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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.

flag	asked Nov 16 2011 at 17:20 javier 2,186●5●16

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+1 as the question is more subtle than it seems... – Alain Valette Nov 16 2011 at 17:41

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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 mathoverflow.net/questions/47369/… – Andreas Thom Nov 16 2011 at 18:28

I cannot / could not believe that this is not true in general. – Martin Brandenburg Nov 17 2011 at 7:06

Alireza Abdollahi · Accepted Answer · 2011-11-16 18:28:12Z UTC

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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. .......

answered Nov 16 2011 at 18:28

Alireza Abdollahi
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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 2011 at 19:06

This completely address my question. Thank you very much, Alireza! – javier Nov 17 2011 at 11:49

Invertible matrices over noncommutative rings

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