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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    $\begingroup$ +1 as the question is more subtle than it seems... $\endgroup$ Nov 16, 2011 at 17:41
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    $\begingroup$ 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/… $\endgroup$ Nov 16, 2011 at 18:28
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    $\begingroup$ I cannot / could not believe that this is not true in general. $\endgroup$ Nov 17, 2011 at 7:06

1 Answer 1


See: R.N. Gupta, Anjana Khurana, Dinesh Khurana, and T.Y. Lam, Rings over which the transpose of every invertible matrix is invertible; J. Algebra 322 (2009), no. 5, 1627–1636 (MR).

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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    $\begingroup$ 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). $\endgroup$ Nov 16, 2011 at 19:06
  • $\begingroup$ This completely address my question. Thank you very much, Alireza! $\endgroup$
    – javier
    Nov 17, 2011 at 11:49

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