1
$\begingroup$

Consider $\mathbb{S}^n$, the set of all $n\times n$ real symmetric matrices. Let $A\in \mathbb{S}^n$ and $A=U\Lambda U^{\top}$ be its spectral decomposition. I want to know how to prove that $\Pi_{\Bbb S_+^n}(A)=U\Lambda ^+U^{\top}$, where $\Lambda^+$ is the $n\times n$ diagonal matrix given by $\Lambda_{ii}^+=\max\{\Lambda_{ii}, 0\}$, for $i=1,...,n$;$\Pi_{S}(x)=\arg\min_{z\in S}\|x-z\|^2_2$.

This projection means that we can project any symmetric matrix on $\mathbb{S}_+^n$ and get the closest positive semi-definite matrix, but how can we prove that it is the closest one?

$\endgroup$
  • $\begingroup$ Show that $\Pi_S(U A U^T) = U \Pi_S(A) U^T$ (as long as $USU^T = S$). Solving the problem for diagonal $A$ should be straightforward. $\endgroup$ – copper.hat Sep 19 '17 at 14:44
  • $\begingroup$ Also think carefully about what norm you're using for "closest." $\endgroup$ – Brian Borchers Sep 19 '17 at 14:53
0
$\begingroup$

$\Pi_{\mathbb{S}_+^n}(A)=\arg\min_{X \in \mathbb{S}_+^n}\|X-A\|^2_F=\arg\min_{X \in \mathbb{S}_+^n}\|\Lambda_X -V^{\top}A V\|^2_F$

$X = V \Lambda_X V^{\top}$ (as $X\in \mathbb{S}^n_+$, so it is diagolalizable), $B=V^{\top}A V$

$\|\Lambda_X -V^{\top}A V\|^2_F = \sum_{i \neq j}b_{ij}^2+\sum_i (\Lambda_{X_{ii}}-b_{ii})^2$

Minimum occurs when $b_{ij}=0$, and $\Lambda_{X_{ii}}=b_{ii}$. Note that $\Lambda_{X_{ii}} \geq 0$, so $\Lambda_{X_{ii}}=\max\{0,b_{ii}\}.$ $B$ is a diagonal matrix. So, $V=U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.