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Does a matrix representation of an operator must be square? Or could be $n\times m$ where $n\neq m$?

MaxGold
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  • What exactly do you mean by "operator"? That term is sometimes used in different ways. – littleO Feb 02 '14 at 12:53
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    If by operator you mean what's meant almost always, i.e. a linear map between a vector space and itself (i.e., an endomorphism), then the answer is yes...when the dimension is finite, of course, otherwise we'd be talking of infinite matrices. – DonAntonio Feb 02 '14 at 12:53
  • If you're asking in the context of operator theory, the answer is no. Howevr, if you mean that the domain and codomain are the same, then the answer is yes and I think you should remove the tag "operator-theory" – user40276 Feb 02 '14 at 13:45
  • Indeed I was talking about a finite dimension. And yes between a vector space and it self. Tag for "operator theory" removed. – MaxGold Feb 02 '14 at 15:51

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A linear operator from $m$-dimensional vector space into $n$-dimensional vector space can be represented by a matrix of size $n\times m$, that is $n$ rows and $m$ columns.

When $n=m$ (in particular, if the domain and codomain are the same space) this is a square matrix.