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As we have eigenspace , I have also read that eigenvector can not be a zero vector , so how has it been possible to say eigenspace is there without null vector .

abstract
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  • Let $v$ be the zero vector , then $Av=tv$ for any $t$ – QED Dec 10 '17 at 13:49
  • You define the eigenspaces for $A$ corresponding to a scalar $\lambda$ as the null space of $A-\lambda I$. This is a vector space, and it is nontrivial if and only if $\lambda$ is an eigenvalue. In that case, one easily checks that every vector in the eigenspace, except for 0, is an eigenvector. It is just a matter of definitions (the point being that definitions are chosen to help us, not to become obstacles. Eigenvectors and eigenspaces serve different purposes). – Andrés E. Caicedo Dec 10 '17 at 14:17
  • sure but i did not check unfortunately – abstract Dec 10 '17 at 15:25

2 Answers2

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The eigenspace contains the zero vector, but the zero vector is often not considered to be an eigenvector (e.g. because if it were, then every linear map would have all elements of the base field as eigenvalues, since $\lambda\cdot0=0$ for all $\lambda$...)

Daniel Robert-Nicoud
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  • Sir If zero vector is not an eigenvector so how can we keep it in the eigenspace where all vectors are eigenvectors . – abstract Dec 10 '17 at 13:49
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    @AbdulFatahRajri It is simply not true that every vector in an eigenspace is an eigenvector! If $\lambda$ is an eigenvalue the corresponding eigenspace is the set of all corresponding eigenvectors, plus the zero vector. – David C. Ullrich Dec 10 '17 at 14:31
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An eigenvector is nonzero by definition. I like the following reason: Let $T: V \to V$ be linear. Then, we will say that a one-dimensional eigenspace $W \subseteq V$ is a one-dimensional subspace such that $T|_W = \lambda \,\mathrm{id}_W$ for some scalar $\lambda$. This definition makes it clear that a zero-dimensional eigenspace is always uninteresting.

Then, we make a secondary definition of eigenvector, which is an element whose span is a one-dimensional eigenspace.

Joppy
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