# Show the matrix commutes with companion matrix is a polynomial

Let $$A$$ be a linear transform on $$n$$-dimensional $$V$$ over a field $$F$$. Under a basis $$\alpha_1, \cdots, \alpha_n$$, the matrix representation of $$A$$ is as follows: $$A = \begin{bmatrix} 0 & 0 & \dots & 0 & -a_0 \\ 1 & 0 & \dots & 0 & -a_1 \\ 0 & 1 & \dots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{bmatrix}.$$ Let $$C(A):= \{T: T\text{ is a linear transform on V and } TA = AT \}$$, and let $$F[A]$$ denotes all the polynomials in $$A$$. Show that: $$C(A) = F[A]; \dim(C(A)) = n.$$

First of all, the minimal polynomial $$m(\lambda)$$ of $$A$$ is the same as its characteristic polynomial $$f(\lambda)$$, namely $$m(\lambda) = f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots a_0$$. Thus, plugging in $$A$$, we see that all $$A^{k}$$ with $$k \geq n$$ could be expressed by $$I, A, A^2, \cdots, A^{n-1}$$. So $$\dim F[A] \leq n$$. If $$\dim F[A] < n$$, say $$k_0 I + k_1 A + \cdots + k_r A^r = 0$$ with $$r < n$$ and some $$k_j \neq 0$$, then we have that $$g(\lambda) = k_0 + k_1 \lambda + \cdots + k_r \lambda^r$$ is another polynomial with $$g(A) = 0$$. By the definition of minimal polynomial, we must have that $$r \geq n$$, a contradiction. So $$\dim(F[A]) = n$$, and it remains to show the first equality $$C(A) = F[A]$$.

Also, one could see that $$F[A] \subseteq C(A)$$. But I am not sure how to show the other direction. Could someone give me a hint?

Define a linear map $$\Psi \colon C(A) \rightarrow V$$ by $$\Psi(T) = T\alpha_1$$. Let's show that this map is an isomorphism. First, note that $$A^i \in C(A)$$ for all $$i \in \mathbb{N}_0$$ and
$$\Psi(I) = \alpha_1, \psi(A) = A\alpha_1 = \alpha_2, \cdots, \psi(A^{n-1}) = A^{n-1}\alpha_1 = \alpha_n.$$
This shows that $$\Psi$$ is surjective. Next, let's assume that $$\psi(T) = T\alpha_1 = 0$$. Then $$T\alpha_2 = T(A\alpha_1) = A(T\alpha_1) = A(0) = 0, \\ T\alpha_3 = T(A\alpha_2) = A(T\alpha_2) =0, \\ \vdots,\\ T\alpha_n = T(A\alpha_{n-1}) = A(T\alpha_{n-1}) = 0 \\$$
which shows that $$T \equiv 0$$. This shows that $$\Psi$$ is injective. Hence, $$\dim C(A) = n$$ and since $$\dim F[A] = n$$ and $$F[A] \subseteq C(A)$$, we deduce that $$F[A] = C(A)$$.
• Yes I think this is excellent! But how do we come up with such $\Psi$? – mathdoge Jun 5 at 12:00
• The matrix $A$ acts on your basis by $\alpha_1 \mapsto \alpha_2 \dots \mapsto \alpha_n$ (where $x \mapsto y$ means that $A$ sends $x$ to $y$). By applying $T$ and using the fact that $A$ and $T$ commute, you can see that we also have $T\alpha_1 \mapsto T\alpha_2 \dots \mapsto T\alpha_n$. From here you can already see that if we know $T\alpha_1$ and $A$, we also know $T\alpha_i$ for all $2 \leq i \leq n$. – levap Jun 5 at 12:06