TITLE: Finding all matrices that satisfy a 'wrong property'
QUESTION [6 upvotes]: I was reminded of a very usual highschool algebra question of a 'wrong property' where students usually just 'distribute squares' like: $(x+y)^2 = x^2+y^2$. Clearly this is wrong and that if we solve this equation anyway, then we arrive at $2xy = 0.$ This means that the solution set for which this 'wrong property' holds is when you have any real number $x$ and $y=0$, or both are zero.
Coming to linear algebra, we have a quite similar common misconception that $(\mathbf{A} + \mathbf{B})^{-1} = \mathbf{A}^{-1} + \mathbf{B}^{-1}$. My question is, can we find all $2\times 2$ matrices that satisfy this equation?
I tried solving this, but it is proving a bit difficult, so I made some assumptions like let $\mathbf{A}$ be the identity matrix $\mathbf{I}_2$, can we find all $\mathbf{B}$ that satisfies this 'wrong property'?

REPLY [3 votes]: Lets work with $n\times n$ matrices, and later focus on $n=2.$ It is easy to prove that $A^{-1}+B^{-1}$ is really the inverse of $A(A + B) ^{−1}B,$ not $A+B.$ Also $$(A+B)(A^{-1}+B^{-1})=I \iff AB^{-1}+BA^{-1}+I=0.$$
Theorem
There are two square matrices satisfying this condition if and only if their dimension $n$ is even. Moreover there is a one to one correspondence between complex structures on $\mathbb{R}^n$ and $AB^{-1}.$
Proof:
Let $C=AB^{-1}$ be a non-identity matrix, then the last condition is equivalent to saying that $$C+C^{-1}+I=0 \iff C^3=I.$$ Now, $J=\frac{1}{\sqrt{3}}(I+2C)$ has the property that $J^2=-I,$  i.e., $J$ is a complex structure on $\mathbb{R}^n.$ Then $\det(J)^2=(-1)^n$ implies that $n$ must be even.
Conversely, if $n=2m$ is even, then we can define a complex structure $J: \mathbb{R}^{2m}\to \mathbb{R}^{2m}$  by picking a basis $\{e_1, e_2,\cdots, e_m, e_{m+1},\cdots, e_{2m}\}$ and mapping  $J(e_r)=e_{m+r}$ and $J(e_{m+r})=-e_r$ for all $r=1, 2, \cdots, m.$ So, $J^2=-I$ and the previous equation $C=\frac{1}{2}(\sqrt{3}J-I)$ defines a matrix $C$ with the desire property.
In fact, this proof can use as an algorithm to explicitly compute such matrices $C,$ and hence $A=CB$ for any invertible matrix $B,$ in any even dimension $n.$
Added: We can parametrize all complex structures on $\mathbb{R}^2$ by $$J_{\theta, \lambda}=\begin{pmatrix}
\sinh \theta & -\lambda\cosh \theta \\
\dfrac{1}{\lambda} \cosh \theta & -\sinh \theta
\end{pmatrix},\qquad \lambda\neq0.$$ Hence there is a complete description of all $2\times 2$ matrices with this wrong property.