# Finding all matrices that satisfy a 'wrong property'

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'?

• 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 $(A + B) \times \left(A^{-1} + B^{-1}\right) = I$? Mar 18, 2021 at 3:25
• 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. Mar 18, 2021 at 3:35

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.

• Cool stuff! Is it obvious that this $C$ will be invertible? Mar 18, 2021 at 4:20
• @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. Mar 18, 2021 at 4:35
• Of course, good point! I really like your answer! Mar 18, 2021 at 4:39
• may I know where $J = \dfrac{1}{\sqrt{3}}(I+2C)$ came from?
– cgo
Mar 18, 2021 at 6:42
• Nice to see the symplectic matrix make an all-of-a-sudden appearance in this problem!
– cgo
Mar 18, 2021 at 6:51