Prove that if $A, B \in M_2(\mathbb{R}) $, $AB=BA$, $\det(A+iB) =0$ and $4 \det A > (Tr A) ^2$, then $A^2+B^2=O_2$. From $\det (A+iB) =0$ I got that $\det A=\det B$ and $Tr A\cdot Tr B=Tr(AB) $, but I don't know how to use this.
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determinant of sum of 2x2 matrices: https://math.stackexchange.com/questions/673934/expressing-the-determinant-of-a-sum-of-two-matrices $(A + B)^2 = A^2 + AB + BA + B^2$ – R zu Oct 28 '18 at 17:43
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How is this relevant to the problem? – user69503 Oct 28 '18 at 17:45
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Hint: if $C$ is a real 2-by-2 matrix with an eigenvalue $i$, then $C^2=-I$. – user1551 Oct 28 '18 at 18:30
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How to use this? – user69503 Oct 28 '18 at 19:32
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Since $A$ is a real square matrix, its determinant and its trace are real numbers. As $$4\,\det(A)>\big(\text{trace}(A)\big)^2\geq 0\,,$$ we have $\det(A)\neq 0$, so $A$ is invertible. Take $C:=A^{-1}\,B$. Then, $C$ is a real matrix such that $$\det(I+\text{i}\,C)=0\,,$$ where $I$ is the $2$-by-$2$ identity matrix. Use this to prove that $\text{i}$ is an eigenvalue of $C$, and then apply user1551's hint. With $C^2=-I$ and $AB=BA$, show that $A^2+B^2=0$.
Batominovski
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Because $\det(A+\text{i},B)=\det(A),\det\left(I+\text{i},A^{-1},B\right)$. – Batominovski Oct 28 '18 at 20:40
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