Homework 8

1. We could compute the Smith normal form for each matrix and discover that they have the same Smith normal form. This would let us conclude that $A$ and $B$ have the same invariant factors, hence the same minimal polynomial, characteristic polynomial and rational canonical form. However, this problem is about proving those last three facts without computing invariant facto

   So instead we directly compute characteristic polynomials using determinants. We first compute

   $\begin{array}{rl}{c}_A(x)& =det(x{I}_3-A)\\ & =\left|\begin{array}{ccc}x& 4& -85\\ -1& x-4& 30\\ 0& 0& x-3\end{array}\right|\\ & =(x-3)\left|\begin{array}{cc}x& 4\\ -1& x-4\end{array}\right|\\ & =(x-3)(x^2-4x+4)\\ & =(x-3)(x-2{)}^2,\end{array}$

   and similarly

   $\begin{array}{rl}{c}_B(x)& =det(x{I}_3-B)\\ & =\left|\begin{array}{ccc}x-2& -2& -1\\ 0& x-2& 1\\ 0& 0& x-3\end{array}\right|\\ & =(x-2)(x-2)(x-3).\end{array}$

2. The minimal polynomial of $A$ needs to divide the characteristic polynomial and have the same roots. Since the characteristic polynomial of $A$ is$c_A(x)=(x-2{)}^2(x-3)$, this leaves only two possibilities: either $m_A(x)=(x-2)(x-3)$ or $m_A(x)=(x-2{)}^2(x-3)$. To determine which it is, we compute

   $\begin{array}{rl}(A-2I)(A-3I)& =\left[\begin{array}{ccc}-2& -4& 85\\ 1& 2& -30\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-3& -4& 85\\ 1& 1& -30\\ 0& 0& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}2& 4& -50\\ -1& -2& 25\\ 0& 0& 0\end{array}\right]\end{array}$

   Since $(A-2I)(A-3I)\ne 0$ we have $m_A(x)\ne (x-2)(x-3)$ and so (by our above discussion) $m_A(x)=(x-2{)}^2(x-3)$.

   Since $B$ has the same characteristic polynomial as $A$, the same logic applies. We compute

   $\begin{array}{rl}(B-2I)(B-3I)& =\left[\begin{array}{ccc}0& 2& 1\\ 0& 0& -1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}-1& 2& 1\\ 0& -1& -1\\ 0& 0& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}0& -2& -2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\end{array}$

   As above, the fact that $(B-2I)(B-3I)\ne 0$ is enough to conclude that $m_B(x)=(x-2{)}^2(x-3)$.

3. For a $3\times 3$ matrix with minimal polynomial $m(x)=(x-2{)}^2(x-3)$ and characteristic polynomial $c(x)=m(x)$, the only possible list of invariant factor(s) is $a_1(x)=(x-2{)}^2(x-3)=x^3-7x^2+16x-12$. The corresponding rational canonical form matrix is

   $R=\left[\begin{array}{ccc}0& 0& 12\\ 1& 0& -16\\ 0& 1& 7\end{array}\right].$

We first recall the following facts about invariant factors. For an $n\times n$ matrix $A$ over a field $F$, the invariant factors of $A$ are polynomials $a_1(x),a_2(x),\dots ,a_m(x)\in F[x]$ that satisfy the following properties:

• Each is a monic nonconstant polynomial
• Each $a_i(x)$ divides $a_{i+1}(x)$ in $F[x]$
• The largest invariant factor is the minimal polynomial: $a_m(x)=m_A(x)$
• The product of the invariant factors is the characteristic polynomial $c_A(x)$, which is degree $n$

Lastly, two $n\times n$ matrices over $F$ are similar if and only if they have the same lists of invariant factors. As a consequence, if $A$ and $B$ are similar then they always have the same minimal and characteristic polynomials.

1. Suppose $A$ is any $2\times 2$ matrix over a field $F$. The minimal polynomial $m_A(x)$ is at least degree 1 but at most degree 2 (since it divides the characteristic polynomial $c_A(x)$, which is degree 2). If $m_A(x)$ is degree 1, then $A$ must have exactly two invariant factors $a_1(x)$ and $a_2(x)=m_A(x)$ which must both be linear. Since $a_1(x)$ divides $a_2(x)$, we must in fact have $a_1(x)=a_2(x)=m(x)=x-\lambda$. Then the rational canonical form of $A$ is

   $D=\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]=\lambda I_2.$

   Since such scalar matrices commute with all $2\times 2$ matrices under product, they are fixed by conjugation; i.e., $PD{P}^{-1}=D$ for every invertible $2\times 2$ matrix $P$. But $A$ is similar to $D$, so this implies we must actually have $A=D$; i.e., $A$ is a scalar matrix.

   So, if $A$ is not a scalar matrix then its minimal polynomial $m_A(x)$ must be quadratic and so must equal its characteristic polynomial. The matrix $A$ thus has exactly one invariant factor, namely $a_1(x)=m(x)=c(x)$. It now follows that if $A$ and $B$ are two non-scalar matrices with the same characteristic polynomial, then they have the same (single) invariant factor and hence are similar.

2. Suppose $A$ and $B$ have the same minimal and characteristic polynomials. We consider cases according to the degree of their shared minimal polynomial $m(x)$.

   • If $m(x)$ is degree 3, then we must have $m(x)=c(x)$ and so both matrices have the same single invariant factor $a_1(x)=m(x)=c(x)$. They are therefore similar.
   • If $m(x)$ is degree 2, then both $A$ and $B$ must have exactly two invariant factors, say $a_1(x)$ and $a_2(x)=m(x)$ for the matrix $A$ and $b_1(x)$ and $b_2(x)=m(x)$ for $B$. But we also have $a_1(x)m(x)=c(x)$ and similarly $b_1(x)m(x)=c(x)$, hence by cancellation (of the nonzero element $m(x)$ in the integral domain $F[x]$) we must have $a_1(x)=b_1(x)$. Thus, $A$ and $B$ have the same two invariant factors and are therefore similar.
   • If $m(x)$ is degree 1, then $A$ and $B$ both have the three invariant factors $a_1(x)=a_2(x)=a_3(x)=m(x)$. Thus, $A$ and $B$ have the same invariant factors and hence are similar.

3. There are many examples. For a specific example, let $A$ be a matrix that has invariant factors $a_1(x)=x-1$, $a_2(x)=x-1$, and $a_3(x)=(x-1{)}^2$, and let $B$ be a matrix with invariant factors $b_1(x)=(x-1{)}^2$ and $b_2(x)=(x-1{)}^2$. Then $A$ and $B$ both have minimal polynomial $m(x)=(x-1{)}^2$ and characteristic polynomial $c(x)=(x-1{)}^4$, but $A$ and $B$ are not similar (since they have different lists of invariant factors). Two such matrices (already in rational canonical form!) are

   $A=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& -2\\ 0& 0& 1& 2\end{array}\right]\text{and}B=\left[\begin{array}{cccc}0& -2& 0& 0\\ 1& 2& 0& 0\\ 0& 0& 0& -2\\ 0& 0& 1& 2\end{array}\right].$

1. Recall that the matrix $A$ endows ${\mathbf{Q}}^3$ with the structure of a $\mathbf{Q}[x]$-module, where $x$ acts as multiplication by $A$ (when using the standard basis for ${\mathbf{Q}}^3$). The minimal polynomial $m_A(x)$ is the monic generator for the annihilator $I_A$ of the torsion $\mathbf{Q}[x]$-module ${\mathbf{Q}}^3$. Now observe that $A^6=I_3$ if and only if $A^6-I_3=0_3$, which in turn is true if and only if $x^6-1$ is in the annihilator $I_A=(m_A(x))$; i.e., if and only if $m_A(x)$ divides $x^6-1$ in $\mathbf{Q}[x]$.

2. By part (1), the polynomial $m_A(x)$ is a nonconstant monic factor of $x^6-1$. It is also at most a cubic, since it divides the characteristic polynomial $c_A(x)$ (which is a cubic). In light of the irreducible factorization of $x^6-1$ in $\mathbf{Q}[x]$, the only possibilities are therefore

• $m_A(x)=x-1$
• $m_A(x)=x+1$
• $m_A(x)=x^2-x+1$
• $m_A(x)=x^2+x+1$
• $m_A(x)=(x-1)(x+1)$
• $m_A(x)=(x-1)(x^2-x+1)$
• $m_A(x)=(x-1)(x^2+x+1)$
• $m_A(x)=(x+1)(x^2-x+1)$
• $m_A(x)=(x+1)(x^2+x+1)$

3. There are at most three invariant factors $a_1(x)$, $a_2(x)$, $a_3(x)$, each must divide the next, each is nonconstant and monic, the largest is $m_A(x)$, and their product is the cubic $c_A(x)$. We run through the nine possibilities in part (2) and for each give the possible invariant factors:

• If $m_A(x)=x-1$, then the only possible list of invariant factors is

$x-1,x-1,x-1$

• If $m_A(x)=x+1$, then the only possible list of invariant factors is

  $x+1,x+1,x+1$

• If $m_A(x)=x^2-x+1$, there is no possible list of invariant factors. Thus this situation cannot occur.

• If $m_A(x)=x^2+x+1$, there is no possible list of invariant factors. Thus this situation cannot occur.

• If $m_A(x)=(x-1)(x+1)$, then there are two possible lists of invariant factors:

  $x-1,(x-1)(x+1)\text{or}x+1,(x+1)(x-1)$

• In each of the last four possibilities (for which