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Let $A$ be a Hermetian matrix having all district eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$. If $X_1, X_2,\ldots X_n$ are corresponding eigenvectors then show that the $n\times n$ matrix $C$ whose $k^{th}$ column consists of the vector $X_k$ is nonsingular.

I need to show that matrix $C$ in the question (above or image) is non singular.

snulty
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Gcap
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    Hint: Vectors having pairwise distinct eigenvalues are linearly independent. What does this tell you about the (column) rank of your matrix? – Mr. Chip Aug 09 '16 at 11:50
  • Related: http://math.stackexchange.com/questions/29371/how-to-prove-that-eigenvectors-from-different-eigenvalues-are-linearly-independe – Martin Sleziak Aug 09 '16 at 11:58
  • I found from other problems that Eigen vectors corresponding to two distinct eigenvalues of hermitian matrix are mutually orthogonal. Then normalising the eigenvectors a unitary square matrix is formed which is non singular with nonzero determinant. (Is solving the main question on similar lines all right?) – Gcap Aug 10 '16 at 12:27

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