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For a symmetric matrix, the SVD is $$A = USV^T$$ and, since $A = A^T$, we have $$USV^T = VSU^T$$ $(S = S^T)$, but how to prove $U=V$?

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  • In general, we can't take $U = V$; that only happens if $A$ additionally has non-negative eigenvalues. – Ben Grossmann Nov 10 '16 at 14:09
  • Take $A$ to be $n\times n$ zero matrix. Then $S=A$ and any unitary $U,V$ give a possible SVD without $U=V$. – A.Γ. Nov 10 '16 at 14:18

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