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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

2 Answers2

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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
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The argument I found essentially copies the format of Theorem 5.22 from Axler (4th edition) about the "existence, uniqueness, and degree of minimal polynomial[s]".

By induction if dim $V = 0$, then every map is the $0$ map and can be represented by the minimal polynomial $I$. We can also directly show the result for the case of dim $V = 1$: note that every map $T$ that can be represented via a rational coefficient matrix from some basis of $V$ to itself is simply of the form $T(x*b) = k*x*b$ where $k$ is a rational scalar, and $b$ is the basis vector for $V$. Then the minimal polynomial is indeed rational as it is simply "$T - kI$" (note that the minimal polynomial cannot be of lower degree, i.e. degree $0$, as the only constant polynomial that maps every vector to $0$ is the polynomial "$0$", and that is not a monic polynomial). Now comes the induction step that will mimic 5.22:

Assume that the result holds for all $\text{dim}\: V < n$. Now let $\text{dim}\: V = n$, and consider an appropriate $T$. Let $v \neq 0$, and consider the smallest positive integer "$k$" s.t. $\{v,T(v),...,T^k(v)\}$ becomes linearly dependent: $$c_k*T^k(v) = c_{k-1}*T^{k-1}(v) + ... + c_1*T^1(v)+ c_0*I(v)$$

Where $c_k$ may not be $0$, otherwise the assumption of having a minimal value for "$k$" is wrong. So then divide through by "$c_k$" and note we have a matrix equation:

$$T^k(v) = [T^{k-1}(v),...,v] [c'_{k-1},..., c'_(0)]^T$$

where $c'_i = \frac{c_i}{c_k}$, and where the LHS is an $n \times 1$ vector, and the RHS is an $n \times k$ matrix (call it $M$, that is, let $M = [T^{k-1}(v),...,v]$) multiplied by a $k \times 1$ vector. Now, we know that $k \leq n$ by the fact that $\text{dim}\: V = n$.

Also, we may now interpret $M$ as a matrix over the field $\mathbb{Q}$, and note that because its columns are independent over $\mathbb{R}$, they are also independent over $\mathbb{Q}$. Now we know by the independence of the columns of $M$ that $k = \text{rank of the columns of } M = \text{rank of the rows of }M$ (as the column rank of a matrix always equals its row rank).

Now, since the rows of M have rank $k$, it follows that we can find a matrix $L$ over $\mathbb{Q}$ s.t. $LM$ is the $k \times k$ identity matrix. Now on the LHS, $LT^k(v)$, whatever it may be, is a vector of rational numbers, and so it follows that "$[c'_{k-1},..., c'_(0)]^T$" is a vector of rational numbers.

Finally we now know that $$-T^k(v) + c'_{k-1}T^{k-1}(v) + ... + c'_1T^1(v)+ c'_0I(v)$$ is a rational polynomial, which we denote by "$p(T)$". Now $p(T)$ when applied to $V$ has a kernel with $\text{dim ker } p(T) \geq k$, and we now copy the idea of the proof from 5.22 to finish the induction step, i.e. to argue that the minimal polynomial of $T$ is equal to $p(T)q(T)$ where $q(T)$ is the minimal polynomial of $T$ over the smaller subspace $\text{range } p(T)$, and which, by our inductive hypothesis is a rational polynomial.

We are done....

Just_a_fool
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