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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

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    $\begingroup$ 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. $\endgroup$
    – Alecos Papadopoulos
    Nov 20, 2017 at 11:55
  • $\begingroup$ Closely related: stats.stackexchange.com/questions/197067. $\endgroup$
    – whuber ♦
    Nov 20, 2017 at 18:43
  • $\begingroup$ 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 $0=A-A$. $\endgroup$
    – whuber ♦
    Nov 20, 2017 at 19:52
  • $\begingroup$ Whan posting homework-like questions please tag it as [self-study] and follow our policies. $\endgroup$
    – Tim ♦
    Nov 27, 2017 at 10:28

1 Answer 1

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$\begingroup$

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

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

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  • $\begingroup$ Oops - have now edited to fix $\endgroup$
    – user185461
    Nov 20, 2017 at 12:25
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    $\begingroup$ A quicker but equivalent way is to start with the original statement and add $B(A+B)^-1B$ to both sides. $\endgroup$
    – Flounderer
    Nov 20, 2017 at 16:10
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    $\begingroup$ Please read carefully this, our policy about answering the homework-like questions is that we don't offer solutions, but hints. $\endgroup$
    – Tim ♦
    Nov 27, 2017 at 10:26

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