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Here $B^T$ denotes the transpose of $B$.

$A$ and $B$ are invertible $3\times 3$ matrices with integer entries.

$A$ is symmetric positive definite with at most two zero entries.

We want the determinants of both $A$ and $B$ be as small as possible (in terms of absolute value)

For a fixed matrix $A$, we want to find all matrices $B$ with small determinants, preferably less than 10, if they exist.

Drew
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2 Answers2

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two completely different questions.........

$$ \left( \begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{array} \right) \left( \begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) $$

You should be aware that, for any symmetric matrix $A$ of integers, there is a rational matrix $B$ of determinant $\pm 1$ such that $B^T A B$ is diagonal. See reference for linear algebra books that teach reverse Hermite method for symmetric matrices

If, as appears, you are writing a computer program to solve the second question: it is more difficult than you think. I suggest that you solve it completely for the 2 by 2 case; I guess this means a working program. You will be able to find numerous references on equivalence of binary quadratic forms over the rationals and over the integers. For the three by three case, start with my introduction.

Community
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Will Jagy
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  • Thank you for your helpful comments. I forgot to add the requirement that $detB > 1$, – Drew Jul 23 '16 at 21:26
  • @Drew in the 3 by 3 case that does not matter at all, since negating $B$ negates its determinant in odd dimension. Still, I do not really know, given how complicated this task actually is, how to advise you. Can you do the 2 by 2 case? – Will Jagy Jul 23 '16 at 21:31
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If A is a symmetric positive definite matrix then $B^TAB$ is a diagonal matrix iff B is an orthogonal matrix i.e. $B^T=B^{-1}$. If $\exists$ such a matrix $B$ then $B^TAB=diag(\lambda_1,\lambda_2,...,\lambda_n)$ where $\lambda_i$ are the eigenvalues of $A$. And the columns of $B$ are the corresponding eigenvectors of $A$. So to find the $B$ s.t. $det(B)<10$ you will need to find the eigenvectors of A.

1D_Particle
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  • $B$ is not necessarily orthogonal. We need all the entries to be integers. – Drew Jul 23 '16 at 14:33
  • You can always multiply the eigenvectors by some scalar such that all the entries are integers. – 1D_Particle Jul 23 '16 at 14:34
  • You will need $B$ to be orthogonal for $B^TAB=diag$ to hold. Infact this only holds because A is positive definite and symmetric. – 1D_Particle Jul 23 '16 at 14:39
  • If $A=\left( \begin{array}{rr} 2 & 1 \ 1 & 5 \end{array} \right) $ and $B=\left( \begin{array}{rr} 1 & -2 \ 1 & 1 \end{array} \right)$ – Drew Jul 23 '16 at 14:55