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Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$


Tried to write the matrices as $$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g&h}$$ and solve $(A+B)^{-1}=A^{-1}+B^{-1}$, But make it too complex. Any more convenient ways?

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  • $\begingroup$ Use the formula for $(A+B)^{-1}$ given here. $\endgroup$ – Dietrich Burde Aug 24 '18 at 12:54
  • $\begingroup$ @ Dietrich Burde . Actually it's $1/(\lambda+\mu)=1/\lambda+1/\mu $ and doesn't have a solution. $\endgroup$ – Jaqen Chou Aug 24 '18 at 12:59
  • $\begingroup$ Yes, I know. Your way is very good, and not too complex when using the formula. $\endgroup$ – Dietrich Burde Aug 24 '18 at 13:14
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You get $I = (A^{-1}+B^{-1})(A+B) = I + A^{-1}B + B^{-1}A + I$. Thus we get:

$$(A^{-1}B) + (A^{-1}B)^{-1} = -I$$ $$(A^{-1}B)^2 + I = -(A^{-1}B)$$

So $A^{-1}B$ satisfies the polynomial $x^2 + x + 1 = 0$. Take any matrix satisfying this polynomial; for example you can take

$$A^{-1}B = \begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$

$$B = A\begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$

Hence you can take any invertible $A$ and produce $B$ of the wanted form.

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From $(A+B)^{-1}=A^{-1}+B^{-1}$ we can get $$A^{-1}B+B^{-1}A=-I \quad (*)$$ now let's chose $A=I$ so (*) become $$B+B^{-1}=-I$$ If we choice $B$ a diagonale matrix, then

$$B=\pmatrix {\alpha & 0\\0&\beta};B^{-1}=\pmatrix {\frac{1}{\alpha} & 0\\0&\ \frac{1}{\beta}} $$ Then every thing is done by solving $$\alpha + \frac{1}{\alpha}=-1$$

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Using the formula from

Inverse of the sum of matrices

we can take, for arbitrary $a,b$ with $a+b\neq 0$, $$ A=\begin{pmatrix} -1 & 1 \cr 0 & 2 \end{pmatrix},\quad B=\begin{pmatrix} a & b \cr \frac{2(a^2-a+1)}{a+b} & \frac{-2(b-ab+1)}{a+b} \end{pmatrix}. $$ All of $A$, $B$ and $A+B$ have determinant $-2$. It is easy to check that $(A+B)^{-1}=A^{-1}+B^{-1}$. So your way is not too "complex".

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