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As we know, SVD decomposites any matrix $M$ into the form: $$M=U\Sigma V^*,$$ where $U$ and $V$ are normally different.

In here Wikipedia says that a matrix A is normal if and only if $U=V$. But in the very same article, it raised an example of a normal matrix $$M=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}=U\Sigma V^*,$$ where $$U= \left( \begin{array}{ccc} -0.57735 & 0.816497 & 0. \\ -0.57735 & -0.408248 & -0.707107 \\ -0.57735 & -0.408248 & 0.707107 \\ \end{array} \right),$$ but $$V=\left( \begin{array}{ccc} -0.57735 & 0.408248 & 0.707107 \\ -0.57735 & 0.408248 & -0.707107 \\ -0.57735 & -0.816497 & 0. \\ \end{array} \right),$$ according to mathematics software. They are not equal, why is that?

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2 Answers

up vote 2 down vote accepted

Let $M = U \Sigma V^*$ be an SVD of $M$. Suppose that $V=U$. Then $M$ is symmetric positive-definite. Conversely, if $M$ is symmetric positive semi-definite, then any unitary eigendecomposition $M=U \Lambda U^*$ in which the diagonal entries of $\Lambda$ are ordered from maximal to minimal, is an SVD of $M$.

Hence $U=V$ in the SVD if and only if $M$ is symmetric positive-semidefinite.

Also, the wiki article you reference, does not say that $U=V$ if and only if $M$ is normal.

In your example $M$ is not even symmetric hence it can't possibly be the case that $U=V$.

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Read the Wiki page again. It says $M$ is normal if and only if $M=U\Lambda U^\ast$ for some diagonal matrix $\Lambda$ and some unitary matrix $U$. Note that $\Lambda$ may be non-real. However, in a SVD $M=U\Sigma V^\ast$, $\Sigma$ is a real diagonal matrix. So, the two decompositions are not in conflict with each other.

Actually, as pointed out in Manos' answer, if $U=V$ in the SVD of $M$, then $M$ is not just normal, but it must be positive semidefinite.

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