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Let $A\in M_m(R)$ be an invertible square matrix over a noncommutative ring $R$. Is the transpose matrix $A^t$ also invertible? If it isn't, are there any easy counterexamples?

The question popped up while working on a paper. We need to impose that the transpose of certain matrix of endomorphisms is invertible, and we wondered if that was the same as asking if the matrix is invertible.

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+1 as the question is more subtle than it seems... – Alain Valette Nov 16 '11 at 17:41
It is a classical observation due to Nathan Jacobson that a division ring such the set of invertible matrices is closed under transposition has to be a field, i.e. commutative. See… – Andreas Thom Nov 16 '11 at 18:28
I cannot / could not believe that this is not true in general. – Martin Brandenburg Nov 17 '11 at 7:06
up vote 42 down vote accepted

See the following paper:

R.N. Gupta, Anjana Khurana, , Dinesh Khurana, T.Y. Lam, Rings over which the transpose of every invertible matrix is invertible Journal of Algebra Volume 322, Issue 5, 1 September 2009, Pages 1627-1636

Abstract We prove that the transpose of every invertible square matrix over a ring $R$ is invertible if and only if $R/\text{rad}(R)$ is commutative. .......

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Note that the $\Longleftarrow$ direction is trivial (once you know that the Jacobson radical of a matrix ring is the ideal of all matrices which have all of their entries lying in the Jacobson radical of the ring itself). – darij grinberg Nov 16 '11 at 19:06
This completely address my question. Thank you very much, Alireza! – javier Nov 17 '11 at 11:49

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