I feel pretty stupid for doing this, but here goes anyway. Earlier today I asked: Find an $n\times n$ integer matrix with determinant 1 and $n$ distinct eigenvalues. As it turns out, for my problem I need an $n$-by-$n$ matrix with integer entries, determinant $1$ and $n$ distinct strictly positive eigenvalues (where $n$ is an arbitrary positive integer).

I tried modifying the answer given above (a very easy one, by the way) by attempting to transform the $n$'th roots of unity to distinct numbers on the positive axis. (If you add a root of unity to its conjugate, you end up in $\mathbb{R}$) but whatever I do, my transformation ends up having determinant $0$. What to do, what to do...

(When things like this occur, i.e., having a question answered before finding out that it was the wrong question, what is the usual procedure?)


Let $(a_i)_i$ be distinct positive real s.t. $\Pi_ia_i=1$ and $P(x)=\Pi_i(x-a_i)$ is a polynomial with coefficients in $\mathbb{Z}$. Choose for $A$ the companion matrix associated to $P$.

EDIT 1: OK Bryder, I understand.

a) For companion matrices, see http://en.wikipedia.org/wiki/Companion_matrix

b) Instead of choosing the $(a_i)$, it is easier to choose any $P\in\mathbb{Z}[x]$ of degree $n$, with $n$ distinct positive roots, in the form $x^n+b_{n-1}x^{n-1}+\cdots+b_1x\pm 1$.

EDIT 2. If you want an explicit polynomial $P$, do as follows: if $n=2p$ is even, then $P_{2p}=(x^2-3x+1)(x^2-4x+1)(x^2-5x+1)\cdots$ and, if $n=2p+1$ is odd, then $P_{2p+1}=P_{2p}\times (x-1)$.

  • $\begingroup$ What do you mean by companion matrix? And how would you go about choosing $(a_i)_i$? $\endgroup$ – Bryder Apr 14 '14 at 21:19
  • $\begingroup$ cf. math.stackexchange.com/questions/753485/… How many times have you asked the question? $\endgroup$ – loup blanc Apr 14 '14 at 21:22
  • $\begingroup$ As mentioned above, I made a mistake in my original phrasing of the problem. As I accepted an answer before I discovered my mistake, I thought it would be better to ask a new question (with an apology, of course). $\endgroup$ – Bryder Apr 14 '14 at 21:28
  • $\begingroup$ Wow, that is freakishly smart. Never heard about companion matrices before. Thanks! $\endgroup$ – Bryder Apr 15 '14 at 6:45

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