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.