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Suppose $T \in \mathcal{L}(V)$ is such that with respect to some basis of $V$, all entries of the matrix of $T$ are rational numbers. Explain why all coefficients of the minimal polynomial of $T$ are rational numbers.

My hunch is that it is because the rationals are closed under addition and multiplication, but I don't know how to formalize this into a rigorous proof. Any help is appreciated!

Aphelli
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    Hi, welcome to MSE! You may find this answer and the ideas in it helpful. – Izaak van Dongen Mar 05 '24 at 22:25
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    Let $M$ be the matrix of $T$, $d$ be the degree of its min. poly. and $$A=\pmatrix{\text{vec}(I),\text{vec}(M),\text{vec}(M^2),\ldots,\text{vec}(M^d)},$$ where $\text{vec}(X)$ denotes vectorisation. Then $A$ has rational entries. Also, over some field containing $\mathbb Q$, the first $d$ columns of $A$ are linearly independent and the last one is a linear combination of the others. Hence $Ax=0$ for some nonzero vector $x$ with rational entries, because $x$ can be obtained from Gaussian elimination and backward substitution. – user1551 Mar 05 '24 at 23:01

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The trick here is essentially this lemma.

Lemma: Let $k$ be a subfield of $K$, and let $v_1, \ldots, v_n$ be $k$-independent elements of $k^m$. Then $v_1, \ldots, v_n$ are $K$-independent elements of $K^n$.

There are several ways this can be proved. Here is one. Fix a $k$-linear isomorphism $\phi : k^n \to k^n$ such that $\phi(v_i) = e_i$. Then $\phi$ extends to a $K$-linear isomorphism $\Phi : K^n \to K^n$ such that $\Phi(v_i) = e_i$ (just using the matrix of $\phi$ and of its inverse); thus, the $v_i$s are $K$-independent.

Now identify $\mathcal{L}(V)$ with $K^{n^2}$ in the obvious way (here $k = \mathbb{Q}$, $K = \mathbb{R}$); the operator $M$, being a matrix with rational coefficients, is identified with an element of $k^{n^2}$, as are all powers of $M$. Let $m$ be the lowest degree of a polynomial $P \in k[x]$ such that $P(M) = 0$. Then $M^0, \ldots, M^{m - 1}$ are $k$-independent elements of $k^{n^2}$, and hence are $K$-independent elements of $K^{n^2}$, so there is no polynomial $Q$ of degree $<m$ such that $Q(M) = 0$. Thus, $P$ is indeed minimal, and has coefficients in $k$.

Mark Saving
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  • Thank you for your in-depth response! Since this question is from a linear algebra textbook, is there any way to prove it without using abstract algebra? –  Mar 06 '24 at 00:06
  • @TheAlgebraist this answer is not using anything past linear algebra, just plug in $k = \mathbb{Q}$, $K = \mathbb{R}$ everywhere. – ronno Mar 06 '24 at 07:57
  • @TheAlgebraist Everything I wrote should be more than doable using Axler’s book; the only tricky part is showing $\phi$ exists. To do this, you need merely complete $v$ to a basis of $k^n$ and make $\phi$ the change of basis map. To extend $\phi$ and $\phi^{-1}$ to $\Phi$ and $\Phi^{-1}$, just define $\Phi(e_i) = \phi(e_i)$ and $\Phi^{-1}(e_i) = \phi^{-1}(e_i)$; it is then easy to show that $\Phi$ and $\Phi^{-1}$ are inverses. If you would like me to cite particular Axler lemmas, I can, but it will take a bit more time. – Mark Saving Mar 07 '24 at 00:07