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Is there a standard name for a linear operator $T$ on a finite dimensional vector space satisfying $T^n=T^{n+1}$ for some $n\geq 1$ or, equivalently, $T$ is a similar to a direct sum of a nilpotent matrix and an identity matrix? I am not looking so much for name suggestions, but rather for a generally accepted terminology from the literature.

Added Motivation. In Kovacs proof that the complex algebra of the monoid of $n\times n$-matrices over a finite field is semisimple a key step is to show that the ideal of the monoid algebra spanned by the singular matrices is a unital ring. He shows that the identity is a linear combination of matrices satisfying the above property. He calls such matrices semi-idempotent. But I believe he invented the name.

Being a semigroup theorist I don't like math terms involving "semi" and so in my book I would prefer another term, preferably one in use in the matrix theory literature.

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The closest term I know is "idempotent", though that's only for $n=1$.

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    $\begingroup$ But if n>1, it need not be idempotent. It could be nilpotent. Laci Kovacs calls them semi-idempotent by but I think he invented the term. $\endgroup$ – Benjamin Steinberg Feb 18 '16 at 23:28
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    $\begingroup$ It's a shame that 'idempotent' came first, and was used only in the sense that you mention; it seems like it would be the perfect word for @BenjaminSteinberg's situation if it weren't already taken. I know that the question specifically doesn't ask for suggestions, but it's hard to avoid imagining them: 'eventually idempotent'? (That sounds like it's saying that some $T^n$ is idempotent, though.) 'Stabilipotent'? $\endgroup$ – LSpice Feb 19 '16 at 0:33
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    $\begingroup$ @LSpice Eventually idempotent is a name I have been considering if an official name doesn't come up. A good joke name is demi-potent since it is an anagram of idempotent $\endgroup$ – Benjamin Steinberg Feb 19 '16 at 0:45
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I would call it projectipotent :)

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I would say "unipotent" is the standard term.

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    $\begingroup$ Really? Wikipedia says a unipotent matrix $T$ would have $1$ as its only eigenvalue. But the OP's matrices might have both $0$ and $1$. $\endgroup$ – Robert Israel Feb 18 '16 at 23:02
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    $\begingroup$ In my experience, "unipotent" means "characteristic polynomial $(T-1)^n$", so $\mathrm{Id}+\mathrm{nilpotent}$. $\endgroup$ – David E Speyer Feb 18 '16 at 23:02
  • $\begingroup$ @DavidSpeyer If you read the OP's question, that is exactly how he characterizes his operators :) $\endgroup$ – Igor Rivin Feb 18 '16 at 23:14
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    $\begingroup$ @IgorRivin if you read the OP's question, he said direct sum! $\endgroup$ – Allen Knutson Feb 18 '16 at 23:15
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    $\begingroup$ Yes it is direct sum not sum. Only the identity can be unipotent and have my property. $\endgroup$ – Benjamin Steinberg Feb 18 '16 at 23:29

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