# Solve for $D$ in $R^{T}DSDR = Id$

Given that $R$ is a rectangular matrix, $D$ is a diagonal, square matrix and $S$ being a square matrix along with the fact that both $D$ and $S$ are invertible. $S$ in this specific case can be assumed to be symmetric, positive-definite.

$R$ is actually the first few columns of the discrete Fourier Transform matrix. Thus, $R^{T}R=Id$. Note $RR^{T}$ need not be $Id$. Also, $R$ has both dimensions of the same order (but with more rows than columns). $R$ also has a full row rank.

I would like to obtain $D$ in terms of the given matrices, which are $R$ and $S$

• How can $R$ be invertible if it's rectangular? – Wolfgang Bangerth Feb 16 '16 at 12:30
• Also, is $S$ symmetric, positive definite, invertible? Is $R$ a tall or a wide rectangular matrix? – Wolfgang Bangerth Feb 16 '16 at 12:31
• I have edited the question to reflect changes. $S$ is indeed symmetric, positive-definite. Both dimensions of $R$ are of the same order but with more rows than columns. – gpavanb Feb 16 '16 at 15:18
• $D$ will not be unique in general. Think for example of the case where $S=id$ and $R$ is a single column. – Jan Weidner Feb 16 '16 at 18:47
• Thanks. In case where $S=Id$, and $R$ is a single column, i.e., a vector of ones because it is the first column of the Discrete Fourier Transform matrix, we are left with $\sum_{i} D_{i}^{2} = 1$ and the solution is indeed not unique. – gpavanb Feb 16 '16 at 19:22