1
$\begingroup$

Let

$$ G =\begin{pmatrix} \Sigma & \Sigma - \text{diag} \lbrace s \rbrace \\ \Sigma - \text{diag} \lbrace s \rbrace & \Sigma \end{pmatrix} $$

where $s \in \mathbb{R}_{\geq 0}^p$ is some nonnegative vector, and $\Sigma$ is a positive definite covariance matrix.

I need to argue that $G$ is positive semidefinite using the Schur complement. That is, $G$ is positive semidefinite iff the Schur complement $$ 2\text{diag} \lbrace s \rbrace-\text{diag} \lbrace s \rbrace \Sigma^{-1} \text{diag} \lbrace s \rbrace $$ is positive semidefinite. However I am not sure how to prove this.

$\endgroup$
  • $\begingroup$ Is $s \geq 0_p$? $\endgroup$ – Rodrigo de Azevedo Jun 21 '18 at 20:08
  • $\begingroup$ It is not possible to say anything without more information on $s$. Take e.g. $\begin{bmatrix}1 & 1-s\\1-s & 1\end{bmatrix}$. It is psd iff $0\le s\le 2$. $\endgroup$ – A.Γ. Jun 22 '18 at 6:18
  • $\begingroup$ Yes, that is correct - s is a nonnegative vector. $\endgroup$ – mathstudent Jun 22 '18 at 6:29
  • $\begingroup$ We can assume that $\Sigma$ is invertible. And since it is a covariance matrix $\Sigma$ is positive semidefinite and symmetric. $\endgroup$ – mathstudent Jun 22 '18 at 8:13
1
$\begingroup$

You can write down the nonstrict version of the Schur complement formula

$$ G\succeq 0 \iff \left\{\begin{array}{cc}\Sigma\succeq 0\\ \Sigma-(\Sigma-\textrm{diag}(s))\Sigma^{\dagger}(\Sigma-\textrm{diag}(s))\succeq0\\ (I-\Sigma\Sigma^\dagger)(\Sigma-\textrm{diag}(s))=0\end{array}\right. $$ The second condition expands to $$ \Sigma-(\Sigma-\textrm{diag}(s))\Sigma^{\dagger}(\Sigma-\textrm{diag}(s))= \textrm{diag}(s)\Sigma^{\dagger}\Sigma + \Sigma\Sigma^{\dagger}\textrm{diag}(s)-\textrm{diag}(s)\Sigma^{\dagger}\textrm{diag}(s) $$ if $\Sigma$ is also invertible then you are done.

$\endgroup$
  • $\begingroup$ Thank you for the answer. However I am not sure that I understand the last step. Why is the last matrix positive semidefinite if $\Sigma$ is invertible? $\endgroup$ – mathstudent Jun 22 '18 at 10:07
  • 1
    $\begingroup$ @mathstudent $\Sigma\Sigma^\dagger$ will be $I$ matrix. $\endgroup$ – percusse Jun 22 '18 at 10:21

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.