Finding the subspace generated by eigenvectors of eigenvalue $\lambda$

Question

I've been asked to find a basis of the subspace generated by the eigenvectors of eigenvalue $\lambda$ plus the vector $0$. I know how to find the eigenvalues and the subset of the eigenvectors. The problem is that "plus the vector $0$". I don't understand this restriction because the vector $0$ is in every subspace. Am I missing something?

Answer 1

My conjecture is that whoever proposed that in problem had this in mind:

Given a scalar $\lambda$, find a basis of the vector space of all eigenvectors with eigenvalue $\lambda$ together with the $0$ vector.

Indeed, if you replace the set of all eigenvectors with eigenvalue $\lambda$ with the space spanned by them, it becomes needless to add the $0$ vector.