Alternate representation of the Kalman Filter recursions for mean and variance in LGSSM I'm reading the thesis Variational Algorithms for Approximate Bayesian Inference. In section 5.3 the author gives a variational treatment to a linear Gaussian state-space model (LGSSM):
\begin{align}
{\bf x}_t &= A {\bf x}_{t-1} + {\bf w}_t, && {\bf w}_t \sim \mathcal N({\bf 0}, Q) \\
{\bf y}_t &= C {\bf x}_t + {\bf v}_t, && {\bf v}_t \sim \mathcal N({\bf 0}, R),
\end{align}
where ${\bf x}_t$ is $k \times 1$ vector of hidden states, $A$ is the $k \times k$ state dynamics matrix, ${\bf y}_t$ is the $p \times 1$ observation vector, and $C$ is the $p \times k$ observation matrix. $Q$ and $R$ are covariance matrices (symmetric and invertible) and  of size $k \times k$ and $p \times p$, respectively.
The author further assumes that $Q = I_k$, the identity matrix of size $k$. In section 5.3.3. on page 175 the author derives the recursive equations for the Kalman Filter: 
\begin{align}
\Sigma_t &\equiv \text{var}(x_t |y_{1:t}) = \left[ I_k + C^{\top} R^{-1} C - A \Sigma_{t-1}^* A^{\top}  \right]^{-1} \tag{1} \\
\mu_t &\equiv E(x_t | y_{1:t}) = \Sigma_t \left[ C^{\top} R^{-1} y_t + A \Sigma_{t-1}^* \Sigma_{t-1}^{-1} \mu_{t-1} \right], \tag{2}
\end{align}
where $$\Sigma_{t-1}^* = \left[ \Sigma_{t-1}^{-1} + A^{\top} A \right]^{-1},$$
and $(\cdot)^{\top}$ denotes matrix transpose. Note that $\Sigma_t$ and $\Sigma_{t-1}$ are both $k \times k$ matrices, hence $\Sigma_{t-1}^*$ is also $k \times k$.
When I derived the recursion equations on my own I came up with what is below. It matches what I have seen in many articles regarding this topic:
\begin{align}
\Sigma_t &\equiv \text{var}(x_t |y_{1:t}) = \Sigma_{t|t-1} - \Sigma_{t|t-1} C^{\top} \left[ C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C \Sigma_{t|t-1} \tag{3}  \\
\mu_t &\equiv E(x_t | y_{1:t}) = \mu_{t|t-1} + \Sigma_{t|t-1} C^{\top} \left[C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} (y_t - C \mu_{t|t-1}), \tag{4}
\end{align}
where $$\mu_{t|t-1} = A \mu_{t-1} \quad \text{ and } \quad \Sigma_{t|t-1} = A \Sigma_{t-1} A^{\top} + I_k.$$

Does anyone see how equations $(1)$ and $(3)$ are equivalent? Or how equations $(2)$ and $(4)$ are equivalent?

I believe it boils down to matrix algebra (the inversion lemma, to be specific). I specified values for $A$, $C$, $R$, $\Sigma_{t-1}$, $y_t$, and $\mu_{t-1}$ in MATLAB and $(1)$ and $(3)$ seem to agree. 
 A: I've been working on this for at least 3 days now and it has finally clicked, so my sincerest apologies for anyone who spent some time working on this. 
I have working with the document Matrix Inversion Identities. In particular, the lemma that states that 
\begin{align}
(A + BCD)^{-1} &= A^{-1} - A^{-1} B (C^{-1} + D A^{-1} B)^{-1} D A^{-1} \\
&= A^{-1} - A^{-1} B (D A^{-1} B + C^{-1})^{-1} D A^{-1}
\end{align}
for invertible matrices $A$ and $C$. 
My mistake that has cost me so much time was that I kept trying to use the lemma above to expand the far RHS of equation $(3)$ into that of equation $(1)$. Instead of doing that, the trick is to use the lemma above to condense the far RHS of equation $(3)$ into that of equation $(1)$.
$$\Sigma_{t|t-1} - \Sigma_{t|t-1} C^{\top} \left[ C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C \Sigma_{t|t-1} = \left[ \Sigma_{t|t-1}^{-1} + C^{\top} R^{-1} C \right]^{-1}$$
Lastly, using the lemma again we get that 
\begin{align}
\Sigma_{t|t-1}^{-1} &= \left[ A \Sigma_{t-1} A^{\top} + I_k \right]^{-1} \\
&= \left[ I_k + A \Sigma_{t-1} A^{\top} \right]^{-1} \\
&= I_k - A \left(\Sigma_{t-1}^{-1} + A^{\top} A \right)^{-1}A^{\top} \\
&= I_k - A \Sigma_{t-1}^* A^{\top},
\end{align}
which shows that equation $(1)$ and $(3)$ are equivalent. 

Manipulating equation $(2)$ a bit gives the following:
$$ \mu_t = \Sigma_t C^{\top} R^{-1} y_t + \Sigma_t A \Sigma_{t-1}^* \Sigma_{t-1}^{-1} \mu_{t-1} $$
Manipulating equation $(4)$ a bit gives the following:
$$\mu_t = \Sigma_{t|t-1} C^{\top} \left[C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} y_t + \left( I_k - \Sigma_{t|t-1} C^{\top} \left[C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C \right) \mu_{t|t-1}
$$
So we need to show that 
$$\Sigma_t C^{\top} R^{-1} = \Sigma_{t|t-1} C^{\top} \left( C \Sigma_{t|t-1} C^{\top} + R \right)^{-1} \tag{5}$$
and 
$$\Sigma_t A \Sigma_{t-1}^* \Sigma_{t-1}^{-1} =  I_k - \Sigma_{t|t-1} C^{\top} \left[C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C. \tag{6}$$
Proof of $(5)$:
\begin{align}
\Sigma_t C^{\top} R^{-1} &= \Sigma_{t|t-1} C^{\top} R^{-1} - \Sigma_{t|t-1} C^{\top} \left( C \Sigma_{t|t-1} C^{\top} + R \right)^{-1} C \Sigma_{t|t-1} C^{\top} R^{-1} \\
&= \Sigma_{t|t-1} C^{\top} R^{-1} - \Sigma_{t|t-1} C^{\top} \left( C \Sigma_{t|t-1} C^{\top} + R \right)^{-1} \left( C \Sigma_{t|t-1} C^{\top} + R - R \right) R^{-1} \\
&= \Sigma_{t|t-1} C^{\top} R^{-1} - \Sigma_{t|t-1} C^{\top} R^{-1} + \Sigma_{t|t-1} C^{\top} \left( C \Sigma_{t|t-1} C^{\top} + R \right)^{-1} \\
&= \Sigma_{t|t-1} C^{\top} \left( C \Sigma_{t|t-1} C^{\top} + R \right)^{-1}
\end{align}
Proof of $(6)$:
\begin{align}
\Sigma_t A \Sigma_{t-1}^* \Sigma_{t-1}^{-1} &= \Sigma_t A \left[ \Sigma_{t|t-1} - \Sigma_{t|t-1} A^{\top} \left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} A \Sigma_{t|t-1} \right] \Sigma_{t|t-1}^{-1} \\
&= \Sigma_t A \left[ I_k - \Sigma_{t|t-1} A^{\top} \left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} A \right] \\
&= \Sigma_t \left[ A - A \Sigma_{t|t-1} A^{\top} \left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} A \right] \\
&= \Sigma_t \left[ A - \left(-I_k + I_k + A \Sigma_{t|t-1} A^{\top} \right) \left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} A \right] \\
&= \Sigma_t \left[ A - \left(-\left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} A + A \right) \right] \\
&= \Sigma_t \left( I_k + A \Sigma_{t|t-1} A^{\top} \right)^{-1} \\
&= \Sigma_t \Sigma_{t|t-1}^{-1} \\
&= \left(\Sigma_{t|t-1} - \Sigma_{t|t-1} C^{\top} \left[ C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C \Sigma_{t|t-1} \right) \Sigma_{t|t-1}^{-1} \\
&= I_k - \Sigma_{t|t-1} C^{\top} \left[ C \Sigma_{t|t-1} C^{\top} + R \right]^{-1} C.
\end{align}
This shows that the representations of the Kalman Filter recursions are equivalent. 
I probably can shorten things somewhere, but this is good enough for me! 
