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A quite popular exercise in linear algebra is the following (or very related exercises, see for example https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 and https://math.stackexchange.com/questions/3109173/ab-ba-invertible-and-a2b2-ab-then-3-divides-n):

Let $K$ be a field of characteristic different from 3 and $X$ and $Y$ two $n\times n$-matrices with $X^2+Y^2+XY=0$ and $XY-YX$ invertible. Then 3 divides $n$.

A (representation-theoretic) proof can be given as in the answer of Mariano Suárez-Álvarez in https://math.stackexchange.com/questions/299651/square-matrices-satisfying-certain-relations-must-have-dimension-divisible-by-3 .

Question: Is this also true for fields of characteristic 3?


edit: So it turned out that the result holds for any field. A bonus question might be to find a proof that works independent of the characteristic of the field.

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    $\begingroup$ A simple observation in characteristic 3 is that the hp is equivalent to (X-Y)^2 = [X,Y] invertible. So in particular X = M +Y where M is invertible. Substituting M yields M^2 = [M,Y] for some Y. This is a bit stronger then tr M^2 = 0. $\endgroup$ – Andrea Marino Jun 3 '19 at 18:19
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    $\begingroup$ Following Andrea's simplification. By multiplying both on the left and right by $M^{-1}$ we get $I=[Y,M^{-1}]$. Taking traces of both sides gives $n=0$, which means $3|n$ since we are in characteristic 3. $\endgroup$ – Gjergji Zaimi Jun 3 '19 at 19:27
  • $\begingroup$ This is a very beautiful solution. It would be nice to turn it into an answer. $\endgroup$ – Mare Jun 3 '19 at 19:56
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    $\begingroup$ Great job Gjergji! We did it. Who's going to write it? $\endgroup$ – Andrea Marino Jun 3 '19 at 21:06
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    $\begingroup$ @LSpice - no it is not required. The other answer next to Mariano's essentially works just fine whenever there exists a primitive cube root of unity. $\endgroup$ – Vladimir Dotsenko Jun 4 '19 at 16:26
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Assembling the comments, I write the entire solution.

STEP 1: the first hypothesis in characteristics 3 is equivalent to $(X-Y)^2 = [X,Y]$.

Indeed, note that

$$(X-Y)^2 - [X,Y] = X^2-XY-YX+Y^2- XY+YX = X^2-2XY+Y^2 = X^2+Y^2+XY$$

Why did I make this calculation? Note that $X^2+Y^2+XY$ resembles a factor of $X^3-Y^3$, which would be equal (in commutative case) to $(X-Y)^3$ in char 3. To get there we would need a $X-Y$ more, so we can guess that $(X-Y)^2$ and $X^2+Y^2+XY$ will be equal up to some commutators!

STEP 2: getting the final result.

We have that $(X-Y)^2 = [X,Y]$ is invertible by the second hypothesis. Thus in particular $X-Y$ is invertible. This allows us to substitute $X=M+Y$ with M invertible, obtaining

$$M^2 = [M+Y,Y] = [M,Y]$$

The only relevant information we have about arbitrary commutators is that the trace is zero. We hence would like to have some matrix with easy trace (like the identity) to compare witha commutator. To do this, let's multiplicate on both the left and the right for $M^{-1}$:

$$ I = M^{-1}[M,Y]M^{-1} = M^{-1}(MY-YM)M^{-1} = YM^{-1}-M^{-1}Y = [Y,M^{-1}]$$ Taking traces we get $n=0$. Being in characteristics 3, this gives $3 \mid n$.

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Let me start from $M^2=[M,Y]$ with $M\in {\bf GL}_n(k)$, as suggested by Andrea. Wlog, I assume that the characteristic polynomial of $M$ splits over $k$, and I decompose $k^n$ as the direct sum of characteristic subspaces $E_\mu=\ker(M-\mu I_n)^n$. It is enough to prove that the dimension of each $E_\mu$ is a multiple of $3$. To this end, observe that $Y$ acts over $E_\mu$. In details, let $x$ be an eigenvector, $Mx=\mu x$. Then $(M-\mu)Yx=\mu^2x$. Because $M$ is invertible, $\mu\ne0$ and therefore $$Yx\in\ker(M-\mu)^2\setminus\ker(M-\mu).$$ Likewise $$(M-\mu)Y^2x=2\mu^2Yx+2\mu^3x,$$ hence $Y^2x\in\ker(M-\mu)^3\setminus\ker(M-\mu)^2$. Eventually, using again ${\rm char}(k)=3$, we have $$(M-\mu)Y^3x=0.$$ We deduce that a basis $E_\mu$ is obtained by taking a basis $\cal B$ of $\ker(M-\mu)$, and adjoining the vectors of $Y\cal B$ and $Y^2\cal B$ ; all of them are linearly independent, as seen above. Thus $\dim E_\mu=3\dim\ker(M-\mu)$.

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