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The matrices $A,B,C$ all $2\times 2$ dimensions so: $$A^2 +B^2 +C^2 =AB+BC+CA.$$ Prove that $$(A^2 +B^2 +C^2 -BA-CB-AC)^2=O_2$$

Can someone help me with this? Thank you.

AlexG
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    Seems relevant: http://math.stackexchange.com/questions/487317/if-a2-2abb2-0-then-a-b2-0-is-true-how-about-n-th-degree-case?rq=1 – yurnero Dec 15 '16 at 13:10

2 Answers2

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Set $X=A-B,Y=B-C$, the condition is equivalent to $X^2+YX+Y^2=0$. Hence $A^2+B^2+C^2-BA-CB-AC=X^2+XY+Y^2=XY-YX$, whose trace is obviously $0$. We will frequently use the following special case of Hamilton-Cayley theorem:

Let $M$ be a $2\times 2$ matrix, then $$M^2=(\operatorname{tr}M)M-(\det M)I$$

For example, $(XY-YX)^2=-(\det(XY-YX))I$, so we only need to prove that $XY-YX$ is not invertible. Note that $X^2+YX+Y^2=0$ implies
$$0=(X^2+YX+Y^2)X-Y(X^2+YX+Y^2)=X^3-Y^3$$ So $X^3=Y^3$. Then one may obtain $$X(YX+Y^2)=-X^3=-Y^3=(X^2+YX)Y=Y(X^2+YX)$$ or equivalently $$(XY-YX)X=Y^2X-XY^2,\hspace{1em}Y(XY-YX)=YX^2-X^2Y$$ Now Hamilton-Cayley implies $Y^2X-XY^2=(\operatorname{tr}Y)(YX-XY)$, so $$(\det X)\det(XY-YX)=(\operatorname{tr}Y)^2\det(YX-XY)=(\operatorname{tr}Y)^2\det(XY-YX)$$ Similarly, $$(\det Y)\det(XY-YX)=(\operatorname{tr}X)^2\det(YX-XY)=(\operatorname{tr}X)^2\det(XY-YX)$$ Now we suppose on the contrary that $XY-YX$ is invertible, then $\det(XY-YX)\not=0$, hence $(\operatorname{tr}X)^2=\det Y,(\operatorname{tr}Y)^2=\det X$. But $X^3=Y^3$ implies $\det X=\det Y$, thus $$(\operatorname{tr}X)^2=(\operatorname{tr}Y)^2=\det X=\det Y$$ Applying Hamilton-Cayley again, we see that $\det X=0$ would result in $X^2=Y^2=0$, hence $XY=0$ and $(XY-YX)^2=(YX)(YX)=Y(XY)X=0$. So we may assume $\det X\not=0$. In that case we have \begin{align} X^3=X\cdot X^2 & =X((\operatorname{tr}X)X-(\det X)I) \\ & =(\operatorname{tr}X)X^2-(\det X)X \\ & =(\operatorname{tr}X)((\operatorname{tr}X)X-(\det X)I)-(\det X)X\\ & =-(\operatorname{tr}X)^3I \end{align} For the same reason, $Y^3=-(\operatorname{tr}Y)^3I$. Now $X^3=Y^3$ yields $\operatorname{tr} X=\operatorname{tr} Y$. Let $\lambda=\operatorname{tr} X\not=0$. Then $0=X^2+YX+Y^2=\lambda(X+Y)+YX-\lambda^2 I$. On the other hand, $$0=(X^2+YX+Y^2)X=-\lambda^3 I+Y(\lambda X-\lambda^2 I)+(\lambda Y-\lambda^2 I)X=\lambda(2YX-\lambda(X+Y)-\lambda^2I)$$ "Solving" these two equations we get $$X+Y=\frac 1 3\lambda I,\hspace{1em} YX=\frac 2 3\lambda^2 I$$ But $\det(YX)=(\det Y)(\det X)=\lambda^4$, while $\det(2\lambda^2I/3)=4\lambda^4/9$, contradiction. Now we're done.

Cave Johnson
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Hints. As indicated by another answer here, if you put $X=A-B$ and $Y=B-C$, the given condition becomes $X^2 + YX + Y^2 = 0$ and the hypothesis you want to prove becomes $(XY - YX)^2 = 0$. Since $X$ and $Y$ are $2\times2$, Cayley-Hamilton theorem dictates that $X^2$ is a linear combination of $X$ and $I$ and the analogous holds for $Y^2$. So, it suffices to prove the following proposition:

Proposition. Suppose $X,Y$ are $2\times2$ matrices over any field and $YX=aX+bY+cI$ for some scalars $a,b,c$. Then $XY-YX$ is nilpotent.

Sketch of proof. Let $X'=X-bI$ and $Y'=Y-aI$. Then $XY-YX=X'Y'-Y'X'$ and $$ Y'X'=(Y-aI)(X-bI)=(ab+c)I. $$ Now:

  • If $ab+c\ne0$, what is the relationship between $X'$ and $Y'$?
  • if $ab+c=0$, then $Y'X'=0$. Can you calculate $(X'Y'-Y'X')^2$ directly?
user1551
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  • Very nice answer! I tried to work out with the magic $AB=I\implies BA=I$ but failed. I'm happy to see that the actual answer to the queston finally arises, instead of boring calculation like mine, – Cave Johnson Dec 19 '16 at 04:58
  • @CaveJohnson This is somewhat similar to exercises like "A+B=AB implies that A and B commute". Just add an auxiliary term and then factorise. No big deal. The truly crucial part, which requires good observation, is to recognise that one should simplify the equations in three variables A,B,C to equations in two variables X,Y. And this is your contribution. – user1551 Dec 19 '16 at 05:34