Therefore it is NOT in the kernel of the linear transformation.
But, since you said "I have found the kernel which is 0" in your first post, I think you were asking in general, not for this particular transformation. It is not possible for the kernel of a linear transformation to be a single non-zero vector!
The kernel of a linear transformation, A, is the set of all vectors, v, such that Av= 0. That is not true for (2, 0). Further, if the kernel is not "trivial", that is, if Av= 0 for any non-zero v, it cannot consist of a single vector! If v is a non-zero vector such that Av= 0 then A(xv)= xAv= 0 for any number x so there must exist an infinite number of vectors in the kernel. The kernel of a linear transformation is a subspace, possibly trivial, which is why it has a dimension.