The eigenspace corresponding with the eigenvalue zero is the same as the null space of the original matrix. All vectors in the null space are linearly independent so the eigenvectors of zero are also independent.
Is this conclusion right?
You're right that the eigenspace for 0 is the kernel of the transformation.
The last sentence is not correct, however. Certainly not all vectors in the nullspace are linearly independent. If $Ax=0$ so that $x$ is in the nullspace, then $2x$ is another vector in the nullspace that is linearly dependent with $x$.
I think you have some thinking to do about what linear independence means. One way to think about a collection of vectors being linearly independent is that you can't express one of those vectors as a nontrivial linear combination of the others.