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Let $A$ and $B$ be two $n \times n$ non-commuting matrices with complex elements such that $A^2 + B^2 = 2AB$. Prove that there is at least one $m \leq \frac{n+1}{2}$ so that $(AB-BA)^m = O$, where $O$ is the null matrix.

Sungjin Kim
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C_M
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    Hint: try to show that $AB-BA=(A-B)^2$. – Mosquite Mar 14 '17 at 20:13
  • Please trying using MathJax formatting (it's the same syntax as LaTeX). It makes for a more readable text, attracting more attention. – mlc Mar 14 '17 at 20:13
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    I know that $AB-BA=(A-B)^2$, but how does it help me?($(A-B)^2 = A^2 - AB - BA + B^2, so AB-BA=(A-B)^2$) – C_M Mar 14 '17 at 20:24
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    I propose a change in your title in order to attract attention and to stick to the real issue '"A difficult problem..." could be the title of at least half of the questions ... and is not informative at all. – Jean Marie Mar 14 '17 at 20:24

1 Answers1

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Indeed, @Mosquite's comment is extremely helpful. The main idea is to use the argument in Show that a matrix is nilpotent. Let $C=AB-BA = (A-B)^2$ and we will prove that $\mathrm{tr}(C^k)=0$ for $k\geq 1$.

From $C=(A-B)^2$, we see that $C$ commutes with $A-B$. Thus, $C^k$ commutes with $A-B$ for any $k\geq 1$.

Since $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, we have $\mathrm{tr}(C)=0$. Suppose that for some $k\geq 2$ we proved that $\mathrm{tr}(C^{k-1})=0$. Then by $\mathrm{tr} (XY)= \mathrm{tr} (YX)$, we have $$ \begin{align} \mathrm{tr}(C^k)&= \mathrm{tr} C^{k-1}( AB-BA) \\ &= \mathrm{tr} C^{k-1}( (A-B)B+B^2-BA)\\ &=\mathrm{tr} ( (A-B) C^{k-1} B + C^{k-1}B(B-A) )\\ &=\mathrm{tr}( C^{k-1}B(A-B) + C^{k-1}B(B-A)) =0. \end{align} $$

Therefore, the claim that $\mathrm{tr}(C^k)=0$ for $k\geq 1$ is proved. Consequently, $C$ is nilpotent so that $A-B$ is also nilpotent. Thus, $(A-B)^n=O$. Then for a least integer $m$ with $2m\geq n$, we have $$ (AB-BA)^m = (A-B)^{2m} = O. $$ This $m$ satisfies $m\leq \frac{n+1}2$. So, we are done.

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Sungjin Kim
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