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Given the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$, how would I find a real orthogonal matrix $P$ such that $PAP^t$ is a diagonal matrix?

I've found the eigenvalues $0, \dfrac{9\pm\sqrt{105}}{2}$, but I don't know how to proceed from here.

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There exists a basis of eigenvectors since $A$ is symmetric. Form an orthonormal basis of eigenvectors. Form the matrix whose columns are the basis vectors, as your $P$.

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  • $\begingroup$ How do you form this orthonormal basis of eigenvectors? $\endgroup$ – user386867 Feb 21 at 18:18
  • $\begingroup$ For each eigenvalue $\lambda$, find the kernel of $A-\lambda I$. Get three linearly independent eigenvectors, then use Gram-schmidt. $\endgroup$ – Chris Custer Feb 21 at 18:36

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