# How to prove that the Schur complement of symmetric, positive-definite matrix is positive-definite?

We have $$A \in \mathbb{R}^{n \times n}$$ which is symmetric and positive-definite. Also, $$A$$ is a block matrix:

$$A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix}$$

I have managed to show that both $$A_{11}$$ and $$A_{22}$$ are symmetric and positive-definite. Also, it is easy to show $$S = A_{22} -A_{12}^T A_{11}^{-1} A_{12}$$ (Schur complement) is symmetric. What I cannot do is to show that $$S$$ is positive definite as well.

## 2 Answers

Let $$S$$ be the Schur complement. For any (column) vector $$v$$ define $$\tilde{v}=\begin{bmatrix}-A_{11}^{-1}A_{12}v\\v\end{bmatrix}.$$ Then $$\tilde{v}^TA\tilde{v} = v^TSv$$.

Show that $$A$$ $$A = \begin{bmatrix}I \\ A_{12}^\top A_{11}^{-1} & I\end{bmatrix} \begin{bmatrix}A_{11} \\ & A_{22} - A_{12}^\top A_{11}^{-1} A_{12}\end{bmatrix} \begin{bmatrix}I & A_{11}^{-1} A_{12} \\ & I\end{bmatrix} =: NDN^\top.$$

Show that $$N$$ is invertible.

$$N^{-1} = \begin{bmatrix}I \\ -A_{12}^\top A_{11}^{-1} & I\end{bmatrix}.$$

Show that consequently, $$D$$ is positive-definite.

$$D = N^{-1} A (N^{-1})^\top$$ so for any $$v \ne 0$$ we have $$v^\top D v = v^\top N^{-1} A (N^{-1})^\top v = w^\top A w > 0$$ where $$w = (N^{-1})^\top v$$.

Show that $$A_{22} - A_{12}^\top A_{11}^{-1} A_{12}$$ is positive-definite.

• D being positive-definite is a consequence of N being invertible? How is that? Commented Nov 5, 2019 at 19:47
• @Uppermost See my edit Commented Nov 5, 2019 at 20:25
• Got it, thanks a lot! Commented Nov 6, 2019 at 8:36