[proofplan] We compute the posterior density of the latent factor $F$ after observing $X=x$. Since $F$ is Gaussian and the conditional model $X\mid F=f$ is Gaussian with mean $\Lambda f$ and covariance $\Psi$, the conditional density is proportional to the product of two Gaussian densities. Completing the square identifies the conditional law as a Gaussian with precision matrix $I_k+\Lambda^\top\Psi^{-1}\Lambda$. The Woodbury identities then rewrite the resulting mean and covariance in the stated $\Sigma^{-1}$ form. [/proofplan] [step:Compute the covariance of the observed vector] Since $\Psi$ is positive diagonal, it is positive definite. For every nonzero vector $u \in \mathbb{R}^p$, \begin{align*} u^\top \Sigma u &= u^\top \Lambda\Lambda^\top u + u^\top \Psi u\\ &= |\Lambda^\top u|^2 + u^\top \Psi u\\ &>0. \end{align*} Thus $\Sigma$ is positive definite. Because $F$ and $\varepsilon$ are independent, centred, and have covariance matrices $I_k$ and $\Psi$, respectively, \begin{align*} \operatorname{Cov}(X) &= \operatorname{Cov}(\Lambda F+\varepsilon)\\ &= \Lambda \operatorname{Cov}(F)\Lambda^\top+\operatorname{Cov}(\varepsilon)\\ &= \Lambda\Lambda^\top+\Psi\\ &= \Sigma. \end{align*} Also, \begin{align*} \operatorname{Cov}(F,X) &= \mathbb{E}\bigl[F(\Lambda F+\varepsilon)^\top\bigr]\\ &= \mathbb{E}[FF^\top]\Lambda^\top+\mathbb{E}[F\varepsilon^\top]\\ &= \Lambda^\top, \end{align*} because independence and centering give $\mathbb{E}[F\varepsilon^\top]=0$. [/step] [step:Complete the square in the conditional density] Fix $x \in \mathbb{R}^p$. For $f \in \mathbb{R}^k$, the conditional density of $X$ given $F=f$ is proportional to \begin{align*} \exp\left(-\frac{1}{2}(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f)\right), \end{align*} and the density of $F$ is proportional to \begin{align*} \exp\left(-\frac{1}{2}f^\top f\right). \end{align*} Therefore the conditional density of $F$ given $X=x$ is proportional, as a function of $f$, to \begin{align*} \exp\left( -\frac{1}{2} \left[ f^\top f+(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f) \right] \right). \end{align*} Define \begin{align*} A &:= I_k+\Lambda^\top\Psi^{-1}\Lambda,\\ b(x) &:= \Lambda^\top\Psi^{-1}x. \end{align*} The matrix $A$ is positive definite because for every nonzero $v \in \mathbb{R}^k$, \begin{align*} v^\top A v = |v|^2+(\Lambda v)^\top\Psi^{-1}(\Lambda v)>0. \end{align*} Expanding the exponent gives \begin{align*} f^\top f+(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f) &= f^\top A f-2b(x)^\top f+x^\top\Psi^{-1}x\\ &= (f-A^{-1}b(x))^\top A(f-A^{-1}b(x))\\ &\quad +x^\top\Psi^{-1}x-b(x)^\top A^{-1}b(x). \end{align*} The final two terms are independent of $f$. Hence \begin{align*} F\mid X=x \sim \mathcal{N}\left(A^{-1}b(x),A^{-1}\right). \end{align*} [guided] Fix $x \in \mathbb{R}^p$. We want the conditional law of the latent vector $F$ after observing $X=x$. The model says that, conditional on $F=f$, the observation is \begin{align*} X=\Lambda f+\varepsilon, \end{align*} where $\varepsilon \sim \mathcal{N}(0,\Psi)$. Thus the conditional density of $X$ given $F=f$ is proportional to \begin{align*} \exp\left(-\frac{1}{2}(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f)\right). \end{align*} The prior density of $F$ is proportional to \begin{align*} \exp\left(-\frac{1}{2}f^\top f\right). \end{align*} Multiplying the likelihood and prior gives a conditional density for $F$ given $X=x$ proportional to \begin{align*} \exp\left( -\frac{1}{2} \left[ f^\top f+(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f) \right] \right). \end{align*} Define \begin{align*} A &:= I_k+\Lambda^\top\Psi^{-1}\Lambda,\\ b(x) &:= \Lambda^\top\Psi^{-1}x. \end{align*} The role of $A$ is that it is the posterior precision matrix. It is positive definite: if $v \in \mathbb{R}^k$ and $v \ne 0$, then \begin{align*} v^\top A v = |v|^2+(\Lambda v)^\top\Psi^{-1}(\Lambda v)>0, \end{align*} because $\Psi^{-1}$ is positive definite. Now expand the quadratic expression in $f$: \begin{align*} f^\top f+(x-\Lambda f)^\top\Psi^{-1}(x-\Lambda f) &= f^\top f+x^\top\Psi^{-1}x-2x^\top\Psi^{-1}\Lambda f+f^\top\Lambda^\top\Psi^{-1}\Lambda f\\ &= f^\top A f-2b(x)^\top f+x^\top\Psi^{-1}x. \end{align*} Completing the square with the positive definite matrix $A$ gives \begin{align*} f^\top A f-2b(x)^\top f+x^\top\Psi^{-1}x &= (f-A^{-1}b(x))^\top A(f-A^{-1}b(x))\\ &\quad +x^\top\Psi^{-1}x-b(x)^\top A^{-1}b(x). \end{align*} The last two terms do not depend on $f$, so they only affect the normalising constant of the conditional density. Therefore the conditional density is Gaussian with covariance $A^{-1}$ and mean $A^{-1}b(x)$: \begin{align*} F\mid X=x \sim \mathcal{N}\left(A^{-1}b(x),A^{-1}\right). \end{align*} [/guided] [/step] [step:Rewrite the conditional covariance in terms of $\Sigma^{-1}$] We prove \begin{align*} A^{-1}=I_k-\Lambda^\top\Sigma^{-1}\Lambda. \end{align*} Since $\Sigma=\Psi+\Lambda\Lambda^\top$, direct multiplication gives \begin{align*} \Sigma\Psi^{-1}\Lambda &=(\Psi+\Lambda\Lambda^\top)\Psi^{-1}\Lambda\\ &=\Lambda+\Lambda\Lambda^\top\Psi^{-1}\Lambda\\ &=\Lambda(I_k+\Lambda^\top\Psi^{-1}\Lambda)\\ &=\Lambda A. \end{align*} Multiplying on the left by $\Sigma^{-1}$ and on the right by $A^{-1}$ yields \begin{align*} \Psi^{-1}\Lambda A^{-1}=\Sigma^{-1}\Lambda. \end{align*} Now multiply $I_k-\Lambda^\top\Sigma^{-1}\Lambda$ by $A$: \begin{align*} (I_k-\Lambda^\top\Sigma^{-1}\Lambda)A &= A-\Lambda^\top\Sigma^{-1}\Lambda A\\ &= A-\Lambda^\top\Psi^{-1}\Lambda\\ &= I_k. \end{align*} Thus $I_k-\Lambda^\top\Sigma^{-1}\Lambda$ is a right inverse for $A$. Since $A$ is invertible, it equals $A^{-1}$. [/step] [step:Rewrite the conditional mean in terms of $\Sigma^{-1}$] Using $\Psi^{-1}\Lambda A^{-1}=\Sigma^{-1}\Lambda$ and transposing both sides, we obtain \begin{align*} A^{-1}\Lambda^\top\Psi^{-1}=\Lambda^\top\Sigma^{-1}, \end{align*} because $A$, $\Psi$, and $\Sigma$ are symmetric. Therefore \begin{align*} A^{-1}b(x) &=A^{-1}\Lambda^\top\Psi^{-1}x\\ &=\Lambda^\top\Sigma^{-1}x. \end{align*} Combining this identity with the covariance identity from the previous step and the conditional Gaussian law found above gives \begin{align*} \mathbb{E}[F\mid X=x] &= \Lambda^\top\Sigma^{-1}x,\\ \operatorname{Cov}(F\mid X=x) &= I_k-\Lambda^\top\Sigma^{-1}\Lambda. \end{align*} These are exactly the stated conditional moments. [/step]