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I just learned singular value decomposition that a matrix $A$ can be decomposed as $A=U^* \Sigma V$ where $U,V$ are unitary matrices and $\Sigma$ is a diagonal matrix with singular values on the upper part of the diagonal.

By this, $\det A=\det U^* \times \det \Sigma \times \det V= { \pm }\det \Sigma$ is the sum of those real non-negative singular values plus the sign, and has to be real? This looks incredible to me.

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    $\begingroup$ Huh? Who told you that $\det(U^\ast V)=\pm1$? $\endgroup$
    – user1551
    Commented Mar 31, 2017 at 14:48
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    $\begingroup$ The title question: take the $1\times 1$ matrix $A=(i)$. Then $\det(A)=i\not\in \mathbb{R}$. $\endgroup$ Commented Mar 31, 2017 at 14:48
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    $\begingroup$ If U is unitary then |det U| = 1. This does not mean that det U = +/-1. $\endgroup$
    – gandalf61
    Commented Mar 31, 2017 at 14:52

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You're confusing the definition of unitary matrix. For matrices over $\mathbb{R}$ it is the case that unitary matrices have determinate $\pm 1$, but for matrices over $\mathbb C$ that is no longer true. In general, a matrix $U$ is unitary if and only if $UU^\ast=I$. In $\mathbb C$, unitary matrices can have det $i$ or any other complex number with norm $1$.

As an obvious check, you can see your conclusion is wrong because $[i]$ is a matrix over $\mathbb{C}$ and clearly doesn't have real determinant. For a less degenerative case, take the identity matrix and change $a_{0,0}$ to $i$. The resulting matrix has determinant $i$.

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