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Let $A \in \mathcal{M}_{n\times n}(\mathbf{R})$ and $b \in \mathbf{R}^n$ with $n\geqslant2$ be given.

Now should $(A,b)$ satisfy the Kalman rank condition \begin{equation} \text{rank}[b \,\, Ab \,\, \ldots \,\, A^{n-1}b] = n, \end{equation} or equivalently \begin{equation} \text{span}\{b, Ab, \ldots, A^{n-1}b\} = \mathbf{R}^n, \end{equation} then it is known that there exists $P \in \mathrm{GL}_n(\mathbf{R})$ such that \begin{equation} A = P\mathbf{A}P^{-1} \quad \text{ and } \quad b = Pe_n, \end{equation} where $e_n = [0, \ldots, 1]^T$ denotes the vector of the canonical basis of $\mathbf{R}^n$, whereas $\mathbf{A}$ is the companion matrix of $A$ (as defined in https://en.wikipedia.org/wiki/Companion_matrix): \begin{equation} \mathbf{A} = \begin{bmatrix} 0 & 1 & 0 &\ldots & 0\\ 0 & 0 & 1 &\ldots & 0 \\ \vdots & & & \ddots &0\\ -a_n &\ldots &\ldots & \ldots & -a_1 \end{bmatrix}, \end{equation} where $a_1, \ldots, a_n$ are the coefficients of the characteristic polynomial of $A$: \begin{equation} \chi_A(x) = x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n. \end{equation}

My question is:

Is the matrix $P$ unique?

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Yes. Suppose there was another such matrix. Since $P$ is invertible, it could be written as $QP$ for some $n \times n$ matrix $Q$. Thus $A Q P = Q P {\bf A} = Q A P$, so $Q$ commutes with $A$, and $Q P e_n = b$. But then $A^k b = A^k Q P e_n = Q A^k P e_n = Q A^k b$. Since the vectors $A^k b$ span $\mathbb R^n$, we conclude that $Q = I$.

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