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Let $$A=\left( \begin{array}{ccc}A_{11}&A_{12}\\A_{21}&A_{22}\end{array}\right)\in R^{n\times n}$$ be a symmetric positive definite matrix with blocks $A_{ij}\in\mathbb{R}^{n_i\times n_j},i,j=1,2$,$n_1+n_2=n$,Let $B_{22}\in\mathbb{R}^{n_1\times n_2}$ be a symmetric positive definite matrix. Prove that the matrix $$\left( \begin{array}{ccc}A_{11}&A_{12}\\A_{21}&B_{22}+A_{21}A_{11}^{-1}A_{12}\end{array}\right)$$is symmetric positive definite.

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Note that since $A_{11}$ is nonsingular (in fact SPD), we can do the following factorisation: $$ \begin{bmatrix}A_{11}&A_{12}\\A_{21}&X\end{bmatrix} =\begin{bmatrix}I&0\\A_{21}A_{11}^{-1}&I\end{bmatrix} \begin{bmatrix}A_{11}&0\\0&X-A_{21}A_{11}^{-1}A_{12}\end{bmatrix} \begin{bmatrix}I&A_{11}^{-1}A_{12}\\0&I\end{bmatrix}. $$ Hence it is SPD iff the Schur complement $X-A_{21}A_{11}^{-1}A_{12}$ is SPD. With $X=B_{22}+A_{21}A_{11}^{-1}A_{12}$, one has $X-A_{21}A_{11}^{-1}A_{12}=B_{22}$. Therefore since $B_{22}$ is SPD, the matrix $$\begin{bmatrix}A_{11}&A_{12}\\A_{21}&B_{22}+A_{21}A_{11}^{-1}A_{12}\end{bmatrix}$$ is SPD as well.

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