TITLE: Similarity to companion matrix, uniqueness
QUESTION [1 upvotes]: 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?

REPLY [2 votes]: 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 $Q = I$.
{"set_name": "stack_exchange", "score": 1, "question_id": 384067}