Finding matrices with $A_1^{-1}+A_2^{-1}+\dots+A_k^{-1}=(A_1+A_2+\dots+A_k)^{-1}$

Question

Prove that for any $n, k\geq2$ there exist nonsingular nondiagonal matrices $A_1, A_2, \dots, A_k \in M_n(\mathbb{R})$ such that $$A_1^{-1} + A_2^{-1} + \dots + A_k^{-1} = \left( A_1 + A_2 + \dots + A_k \right)^{-1}$$

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For $n=2$, if we have $A^{-1} + B^{-1} = (A+B)^{-1}$, if I am not mistaken, we can prove that $$\det(A)=\det(B)=\det(A+B)$$ if $A,B\in M_n(\mathbb{R})$, so it would be natural to consider that $\det(A_1)=\det(A_2)=\dots=\det(A_k)$, but from here I don't have any idea what should I do. What should we do?

Answer 1

If $k$ is odd, then it is easy: take $A_{2m} = -A_{2m+1}$ for every $m\ge 1$ so that the whole expression reduces to $A_1^{-1} = A_1^{-1}$ that is true for every $A_1$ invertible.

If $k$ is even, then using the same trick as before, what's left is $$ A_1^{-1} + A_2^{-1} = (A_1+A_2)^{-1} $$ So if you solve it for $k=2$, you solve it for every $k$.

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Let's prove that $n=3$ and $k=2$ produces an absurd. It turns out that $$ 2I + A_1A_2^{-1} + A_2A_1^{-1} = I $$ and if $X = A_1A_2^{-1}$ then $X^{-1} = A_2A_1^{-1}$ and $$ X^2 + X + I = 0. $$ This means that the minimal polynomial of $X$ is $x^2+x+1$, so the characteristic polynomial must be $(x^2+x+1)^m$ for some $m$, but then $n=2m=3$, that is impossible.

Answer 2

HINT

If the defining identity is left-right multiplied by $P^{-1}$ and $P$ then $$ \eqalign{ & P^{\, - 1} \left( {{I \over {A_{\,1} }} + {I \over {A_{\,2} }} + \cdots + {I \over {A_{\,k} }}} \right)P = \left( {{I \over {PA_{\,1} P^{\, - 1} }} + \cdots + {I \over {PA_{\,k} P^{\, - 1} }}} \right) = \cr & P^{\, - 1} {I \over {A_{\,1} + A_{\,2} + \cdots + A_{\,k} }}P = {I \over {PA_{\,1} P^{\, - 1} + \cdots + PA_{\,k} P^{\, - 1} }} \cr} $$ thus $\{PA_kP^{-1}\}$ also satisfies the identity.