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

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?

• You don't need the "plus the vector zero". It is enough with " the subspace generated by the eigenvectors of a single eigenvalue $\;\lambda\;$" . That automatically will contain the zero vector. Feb 5, 2018 at 16:04
• This is technical but crucial, the vector zero is not an eigenvector and thus the set of all eigenvectors for the eigenvalue $\lambda$ is not a vector space! (it will become one, once you add the zero vector). Feb 5, 2018 at 16:07
• If it said "comprising" rather than "generated by", it would make sense as eigenvectors are usually defined to be non-zero. Feb 5, 2018 at 16:07

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.