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Is it possible to have a matrix for which eigen vectors won't change by changing the eigen values? Please help me! I am seraching for the answer to this question.

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    $\begingroup$ What do you mean by changing the eigenvalues? $\endgroup$ – user63181 Aug 19 '14 at 15:52
  • $\begingroup$ I mean the same eigen vector for all the eigenvalues...if such a matrix exists is there any particular form of that matrix? $\endgroup$ – Anupam Aug 19 '14 at 15:54
  • $\begingroup$ Two different eigenvalues have different eigenvectors. More precisely $\ker(A-\lambda_1I)\cap \ker(A-\lambda_2I)=0$ for any eigenvalues $\lambda_1\neq \lambda_2$. $\endgroup$ – Quang Hoang Aug 19 '14 at 16:01
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    $\begingroup$ @Frunobulax, sure. However in Linear Algebra, $0$ also denotes the trivial vector space, i.e. $\{0\}$. $\endgroup$ – Quang Hoang Aug 19 '14 at 16:14
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    $\begingroup$ @Frunobulax: It's just a notation, doesn't mean a set. It makes sense when you consider the direct/tensor products $0\oplus M=M$ or $0\otimes M=0$. $\endgroup$ – Quang Hoang Aug 19 '14 at 16:27
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No, eigenvectors for different eigenvalues are not only different but also linear independent.

See here.

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  • $\begingroup$ To look at it another way, given a matrix $A$ with eigenvector $u\neq 0$, the corresponding eigenvalue is uniquely determined by the definition: $$ Au = \lambda u $$ Further, thus while two different eigenvectors $u,v$ might share a common eigenvalue, two different eigenvalues cannot share a common eigenvector. $\endgroup$ – hardmath Aug 19 '14 at 16:07

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