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

  • 2
    $\begingroup$ 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

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 $Av=\lambda v$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.