3
$\begingroup$

Let $A$ be some matrix over $\mathbb{Q}$ (then it's also over $\mathbb{R}$). Suppose $A$ is invertible over $\mathbb{R}$ (that is, $A^{-1}$ is over $\mathbb{R}$). Prove that $A^{-1}$ is also over $\mathbb{Q}$.

I know that I have to prove that $A^{-1}$ contains no irrational numbers but I fail to do so. I would appreciate any suggestions.

$\endgroup$
3
  • $\begingroup$ Are we supposed to assume that the elements of $A$ are all rational numbers? Is that implied by the expression "some matrix over $\mathbb{Q}$? $\endgroup$
    – Mico
    Nov 9, 2014 at 13:19
  • $\begingroup$ @Mico - It just means that $A$ doesn't have irrational numbers. $\endgroup$
    – user191052
    Nov 9, 2014 at 14:36
  • $\begingroup$ @user191052 "Doesn't have irrational numbers" is the same as "Has only rational numbers". $\endgroup$ Nov 9, 2014 at 15:45

3 Answers 3

4
$\begingroup$

You can calculate an inverse of a matrix in multiple ways. (Adjuncts, extend it with identity matrix and Gauss eleminate it, LUP decompositions,...). But in the end, all you will be doing is taking sums and products of fractions so the outcomes can never be irrational. (Sum or products of fractions will still be fractions)

$\endgroup$
1
  • $\begingroup$ Intuitively, I understand it. I just wanted a more formal proof. But thanks! $\endgroup$
    – user191052
    Nov 9, 2014 at 11:52
2
$\begingroup$

If $A$ is an invertible matrix with entries in $ℚ$, what two things can you say about its determinant? Think of Cramer's Rule.

$\endgroup$
5
  • $\begingroup$ Yes, I thought about determinants. In particular, we know that if $A$ is invertible then $\det{A} \neq 0$ and also $\det{A} \in \mathbb{Q}$. Then $\det{A^{-1}}=\frac{1}{\det{A}}$. Thus $\det{A^{-1}} \in \mathbb{Q}$ and $A^{-1}$ is over $\mathbb{Q}$. However I'm not sure about this proof. Or is it correct? $\endgroup$
    – user191052
    Nov 9, 2014 at 11:49
  • $\begingroup$ @user131052: $\det(A^{-1})\in\mathbb Q$ does not imply $A^{-1}$ is over $\mathbb Q$. Take $\begin{pmatrix}\sqrt 2 & 0\\ 0 & \sqrt 2\end{pmatrix}$ for $A^{-1}$, for example. What k.stm suggests is to use this fact together with Cramer's formula for inverses. $\endgroup$ Nov 9, 2014 at 12:57
  • 1
    $\begingroup$ @rosebud - so basically what you mean is that because $\frac{1}{\det(A)} \in \mathbb Q$ and because $\mathrm{adj}(A)$ is over $\mathbb Q$ then $A^{-1}$ is over $\mathbb Q$? $\endgroup$
    – user191052
    Nov 9, 2014 at 15:02
  • $\begingroup$ @user191052 Yes. This also generalizes to arbitrary rings. $\endgroup$
    – k.stm
    Nov 9, 2014 at 16:18
  • 1
    $\begingroup$ @k.stm: To arbitrary commutative rings. $\endgroup$ Nov 9, 2014 at 17:01
2
$\begingroup$

I think finding inverse of a matrix by adjoint method gives the answer to your question; because determinant function takes values in the field of rational functions and the adjoint of a matrix also takes values in the field of rational numbers( while we are computing the adjoint of a matrix, we just multiply and add the elements in the field of rational numbers that is we do computations in a field so the adjoint of a matrix (over the field of rational numbers) will be a matrix over the field of rational numbers.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .