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Supposing $A$ is an $m \times n$ matrix where $m > n$ and $A$ has full column rank. I want to find a $C$ (an $m \times m$ matrix) such that $A^TCA$ is a diagonal matrix and also that the maximum singular value of $C$ is the smallest possible.

EDIT: $C$ also has to satisfy: $C= VDV^T$ where $V$ is an orthogonal matrix and $D$ is diagonal with positive entries on the diagonal.

Thanks

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    @m_l: WLOG $C$ can be considered just to be a symmetric invertible matrix, as any such $C$ would have a decomposition $VDV^T$ (if I am not mistaken). $C$ has entries in $\mathbb{R}$.2012-04-13

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There is no $C \in \mathbb{R}^{m \times m}$ with minimal singular values.

By the restriction $C = VDV^{tr}$, $C$ must be symmetric and positive definite. Now suppose $C \in \mathbb{R}^{m \times m}$ symmetric and positive definite such that $A^{tr}CA$ is diagonal. As Rahul pointed out in his comment, consider $\widetilde{C} := \frac{1}{2}C$. Then $\widetilde{C}$ is symmetric and positive definite and $A^{tr}\widetilde{C}A = \frac{1}{2} A^{tr}CA$ is diagonal, but the singular values of $\widetilde{C}$ are strictly smaller than those of $C$.

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    I think so. Find $B \in \mathbb{R}^{m \times m}$ such that $BA$ has the identity matrix as its first $n \times n$ submatrix and all rows below contain only 0. Then $C := B^{tr}B$ should be what you are looking for.2012-04-13