Let $A$ and $B$ $\in M^{n\times n}$ be matrices with $n$ distinct eigenvalues $\{\lambda_1,\ldots ,\lambda_n\}$ and corresponding eigenvectors $\{v_1,v_2,\ldots,v_n\}$. Is it necessary that $A$ and $B$ are equal?
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2 Answers
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Yes. Since there are $n$ distinct eigenvalues the corresponding eigenvectors form a basis for $\mathbb R^n$. So knowing the eigenvectors and eigenvalues completely determines what the linear transformation corresponding to the matrix is.
answered Apr 19, 2016 at 13:29
hmakholm left over Monicahmakholm left over Monica
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For a 2x2 matrix you can visualize the answer Henning Makholm using the eigencircle of the matrix:
If you scroll up from the bookmark Eigencircles of special transformations,
You will see that if both eigenvalues and eigenvectors are fixed the transformation is fully defined.
The two green vectors$v_{A1}$ and $v_{A2}$ are the eigenvectors of A in the inserted figure.
you can also go straight the articles on eigencircles:
Englefield, M. J., & Farr, G. E. (2006). Eigencircles of 2 x 2 Matrices. Mathematics Magazine Vol. 79 Oct.,2006, 281-289.
