Rank of sum of matrices considering the Moore-Penrose pseudoinverse

Question

Consider a collection of matrices $A_i\in\mathbb{R}^{n\times n}$ which satisfy $rank(\sum_{i=1}^{n}A_i)=m\leq n$. Then the question is whether $rank(\sum_{i=1}^{n}A_i^{\dagger}A_i)\geq m$ holds? Specially, the equality holds if $m=n$. Here, $A_i^{\dagger}$ is the Moore-Penrose pseudoinverse of $A_i$, and $A_i$'s have no relation.

Answer 1

Note that if $A_i$'s are independent, then for $i\ne j$, $\mathcal{R}(A_i)\cap\mathcal{R}(A_j)=\{\mathbf{0}\}$, which implies $$\mathcal{R}(\sum_{i=1}^k A_i)=\sum_{i=1}^k\mathcal{R}(A_i).$$

Also note that for any generalized inverse $G$ of $A$,we have $$\mathrm{rank}(GA)\le\mathrm{rank}(A)=\mathrm{rank}(AGA)\le\mathrm{rank}(GA) \implies\mathrm{rank}(GA)=\mathrm{rank}(A)$$ Now obviously we have $\mathcal{R}(GA)\subseteq\mathcal{R}(A)$, but $\mathrm{dim}(\mathcal{R}(GA))=\mathrm{rank}(GA)=\mathrm{rank}(A)=\mathrm{dim}(\mathcal{R}(A))$, hence $$\mathcal{R}(GA)=\mathcal{R}(A)$$

This implies for $i\ne j$, $\mathcal{R}(A_i^{\dagger}A_i)\cap\mathcal{R}(A_j^{\dagger}A_j)=\mathcal{R}(A_i)\cap\mathcal{R}(A_j)=\{\mathbf{0}\}$. Hence we have $\mathcal{R}(\sum_{i=1}^kA_i^{\dagger}A_i)=\sum_{i=1}^k\mathcal{R}(A_i^{\dagger}A_i)=\sum_{i=1}^k\mathcal{R}(A_i)=\mathcal{R}(\sum_{i=1}^k A_i)$. Thus $$\mathrm{rank}(\sum_{i=1}^kA_i^{\dagger}A_i)=\dim(\mathcal{R}(\sum_{i=1}^kA_i^{\dagger}A_i))=\dim(\mathcal{R}(\sum_{i=1}^k A_i))=\mathrm{rank}(\sum_{i=1}^k A_i)$$