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

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.

  • $\begingroup$ Is $s \geq 0_p$? $\endgroup$ Commented Jun 21, 2018 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.
    Commented Jun 22, 2018 at 6:18
  • $\begingroup$ Yes, that is correct - s is a nonnegative vector. $\endgroup$ Commented Jun 22, 2018 at 6:29
  • $\begingroup$ We can assume that $\Sigma$ is invertible. And since it is a covariance matrix $\Sigma$ is positive semidefinite and symmetric. $\endgroup$ Commented Jun 22, 2018 at 8:13

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jun 22, 2018 at 10:07
  • 1
    $\begingroup$ @mathstudent $\Sigma\Sigma^\dagger$ will be $I$ matrix. $\endgroup$
    – percusse
    Commented Jun 22, 2018 at 10:21

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