# Positive semidefinite matrix using Schur Complement

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.

• Is $s \geq 0_p$? – Rodrigo de Azevedo Jun 21 '18 at 20:08
• 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 $0\le s\le 2$. – A.Γ. Jun 22 '18 at 6:18
• Yes, that is correct - s is a nonnegative vector. – mathstudent Jun 22 '18 at 6:29
• We can assume that $\Sigma$ is invertible. And since it is a covariance matrix $\Sigma$ is positive semidefinite and symmetric. – mathstudent Jun 22 '18 at 8:13

## 1 Answer

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.

• 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? – mathstudent Jun 22 '18 at 10:07
• @mathstudent $\Sigma\Sigma^\dagger$ will be $I$ matrix. – percusse Jun 22 '18 at 10:21