ac.commutative algebra - Filtration over tensor product

nt.quantity principle – Deciding unimodular versus a eccentric matrix retort

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nt.quantity principle – Deciding unimodular versus a eccentric matrix

$L$ and $R$ are matrices in ${0,1}^{ntimes n}$ given to you and one in every of $L$ and $R$ is eccentric and the opposite is unimodular on the identification as a biadjacency of a bipartite graph it has one consummate matching.

Is there a unostentatious path to establish which of $L$ or $R$ is unimodular with out computing determinants? By less complicated I meant to graze if there’s path to establish in $NC^1$ (https://complexityzoo.bag/Complexity_Zoo:N#nc1) or equally in $5$$PBP$ (https://complexityzoo.bag/Complexity_Zoo:P#kpbp). The drawback is in $NC^2$ (https://complexityzoo.bag/Complexity_Zoo:N#nc2).

If wanted I can present a foundation $M$ for a lattice in $n$ dimensions in $mathbb R^n$.

The drawback is contained in $C_=L$ (https://complexityzoo.bag/Complexity_Zoo:C#cequalsl) as it’s testing for singularity however it’s less complicated for the reason that non-eccentric matrix has determinant $pm1$.

It is in $oplus L$ (https://complexityzoo.bag/Complexity_Zoo:Symbols#parityl) however it’s less complicated for the reason that covenant is we’re constraining to eccentric matrices if determinant is $0bmod 2$ and to unimodular matrices if determinant is $1bmod 2$.

It is in $UL$ by developing a matrix whose determinant is a sq. of given matrix (https://complexityzoo.bag/Complexity_Zoo:U#ul) however it’s less complicated since $UL$ shouldn’t be recognized to amass full issues.

So wanting on the zoo https://www.math.ucdavis.edu/~greg/zoology/diagram.xml it seems the issue is rapid to $L$

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