Polar decomposition, positive-definiteness

Question

The polar decomposition says that any matrix $A$ can be decomposed as $A = WP$ where $W$ is unitary, and $P$ is positive-semidefinite.
In proving the existence of a polar decomposition, one uses the fact that $V\Sigma V^{*} $ obtained from the singular value decomposition of $A$ is positive-semidefinite. This also uses the fact that $\Sigma$ is positive-semidefinite.
However, $\Sigma$ is not even a square matrix, how can it be positive-semidefinite?

Answer 1

A unitary matrix is square, by definition. A positive-definite matrix is Hermitian, so also square by definition.

Thus, one can talk about the polar decomposition only for square matrices. And then $\Sigma$ is square too.

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If $A$ is any square matrix, it has a SVD $A=U\Sigma V^*$, where $U$ and $V$ are unitary. Then, setting $W=UV^*$ and $P=V\Sigma V^*$, we have that $A=WP$, $W$ is unitary, and $P$ is positive-semidefinite, because $\Sigma$ is diagonal with non negative entries by definition.

Answer 2

In the singular value decomposition, $\Sigma$ is not positive semi-definite as a matrix. It is positive semi-definite in the sense that its diagonal entries are real and non-negative.