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Given a rectangular matrix $X\in\mathbb{R}^{d\times p}$, $d>p$, and a diagonal matrix $D\in\mathbb{R}^{d\times d}$ with positive diagonal entries, and property $$X^TDX=I,$$ with $I\in\mathbb{R}^{p\times p}$ being the identity matrix, could some property be derived on SVD of $X$ (ie, singular values and singular values)?

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I assumed the diagonal matrix was supposed to be called $D$, and editied your question accordingly. – Harald Hanche-Olsen Oct 10 '12 at 10:51
It is tempting to multiply the equation by a pseudo-inverse of $X$ on the right, and its transpose on the left. In the general case, the projection on the range of $X$ will not commute with $D$, however, so this does not settle the question. But it might be a beginning. – Harald Hanche-Olsen Oct 10 '12 at 10:58
This might be helpful...(but I am not sure), your equation implies $Y^tY=I$ with $Y=\sqrt{D}X$. So, we get the singular values of $Y$. Can we get the singular values from $X$, then? – Tapu Oct 10 '12 at 12:01
@Tapu yes, singular values of $X$ are inverse of square roots of diagonal entries of $D$. Please organize your answer. – user506901 Oct 10 '12 at 14:39

All right, as I said in comment, your equation can be written as $$Y^tY=I$$ with $Y=\sqrt{D}X$. This shows that the eigenvalues of $Y^TY$, i.e., the singular values of $Y$ are $1$ ($p$ times) and $0$ ($d-p$ times). Now it is possible to find the singular values of $X$.

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