An element a of a ring R is nil-clean provided that there exists an idempotent element e ∈ Id(R) and a nilpotent b ∈ Nil(R) such that a = b + e, a ring is called nil-clean if each element a ∈ R is nil-clean. I would like to have some examples of nontrivial unit (invertible) 2×2 matrices nil-clean (over real numbers).
If you consider the matrix a=1/2[1-cos(t), 0; sin(t), 1+cos(t)]. Then e=1/2[1-cos(t), sin(t); sin(t), 1+cos(t)] is idempotent and b=1/2[0, -sin(t); 0, 0] is nilpotent. Also, a is invertible if 1-cos^2(t) is non zero and it is a nontrivial unit matrix
@Cenap,
May be he considers them in M2 (IR). So, just that matrix is an example of nil-clean element, not the ring M2 (IR).
yes u r right hanifa
ofcourse it is one element in M_2(R) no algebraic structure there is no addition on these type mtricess how can we get ring sructure
this is related to jordan canonicsl formmm dea profesor wiwat
plz think unipotent matricess
@Cenap
ok, I understand you now.
We can find just nil-clean elements over IR, but a set of those elements to be a ring, it is not easy, may be it doesn't exist.
Even canonical Jordan form can't give result, because of the different bases
Dear Hanifa and Cenap,
Sorry. I forgot that it should satisfy the closure condition. Let me argue in a different way. If there are some errors, let me know.
Suppose a ∈ R be a nontrivial unit, then a = b + e, where b is nilpotent and e is idempotent. Since e is idempotent and 2x2. There are two possibilities (1) e is diagonal with 1 or 0 (2) trace(e)=1. Since b is nilpotent trace(b)=0. Also trace(a)=trace(b)+trace(e). Therefore, trace(a)=1. Therefore eigenvalues sum is 1. But a unit is nil clean iff it is unipotent. So, a is unipotent. a is unipotent iff its characteristic polynomial is a power of t-1 or equivalently all its eigen values are 1. Since trace(a)=1, the characteristic polynomial is t^2-t+1 and hence the eigen value of a is complex, which is a contradiction. So, I claim that there won't be any nontrivial unit. If any flaws in the proof, please post it.
Well done, dear Mariappan. But what is the problem with your example above?
I think I found the problem.
In the definition of a nil-clean element, it requires that the idempotent and the nilpotent commute and it uses the terminology "strongly nil-clean". In Mariappan's example, they do not.
For the definition, see: http://arxiv.org/pdf/1310.0416.pdf
for nil-clean matrices and it requires commutativity of the idempotent and the nilpotent, see the introduction in: http://arxiv.org/pdf/1503.06668.pdf
Dear Hanifa,
Good point.
My aim was to prove that their collections wont form a ring. But, in the middle It diverted from nilclean to strongly nilclean.
There is a flaw in the proof. As per this paper(http://arxiv.org/pdf/1310.0416.pdf), a unit is strongly nil-clean iff it is unipotent.
The first paper(http://www.sciencedirect.com/science/article/pii/S0021869313001075) of nil-clean shows that A=[1 -2 ; 1 0] is a nil-clean element.
So, nontrivial unit nil-clean element exists.
If we replace nil-clean by strongly nil-clean, the above post is valid.
Claim: Collection of nil-clean elements won't form a ring.
Following the same method we obtain the following
trace(a)=trace(b)+trace(e)=1. Now trace(a+a)=trace(2a)=2trace(a)=2. So, a + a is not nil-clean and hence not closed under addition.
So dear Mariappan, the conclusion is:
1) nontrivial nil-clean 2x2 matrices over IR exist ( we can take your example, or the example of the paper).
2) nontrivial strongly nil-clean 2x2 matrices over IR do not exist as you proved in the first page ( in your proof, we correct to: a unit is strongly nil-clean iff it is unipotent).
3) The set of nil-clean 2x2 matrices over IR is not closed under addition.
Dear Panchatcharam, Cenap and Hanifa,
Thank you so much for your kindly comments and advice.
Regards,
Wiwat Wanicharpichat
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