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I know that if A is an nxn intger matrix, with A = XRX^-1 where R is the rcf of A. Then R is also an integer matrix.

My question is Does X and X^- are integer matrices as well ?

user593805
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There's no reason for the similarity matrices to be integer matrices. The most natural construction is to start with some fairly arbitrary $v_1$ and define $v_2=Av_1$, $v_3=Av_2$, and so on until we run out of room and they stop being linearly independent; the $v_i$ will be columns of $X$. While starting with an integer vector $v_1$ will make the other $v_i$ integer vectors, there's no control of the determinant, and we will be dividing by something when it comes time to take the inverse $X^{-1}$.

As an example, I constructed a random (entries discrete uniform between $-5$ and $5$) $5\times 5$ matrix: $$A=\begin{pmatrix}4&-4&-2&-3&2\\0&-4&1&-5&-3\\4&5&4&-5&-3\\-3&-3&-2&0&4\\3&-5&3&-5&2\end{pmatrix},\quad X=\begin{pmatrix}1&4&23&162&1749\\0&0&10&23&853\\0&4&38&319&2665\\0&-3&-32&-355&-2725\\0&3&45&383&3869\end{pmatrix}$$ Then $X$ has determinant $183716$, so its inverse certainly isn't an integer matrix.

Incidentally, for this example, we get $R=\begin{pmatrix}0&0&0&0&-1349\\1&0&0&0&1531\\0&1&0&0&-548\\0&0&1&0&71\\0&0&0&1&6\end{pmatrix}$.

jmerry
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  • I have a follow up question. Is it always with rational entries ? Does X, X^-1 € M(n, Q) ? – user593805 Feb 18 '19 at 11:08
  • There are many choices of $X$; it depends on what we "seed" it with. There's always a way to do it with rational numbers, at least - rational canonical form stays entirely inside the field that the elements come from, unlike eigenvalue-based forms that depend on extending to an algebraically closed field. – jmerry Feb 18 '19 at 11:14
  • Omg !!! This really helps . Thnk u so much. Do you have a paper or book that will prove that is true ? Can you give me some references..? Thank u so much. – user593805 Feb 18 '19 at 11:17
  • No reference here. I'm pretty sure I didn't learn this stuff from a textbook; I recall being annoyed that it wasn't in the textbook when I actually got around to taking an upper-division linear algebra course. The full canonical form has a lot of fiddly bits, but it's actually very easy to prove in the one-block case when the minimal polynomial has full degree - I found $X$ in my example by just picking a vector and repeatedly multiplying by $A$ to get the columns. – jmerry Feb 18 '19 at 11:26
  • Ohhh hahaha.. Still, thank you so much. Yes yes. if minimal polynomial is equal to chararacteristic polynomial. Thnk you again. More power hehe – user593805 Feb 18 '19 at 11:30