Help Understanding This Linear Algebra I am working through a paper here and trying to understand all of the linear algebra involved, but I get stuck going from equations 5-6 to equation 7. I will skip the bold matrix notation because everything is a matrix/vector
Equation 5
$$
p(y|X, \bar{X}, \bar{f}) = \mathcal{N}(y|K_{NM}K_{M}^{-1}\bar{f}, \Lambda + \sigma^2I)
$$
Equation 6
$$
p(\bar{f}|\bar{X}) = \mathcal{N}(\bar{f}|0, K_{M})
$$
Equation 7
The paper says "we find the posterior distribution over pseudo targets $\bar{f}$ using Bayes' rule on (5) and (6)."
$$
p(\bar{f}|\mathcal{D}, \bar{X}) = \mathcal{N}(\bar{f}|K_MQ^{-1}_MK_{MN}(\Lambda + \sigma^2I)^{-1}y, K_MQ_M^{-1}K_M)
$$
where
$$
Q_M = K_M + K_{MN}(\Lambda + \sigma^2I)^{-1}K_{MN}
$$
By saying that they used Bayes' rule, they are saying that they are doing the following...
$$
p(\bar{f}|\mathcal{D}, \bar{X}) = \frac{p(y|X, \bar{X}, \bar{f})p(\bar{f}|\bar{X})}{p(y)}
$$
But I do not see where they can get the denominator from as it is not given. I tried multiplying the likelihood by the prior in the numerator using these Gaussian identities, but I go something totally different, not even close to the definition of $p(\bar{f}|\mathcal{D}, \bar{X})$. How can you combine the equations 5 and 6 with Bayes' rule to get equation 7?
EDIT: WIP to get details final derivation correct
Point 1
I am now very close to getting the full thing, but there is one part that is eluding me, which is probably something trivial. For one, the proportions have different x values, so I think there needs to be some subscript to set the variables apart, like so...
$$ \exp(a(x_1-b)^2) \exp(cx_2^2) \propto \exp((a+c)x_2^2 - 2abx_1).$$
which would mean that $1 \rightarrow 2$ could be represented as...
$$
\begin{align} 
 & \propto \exp(-\frac{1}{2} (y - L \mu)^T \Sigma^{-1}(y - L \mu))\exp(-\frac{1}{2}\mu^T K^{-1} \mu) \\
&\propto \exp(-\frac{1}{2}(\underbrace{-2\mu^T L^T \Sigma^{-1}y + \mu^T(\Sigma^{-1}+K^{-1})\mu}_\text{Terms that do not include $\mu$ are factorized away})) \\
\end{align}
$$
if it is the case that...
$$
\begin{aligned}
a &= \Sigma^{-1} \\
b &= \mu^TL^T \\
c &= K^{-1} \\
x_1 &= y \\
x_2 &= \mu \\ 
\end{aligned}
$$
But I do not see how we would end up with $L^T\Sigma^{-1}L + K^{-1}$ inside of the parentheses at the end.
Point 2
I think the final line of the derivation is missing an inverse at the end, is it correct?
$$ 
\propto \exp(-\frac{1}{2}\left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T \Sigma^{-1} y\right)^T (L^T\Sigma^{-1}L+K^{-1})^{\color{red}{-1}} \left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y\right)).
$$
 A: Preamble
I shall use the common Bayesian statistics trick used in figuring out the posteriors - work with PDF kernels and figure out the normalization constants later. I will also make an intensive use of the proportionality arithmetic, $\propto$, which simplifies the derivations.
The main logic of $\propto$ relation is expressed as:
$$ C_1\exp(C_2 + h(x)) \propto C_3\exp(C_4 + h(x)) \propto \exp(h(x))$$
for any constants $C_i$ and $h(\cdot)$ is some function of $x$.
Obviously $f(x) \propto h(x)$ implies $h(x) \propto f(x)$.
The following relations are relatively not obvious but will be useful below:
$$ \exp(a(x-b)^2) \propto \exp(ax^2 - 2ab x) \propto \exp(ax(x-2b)),$$
$$ \exp(a(x-b)^2) \exp(cx^2) \propto \exp((a+c)x^2 - 2ab x).$$
Tedious algebra
So, we have normally distributed data $y|\mu \sim \mathcal{N}(L \mu, \Sigma), \mu\in\mathbb{R}^n$ and normally distributed mean parameter $\mu \sim \mathcal{N}(0, K)$. The posterior density of $\mu$ is proportional to the product of PDF kernels of the two:
\begin{align} 
  \overline{p}(\mu|y) \quad & \propto \exp(-\frac{1}{2} (y - L \mu)^T \Sigma^{-1}(y - L \mu))\exp(-\frac{1}{2}\mu^T K^{-1} \mu) \\
&\propto \exp(-\frac{1}{2}(\underbrace{-2\mu^T L^T \Sigma^{-1}y + \mu^T(L^T \Sigma^{-1}L+K^{-1})\mu}_\text{Terms that do not include $\mu$ are factorized away})) \\
&\propto \exp(-\frac{1}{2} \underbrace{\mu^T(L^T\Sigma^{-1}L+K^{-1})}_\text{Factorized inside $\exp$} \left(-2 (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y + \mu\right))\\
&\propto \exp(-\frac{1}{2}\left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T \Sigma^{-1} y\right)^T (L^T\Sigma^{-1}L+K^{-1}) \left(\mu - (L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y\right)).
\end{align}
Careful examination of the last line reveals the PDF kernel of a normal rv
$$\mu|y \sim \mathcal{N}((L^T\Sigma^{-1}L+K^{-1})^{-1} L^T\Sigma^{-1} y, (L^T\Sigma^{-1}L+K^{-1})^{-1}). $$
Using the Woodbury identity one can rewrite the posterior covariance with an eye on simplifying it in the context of the paper:
$$ 
(L^T\Sigma^{-1}L+K^{-1})^{-1} = K - K L^T(\Sigma + L K L^T)^{-1} L K
$$
In the context of the paper,
\begin{align}
\mu &= \overline{f}, \\  \Sigma &= \Lambda + \sigma^2 I, \\  K &= K_M,\\
L &= K_{NM}K_M^{-1}, \\ KL^T &= K_{MN}, \\  L K L^T &= K_{NM} K_M^{-1} K_{MN},
\end{align}
and the posterior covariance matrix equals
\begin{align} 
&K_M - K_{MN} (\Lambda + \sigma^2I + K_{NM} K_M^{-1} K_{MN} )^{-1} K_{NM} \\
=& K_M (K_M^{-1} - K_M^{-1} K_{MN} (\Lambda + \sigma^2I + K_{NM} K_M^{-1} K_{MN} )^{-1} K_{NM} K_M^{-1} ) K_M\\
=& K_M (K_M + K_{MN} (\Lambda + \sigma^2 I)^{-1}K_{NM})^{-1} K_M \\
= & K_M Q_M^{-1}K_M. 
\end{align}
(Woodbury identity is used again to get the third line.)
The mean of the posterior is obtained in a similar fashion.
