Solving (A+B)^-1: Is A^-1+ B^-1 the Answer?

[roscany]

Does (A+B)^-1= A^-1+ B^-1?

Thanks!

 

[I like Serena]

Does (A+B)^-1= A^-1+ B^-1?

Thanks!

Hi roscany,

Suppose we pick A = B = I, what do you get then?

 

[Dethrone]

Assume $A$, $B$, and $A+B$ are invertible, otherwise the inverse would not exist. Let $(A+B)^{-1}=A^{-1}+X$, where $X$ is to be determined.

$(A+B)^{-1}=A^{-1}+X$
Multiply both sides by $A+B$:
$I=(A^{-1}+X)(A+B)$
$I=A^{-1}A+A^{-1}B+XA+XB$
$I=A^{-1}B+X(A+B)$
$X(A+B)=-A^{-1}B$
$X=(-A^{-1}B)(A+B)^{-1}$
Recall from above:
$X=(-A^{-1}B)(A^{-1}+X)$
Isolating for $X$...
$X= - (I + A^{-1}B)^{-1} A^{-1} B A^{-1}$

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

Therefore, the original statement is not true in general.

 

[I like Serena]

Assume $A$, $B$, and $A+B$ are invertible, otherwise the inverse would not exist. Let $(A+B)^{-1}=A^{-1}+X$, where $X$ is to be determined.

$(A+B)^{-1}=A^{-1}+X$
Multiply both sides by $A+B$:
$I=(A^{-1}+X)(A+B)$
$I=A^{-1}A+A^{-1}B+XA+XB$
$I=A^{-1}B+X(A+B)$
$X(A+B)=-A^{-1}B$
$X=(-A^{-1}B)(A+B)^{-1}$
Recall from above:
$X=(-A^{-1}B)(A^{-1}+X)$
Isolating for $X$...
$X= - (I + A^{-1}B)^{-1} A^{-1} B A^{-1}$

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

Therefore, the original statement is not true in general.

Wait. (Wait)

Are you saying that $B^{-1}$ does not have to be equal to $-(I + A^{-1}B)^{-1} A^{-1} B A^{-1}$?

Why would that be the case? (Wondering)

 

[Dethrone]

Wait. (Wait)

Are you saying that $B^{-1}$ does not have to be equal to $-(I + A^{-1}B)^{-1} A^{-1} B A^{-1}$?

Why would that be the case? (Wondering)

I am saying that in general,
$(A+B)^{-1}=A^{-1}-(I + A^{-1}B)^{-1} A^{-1} B A^{-1}$
Now, the OP's statement that $(A+B)^{-1}=A^{-1}+B^{-1}$ is true if and only if $B^{-1}=-(I + A^{-1}B)^{-1} A^{-1} B A^{-1}$. Since you were able to point out a counter-example, clearly the RHS cannot be simplified to $B^{-1}$.