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If a matrix has the same number of rows and columns, we call it a square matrix. The analogous concept for linear operators would be operators with the same domain and range, i.e., endomorphisms.

Is there an established adjective that can be added to the word "operator" that denotes this concept?

In other words, what could I fill in for xxx in the following example sentence: "A square matrix is invertible iff it has full rank, but not every full-rank xxx operator is invertible."

Note: In many texts, "operator" already implies that domain and range are the same, but in some texts this is not assumed. Often, this distinction is left implicit. For example, "A course in functional analysis" (Conway 1990) considers bounded operators with different domain and range, while in what is basically the follow up book "A course in operator theory" (Conway 2000), bounded operators are assumed to have same domain and range.


Short summary of suggestions

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    $\begingroup$ Endomorphic? :-) $\endgroup$ – M.G. Jul 28 '19 at 10:45
  • $\begingroup$ @M.G. I find very few mentions of "endomorphic operator" in Google. But if you can post your comment as an answer, people can comment more easily and we can get a feeling for the community opinion. $\endgroup$ – Dominique Unruh Jul 28 '19 at 11:35
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    $\begingroup$ It was more of a joke since the immediate adjective from endomorphism would be endomorphic, but I don't think I've ever heard it being used like this (in mathematics anyway). To be honest, I doubt there is a widely used adjective for this, people just say "endomorphism of this or that". But on the other hand, if I ever see it used, then I'll most likely conclude that the author means an endomorphism. $\endgroup$ – M.G. Jul 28 '19 at 12:13
  • $\begingroup$ I thought “operator” and “self-map” were basically synonyms. In other words, a linear map is a map between two vector spaces, while a linear operator is a map from a vector space to itself. $\endgroup$ – Sam Hopkins Jul 28 '19 at 13:43
  • $\begingroup$ Wikipedia is a little unclear on this point: en.m.wikipedia.org/wiki/Operator_(mathematics) $\endgroup$ – Sam Hopkins Jul 28 '19 at 13:47
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L. Kadison (2012, preprint p. 8) uses endo-operator. (That’s “attributive”, but I guess “predicative” use could still work, in the same way MacLane says an arrow is epi.)

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    $\begingroup$ "Endo-operator" sounds very acceptable to me. (Not important here, but in the jargon I am familiar with, we say an arrow is epic, or that it is an epi, and not that an arrow "is epi". I cannot swear how universal this usage is, but if I found myself saying "is epi", I would regard it as a minor slip.) $\endgroup$ – Todd Trimble Jul 28 '19 at 16:58
  • $\begingroup$ When you refer to using this as an adjective, are you suggesting "the operator is endo" or "the endo operator" (no hyphen)? Without that, this just seems to be a synonym for "endomorphism" (which is just "endo-morphism", after all). $\endgroup$ – LSpice Jul 31 '19 at 13:31
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    $\begingroup$ @LSpice Both (and I agree with your point). $\endgroup$ – Francois Ziegler Jul 31 '19 at 13:39
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In Operator Theory, an operator on the space $X$, is quite commonly used to mean $T:X\to X$, denoting $L(X)$ or $B(X)$ the space of these operators in the same spirit.

Self-map is quite common for Discrete dynamical systems, so I guess linear self-map could do, even though a bit strange. Self-operator seems more natural, but I've never seen it.

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  • $\begingroup$ But this is two nouns and a preposition, whereas the post asks for an adjective. (For the noun, we already have 'endomorphism'.) $\endgroup$ – LSpice Jul 31 '19 at 13:43
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    $\begingroup$ You're absolutely right, but the boldface character made me think that in the lack of a better word, established was somehow preferred to adjective :) $\endgroup$ – Pietro Majer Jul 31 '19 at 13:50

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