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$0 \in \Bbb R^d$ cannot be an eigenvector of $A\in \Bbb R^{d\times d}$. True or false and why? I honestly have no idea

Gono
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    It depends on definition, but by the usual definition, an eigenvector for a matrix $A$ is a nonzero vector $v$ such that $Av=\lambda v$ for some constant $\lambda$ that we call the corresponding eigenvalue. – JMoravitz Dec 06 '17 at 18:22
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    It is a matter of definitions. Eigenvectors are required to be nonzero. Check the convention followed in the book you are using or in the course you are taking. (Eigenspaces, on the other hand, are vector spaces and so 0 belongs to them.) – Andrés E. Caicedo Dec 06 '17 at 18:22
  • The zero vector is generally excluded because it trivially solves the eigenvector equation $Av=\lambda v$ for any $\lambda$. It would be fairly useless as an eigenvector, anyway. It tells you nothing about the behavior of $A$ and doesn’t help you build an eigenspace. – amd Dec 06 '17 at 19:39

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