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'?

  • $\begingroup$ It seems to me that in algebra, $\frac{1}{r} = s \iff sr=1.$ I don't know matrix theory from a giraffe, so, with a lot of salt, what about $(A + B) \times \left(A^{-1} + B^{-1}\right) = I$? $\endgroup$ Mar 18, 2021 at 3:25
  • $\begingroup$ If you multiply on the right by $(\mathbf A + \mathbf B)\mathbf A^{-1} \mathbf B$, you get a quadratic in the matrix $\mathbf M = \mathbf A^{-1} \mathbf B$. Then by completing the square we see that it's a necessary condition that $\tfrac 2{\sqrt 3}(\mathbf M + \tfrac 12 \mathbf I_2)$ (or something similar) squares to $-\mathbf I_2$. But these square roots are fairly well understood. I'm not sure if there's a way to get a necessary and sufficient condition from here, but probably it can generate some examples. $\endgroup$ Mar 18, 2021 at 3:35

1 Answer 1


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 $$(A+B)(A^{-1}+B^{-1})=I \iff AB^{-1}+BA^{-1}+I=0.$$

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.

  • $\begingroup$ Cool stuff! Is it obvious that this $C$ will be invertible? $\endgroup$ Mar 18, 2021 at 4:20
  • 1
    $\begingroup$ @IzaakvanDongen: Good point. It is invertible because of its characteristic polynomial $C^2+C+I=0.$ Also, the simplest case is $$ J=\begin{pmatrix}0&-1\\1&0\\ \end{pmatrix}$$ and is not hard to compute $C$ and check its invertibility. $\endgroup$
    – Bumblebee
    Mar 18, 2021 at 4:35
  • $\begingroup$ Of course, good point! I really like your answer! $\endgroup$ Mar 18, 2021 at 4:39
  • $\begingroup$ may I know where $J = \dfrac{1}{\sqrt{3}}(I+2C)$ came from? $\endgroup$
    – cgo
    Mar 18, 2021 at 6:42
  • $\begingroup$ Nice to see the symplectic matrix make an all-of-a-sudden appearance in this problem! $\endgroup$
    – cgo
    Mar 18, 2021 at 6:51

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