Given a Gaussian distribution $\cal{N}(\mu, \Sigma)$ or $\cal{N}^{-1}(\eta, \Lambda)$, where we have $\Lambda \mu = \eta$, $\Sigma^{-1} = \Lambda$, its marginalization and conditioning can be described as the figure below:

Without explaining the details, we assume we get the final linear equation $Ax = b$ from a SLAM or bundle adjustment problem, where $A$ is a square matrix. Let's put it into an information form. $\Lambda_{cc}$ is a $M \times M$ matrix and $\Lambda_{ll}$ is a $N \times N$ matrix. $M$ is the number of camera keyframes while $N$ is the number of landmarks.

$$ \begin{bmatrix} \Lambda_{cc} & \Lambda_{cl} \\ \Lambda_{lc} & \Lambda_{ll} \end{bmatrix} \begin{bmatrix} \mu_c \\ \mu_l \end{bmatrix}= \begin{bmatrix} \eta_c \\ \eta_l \end{bmatrix} $$

With Schur Complement, we can find $\mu_c$ and $\mu_l$ by inverting two $M \times M$ and $N \times N$ matrices instead of inverting a $(M+N) \times (M+N)$ matrix. We will first find camera positions $\mu_c$, and then find landmark positions $\mu_l$. Thinking it in a probabilistic way, Schur Complement is actually doing following:

$$ p(c, l) = p(c) * p(l|c) $$

It first marginalizes landmarks l to get p(c), then find l conditioned on camera poses c. We assume p(c, l) follows the Gaussian distribution in information form:

$$ p(c, l) \sim (\begin{bmatrix} \eta_c \\ \eta_l \end{bmatrix}, \begin{bmatrix} \Lambda_{cc} & \Lambda_{cl} \\ \Lambda_{lc} & \Lambda_{ll} \end{bmatrix} ) $$

Recall the first figure in this blog, we consider $\alpha$ as $c$ and $\beta$ as $l$. So we the information form based marginalization, we can get $p(c) \sim \cal{N}(\eta, \Lambda)$ as: $$ \begin{align} \eta &= \eta_c - \Lambda_{cl}\Lambda^{-1}_{ll}\eta_l \\ \Lambda &= \Lambda_{cc} - \Lambda_{cl}\Lambda^{-1}_{ll}\Lambda_{lc} \end{align} $$ Then the camera poses will be obtained by solving $\Lambda \mu_c = \eta$, and its corresponding uncertainties are $\Sigma_{cc} = \Lambda^{-1}$.

Once we get $\mu$, we could find landmarks information by: $$ \begin{align} \eta' &= \eta_l - \Lambda_{lc}\mu_c \\ \Lambda' &= \Lambda_{ll} \end{align} $$ Then the landmark positions will be obtained by solving $\Lambda' \mu_l = \eta'$, and its corresponding uncertainties are $\Sigma_{ll} = \Lambda'^{-1}$