 giving me an opportunity to present my research work over here. So, my research topic is image of linear derivations and Matthews our subspaces. So, first of all I will give my notation for my presentation. For me k is a field of characteristic 0, k star is all the non-zero elements, k star is all the non-zero elements in k and kx is the polynomial algebra in n variables over k and further I have defined my a to b of k algebra. What is a k derivation? A k linear map d from a to a such that d of a plus b equals to d of a plus d of b and d of a b equals to b times d of a plus a times d of b, this is called a k derivation of a. Further I will define a ke derivation, again it is a k linear map delta from a to a such that delta of a plus b equals to delta of a plus delta of b and delta of a b is b times delta of a plus a times delta of b minus delta a delta b for all a and b in a. So, this ke derivations they have a very nice structure. So, we can easily classify them because every ke derivation delta is a ke derivation of a if and only a delta is equals to identity minus v for some k andomorphism phi of a. So, using this we can easily classify all the ke derivations of a. I would like to define linear ke derivation, k derivation d of kx is called linear if d of xi is summation of j from 1 to n a ij xj where my aigs are actually sitting inside k, the matrix a is called the associated matrix of the derivation d. And this f this corresponding matrix a actually it is a nilpotent matrix then I can give this derivation as a linear look linear look linear look nilpotent derivation of kx. So, my work is mainly on the linear ke derivations and the linear ke derivations of kx. And I will talk about the Mathew-Zau subspaces. So, this subspaces they have been recently been introduced by Zau in 2010 by a subspace of a ke algebra you always mean a ke linear subspace. So, this is a formal definition of Mathew-Zau subspaces. A k subspace m of a is called the Mathew-Zau subspace of a if the following equivalent conditions holds. If f is in a such that f to the power m belongs to capital M for all m greater than equals to 1 then for every g in a we have gfm is in capital M for all large m. Or the other statement it says that if f belongs to a such that f to the power m belongs to capital M for large m then for every g in a we have gfm belongs to capital M for all large m. So, these are both the equivalent statement for provost spaces of whether a Mathew-Zau subspace or not. So, these are some comments on the Mathew-Zau subspaces. So, it is a natural generalization of notion of ideals in a ring. So, another comment is that every ideal is a Mathew-Zau subspace, but not all the Mathew-Zau subspaces are ideals always. So, I would like to give an example for the same statement. So, I have considered a integral domain R of characteristic 0 and a and a be the algebra of n cross n matrices with entry z in R and a subspace m of a of matrices which are trace 0 matrices. So, if there exist a matrix such that all the powers of it also are having trace 0 then actually it will turn out to be a nilpotent matrix. And we can clearly say that the subspace of nilpotent matrices they will not form a ideal of a, but they actually form a this Mathew-Zau subspace of a. I will talk about the two conjunctures which have been proposed by Zau in 2018. So, the first one it says that the LFED conjecture D is a linear k derivation of kx then the image of the derivation D is a Mathew-Zau subspace of kx. And the second conjecture it says that let D be a linear Rokni nilpotent k derivation of kx then D maps every ideal of kx to a Mathew-Zau subspace of kx. So, I mainly worked on the LFED conjecture for the polynomial algebra in four variables. Also Zau at all they showed that if k bar is a algebraic closure of k and the LFED conjecture it holds for a bar then it also holds for a. So, using this statement we can always work in a overall algebraically closed field. So, before starting with the results I will just give a short definition a set lambda 1 to lambda n where lambda i's are in k is said to be linearly independent over n naught if there exists no non-trivial combination over n naught that equals to 0. So, we have proved the following result that D be a linear k derivation of kx if the eigenvalues of the associated matrix of D are linearly independent over n naught then image of D it always forms a ideal of kx moreover the image of D is actually the ideal generated by x1 to xn. Now, we will talk about the linear k derivations of the polynomial algebra in four variables. So, up to conjugation of matrices we can clearly see that there are 5 possible Jordan forms of a 4 cross 4 matrix. So, corresponding to this I have defined 5 linear k derivations of kx up to conjugation. So, to prove that any key linear image of any linear k derivation forms a Matthewsaw subspace I need it is enough for me to consider only these 5 derivations. So, it has been already been proved by van der Sen and Zhao in 2011 that image of D1 it forms a Matthewsaw subspace of R. So, now we have worked on D2, D3, D4 and D5. We have showed that for D1 that if lambda 1 is in k star and lambda 2 and lambda 3 they both are 0 then image of D2 is an ideal of R generated by x1 and x2. And the next statement which we have showed is that if lambda 2 is in k star and lambda 1 and lambda 3 they both are 0 then image of D2 is an ideal of R generated by x2 and x3. The second statement this is about the derivation D3. So, here we have proved that if lambda 1 is in k star and lambda 2 is 0 then image of D3 is an ideal of R generated by x1, x2 and x3. And if lambda 2 is in k star and lambda 1 is 0 then image of D3 is a Matthewsaw subspace of R. And the third one is like if lambda 1 and lambda 2 they both are 0 then image of D3 is a Matthewsaw subspace of R. So, similar type of statement we have for D4 also. And for D5 we have proved that as there is only one eigenvalue for D5. So, we have proved that if lambda 1 is in k star then image of D5 it forms an ideal of R generated by x1, x2, x3 and x4. So, I would like to also comment that wherever R mentioned that it forms a Matthewsaw subspace of R it is actually not forming an ideal it is just forming a Matthewsaw subspace. And this is a short outline of the proof. So, to prove these statements we have defined varying weights on the k-algebra on R corresponding to these derivations which are mentioned in this table. And this the degrees or weights they helped me a lot to complete my proof. And the second thing which I have used in my proof is that the factorial conjecture which has been proved for homogeneous polynomials in two variables recently by Liu and Sohn in 2020. Now, I will talk about the linear k-e derivations of R. So, again let delta be a linear k-e derivation then delta equals to identity minus phi for a k-andomorphism phi of R. Then again as corresponding to the we discussed that there are five possible Jordanian blocks of the 4 cross 4 matrix corresponding to that we have five different k-andomorphisms of phi of R up to conjugation. So, these are precisely the five different this. Now, if delta let delta is equals to identity minus phi i then up to conjugation a linear k-e derivation of R is conjugate to delta i for some i between 1 to 5. Again to prove that image of delta is a Matthews R image of delta for any k-e derivation of R to prove that it is a Matthews R subspace it is enough to talk about the image of deltas. So, as I said at all in 228 they prove that image of delta 1 is a Matthews R subspace of R. And further in 2022 at all they also derived the result that image of delta 2 it also forms a Matthews R subspace of R. So, we have mainly worked on delta 3 delta 4 and delta 5. So, we have proved that for delta 3 that if lambda 1 p and lambda 2 q product is not equals to 1 for all p and q in n then image of delta 3 it forms a Matthews R subspace of R. And again the second statement it says that if there exists p and q in n there is such that lambda 1 p is equals to 1 and lambda 2 q is equals to 1 then image of delta 3 is a Matthews R subspace of R. Similar kind of statement we have for delta 4 also. So, again overheads with the same conditions we have proved that image of delta 4 it also forms a Matthews R subspace of R. And one more statement we have for delta 5 it says that if lambda 1 p is not equals to 1 for all p in n then image of delta 5 it forms a Matthews R subspace of R. So, to prove this part part overhead this second case and same second case for delta 4 it was not that easy for us we actually defined new automorphisms. Precisely we worked on these automorphisms corresponding to the matrix this new automorphisms and using these matrices we proved that image of delta 3 and image of delta 4 they form a Matthews R subspace of R. Then again using the same weights as we defined for d i's we proved that image of these derivations they form a Matthews R subspace of R. So, that is all and this is the references.