The existence of real SVD 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?
 A: 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.
A: 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.
