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

I have this equality,

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

and the question specifically only states that $A+B$ is nonsingular.

I have looked at this many ways but the only I can see it working is if $A+B$ being nonsingular implies that $A$ and $B$ are nonsingular but I can't see a way of proving this. I have also tried using row equivalence and manipulation of $I=(A+B)(A+B)^{-1}$ but I can't seem to find a way. I need help at this step as I don't understand

• That $A+B$ is non singular does not imply that $A$ and $B$ separately are non. singular. Consider trivially the sum of the Identity matrix with a matrix that has just one "1" in the diagonal and zero everywhere else. Nov 20, 2017 at 11:55
• Closely related: stats.stackexchange.com/questions/197067.
– whuber
Nov 20, 2017 at 18:43
• Notice that $A=1A=(A+B)(A+B)^{-1}$ and $A=A1=A(A+B)^{-1}(A+B)$. Expand each of these into a sum to compute and simplify $0=A-A$.
– whuber
Nov 20, 2017 at 19:52
• Whan posting homework-like questions please tag it as [self-study] and follow our policies.
– Tim
Nov 27, 2017 at 10:28

setting $X=(A+B)^{-1}$,

Consider

$(A-B)X(A+B) = AXA + AXB -BXA - BXB$

$(A+B)X(A-B) = AXA - AXB + BXA - BXB$

Then,

$(A-B)X(A+B) - (A+B)X(A-B) = 2AXB - 2BXA$

so

$(A-B)(A+B)^{-1}(A+B) - (A+B)(A+B)^{-1}(A-B) = 2[A(A+B)^{-1}B - B(A+B)^{-1}A]$

$(A-B)I - I(A-B) = 2(A(A+B)^{-1}B - B(A+B)^{-1}A)$

$0= 2(A(A+B)^{-1}B - B(A+B)^{-1}A)$

so

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

• Oops - have now edited to fix Nov 20, 2017 at 12:25
• A quicker but equivalent way is to start with the original statement and add $B(A+B)^-1B$ to both sides. Nov 20, 2017 at 16:10