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The trace of a $2\times2$ positive integer matrix is a given constant positive value. How many possible choices are there such that the determinant is greater than 0? Each element of matrix is positive.

AJ95
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2 Answers2

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Suppose we're looking for a positive integer matrix with trace $M$. The matrix must have the form $$ \pmatrix{ M - n & a\\ b & n } $$ Where $a,b \geq 1$ and $1 \leq n \leq M-1$. For the determinant to be positive, we must have $$ (M-n)n \geq ab $$ So, we may compute the number of matrices with trace $M$ and a positive determinant as follows: For a given $n$ from $1$ to $M-1$, let $\phi_n(M)$ be the number of products $ab$ such that $ab \leq (M-n)n$ (order matters). Then the total number of matrices for a given $M$ is $$ \sum_{n=1}^{M-1}\phi_n(M) $$ I do not know of any well known function to simplify $\phi$, but at least it is now a number theory problem.

Ben Grossmann
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Infinitely many. Choose any positive integer $m\geq3$, then the matrices $\pmatrix{ m-1 & 1 \cr 1 & 1\cr}$ all have trace $m$ and positive determinant.

EDIT: After the comment pointing out the mistake, I am embarrassed. The answer is finite, bounded below by $m-1$.

  • Set of all matrices with integer (or rational) entries of all possible shapes (including matrices that are square, rectangular, $173\times 29$, singular, of negative determinant) make a set of aleph null cardinality. – P Vanchinathan Apr 05 '14 at 15:55
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    You've only given one example of a matrix with trace $m$ and positive determinant... – ah11950 Apr 05 '14 at 16:12