Are Similar Matrices Over the Reals Also Similar Over the Rationals?

[PingPong]

Homework Statement

Show that two matrices that have rational entries and are similar over the reals are also similar over the rationals. (Hint: Consider the polynomials from the rational canonical form over Q. What happens when we consider A as a real matrix?

Homework Equations

Rational canonical form (we've mostly been dealing with Q) is unique

The Attempt at a Solution

I've not gotten much. I've noticed that if f is the minimal polynomial of A over Q, g is the minimal polynomial of B over Q, and h is the minimal polynomial of A and B over R (it's the same since they're similar) then h divides both f and g over R. I wanted to somehow show that f divides g and reverse the argument to get g dividing f, but I haven't been able to do that.

In terms of the hint, if we look at the polynomials from the RCF of A over Q, they just get split when we consider consider A over R. A friend of mine suggested that this with the fact that the RCF is unique might help, though I'm not seeing how.

Thanks for any help that you can give.

 

[mighty2000]

Thank you for your question. It is a very interesting and important problem in linear algebra. I will try to provide some guidance and hints to help you solve it.

First, let's recall the definition of similar matrices. Two square matrices A and B are said to be similar if there exists an invertible matrix P such that P^-1AP = B. This means that A and B are essentially the same matrix, just written in different bases.

Now, let's consider the rational canonical form (RCF) of A and B over Q. As you mentioned, the RCF is unique, which means that A and B have the same RCF over Q. This RCF can be written as a block diagonal matrix with blocks of the form [Ci], where Ci is a companion matrix of a polynomial pi(x) over Q. This means that the minimal polynomial of A and B over Q is the product of these polynomials pi(x).

Now, let's consider A and B as real matrices. This means that we can view them as elements of R^n×n, where n is the dimension of the matrices. Since A and B are similar over R, they have the same minimal polynomial over R, let's call it h(x). This polynomial h(x) also divides the minimal polynomial of A and B over Q, which is the product of the polynomials pi(x).

Here comes the key idea: since h(x) divides the product of the polynomials pi(x) over Q, it must also divide each of the polynomials pi(x) over R. This is because h(x) is a polynomial over R, and if it divides a polynomial over Q, it must also divide the polynomial over R (think about the definition of divisibility for polynomials).

Now, going back to the definition of similar matrices, we can see that the minimal polynomial h(x) must be the same for both A and B over R. This means that A and B are similar over R, and since they have rational entries, they are also similar over Q.

I hope this helps to guide you in the right direction. Good luck with your proof!