Let $A$ be an $(n+m) \times (n+m)$ symmetric positive definite matrix
$$A=\begin{bmatrix}A_{11} & A_{12}\\ A_{12}^T & A_{22}\end{bmatrix}$$
where $A_{11}$ is an $n \times n$ matrix, $A_{12}$ is an $n \times m$ matrix, and $A_{22}$ is an $m \times m$ matrix. We can factor $A$ into $LL^T$, as follows:
$$A=\begin{bmatrix}L_{11} & 0\\L_{21} & L_{22} \end{bmatrix}\begin{bmatrix}L_{11} & 0\\L_{21} & L_{22} \end{bmatrix}^T \text{,}$$
where $L_{11}$ is an $n \times n$, $L_{21}$ is $n \times m$, and $L_{22}$ is $m \times m$. I want to show that $S=L_{22}L_{22}^T$ where $S=A_{22}-A_{12}^TA_{11}^{-1}A_{12}$ is the Schur complement of $A_{11}$.
I'm stuck trying to unravel how to apply this to block matrices. I'm having a bit of a conceptual hurdle translating my definitions to this problem.
We know that $$ A=\left[\matrix{A_{11} & A_{12}\\ A_{12}^T & A_{22}}\right]=\begin{bmatrix}L_{11} & 0\\L_{21} & L_{22} \end{bmatrix}\begin{bmatrix}L_{11}^T & L_{21}^T\\0 & L_{22}^T \end{bmatrix}= \left[\matrix{L_{11}L_{11}^T & L_{11}L_{21}^T\\ L_{21}L_{11}^T & L_{21}L_{21}^T+L_{22}L_{22}^T}\right]. $$ Since we assume that $A_{11}=L_{11}L_{11}^T$ is invertible then $$ \color{red}{A_{11}^{-1}}=(L_{11}L_{11}^T)^{-1}=\color{red}{(L_{11}^T)^{-1}L_{11}^{-1}}. $$ Now let's see what $S$ is $$ S=\color{blue}{A_{22}}-A_{12}^T\color{red}{A_{11}^{-1}}A_{12}=\color{blue}{L_{21}L_{21}^T+L_{22}L_{22}^T}-L_{21}\underbrace{L_{11}^T\color{red}{(L_{11}^T)^{-1}L_{11}^{-1}}L_{11}}_{=I}L_{21}^T=L_{22}L_{22}^T. $$