Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $A^{-1} + B^{-1} =(A+B)^{-1}$

Question

Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $$A^{-1} + B^{-1} =(A+B)^{-1}$$ then prove that $\operatorname{det}(A)=\operatorname{det}(B)$. Also show that this result is not valid for complex matrices.

I'm not getting any way out to solve this. Can anybody solve the problem?
Thanks for assistance in advance.

Answer 1

By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$\det(U)=1$$or equivalently$$\det(A)=\det(B)$$

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Counter example on $\Bbb C$

Let $A=I_2$ and $b=kI_2$ with $k={-1+i\sqrt 3\over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1\over k}I\\(A+B)^{-1}={1\over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$\det(B)=k^2=-1-k={-1-i\sqrt 3\over 2}\ne 1=\det(A)$$

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Comment

Even on $\Bbb C$, from $U^3=I$ we can conclude that$$\left|\det(A)\right|=\left|\det(B)\right|$$

Answer 2

For $A,B \in \mathcal{M}_n^n(\mathbb{R})$ We've $$(A+B)(A+B)^{-1}=(A+B)(A^{-1}+B^{-1})=I+AB^{-1}+BA^{-1}+I=I\\ \implies AB^{-1}+BA^{-1}+I=O$$ Then post multiplying the equation , once by $B$ and then by $A$ we end up with $$AB^{-1}A=BA^{-1}B=-(A+B)$$ Considering the determinants we get $$\frac{[det(A)]^2}{det(B)}=\frac{[det(B)]^2}{det(A)} \implies [det(A)]^{3}-[det(B)]^{3}=0 \implies det(A)=det(B)$$

If we're working with $\mathcal{M}^n_n(\mathbb{C})$ , the result doesn't hold.

Let's define $\zeta_{n}^{j}$ as the $n$th roots of unity for some odd positive integer $n$ with $1 \le j \le n$ and $j \in \mathbb{Z}^{+}$, then let's define $A=\zeta_n I_{n}$ and $B=\sum_{j=2}^{n}\zeta_{n}^{j}I_n$

Clearly $$A^{-1} +B^{-1}=I\left[\zeta_{n}^{-1} +(\sum_{j=2}^{n}\zeta_{n}^{j})^{-1} \right]=O$$ and $$(A+B)^{-1}=O$$ But we note that $$det(A)=(\zeta_{n})^n =1 \\ det(B)=(\sum_{j=2}^{n}\zeta_{n}^{j})^n=(-\zeta_{n})^n =-1$$

NOTE:Properties of $\zeta_n^{j}$ are used in the arguments without any mention