SVD of symmetric but indefinite matrix

The SVD of the matrix AA is A=UΣVTA = U \Sigma V^T, where A∈Rm×nA\in R^{m\times n} is symmetric positive definite or semi positive definite matrix and UU and VV are square orthogonal matrices.

Does AA has to be positive or semi positive? If A∈Rn×nA\in R^{n\times n} is symmetric but indefinite can we still have a SVD?
Also if A∈Rm×nA\in R^{m\times n} but UU and VV are not orthogonal will AA and UAVUAV still have same singular values?

Basicaly I want to understand if we can have SVD for any matrix and if the case when AA is symmetric positive definite with orthogonal eigenvectos just a special case of SVD?

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If AA is m×nm \times n and m≠nm \ne n, it can’t be symmetric.
– Robert Israel
2 days ago

  

 

In that question it is stated that A∈RnxnA\in R^{nxn}
– Biljana
2 days ago

  

 

In the first line it says m×nm \times n.
– Robert Israel
2 days ago

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1 Answer
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SVD decomposition exists for any real matrices. No other additional requirement is needed. Σ\Sigma share the same size as AA and UU and VV are orthogonal matrices.

As for your another question when UU and VV need not be orthogonal, not true in general, for example if you let U=0U=0 then the singular value is just 00.

When a matrix is symmetric, we know that it has an eigenvalue decomposition, that is you can pick UU and VV to be the same orthogonal matrix.