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I am trying to reconstruct a symmetric 3 x 3 matrix from just its eigenvalues and eigenvectors.

I think the solution involves orthogonalizing two of the eigenvectors using the Gram-Schmidt procedure, but I am not sure why or how. Thanks.

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    $\begingroup$ Normalize the eigenvectors so that they're mutually orthogonal, and assemble them in a matrix $\mathbf V$. Then with the diagonal matrix of eigenvalues $\Lambda$, form $\mathbf V\Lambda\mathbf V^\top$. I omitted a few details that you should fill in. $\endgroup$ Commented May 2, 2015 at 3:01
  • $\begingroup$ You shouldn't orthogonalize eigenvectors unless they have the same eigenvalue, should you? $\endgroup$
    – user856
    Commented May 2, 2015 at 3:11
  • $\begingroup$ Does this answer your question? Reconstruct the original symmetric matrix given Eigen values and the longest Eigen vector $\endgroup$
    – PinkyWay
    Commented May 10, 2020 at 8:08

2 Answers 2

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If the columns of a matrix $P$ are three linearly independent eigenvectors of a matrix $A$ corresponding to the eigenvalues $a,b,c$ respectively, then $P\operatorname{diag}(a,b,c)P^{-1}$ must be equal to $A$, regardless of whether $P$ is orthogonal or not. You don't need any orthogonalisation process. The resulting matrix is automatically symmetric (except that there are some rounding errors, of course, if you use floating point compuatation instead of symbolic computation) when $A$ is symmetric.

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since the matrix is symmetric, you can find three eigenvectors $\{u_1, u_2, u_3\}$ corresponding to the three real eigenvalues $\lambda_1, \lambda_2, \lambda_3$ forming an orthonormal basis for $R^3.$ then $$A = \lambda_1u_1u_1^\top +\lambda_2u_2u_2^\top +\lambda_3u_3u_3^\top . $$

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