I am taking an abstract algebra course and am getting quite confused with the terminology of invariant factors, elementary divisors, and the normal forms. I am asked to compute the rational canonical form of a matrix $A$ whose entries are all $1 \in \mathbb F_p$, for some prime $p$.

I am not asking for an answer, I would much rather be pointed in the correct direction with hints or resources. Here is what I currently believe to understand:

I'm very sure the minimal polynomial of this matrix is $m(x) = x - 1$, since the matrix has all entries of $1$. I thought the minimal polynomial was used in the calculation of the rational canonical form, but am failing to find a resource on how $m(x)$ can be of any help.

I also attempted to compute the Smith normal form of an example $2 \times 2$ matrix $A - xI$ to find it's invariant factors, and arrived at

$$ SNF(A - xI) = \left( \begin{array}{c c} 1 & 0\\ 0 & x(x - 2) \end{array} \right)$$

From that, I believe the invariant factor is merely $x(x - 2)$ and thus the rational canonical form is $$ RCF(A) = \left( \begin{array}{c c} 0 & 0\\ 1 & 2 \end{array} \right)$$

So, my main questions are, is this correct? And, since the entries are in $\mathbb F_p$, how should I go about handling $n\times n$ matrices where $n \ge p$.

  • 4
    $\begingroup$ Only the identity matrix has $x-1$ as its minimum polynomial. To find the actual minpoly of $A$, just compute $A^2$, and look at the result. $\endgroup$
    – quasi
    Mar 8 '17 at 23:41
  • $\begingroup$ OH! Im so foolish. I was wondering what was meant by $x - c$ for constants $c$, so $m(x) = x - 1 = x - 1 \cdot I$. Thank you. $\endgroup$ Mar 8 '17 at 23:46
  • $\begingroup$ So, in my calculations, I have reached the fact that the invariant factors of an $n \times n$ matrix with entries all equal to $1 \in \mathbb F_p$ are $x, x, \dots, x, x(n - x)$, where there are $n - 2$ $x$'s in that list. For $n \ge p$, I'm concerned about what happens to the invariant factors. Do you have any suggestions on how to deal with this? $\endgroup$ Mar 9 '17 at 0:45
  • $\begingroup$ Why not just replace $n$ by $(n\text{ mod }p)$? $\endgroup$
    – quasi
    Mar 9 '17 at 2:46
  • $\begingroup$ Yes, that seems to be the conclusion I reached also. Thank you for your help. $\endgroup$ Mar 9 '17 at 5:30

Checking your results with Maple, for the matrix $$ A = \left( \begin{array}{c c} 1 & 1\\ 1 & 1 \end{array} \right) $$

I get

$$ SNF(A - xI) = \left( \begin{array}{c c} 1 & 0\\ 0 & x(x-2) \end{array} \right) \\[10pt]$$

$$ RCF(A) = \left( \begin{array}{c c} 0 & 0\\ 1 & 2 \end{array} \right)$$

which are almost the same as yours, except for sign changes in the lower right corner entries.

  • $\begingroup$ Ah, I see I made a sign error. Thank you $\endgroup$ Mar 9 '17 at 0:22
  • $\begingroup$ The sign changes do not matter correct? Since we are only looking for uniqueness up to units, where $-1$ is a unit correct? $\endgroup$ Mar 9 '17 at 0:29
  • $\begingroup$ I'm not the one to ask on this. I was just using Maple to check your results for the $2{\,\times\,}2$ case. $\endgroup$
    – quasi
    Mar 9 '17 at 1:00

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