Theorem 5, Section 7.2 of Hoffman’s Linear Algebra

Question

Definition: $A\in M_{n\times n}(F)$ is in rational form if $$A=\begin{bmatrix} A_1& & \\ & \ddots & \\ & & A_r\\ \end{bmatrix}$$ where $A_i$ is companion matrix of non scalar monic polynomial $p_i$, and $p_{i+1}|p_i$, for all $1\leq i\leq r-1$.

---

Let $F$ be a field and let $B$ be an $n\times n$ matrix over $F$. Then $B$ is similar over the field $F$ to one and only one matrix which is in rational form.

Proof: Let $T$ be the linear operator on $F^n$ which is represented by $B$ in the standard ordered basis. As we have just observed, there is some ordered basis for $F^n$ in which $T$ is represented by a matrix $A$ in rational form. Then $B$ is similar to this matrix $A$. Suppose $B$ is similar over $F$ to another matrix $C$ which is in rational form. This means simply that there is some ordered basis for $F^n$ in which the operator $T$ is represented by the matrix $C$. If $C$ is the direct sum of companion matrices $C_i$ of monic polynomials $g_1,…,g_s$ such that $g_{i+1}$ divides $g_i$ for $i=1,…s- 1$,then it is apparent that we shall have non-zero vectors $\beta_1,…,\beta_s$ in $V$ with $T$- annihilators $g_1,…,g_s$ such that $$V=Z(\beta_1;T)\oplus \dots \oplus Z(\beta_s;T)$$ But then by the uniqueness statement in the cyclic decomposition theorem, the polynomials $g_i$ are identical with the polynomials $p_i$ which define the matrix $A$. Thus $C=A$.

Que: How to rigioursly show existence of $\beta_1,…,\beta_s\in V\setminus \{0\}$ such that $T$-annihilator of $\beta_i$ is $g_i$ and $V= Z(\beta_1;T)\oplus \dots \oplus Z(\beta_s;T)$?

Answer 1

Let $C\in M_{n\times n}(F)$ is in rational form. Then $C= \begin{bmatrix}C_1& & \\ &\ddots & \\ & &C_s\\ \end{bmatrix}$, where $C_i\in M_{k_i\times k_i}(F)$ is companion matrix of $g_i=x^{k_i}+\sum_{j=0}^{k_i-1}a_{i,j}x^j$. That is $C_i=\begin{bmatrix} & & &-a_{i,0}\\ 1& & &-a_{i,1}\\ & \ddots & & \vdots\\ & & 1&-a_{i,k_i-1}\\ \end{bmatrix}$. Since $B\backsim C$, we have $\exists \mathcal{B}$ basis of $F^n$ such that $C=[T]_\mathcal{B}$. Let $\mathcal{B}=\{\beta_{1,1},…,\beta_{1,k_1},…,\beta_{s,1},…,\beta_{s,k_s}\}$. Note $|\mathcal{B}|=n$, since $\sum_{i=1}^sk_i=n$. Let $B_i=\{\beta_{i,1},…,\beta_{i,k_i}\}$, for all $i\in J_s$. Then $\mathcal{B}=\bigcup_{i=1}^sB_i$. Fix $i\in J_s$. Since $T(\beta_{i,j})=\beta_{i,j+1}$, $\forall 1\leq j\leq k_i-1$, we have $B_i=\{\beta_{i,1},T(\beta_{i,1}),…,T^{k_i-1}(\beta_{i,1})\}$. Since $T(\beta_{i,k_i})=T^{k_i}(\beta_{i,1})=-a_{i,0}\beta_{i,1}-…-a_{i,k_i-1}T^{k_i-1}(\beta_{i,1})$ and by linear independence of $B_i$, we have $g_i$ is $T$-annihilator of $\beta_{i,1}$.

By theorem 1 section 7.1, $B_i$ is basis of $Z(\beta_{i,1};T)$. Since $\text{span}(\mathcal{B})=V$, we have $V=Z(\beta_{1,1};T)+…+Z(\beta_{s,1};T)$. Since $\mathcal{B}$ is independent, we have $Z(\beta_{1,1};T),…,Z(\beta_{s,1};T)$ are independent. Hence $V= Z(\beta_{1,1};T)\oplus …\oplus Z(\beta_{s,1};T)$.