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?
> 
> 
> 



Similarity to companion matrix, uniqueness
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 $Q = I$.


