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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?

89085731
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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. – J. M. ain't a mathematician Sep 02 '12 at 05:51
  • @J.M. I have updated the question. – 89085731 Sep 02 '12 at 14:01
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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 $\mathbf A$? – J. M. ain't a mathematician Sep 02 '12 at 14:07
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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? – Seyhmus Güngören Sep 02 '12 at 14:07
  • @J.M. sorry,i have not noticed that. – 89085731 Sep 02 '12 at 14:11
  • Here's a related question on math.SE: http://math.stackexchange.com/questions/47764/can-a-real-symmetric-matrix-have-complex-eigenvectors – Kartik Audhkhasi Sep 02 '12 at 14:23
  • @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. – makansij Oct 11 '17 at 04:44
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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. – J. M. ain't a mathematician Oct 11 '17 at 04:59

2 Answers2

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As the comments and the other answer have pointed out, this follows from eigendecomposition. By transposing $A$ if necessary, we may assume that $A$ is tall, i.e. $A$ is $m\times n$ for some $m\ge n$. Let $VSV^T$ be an orthogonal diagonalisation of $A^TA$. Since $A^TA$ is real and positive semidefinite, $V$ can be taken to be real orthogonal and $S$ can be taken as a nonnegative diagonal matrix whose diagonal entries are arranged in decreasing order. Then $(AV)^T(AV)=S$, which implies that the columns of $AV$ are mutually orthogonal. Hence $AV=QS^{1/2}$ for some real $m\times n$ matrix $Q$ with orthonormal columns. Now complete $Q$ to an $m\times m$ real orthogonal matrix $U=\pmatrix{Q&\ast}$ and let $\Sigma=\pmatrix{S^{1/2}\\ 0}$. Then $AV=QS^{1/2}=U\Sigma$. Hence $A=U\Sigma V^T$ and we are done.

user1551
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We only need to show $U$ and $V$ are real. They can be obtained by the eig-decomposition of $AA^*$ and $A^*A$, which in our case are $AA^T$ and $A^TA$. By the spectral theorem of real symmetric matrix we have both $U$ and $V$ real.

yuguaw
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