Questions tagged [idempotents]

For questions about elements which satisfy $x\cdot x=x$ where $\cdot$ is a composition law.

Filter by
Sorted by
Tagged with
0 votes
1 answer
39 views

Incorrect proof for $e$ idempotent $\Rightarrow$ $eA$ projective

I have seen several proofs for $e$ idempotent $\Rightarrow$ $eA$ projective where $A$ is an algebra. I tried something different and produced a proof without using the fact that $e$ is idempotent (so ...
kubo's user avatar
  • 1,640
1 vote
1 answer
44 views

Two comaximal radical ideals, with product the nilradical, are radicals of principal ideals generated by complementary idempotents

Let $\mathfrak a,\mathfrak b\subset R$ be two comaximal radical ideals of the ring $R$, with $\mathfrak{ab}=0$. Then $\mathfrak a=Ra$ and $\mathfrak b=Rb$ for complementary idempotents $a,b\in R$. The ...
Jerry Scott's user avatar
2 votes
1 answer
40 views

Is an idempotent logical expression considered to be in conjunctive normal form?

Is $$A \lor B \lor A$$ technically in conjunctive normal form? Or must we apply the idempotent law for it to be in CNF? $$A \lor B \lor A \equiv A \lor B$$?
justanotheruser's user avatar
2 votes
1 answer
49 views

Number of idempotent functions

I'm supposed to show that the number of functions $\ $$\mathrm{f}$: $[n]$ $\to$ $[n]$ $\ $ such that $\mathrm{f\circ f=f}$ is $$1+\sum _{k=1}^{n}{n \choose k}k^{n-k}$$ But I guess that this result ...
J P's user avatar
  • 333
2 votes
1 answer
84 views

about $\operatorname{tr}(A) = \operatorname{rank}(A)$ for idempotent matrix $A$

I'm trying to prove the above statement, and I had a look at this site. This ends the proof with the following statement the rank is the number of non-zero eigenvalues But, this is what makes me ...
jason 1's user avatar
  • 605
1 vote
1 answer
100 views

Prove $\lVert A \rVert _{p} \geq 1$ for any idempotent matrix $A \neq 0$

As the title states, I'm facing a problem to prove: for any idempotent matrix $A \in \mathbb{C}^{n \times n}$ and $A \neq 0$, $\lVert A \rVert_{p} \geq 1$. Here the p-norm $\lVert A \rVert_{p}$ ...
Yuchun's user avatar
  • 13
3 votes
0 answers
56 views

Category of algebraic vector bundles $VB(X)$ over a scheme $X$, which is not idempotent complete

I have recently come across the category of algebraic vector bundles over a scheme $X$. In short, it is the category of locally free $\mathcal{O}_X$-modules of finite rank. An additive category $\...
Divya's user avatar
  • 533
0 votes
0 answers
134 views

Show that a division ring contains exactly two idempotent elements.

My proof: Suppose that $R$ is a division ring. Since $0a=a0=0$, then $0^2 =0$. If $a \neq 0$ and $a^2 = a$, then the inverse $a^{-1}$ of $a$ exists. So, $a^{-1} (a^2)=a^{-1} (a)$. This implies that $a=...
user avatar
-1 votes
1 answer
60 views

Are idempotents always central in a von Neumann regular ring? [closed]

A ring $R$ is called von Neumann regular if for every $a \in R$; $a=axa$ for some $x \in R$. A ring $R$ called Abelian if every idempotent in $R$ is central. Are von Neumann regular rings abelian?
Cary's user avatar
  • 47
2 votes
1 answer
63 views

Term for $eSe$ where $S$ is a semigroup and $e \in S$ is an idempotent

For a (possibly non-unital) ring $R$ and an idempotent $e \in R$, $eRe$ is a unital ring with identity $e$ and is known as a corner ring. Now, given any semigroup $S$ and any idempotent $e \in S$, $...
Geoffrey Trang's user avatar
1 vote
0 answers
33 views

Directly irreducible ring and central idempotents

My question is mainly concerned with the characterisation of directly irreducible rings as defined on Wikipedia. We say that a (unital, not necessarily commutative) ring $R$ is directly irreducible, ...
Gargantuar's user avatar
1 vote
1 answer
55 views

The natural partial order on the set of idempotents in the semigroup of Boolean relation matrices.

Let $\mathcal{E}_n$ be the set of idempotents in the semigroup $\mathcal{B}_n$ of $n \times n$ Boolean relation matrices. The relation $E \leq F$ iff $EF=FE=E$ is called the natural partial order on $...
geoffrey's user avatar
1 vote
1 answer
94 views

If $R$ is commutative and $I$ is a finitely generated ideal with $I^2=I$, then there exists an idempotent $e\in I$ with $I=Re$

Question Let $R$ be a commutative ring. Let $I$ be a finitely generated ideal. Assume that $I^2=I$. Show that $I$ is a direct summand of $R$. Answer I know that $I=Re$ for some idempotent $e\in I$ and ...
confused's user avatar
  • 499
6 votes
3 answers
304 views

How to prove $\mathbf{1}^\top\mathbf{Q}^+\mathbf{Q}=\mathbf{1}^\top$, where $\mathbf{Q}$ is any element-wise squared correlation matrix?

Let $(X_1,…,X_n)$ be a random vector with $0<\prod_{j=1}^n\text{Var}(X_j)<∞$. Let $\mathbf{Q}=(\mathbf{q}_{1},…,\mathbf{q}_{n})=(ρ_{jk}^2)_{n×n}$, where $ρ_{jk}$ is the Pearson correlation ...
woody's user avatar
  • 87
2 votes
1 answer
71 views

Does there exist an idempotent, pseudo-constant $p$-adic function with an uncountable image?

$f$ is defined to be a pseudo-constant function if $f'(x)=0$. The question simply comes from idly wondering, "What are p-adic idempotent functions like?" Differentiable idempotent functions ...
Merosity's user avatar
  • 2,333
0 votes
0 answers
13 views

conditions for an algebra generated by idempotents to be solvable/nilpotent

Given a finite Lie algebra that is generated by a set of rank-1 idempotents $\{P_{j}\}_{j}$ over the complex field, what are the conditions for it to be solvable (or alternatively nilpotent). A ...
N A McMahon's user avatar
1 vote
0 answers
45 views

Orthogonal idempotents in rings without unity?

Some context In this answer and comments, an argument is given for why an ordered ring with unity can't have nontrivial idempotents. I'm trying to extend the argument for an ordered ring $R$ without ...
Nick F's user avatar
  • 1,153
5 votes
1 answer
101 views

Find all idempotent matrices such that $(A-B)^2 = 0$

Find all idempotent matrices such that $(A-B)^2 = 0$ We can see that the hypotheses imply that $A+B=AB+BA$, and if we multiply by $AB$ on the right, we get $AB+BAB=(AB)^2+BAB$, which also implies ...
Etanche's user avatar
  • 180
2 votes
1 answer
246 views

About idempotent matrices

It seems to me that there do not exist two distinct idempotents matrix such that $(A-B)^2=0$. I have not found a counter-example with $2\times 2$ matrix We can see that the hypotheses imply that $A+B=...
Etanche's user avatar
  • 180
0 votes
0 answers
22 views

(idempotent)Does $rank(A) + rank(A-I_n)=n$ implies $A^2=A$? [duplicate]

Let $A\in M_{n\times n}(R).$ If $rank(A) +rank(A-I_{n})=n$, show that $trace(A)=rank(A)$ I have already known that an idempotent $A^2=A$ implies...$$(1)rank(A) +rank(A-I_{n})=n\qquad(2)trace(A)=rank(...
mlrofcloud's user avatar
15 votes
3 answers
430 views

Does every commutative ring have $2^n$ idempotents?

I've spent a lot of time looking for examples, and I can't find any commutative rings which have a finite number of idempotents other than a power of $2$. Intuitively, adjoining an extra idempotent $a$...
Zoe Allen's user avatar
  • 4,074
0 votes
0 answers
36 views

Idempotents and equitable partitions

Say we have a graph $X$ with $n$ vertices, and $\pi$ is an equitable partition of $X$. Namely, $\pi = \{C_1,\dots,C_r\}$ is a partition of $r$ sets of the vertices of $X$ such that there is a constant ...
Jeff's user avatar
  • 167
6 votes
0 answers
30 views

An element commuting with all non-trivial idempotents is central in a simple ring

Suppose that $R$ is a simple ring that has non-trivial idempotents. I try to prove that if an element $a$ commutes with all the idempotents, then $a$ is in the center of the ring. If we define $[x, y] ...
eeerase's user avatar
  • 147
0 votes
0 answers
137 views

Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$.

Let $A$ be a ring without nonzero nilpotent elements. Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$. I know that ...
Misimiausis's user avatar
2 votes
0 answers
29 views

A nice characterization of commuting idempotent endomaps?

Let $X$ be a set, and let $f\colon X\to X$ be a map. It is idempotent when for each $x\in X$, $f(f(x))=f(x)$. It is equivalent to the requirement that the restriction of $f$ on its image $f(X)$ is the ...
Amaru's user avatar
  • 75
0 votes
1 answer
104 views

Relations, Transitive functions & Idempotent functions

I've read all the stack threads I can find on transitive functions and want to pull together the ideas/questions it leaves me with. This all started because as per this thread I wrongly thought ...
CormJack's user avatar
  • 380
3 votes
1 answer
129 views

How can I prove that this matrix is idempotent?

I have the following matrix $$A=\begin{equation} \begin{pmatrix} 0 & a & -b\\ -a & 0 & c\\ b & -c & 0 \end{pmatrix} \end{equation}$$ I have to prove that $M=A^2+I$ is ...
slow_learner's user avatar
1 vote
0 answers
58 views

Idempotent matrix $X^2=X$ with complex entries

Suppose I have an Idempotent matrix (with complex entries) $X \in \mathbb{C}^{n \times n}$ such that $X^2 = X$ and I want to determine the set of solutions of this equation. So, I started by ...
Ujjwal's user avatar
  • 149
0 votes
0 answers
55 views

Ring generated by all its idempotents

Let $R$ be an associative ring with unity $1$ generated by all its idempotents. Denoted by $R^n$ the additive subgroup of $R$ generated by all elements $a_1a_2\cdots a_n$ for $a_i\in R$. Then, is $R=R^...
Tran Nam Son's user avatar
0 votes
1 answer
71 views

If $\mathcal{C}$ is a cyclic code, with generating idempotent $ε(x)$, then $ε(x)-θ(x)$ is a generating idempotent for $\mathcal{C}_0$

We consider a cyclic code $\mathcal{C}$ over a finite field $\mathbb{F}$, with generating idempotent $ε(x)$ and $\mathcal{C}_0$ a subcode, which contains all the (code)words of zero sum. If $θ(x) = 1/...
user avatar
0 votes
0 answers
30 views

Proving that a polynomial is the generating idempotent of the repetition code of length $n$

First of all, we consider a finite field $\mathbb{F}$, with $|\mathbb{F}|=q$, and a natural number $n$ that is relatively prime to $q$. We also consider the quotient ring $\mathcal{R}_n = \mathbb{F}[...
J.Spi's user avatar
  • 376
3 votes
3 answers
379 views

Assistance with idempotent matrices

I am taking linear algebra for the first time and am struggling with the concept of idempotent matrices. I know that $A = A^2$ is the concept behind it, but I can't seem to understand HOW one would ...
George Innit's user avatar
2 votes
1 answer
92 views

Idempotents in tensor product with a f.g. local Artin algebra

Let $k$ be a field. Let $A$ and $B$ be $k$-algebras (commutative , with unit). Assume that $A$ is finitely generated, local and Artinian. Assume that the set of idempotents in $B$ is finite. (I stress ...
Boogie's user avatar
  • 249
13 votes
3 answers
399 views

How to prove that $AB$ is a projection if $(AB)(BA)=AB$?

I was trying to solve the following problem: Assume $A,B\in M_n\left( \mathbb{C} \right)$,satisfy $$AB^2A=AB.$$ I need to proof $$\left( AB \right) ^2=AB.$$ I tried to use some equivalent substitution ...
fusheng's user avatar
  • 961
2 votes
1 answer
92 views

How to proof there is no idempotent element other than 0 and 1 in a Division Algebra?

If $A$ is a division $K$-algebra. Then I need to proof there is no idempotent element other than $0$ and $1_A$ in $A$. I tried this way : If $0,1_A\neq a\in A$ such that $a^2=a.~$ Now $A$ is division ...
Jayden's user avatar
  • 21
1 vote
3 answers
103 views

Let $e$ be an idempotent of the monoid $M$, $x$, $y$ be two elements of $eMe$. Then, $(eMe)\,x\,(eMe) = (eMe)\,y\,(eMe)$ if and only if $MxM = MyM$ [duplicate]

I am studying semigroups. I saw a Lemma in the text that states: Let $e$ be an idempotent of the monoid $M$, $x$, $y$ be two elements of $eMe$. Then, $(eMe)\,x\,(eMe) = (eMe)\,y\,(eMe)$ if and only ...
CJJ's user avatar
  • 63
7 votes
1 answer
114 views

Does this property of algebra morphisms (related to idempotents) have a name?

Let $F$ be a field. I am in the category of finite-dimensional $F$-algebras. Let $f:A \rightarrow B$ a homomorphism of two of those. The property of $f$ which came up as useful in something I consider ...
Torsten Schoeneberg's user avatar
2 votes
1 answer
76 views

Is it possible to characterize the set of $k$-periodic matrices $\in \mathbb{R} ^{n \times n}$?

A $k$-periodic matrix $M$ is an $\mathbb{R}^{n \times n}$ matrix such that $M^{k+1} = M$ Is it possible to characterize this set of $k$-periodic matrices? My goal is to exploit this characterization ...
Sam's user avatar
  • 999
2 votes
1 answer
67 views

If $A$ has no non-trivial idempotents, then neither does $A/N$

Let $A$ be a commutative, associative, unital, finitely generated algebra over an algebraically closed field $k$. Denote by $N$ the nilradical of $A$, which is the set of all nilpotent elements of $A$ ...
MyWorld's user avatar
  • 2,368
2 votes
2 answers
120 views

Non-commutative abelian rings.

Let $R$ be a ring with unity. An element $e\in R$ is called idempotent if $e^2=e$. Clearly, $0,1$ are idempotents. An element $e\in R$ is called central if $er=re$ for all $r\in R$. Recall that a ring ...
Hussein Eid's user avatar
5 votes
0 answers
119 views

If $M=\bigoplus_{i\in I}M_i$ and $f:M\to M$ is idempotent then $f$ induces an isomorphism between $M_i$ and $f(M_i)$.

Let $R$ be a ring, $M=\bigoplus_{i\in I} M_i$ an $R$-module with $\text{End}(M_i)$ local for all $i\in I$. Then I want to show that if $f:M\to M$ is idempotent (i.e. $f^2=f$) and nonzero, then there ...
14159's user avatar
  • 861
0 votes
0 answers
21 views

Show whether the scalar projection is idempotent

A while ago I asked Show whether the vector projection is idempotent, which made it clear to me that the idempotency of the vector projection could be checked with a simple substitution. Now I am ...
Galen's user avatar
  • 1,800
1 vote
1 answer
119 views

Finding idempotent elements of a $C^{*}$-algebra

Let $T \in B(\mathcal{H})$ (bounded operators on a complex Hilbert space $\mathcal{H}$) and suppose $T$ is normal. Suppose furthermore $\sigma(T)=\{-1\}\cup[2,3]$. Then I want to find a complete list ...
Jacobiman's user avatar
  • 1,031
0 votes
2 answers
115 views

Find the determinant of a sum of idempotent matrices

Let $A, B \in \mathcal{M}_n (\mathbb{R})$ such that $A^2 = A$ and $B^2 = B$. If $\det(2A + B)=0$, prove that $$\det(A + 2B) = 0$$ My attempt: I know that both matrices are diagonalizable and their ...
alien2003's user avatar
  • 1,116
3 votes
1 answer
75 views

Idempotent relations in a ring

Let $(A,+,.)$ be a ring such that, if $x \in A$ with $6x = 0$, then $x=0$. Let $a,b,c \in A$ such that $a-b$ , $b-c$ , $c-a$ are idempotent. Prove that $a=b=c$. Unfortunately, I haven't made any big ...
Michael Goldberg's user avatar
2 votes
1 answer
52 views

Invertible indempotent matrix not equal to identity

On Wikipedia, I found that the only non-singular idempotent matrix is $\mathrm{\mathbf{I}}$. However, the following matrix $$\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 0 \\ 0 & 1 & 1\...
Cordylus's user avatar
1 vote
1 answer
38 views

Idempotent Semigroup $S$ with Equivalence Relation $(a R b) \iff (aba=a), (bab=b)$: $S/R$ is commutative - why?

This is exercise 11.19 (h) in Seth Warner's "Modern Algebra". We are given that $S$ is an idempotent semigroup, that is: $\forall a \in S: aa = a$. Let $R$ be the equivalence relation ...
Prime Mover's user avatar
  • 4,832
0 votes
1 answer
745 views

Let $𝐴 $ be a square matrix of order $𝑛$ such that $A^2=A$ Prove that every $𝑣∈\mathbb R^n$ can be decomposed as $ 𝑣=𝑣_1+𝑣_2$ [closed]

Let $A$ be a square matrix of order $n$ such that $A^2 = A$. Prove that every $v \in \Bbb R^n$ can be decomposed as $v = v_1 +v_2$, where $v_1$ is in the null space of $A$ and $v_2$ is in the column ...
pwop's user avatar
  • 11
2 votes
1 answer
68 views

Conditions for Group given Semigroup with Idempotent Element

Another exercise in Seth Warner's "Modern Algebra" (1965). This one is Exercise 7.15. Let $(S, \circ)$ be a semigroup. Let $(S, \circ)$ have an idempotent element $e$, that is, such that $e \...
Prime Mover's user avatar
  • 4,832
0 votes
0 answers
91 views

Is there general way to prove that for a given function exists formula with a constant amount of operations?

E.g there is a formula for a function that gives a number of idempotent functions for a set with finite size. The solution is $\sum_{k=1}^n{n\choose k}k^{n-k}$ but you need to sum intermediate results....
Andrey Sergeev's user avatar

1
2 3 4 5
10