frac{1}{a_{0i}}& & &  & 0 & 1 \end{pmatrix}.$$
         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
         $$\begin{pmatrix}C_1 & & & \\ & C_2 & &  \\ & & \ddots & \\ & & & C_{r}\end{pmatrix}$$ it follows that the rational canonical form
         of $T^{-1}$ is
         $$\begin{pmatrix}D_1 & & & \\ & D_2 & &  \\ & & \ddots & \\ & & & D_{r}\end{pmatrix}.$$
         This then gives us the decomposition
         $$V_{T^{-1}}\simeq \bigoplus_{i=1}^r k[x]/f_i$$

    \end{enumerate}
\end{enumerate}

\end{document}
