Today, I got stuck at a formula: $$(AA'+BB')(AA'+BB')^+A=A$$ , where $$A$$ and $$B$$ are arbitrary $$m \times n$$ matrices such that $$\mathrm{Col}(A) \cap \mathrm{Col}(B)=\{0\}$$.  After quite some time of brainstorming, I convinced myself.

The keypoint is to prove that $$\mathrm{Col}(A) \subseteq \mathrm{Col}(AA'+BB')$$.

First lets prove that $$A$$ and $$AA'$$ span the same vector space.

By SVD, we have

$$A=UDV'$$, $$U^TU=I_m$$, $$V^TV=I_n$$

And

$$AA'=UD^2U'$$

Note that $$D$$ and $$D^2$$ have nonzero elements at identical positions.  And $$U$$ and $$V'$$ are of full rank, therefore $$Vx , Uy$$ can be arbitrary vectors, and thus $$\mathrm{Col}(DV') = \mathrm{Col}(D^2U')$$.  Therefore, $$\mathrm{Col}(UDV') = \mathrm{Col}(UD^2U')$$, which is exactly $$\mathrm{Col}(A) = \mathrm{Col}(AA')$$.

Next, note $$A$$ and $$B$$ are essentially disjoint, thus by what is just proved, so are $$AA'$$ and $$BB'$$.

Now write $$BB'=PE^2P'$$.  Because both $$U$$ and $$P$$ have full rank, there exists $$P=UV$$.

$$BB' = UVE^2V'U'$$

Comparing

$$AA'=UD^2U'$$ and $$BB' = UVE^2V'U'$$.

It is easily observed that

$$\mathrm{Col}(AA') \cap \mathrm{Col}(BB') = \{0\} \iff \mathrm{Col}(D^2) \cap \mathrm{Col}(VE^2V') = \{0\}$$.  But $$D^2$$

is a diagonal matrix, thus $$VE^2V'$$ has zero columns and rows where $$D$$ has nonzero element.  And hence,

$$\mathrm{Col}(D^2) \subseteq \mathrm{Col}(D^2+VE^2V')$$.

And hence

$$\mathrm{Col}(UD^2U') \subseteq \mathrm{Col}(U(D^2+VE^2V')U')$$

$$\implies \mathrm{Col}(A) = \mathrm{Col}(AA') \subseteq \mathrm{Col}(AA'+BB')$$.

The last step left is to recognized that all columns of $$A$$ are in column space of $$A$$, thus are in $$\mathrm{Col}(AA'+BB')$$.  Thus by a theorem saying that any matrix equation $$Cx=b$$ is consistent if and only if $$CC^+b=b$$.  The proposition is proved.