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Let $F$ be a subfield of the field of complex numbers, and let $A$ be an $n\times n$ matrix over $F$. Let $p$ be the minimal polynomial for $A$. If we regard $A$ as a matrix over $\Bbb{C}$, then $A$ has a minimal polynomial $f$ as an $n\times n$ matrix over $\Bbb{C}$. Use a theorem on linear equations to prove $p=f$. Can you also see how this follows from the cyclic decomposition theorem?

My attempt: Let $A\in M_n(F)$. By theorem 5 section 7.2, $A$ is similar over $F$ to a unique matrix $$P=\begin{bmatrix}P_1& & \\ &\ddots & \\ & &P_r\\ \end{bmatrix}\in M_n(F)$$ where $P_i$ is companion matrix of non scalar monic polynomial $p_i$, and $p_{i+1}|p_i$, for all $1\leq i\leq r-1$. Let $A\in M_n(\Bbb{C})$. By theorem 5 section 7.2, $A$ is similar over $\Bbb{C}$ to a unique matrix $$Q=\begin{bmatrix}Q_1& & \\ &\ddots & \\ & &Q_s\\ \end{bmatrix}\in M_n(\Bbb{C})$$ where $Q_i$ is companion matrix of non scalar monic polynomial $g_i$, and $g_{i+1}|g_i$, for all $1\leq i\leq s-1$. It’s easy to check, $p_1=p$ and $g_1=f$ (here is proof). Since $F\subseteq \Bbb{C}$, we have $P\in M_{n}(\Bbb{C})$ and $A$ is similar over $\Bbb{C}$ to $P$. By uniqueness, $P=Q$. Which implies $r=s$ and $p_i=g_i$. In particular $p_1=g_1$. Hence $p=f$. Is my proof correct?

user264745
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  • You can shorten you proof by saying immediately after the description of $Q$ that $P$ matches that description, so $P=Q$ by uniqueness. – Marc van Leeuwen Mar 19 '23 at 08:01

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The exercise asked first to show this by using a property of linear equations. Since I don't have the book, I cannot give you chapter and verse, however it seems clear what it intended here. A basic fact of linear algebra is that linearly independent family of $l~$vectors in $F^k$ is still linearly independent over$~\Bbb{C}$. To see why this is so, form the $k\times{l}$ matrix over $F$ whose columns are those vectors; the linear independence means that linear homogeneous system with this matrix has trivial (i.e., $0$ dimensional) solution. But that means row reducing the matrix leaves a pivot in every column; this reduction can be performed with coefficients in$~F$ only, but is valid over$~\Bbb{C}$ as well, so the system still has trivial solution over$~\Bbb{C}$.

Now to the problem itself, if the minimal polynomial$~p$ of $A$ over$~F$ has degree$~d$, this implies that the $d$ powers $A^0, A^1, \ldots, A^{d-1}$ are linearly independent over$~F$ (in $F^{n\times n}$), and $p$ expresses a linear dependence of the family obtained by adding $A^d$ to the list. But then the minimal polynomial$~f$ of$~A$ over$~\Bbb{C}$ cannot have degree less than$~d$, and since $p$ still expresses a linear dependence of $A^0,A^1,\ldots,A^d$ over$~\Bbb{C}$, and we can conclude that $f=p$.

Marc van Leeuwen
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