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I have a symmetric matrix A, whose eigenvalues are $\lambda_1 = 6,~ \lambda_2 = 3,~ \lambda_3 = 2$ and eigenvectors are $\vec{v_1} = (1, 1, 1),~\vec{v_2} = (1,1,-1)$. How do I find the third eigenvector $\vec{v_3}$?

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  • $\begingroup$ Eigenvectors of distinct eigenvalues are pairwise orthogonal. $\endgroup$ – xbh Dec 5 '18 at 10:46
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Hint: Since the matrix is symmetric, (and real, I presume) it has an orthogonal basis of eigenvectors.

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  • $\begingroup$ But how do I use this in practice? $\endgroup$ – Henri Södergård Dec 5 '18 at 11:08
  • $\begingroup$ By searching for a vector orthogonal to both $\vec{v_1}$ and $\vec{v_2}$. $\endgroup$ – José Carlos Santos Dec 5 '18 at 11:15
  • $\begingroup$ So, it would be okay to take the cross-product of both vectors, as in $\vec{v_3} = \vec{v_1} \times \vec{v_2}$ and the resulting $\vec{v_3} = (-3,3,0)$ would be what I'm looking for? $\endgroup$ – Henri Södergård Dec 5 '18 at 11:23
  • $\begingroup$ Yes, indeed it is. $\endgroup$ – José Carlos Santos Dec 5 '18 at 11:24

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