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Let $F$ denote a field, and $M$ denote the set of all $n \times n$ matrices over $F$.

Is it true that for all $m \in M$, the following are equivalent?

  1. $m$ is left-cancellative
  2. $m$ is right-cancellative
  3. $m$ is left-invertible
  4. $m$ is right-invertible
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If $M$ is an arbitrary finite-dimensional algebra over a field, then for an element $m \in M$ the four conditions are equivalent, and also to the fifth condition that $m$ is invertible (which is just the conjunction of 1 and 2).

Proof: If $m$ is left-cancellative, then $M \to M, x \mapsto xm$ is an injective linear map, hence an isomorphism. Hence $m$ is left-invertible. The other implications are trivial or follow from symmetry.

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Yes this is true because all four mean that the determinant in nonzero: if the determinant is zero, then the corresponding linear map $f$ fails to be either injective or surjective, hence is non-invertible on the left and right, and counterexamples to left or right cancellativeness are found by taking a nonzero map whose image is in $\ker f$, respectively whose kernel contains $\operatorname{im} f$ (one can do both at the same time, providing a left-and-right annihilator for $f$).

More interestingly this remains true for matrices over any commutative ring such that this property holds in the ring itself (otherwise of course the $1\times1$ matrices would give a counterexample), as is shown in Do these matrix rings have non-zero elements that are neither units nor zero divisors? Moreover, I just saw a recent comment here (by Torsten Schoeneberg) that asks about the noncommutative case, which seems a natural question to ask; it doesn't obviously fail.

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