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I want to find the closest symmetric matrix (using the Frobenius norm, or another matrix norm) to a given symmetric matrix $G$ such that the inner product with a given rank-$1$ matrix $b b^T$ is bigger than a given constant $c$.

Given $G \in \mathcal{M}_{n \times n}$ symmetric, $b \in \mathbb{R}^n$ and $c \in \mathbb{R}$,

$$\begin{array}{ll} \underset{\tilde{G}}{\text{minimize}} & \| \tilde{G} - G \|\\ \text{subject to} & b^T \tilde{G} \ b \geq c\end{array}$$

I think that there should be a relation with the PCA SVD and that somehow the answer should be related to the eigenvectors and eigenvalues of $G$. Thanks.

Rodrigo de Azevedo
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user2784810
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  • Yes. it is symmetric – user2784810 Jul 08 '18 at 15:30
  • this can be phrased as a convex quadratic optimization problem; solving the unconstrained problem and the one with "$=c$" is simple using the KKT conditions. – LinAlg Jul 08 '18 at 17:22
  • You can make your life easier by minimizing $| \tilde{G}-G|_{F}^{2}$. Also note that either $\tilde{G}=G$ will satisfy the constraint and be optimal, or the single linear constraint will be active at optimality. Have you tried to write down the KKT conditions for this problem? – Brian Borchers Jul 08 '18 at 17:55
  • @LinAlg I could use KKT conditions but I'd like to find a general characterization of the solution of this problem. My guess is that I can do an eigendecomposition of the matrix G and move the eigvenvalues according to the weight induced by b, a bit like https://math.stackexchange.com/questions/648809/how-to-find-closest-positive-definite-matrix-of-non-symmetric-matrix – user2784810 Jul 08 '18 at 19:42
  • There is a simle formula that can be derived using the KKT conditions. – Brian Borchers Jul 08 '18 at 22:19
  • @BrianBorchers can u explain? – user2784810 Jul 09 '18 at 08:19

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