Inverse of $A^{-1}+B^{-1}$

Here's a concise proof: $$ \begin{align} A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\ &= B^{-1}AA^{-1} + B^{-1}BA^{-1}\\ &= B^{-1}(A + B)A^{-1} \\ \end{align} $$ So $A^{-1} + B^{-1}$ is the product of invertible matrices and thus is invertible, with inverse equal to $$ \begin{align} (A^{-1} + B^{-1})^{-1} &= (B^{-1}(A + B)A^{-1})^{-1} \\ &= A(A + B)^{-1}B \\ \end{align} $$


\begin{equation} (A^{-1} + B^{-1}) A (A+B)^{-1} B = A^{-1}A(A+B)^{-1}B + B^{-1} A (A+B)^{-1} B \end{equation} by the distributive law. Now here comes the trick: Write $A = (A+B) - B$ to get

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


Here's half of the solution. $$A(A+B)^{-1}B=((A+B)A^{-1})^{-1}B=(I+BA^{-1})^{-1}B$$ (Note that inversion reverses the order of products.) What happens if you do the same thing with $B$? We then get $$(B^{-1}(I+BA^{-1}))^{-1}=(B^{-1}+A^{-1})^{-1}$$ We didn't even have to multiply anything out!

Tags:

Matrices

Related

Can I think of Algebra like this? We can define the derivative of a function whose domain is a subset of rational numbers? To show that group G is abelian if $(ab)^3 = a^3 b^3$ and the order of $G$ is not divisible by 3 Why is it that $\mathscr{F} \ne 2^{\Omega}$? Homotopy classes of maps from projective plane to projective plane Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$ Writing an integer as a sum of two square in many ways, with consecutive arguments Infinite number of decimal places? For $X$ contractible, deformation retract of $CX$ onto $X$. Why is the Pontraygin dual a locally compact group? Structure of groups of order $pq$, where $p,q$ are distinct primes. Is an arbitrary number of the form xyzxyz divisible by 7, 11, 13?

Recent Posts