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I calculated the singular value decomposition (SVD) of a square matrix, $A = U\Sigma V^{\intercal}$. I noticed that the unitary matrices $U$ and $V^{\intercal}$ satisfy $U = -V^{\intercal}$. What property of the original matrix $A$ leads to this outcome?

Edit: I think it might be that $A$ is a symmetric matrix, based on this answer.

Fab von Bellingshausen
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  • Was the matrix of interest antisymmetric? – A rural reader Jun 27 '21 at 01:58
  • If it were $U=-V$ instead, then $A=U \Sigma V^\top = U (-\Sigma) U^\top$ so $A$ is symmetric with nonpositive eigenvalues. However, $U=-V$ is not $U=-V^\top$, so this is a different condition. – angryavian Jun 27 '21 at 02:02

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