hello;
Consider the linear polynomial u:Rn[X]--Rn[X] associated with its second derivative. We immediately obtain:
Ker(u)=R1[X]
*******
My question is how did they immediately obtained
Ker(u)=R1[X]
(R1[X] is a polynomial of degree 1)

Question

hello;
Consider the linear polynomial u:Rn[X]-->Rn[X] associated with its second derivative. We immediately obtain:
Ker(u)=R1[X]
*******
My question is how did they immediately obtained
Ker(u)=R1[X]
(R1[X] is a polynomial of degree 1)

anonymous · Answer

could anyone help me??
i need the answer

anonymous · Answer

I hope I understand correctly. As far as I understand this is linear transofmration from R^n[x] ro R^n[x] assosicated with the action of second dervative. if so then think about it, the R^1[x] are all the polynomials which at most one degree variables. (X, 2x....) if you take a second derviatve of these each one of the R^1[x] you get zero. This is why the R^1[x] is in kerT. 

If I have u in R^1[x] and I take its second dervative I get zero, then it's this whole space is spanned by KerT. understand?

anonymous · Answer

thank you so much...your answer really helped me

anonymous · Answer

wow, that's help me too.
thx a lot

anonymous · Answer

I'm glad to hear  :)

anonymous · Answer

even if you find the null space your will reach to the same answer so , the shortcut is to say R1(x)