show that $(A'A + B'B)^{-1}$ is a g inverse of A'A

Question

Notation: ' is for transpose, C() is for column space

Here's what I've been given/figured out so far

• A is an n by m matrix

• rank(A) = rank(A'A) = r

• B is an s by m matrix

• rank(B) = rank(B'B) = m-r

• C(A') $\cap$ C(B') = {0} (which also implies C(A'A) $\cap$ C(B'B) = {0} right?)

• A'A + B'B is of full rank = m

Using this, I have to show that A'A$(A'A+B'B)^{-1}$A'A = A'A

My guess is to start with something that is obviously true like...

$$A'A(A'A+B'B)^{-1}(A'A+B'B)=A'A$$

and then expand that out so it reads

$$A'A(A'A+B'B)^{-1}A'A + A'A(A'A+B'B)^{-1}B'B = A'A$$

In which case all that's left to do is show that $A'A(A'A+B'B)^{-1}B'B = 0$ which I figure is possible by showing that the rank of said matrix product is equal to 0, but I've yet to figure out just how to pull that off, and thus I'm holding out for a hero here on the exchange...

Answer 1

As you have already got, $C(A'A)\subseteq C(A')$ and $C(B'B)\subseteq C(B')$. So $C(A')\cap C(B')=\{0\}$ implies $C(A'A)\cap C(B'B)=\{0\}$

In the same way, you can prove that $$C(A'A(A'A+B'B)^{-1}BB')\subseteq C(A'A)$$ and $$C(A'A(A'A+B'B)^{-1}BB')\subseteq C(B'B)$$ (Using the symmetric of $A'A$ and $B'B$).

Thus $$C(A'A(A'A+B'B)^{-1}B'B)\subseteq C(A'A)\cap C(B'B) = \{0\}$$ ie. $A'A(A'A+B'B)^{-1}B'B = 0$