Is it true that an nxn symmetric matrix has n linearly independent eigenvectors even for non-distinct eigenvalues? How can we show it rigorously? Basically, I want to prove that if an nxn symmetric matrix has eigenvalue 0 with multiplicity k, then its rank is (n - k). If we can prove that there exist k linearly independent eigenvectors which solve Ax = 0, then we can use the rank-nullity theorem to directly show the result, right?(adsbygoogle = window.adsbygoogle || []).push({});

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# Eigenvectors of a symmetric matrix.

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