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I have known that if matrix A $\in C^{m*n}$,then there exists a SVD $A=U\Sigma V$.My question is if A is real,does there exist SVD which U and V real?

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    I'm not sure I understand. For an arbitrary complex matrix, you can set things up so that the matrix of left and right singular vectors satisfy an orthogonality relation, and the diagonal matrix of singular values are entirely real. If the original matrix is real, then the left and right singular vectors can all be made real.2012-09-02
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    @J.M. I have updated the question.2012-09-02
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    Alright, first things first: are you aware of the relationship between the eigendecompositions of $\mathbf A^\top\mathbf A$ and $\mathbf A\mathbf A^\top$, and the singular value decomposition of $\mathbf A$?2012-09-02
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    @89085731 yes if $A$ is real then all $U$ $V$ and $S$ are real. If $A$ is hermitian then $S$ is still real. Does this answer your question?2012-09-02
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    @J.M. sorry,i have not noticed that.2012-09-02
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    Here's a related question on math.SE: http://math.stackexchange.com/questions/47764/can-a-real-symmetric-matrix-have-complex-eigenvectors2012-09-02
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    @J.M.isnotamathematician what is the conclusion on this? I follow you about the eigendecompositions, but here the OP is asking about SVD is which slightly more complicated.2017-10-11
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    @Hunle, the point was to observe and use relationships between the symmetric eigendecomposition of the cross-product matrices and the singular value decomposition of the original matrix. If anything is still unclear, consider elaborating in a new question.2017-10-11

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