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As mentioned above: Is it true that every eigenvalue has at least one eigenvector?

Or is it possible that while trying to find the basis of a specific eigenspace, i will get only the zero vector (means there are no eigenvectors corresponding to this eigenvalue)?

Thank you

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    $\begingroup$ By definition, $\lambda$ is an eigenvalue of $f$ if (and only if) there is a $v\neq 0$ with $f(v) = \lambda v$. $\endgroup$ – Daniel Fischer Jun 19 '14 at 11:05
  • $\begingroup$ You could in theory define an eigenvalue to be a root of the characteristic polynomial of $f$. Michael's answer provides half of the proof that these two definitions are equivalent. $\endgroup$ – mdp Jun 19 '14 at 11:09
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I assume you are talking of eigenvalues and eigenvectors of an $n\times n$ square matrix $A$.

Define an eigenvalue to be a root of the polynomial $|\lambda I-A|=0$. Then $\lambda I-A$ has determinant $0$. So when you row-reduce it, there will be a row of zeros. There will be at most $n-1$ pivots, so one column lacks a pivot - it is a free variable. So there are non-zero solutions to $(\lambda I-A)v=0$, or $Av=\lambda v$.

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