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A matrix with entries in an Euclidean Domain is invertible if and only if the matrix can be row reduced to the identity matrix.

Is the above statement true?

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closed as off-topic by user26857, Misha Lavrov, C. Falcon, Leucippus, Juniven Apr 22 '17 at 1:47

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  • $\begingroup$ Any references of your "theorem"? $\endgroup$ – Xam Apr 14 '17 at 19:56
  • $\begingroup$ @Xam It is not a theorem though. I just want to know if this statement works under Euclidean domain since it works in basic linear algebra. $\endgroup$ – Kenneth.K Apr 14 '17 at 20:11
  • $\begingroup$ HINT: The only problem is that at each step of the Gaussian elimination, the considered element in the diagonal is not a divisor of the elements below. Now, encoding the Euclidean algorithm for calculating the greatest common divisor, this problem can be solved. $\endgroup$ – Josué Tonelli-Cueto Apr 15 '17 at 0:14
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By the Smith normal form for Euclidean domains, for any given matrix $ X $, there is $ A $ and $ B $; both products of matrices corresponding to invertible elementary row/column operations, such that $ X = ADB $ for a diagonal matrix $ D $. (Stated this way, the Smith normal form relies essentially on the Euclidean algorithm, so is not true for general principal ideal domains.) If $ X $ is invertible, it follows that the nonzero entries of $ D $ are all units, which we may wlog assume to be $ 1 $. Therefore; $ X = AB = (AB)I $, where $ AB $ is a product of matrices corresponding to invertible elementary row operations, and therefore $ (AB)^{-1} X = I $. Thus, the answer to the question is positive.

Things are not as trivial as Josué Tonelli-Cueto surmises in the comments; as using the Euclidean algorithm during row reduction may cause elements which have been reduced to zero to become nonzero again. The Smith normal form, however, gives a straightforward way of answering the question. (Note that the algorithm for producing the Smith normal form also allows for column reduction, so the problems we may run into are resolved more easily.)

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