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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?

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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 $F[A] = C(A)$.

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  • $\begingroup$ Yes I think this is excellent! But how do we come up with such $\Psi$? $\endgroup$ – mathdoge Jun 5 at 12:00
  • $\begingroup$ 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 $2 \leq i \leq n$. $\endgroup$ – levap Jun 5 at 12:06
  • $\begingroup$ Thank you a lot! I used this method to deal with other questions successfully. $\endgroup$ – mathdoge Jun 5 at 12:42

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