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I suspect this is true since it holds in the case over a field. But suppose $A\in M_{m\times n}(R)$ where $R$ is a PID. Does it still hold that $A$ and $A^{T}$ have the same invariant factors?


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up vote 2 down vote accepted

The answer is yes. Indeed, if $d_1,\dots,d_r$ are the invariant factors, and if $d_i$ divides $d_{i+1}$ for $1\le i < r$, then $d_1\cdots d_i$ is the gcd of the $i$-minors of any presentation matrix.

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Thanks, but why does that imply they have the same invariant factors? – Keith Richards Oct 29 '11 at 9:44
Dear @Keith: You're welcome. Do you agree that a matrix and its transpose have the same $i$-minors? – Pierre-Yves Gaillard Oct 29 '11 at 10:10
I'd never thought of this, but I think I've convinced myself of this fact now. – Keith Richards Oct 29 '11 at 10:46

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