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Can you guys help me out with the following problem:

Problem: Find conditions that the matrices $A$ and $B$ have to satisfy in order for the following to be valid: $(A+B)^{-1} = A^{-1} + B^{-1}$.

My solution. The first condition is that $A$ and $B$ have to have the same dimensions. Then by expanding the two equalities: $(A+B) \cdot (A^{-1} + B^{-1})=I $ and $(A^{-1} + B^{-1}) \cdot (A+B)=I$, I deduced the second condition: $A$ and $B$ have both to be invertible, and third condition is: $A^{-1} \cdot B=B \cdot A^{-1}$. Is my answer correct?

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    @MichaelHardy No$I$understand that. I just showed that Zi2018 Alphas' answer is false by counter-example, since (s)he asks if his answer is correct.2011-11-01

2 Answers 2

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To elaborate slightly on @MichaelHardy's comment, from:

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

consider as well that:

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

So when we are talking about all possible solutions, these can be generated by finding $M$ such that:

$M + M^{-1} = -I$

i.e. an invertible matrix $M$ such that:

$M^2 + M + I = 0$

Then if $M$ has the corresponding characteristic root(s), any invertible matrix $B$ can be paired with an invertible matrix $A$ satisfying the desired relation via:

$A = MB$

so that $M = A\;B^{-1}$ and $M^{-1} = B\;A^{-1}$.

Added: A few more words about $M$ and its characteristic roots are useful.

Because it satisfies the polynomial above without repeated roots, complex matrix $M$ must be diagonalizable (similar to a diagonal matrix) with eigenvalues in $\{ (-1 \pm i\sqrt{3})/2 \}$, the nontrivial cube roots of unity. So to construct all such $M$, choose how many of each root so that combined we have $n$ the dimension of $M$, and group them along the diagonal of $D$ having those with positive imaginary part before those with negative imaginary part (for the sake of definiteness). Then take for arbitrary invertible complex matrix $P$ the similarity transformation:

$M = P\; D \; P^{-1}$

If we wanted to restrict $M$ to real matrices, then its eigenvalues must occur in conjugate pairs. Thus the dimension $n$ must be even and the "real Jordan canonical form" must have diagonal blocks $ \bigl(\begin{smallmatrix} \frac{-1}{2}&\frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{2}&\frac{-1}{2} \end{smallmatrix} \bigr)$. Replace $D$ with such a block diagonal matrix and $P$ with an arbitrary invertible real matrix in our above similarity recipe, and you get the construction of all real solutions $M$.

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For third condition you have that:

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

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

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

$\Rightarrow AA^{-1}+AB^{-1}+BA^{-1}+BB^{-1}=A^{-1}A+A^{-1}B+B^{-1}A+B^{-1}B \Rightarrow$

$\Rightarrow AB^{-1}+BA^{-1} =A^{-1}B+B^{-1}A$

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    And, to test if these 3 conditions are enough, you assume all of them are true and see if you can prove the original statement2011-10-29