I think d option is correct
The zero vector by convention is not an Eigen vector, much in the same way that 1 is not a prime number. If we let zero be an eigen vector, we would have to repeatedly say "assume v is a nonzero Eigen vector such that..." since we aren't interested in the zero vector. The reason being that v=0 is always a solution to the system Av=λv
By definition of Eigen vector : a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector —called also characteristic vector
So Option A , B , C are incorrect
However, the eigenspace associated to an eigenvalue always contains the zero vector. .