Invertible Matrices

• Oct 20th 2009, 05:25 PM
ginarific
Invertible Matrices
Hello!

This is one problem I've been tackling for a while:

If A, B, and A+B are all invertible matrices of order "n", prove that $\displaystyle A^{-1} + B^{-1}$ is invertible and that:

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

First, is there a generic formula for $\displaystyle A^{-1} + B^{-1}$? (If so, our instructor never taught us one). I'm also assuming that we can't assume AB is commutable, because that could make everything so much easier. ^_^;

Anyway. Any helpful hints might help. Basically what I've tried to do so far is equate the "middle" and RHS statements by multiplying both sides by inverses to isolate $\displaystyle (A+B)^{-1}$, but I don't really see how that 'proves' anything ...
• Oct 20th 2009, 10:39 PM
tonio
Quote:

Originally Posted by ginarific
Hello!

This is one problem I've been tackling for a while:

If A, B, and A+B are all invertible matrices of order "n", prove that $\displaystyle A^{-1} + B^{-1}$ is invertible and that:

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

First, is there a generic formula for $\displaystyle A^{-1} + B^{-1}$? (If so, our instructor never taught us one). I'm also assuming that we can't assume AB is commutable, because that could make everything so much easier. ^_^;

Anyway. Any helpful hints might help. Basically what I've tried to do so far is equate the "middle" and RHS statements by multiplying both sides by inverses to isolate $\displaystyle (A+B)^{-1}$, but I don't really see how that 'proves' anything ...

How do you show a matrix is invertible? Find a matrix that multiplied by the first one gives you the identity one!
Well, let us show that $\displaystyle A^{-1}+B^{-1}$ is the inverse of the matrix $\displaystyle A(A+B)^{-1}B$:

$\displaystyle A(A+B)^{-1}B(A^{-1}+B^{-1})=A\left[(A+B)^{-1}B+(A+B)^{-1}A\right]A^{-1}=$$\displaystyle A(A+B)^{-1}(B+A)A=I$ and we're done!

Tonio