Moore-Penrose pseudoinverse: product on left with another matrix

The following problem comes from studying the conditional expectation of a multivariate normal distribution. Let $$n\ge2$$ be an integer, and let $$\Sigma$$ be a positive semidefinite, symmetric $$n\times n$$ matrix of real numbers partitioned as $$\Sigma=\begin{pmatrix}\Sigma_{a,a}&\Sigma_{a,b}\\\Sigma_{b,a}&\Sigma_{b,b}\end{pmatrix},$$ where $$\Sigma_{a,a}$$ is $$1\times1,$$ $$\Sigma_{a,b}$$ is $$1\times(n-1)$$ and $$\Sigma_{b,b}$$ is $$(n-1)\times(n-1).$$ Is it true that $$\Sigma_{a,b}=\Sigma_{a,b}\Sigma_{b,b}^+\Sigma_{b,b}?$$ (Here, $$\Sigma_{b,b}^+$$ is the Moore-Penrose pseudoinverse.)

In the CrossValidated post "Conceptual proof that conditional of a multivariate Gaussian is multivariate Gaussian", someone claims this result. The result is clearly true if $$\Sigma_{b,b}$$ is invertible, in which case $$\Sigma_{b,b}^+=\Sigma_{b,b}^{-1}.$$ In addition, I have tried two examples, $$\Sigma=0$$ and $$\Sigma=\begin{pmatrix}\!\!\begin{array}{c|cc}1&1&0\\\hline1&1&0\\0&0&0\end{array}\!\!\end{pmatrix},$$ and in both cases we have $$\Sigma_{a,b}-\Sigma_{a,b}\Sigma_{b,b}^+\Sigma_{b,b}=0,$$ as desired. However, I could not prove the result in generality using the definition or the properties listed in the Wikipedia page.

In the context of this problem, we partition a multivariate Gaussian vector into two subvectors $$X_a$$ and $$X_b$$. The goal is to find a matrix $$C$$ such that $$Z:=X_a- C X_b$$ is uncorrelated with $$X_b$$, so that the equality $$\Sigma_{a,b}=C\,\Sigma_{b,b}\tag1$$ holds. The general solution is to take $$C:=\Sigma_{a,b}\Sigma_{b,b}{}^+,$$ where $$\Sigma_{b,b}{}^+$$ is the Moore-Penrose inverse of $$\Sigma_{b,b}$$.
The reason why this works, even when $$\Sigma_{b,b}$$ is not of full rank, depends crucially on the fact that we are dealing with multivariate Gaussians, in which case $$\Sigma_{b,b}$$ and $$\Sigma_{a,b}$$ have a special form. Recall that every multivariate Gaussian vector is an affine transformation of some vector $$Z$$ of independent standard Gaussians. We can then write the subvectors $$X_a$$ and $$X_b$$ in the form $$X_a = AZ + \mu_a$$, $$X_b = BZ + \mu_b$$, with $$A$$ and $$B$$ matrices of constants. Since the covariance matrix of $$Z$$ is the identity, we calculate $$\Sigma_{a,b} = AB^T$$ and $$\Sigma_{b,b}=BB^T$$.
Using properties of Moore-Penrose inverses, we find $$\Sigma_{b,b}{}^+=(BB^T)^+=(B^T)^+B^+$$ and verify (1): $$C\,\Sigma_{b,b}=\Sigma_{a,b}\Sigma_{b,b}{}^+\Sigma_{b,b} =AB^T(B^T)^+\underbrace{B^+BB^T}_{B^T} =A\underbrace{B^T(B^T)^+B^T}_{B^T}=AB^T=\Sigma_{a,b}.$$
The definition of the Moore-Penrose pseudoinverse of $$\Sigma_{b,b}$$ is $$\Sigma_{b,b}^+ = (\Sigma_{b,b}^\top \Sigma_{b,b})^{-1}\Sigma_{b,b}^\top$$. This gives $$\begin{equation*} \Sigma_{a,b}\Sigma_{b,b}^+\Sigma_{b,b} = \Sigma_{a,b}(\Sigma_{b,b}^\top\Sigma_{b,b})^{-1}\Sigma_{b,b}^\top \Sigma_{b,b} = \Sigma_{a,b}, \end{equation*}$$ as desired. Note that this result is really more general than the context of covariance matrices: given $$A\in\mathbb{R}^{m\times n}$$ and $$B\in\mathbb{R}^{p\times n}$$ with $$\text{rank}(B)=n$$, then $$AB^+B = A(B^\top B)^{-1}B^\top B = A$$.
• What if $\Sigma_{b,b}$ is non-invertible? The equation $\Sigma_{b,b}^+=(\Sigma_{b,b}^T\Sigma_{b,b})^{-1}\Sigma_{b,b}^T$ only works if $\Sigma_{b,b}$ has full rank (i.e., is invertible). Mar 26 '20 at 19:20