6
$\begingroup$

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'?

$\endgroup$
2
  • $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.