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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?

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    $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Feb 5, 2018 at 16:04
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    $\begingroup$ 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). $\endgroup$
    – Yanko
    Feb 5, 2018 at 16:07
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    $\begingroup$ If it said "comprising" rather than "generated by", it would make sense as eigenvectors are usually defined to be non-zero. $\endgroup$
    – Rob Arthan
    Feb 5, 2018 at 16:07

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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.

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