Determinant of a matrix over the integers mod n

  • Thread starter jdstokes
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  • #1
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Hi,

I'm curious if the following statement is true for all prime numbers n,

[itex]\det_{\mathbb{Z}_n}M = (\det_{\mathbb{R}}M)\mod n[/itex]

where [itex]\det_F M[/itex] is the determinant of M over the field F.

Thanks.

James
 
Last edited:

Answers and Replies

  • #2
AKG
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Z isn't a field. But if you mean that if you take a matrix, and compute its determinant by multiplying, adding, and subtracting the entries as integers and then reduce mod n, versus if you do all the arithmetic mod n, then the answer is 'yes'.
 
  • #3
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Thanks for the correction, I guess what I meant to say was [itex]\mathbb{R}[/itex].
 
  • #4
mathwonk
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no, what you meant to saY WAS THATyour notation denoted the determinant over the ring F.

since you cannot consider elements of R as if they were in Z/n unless they are integers.
 

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