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Let $A,B$ be $4 \times 4$ matrices with real elements such that $A \ne B$, $tr(a) \ne 0$, $A^2-2B+I_4=O_4$ and $B^2-2A+I_4=O_4$. Show that $A+B=-2I_4$.

My try: I got an expression of $B$ oin function of $A$, substituded and got $A^4-2A^2-8A+5I_4=O_4$ and I am stuck.

M. Stefan
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1 Answers1

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Substitute the first equation into the four times the second, you get $(A^2 + I)^2 + 4(-2A + I) = 0$. Simplify to get $(A-I)^2(A^2+2A+5I)=0$. Argue that the minimal polynomial of $A$ is ...

  • neither $x-1$ nor $(x-1)^2$, otherwise the condition $A\ne B$ will be violated;
  • neither $(x-1)(x^2+2x+5)$ nor $(x-1)^2(x^2+2x+5)$, otherwise the condition $\operatorname{tr}(A)\ne0$ will be violated.

Hence the minimal polynomial of $A$ is $x^2+2x+5$. Now use the condition $A^2-2B+I=0$ to conclude that $A+B=-2I$.

user1551
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