Prove that if $A$ and $B$ are $3\times 3$ matrices over the field $F$, a necessary and sufficient condition that $A$ and $B$ be similar over $F$ is that they have the same characteristic polynomial and the same minimal polynomial.
My attempt: $(\Rightarrow)$ It’s trivial, and holds for $n\times n$ matrices. $(\Leftarrow)$Let $f_A$ and $f_B$ be characteristic polynomial of $A$ and $B$, respectively. Let $m_A$ and $m_B$ be minimal polynomial of $A$ and $B$, respectively. We first prove for $n=2$. Suppose $A,B\in M_{2\times 2}(F)$. We claim $A\backsim B$$\iff$$m_A=m_B$. Suppose $m_A=m_B=m$. Clearly degree of minimal polynomial can’t be zero. Case (1): $\deg (m)=1$. Then $m=x-c$. By example 2 section 7.2, $A,B\backsim \begin{bmatrix} c& 0\\ 0&c\\ \end{bmatrix}$. Since $\backsim$ is an equivalence relation, we have $A\backsim B$. Case (2): $\deg (m)=2$. Then $m=x^2+c_1x+c_0$. By example 2 section 7.2, $A,B\backsim \begin{bmatrix} 0& -c_0\\ 1& -c_1\\ \end{bmatrix}$. Since $\backsim$ is an equivalence relation, we have $A\backsim B$.
Following is slight generalisation of example 2. Suppose $\dim (V)=3$ and $T:V\to V$ be a linear operator. Let $m$ and $f$ be minimal and characteristic polynomial of $T$, respectively. Case (1): $\deg (m)=3$. Then $m=x^3+c_2x^2+c_1x+c_0$. By Cayley-Hamilton theorem, $m=f$. By theorem 3 corollary 2 section 7.2, $T$ has a cyclic vector. That is $\exists \alpha \in V$ such that $V=Z(\alpha ;T)$. By theorem 2 section 7.1, $\exists \mathcal{B}$ basis of $V$ such that $[T]_\mathcal{B}$ is companion matrix of $m$. So $[T]_\mathcal{B}=\begin{bmatrix} 0&0&-c_0\\ 1&0&-c_1\\ 0&1&-c_2\\ \end{bmatrix}$. Case (2): $\deg (m)=2$. Then $m=x^2+c_1x+c_0$. By cyclic decomposition theorem, $\exists \alpha_1,…,\alpha_r\in V\setminus \{0\}$ with respective $T$-annihilators $p_1,…,p_r$ such that $V=Z(\alpha_1;T)\oplus \cdots Z(\alpha_r;T)$, and $p_{i+1}|p_i$, and $p_1\cdots p_r=f$, and $p_1=m$. So $\deg (p_1)=\deg (m)=2$. By theorem 1 section 7.1, $\dim Z(\alpha_1,T)=\deg (p_1)=2$. Which implies $r=2$ and $\dim Z(\alpha_2;T)=\deg (p_2)=1$. Let $p_2=x-d$. So $\exists \mathcal{B}$ basis of $V$ such that $[T]_B= \begin{bmatrix} 0&-c_0&0\\ 1&-c_1&0\\ 0&0&d\\ \end{bmatrix}$. Case (3): $\deg (m)=1$. Then $m=x-c$ and $T=cI$. So $[T]_\mathcal{B}= \begin{bmatrix} c&0&0\\ 0&c&0\\ 0&0&c\\ \end{bmatrix}$, for any basis $\mathcal{B}$ of $V$. By theorem 5 section 7.2, every $3\times 3$ matrix over $F$ is similar to exactly one matrix of the types
$$\begin{bmatrix} 0&0&-c_0\\ 1&0&-c_1\\ 0&1&-c_2\\ \end{bmatrix}, \quad \begin{bmatrix} 0&-c_0&0\\ 1&-c_1&0\\ 0&0&d\\ \end{bmatrix}, \quad \begin{bmatrix} c&0&0\\ 0&c&0\\ 0&0&c\\ \end{bmatrix}.$$
We show for $n=3$. Suppose $A,B\in M_{3\times 3}(F)$, $f_A=f_B=f$ and $m_A=m_B=m$. Case (1): $\deg (m)=1$. Then $m=x-c$. By above result, $A,B\backsim \begin{bmatrix} c&0&0\\ 0&c&0\\ 0&0&c\\ \end{bmatrix}$. Since $\backsim$ is an equivalence relation, we have $A\backsim B$. Case (2): $\deg (m)=2$. Then $m=x^2+c_1x+c_0$. By above result, $A\backsim \begin{bmatrix} 0&-c_0&0\\ 1&-c_1&0\\ 0&0&d\\ \end{bmatrix}$ and $B\backsim \begin{bmatrix} 0&-c_0&0\\ 1&-c_1&0\\ 0&0&d’\\ \end{bmatrix}$ such that $m(x-d)=f_A$ and $m(x-d’)=f_B$. Since $f_A=f_B$, we have $d=d’$. Since $\backsim$ is an equivalence relation, we have $A\backsim B$. Case (3): $\deg (m)=3$. Then $m=x^3+c_2x^2+c_1x+c_0$. By above result, $A,B\backsim \begin{bmatrix} 0&0&-c_0\\ 1&0&-c_1\\ 0&1&-c_2\\ \end{bmatrix}$. Since $\backsim$ is an equivalence relation, we have $A\backsim B$. Is my proof correct?
Best Answer
Two long for a comment. Just to explain why I think all the "stuff before “By theorem 5 section 7.2” is useless.
By theorem 5 section 7.2, $A$ is similar to a matrix of one the following three forms, which we shall call canonical: $$\begin{pmatrix} 0&0&-c_0\\ 1&0&-c_1\\ 0&1&-c_2\end{pmatrix}, \quad \begin{pmatrix} 0&-c_0&0\\ 1&-c_1&0\\ 0&0&d\end{pmatrix},\quad\begin{pmatrix} c&0&0\\ 0&c&0\\ 0&0&c\end{pmatrix}$$ where in the second form, $X-d\mid X^2+c_1X+c_0.$
The characteristic polynomials of these three matrices are respectively $$X^3+c_2X^2+c_1X+c_0,\quad(X^2+c_1X+c_0)(X-d),\quad(X-c)^3$$ and the minimal polynomials are also easily computed: $$X^3+c_2X^2+c_1X+c_0,\quad X^2+c_1X+c_0,\quad X-c.$$ Therefore, the characteristic and minimal polynomials of $A$ determine completely which canonical form it is similar to. Hence two matrices having the same characteristic and minimal polynomials are similar.