Singular value decomposition (SVD) of matrix R from reduced QR decomposition

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I have a matrix $A=QR$, after performing QR decomposition.

Why is it that performing a singular value decomposition on $A$ gives the same answer for $\Sigma$ and $V$ when performing one on the matrix $R$ from the QR decomposition? I'm using the form $A=U\Sigma V^T$ for the SVD

Is it to do with the orthogonality of the matrix $Q$? Regardless, could somebody explain or at least give a hint as to why this happens?

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The matrix $R$ contains the same information about $A$ as the matrices $\Sigma$ and $V$ do. Specifically, if $$A = QR = U\Sigma V^T,$$ then $$ A^T A = (U \Sigma V^T)^T (U \Sigma V^T) = V \Sigma^2 V^T \\ $$ and $$ A^TA = (Q R)^T Q R = R^T Q^T Q R$$ As you suspected, it is the orthogonality of the matrices involved, i.e., $U$ and $Q$ which is critical.