Linear Algebra: Prove existence of polar form for any $n\times n $ matrix

I have a problem I don't quite understand:

Prove that any $n\times n$ matrix $A$ can be divided "polarly" (translated from Swedish "polärt") in the form $A=PQ$, where $Q$ is an orthogonal matrix and $P$ a positive semidefinite (or positive definite) matrix. This division can be interpreted as the matrix $A$ first rotates vectors and then stretches them with $P$. This can be used in e.g. theory for deformation of material. Guide: Use SVD of $A=U\Sigma V^T$ and that e.g. $A$ can be written as $A=(UU^T)A$

I don't know where to start. I suppose that P has to be a symmetrical matrix with non-negative eigenvalues if it is to be positive semi- or definite matrix. The singular values of $\Sigma $ can be any real number and $U$ is just a rotation matrix so $P=U\Sigma$ is not necessarily a positive semi- or definite matrix, neither is it necessarily symmetrical. Have misinterpreted what positive semi- and definite matrix means?

Since it is an $n\times n$ matrix, do I just say that you can express any matrix $A=[A]_U=U^TA$ and that for all negative singular values in $\Sigma$ you make them positive and multiply every row vector in $V^T$ corresponding to that singular value with $-1$?

Solution 1:

Using the singular value decomposition of $A$, we have:

$$A = U \Sigma V^T = U \Sigma U^T U V^T = P Q$$

where $P = U \Sigma U^T$ and $Q = UV^T$. Being the product of orthogonal matrices, $Q$ itself is orthogonal. Since $\Sigma$ is positive semi-definite and $P$ is unitarily equivalent to $\Sigma$, it follows that $P$ is positive semi-definite too.