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Let $A, C \in M_{2}(\mathbb{R})$ such that $A = AC-CA$. Prove that $A^{2}=0$

My attempt.

Note that $$ \operatorname{tr}(A) = \operatorname{tr}(AC)-\operatorname{tr}(CA) = \operatorname{tr}(AC)-\operatorname{tr}(AC) = 0. $$ So the characteristic polinomial of $A$ has the form $$ p_{A}(t)=t^{2}+(\operatorname {det}A) . $$ This gives, by the Cayley-Hamilton theorem $$ A^{2}+(\operatorname{det}A)I = 0, $$ that is $$ A^{2} = -(\operatorname{det}A)I. $$ Edit: By the equality above we have $\operatorname{tr}(A^{2})=-2\det A$, but $$ \operatorname{tr}(A^{2}) = \operatorname{tr}(AC-CA)(AC-CA)\\ = \operatorname{tr}((AC)^{2}-((AC)(CA))-((CA)(AC))+(CA)^{2})\\ = \operatorname{tr}((AC)^{2})-\operatorname{tr}((A(C^{2}A))-\operatorname{tr}(C(A^{2}C))+\operatorname{tr}((AC)^{2})\\ = \operatorname{tr}((AC)^{2})-\operatorname{tr}((A^{2}C^{2}))-\operatorname{tr}(A^{2}C^{2})+\operatorname{tr}((AC)^{2})\\ =0. $$ Thus the equation above implies that $\det A = 0$, giving $A^{2}=0$

MrGran
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  • It looks to me like all you're using about the commutator is that its trace is $0$. It's not true that the square of any real $2 \times 2$ matrix of trace $0$ is $0$, so your proof is incorrect. – Robert Israel Jun 07 '24 at 17:37
  • I’m using that hypothesis to see how the characteristic polynomial looks like. Since $A$ Is a 2 by 2 matrix, its characteristic polynomial is $t^{2} - (\operatorname{tr} A) t + \det A = t^{2} + \det A$ – MrGran Jun 07 '24 at 17:45
  • And that's all your using it for. For example, $A = \pmatrix{0 & 1\cr 1 & 0\cr}$ has determinant $-1$ and its characteristic polynomial is $t^2 - 1$. – Robert Israel Jun 07 '24 at 17:48
  • Your statement that $D = \pmatrix{\sqrt{\det A} & 0\cr 0 & -\sqrt{\det A}\cr} $ is wrong. – Robert Israel Jun 07 '24 at 17:52
  • actually, a better duplicate is https://math.stackexchange.com/questions/227984/if-a-and-ab-ba-commute-show-that-ab-ba-is-nilpotent since $n=2$, the result follows by Cayley-Hamilton – user8675309 Jun 07 '24 at 18:04
  • I´ve just edited my post. I think it is better now. Thank you for your feedback – MrGran Jun 08 '24 at 09:53
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    it's a lot easier to directly compute $\text{trace}\big(A^2\big)=\text{trace}\big(A(AC-CA)\big)=\text{trace}\big(A^2C\big)-\text{trace}\big(ACA\big)=\text{trace}\big(A^2C\big)-\text{trace}\big(A^2C\big)=0$ – user8675309 Jun 09 '24 at 19:25
  • Yes, I realized about that after I wrote it. – MrGran Jun 10 '24 at 07:32
  • By taking trace on both sides of $A=AC-CA$, we get $\operatorname{tr}(A)=0$. If $A$ is invertible, we would have $I=C-A^{-1}CA$, which is absurd because the LHS has trace $2$ but the RHS is traceless. Hence $A$ must be singular and $\det(A)=0$. The characteristic polynomial of $A$ is then $x^2$, and by Cayley-Hamilton theorem, we obtain $A^2=0$. – user1551 Jul 29 '24 at 11:13

2 Answers2

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Suppose that $\det A \neq 0$. Then the polynomial $p(t) = \det(C-tI) - \det C$ with $\deg p = 2$ vanishes on $\{-1,0,1\}$, which is absurd.

Alan
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Here is another solution. A general result by Jacobson is, that if $A$ and $AC-CA$ commute, then $AC-CA$ is nilpotent. This yields in your case that $A=AC-CA$ is nilpotent. Since $A$ has size $2$, it follows that $A^2=0$.

Reference:

$AB-BA$ is a nilpotent matrix if it commutes with $A$

Dietrich Burde
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    So my solution is correct? Also, thank you for that reference, I didn´t know about that result from Jacobson. – MrGran Jun 07 '24 at 17:29