Is this statement true or not? ∃ A∈M_3 (R) ∣ tr(A)=0 and A^2 +^t A=I


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Question Number 143774 by TheHoneyCat last updated on 18/Jun/21

$Is this statement t r u e or not ?$ $\exists A \in M_{3} (R) ∣ tr (A) = 0 and A^{2} +^{t} A = I_{3}$
Commented by TheHoneyCat last updated on 18/Jun/21
Oh, I actually found the answer. �� sorry. Should I send it?
Commented by ArielVyny last updated on 18/Jun/21

$y e s$
	Answered by TheHoneyCat last updated on 19/Jun/21
	$let A be a matrix such that: A^2 +^t A=I_3 −^t A=A^2 −I_3 so −A=^t A^2 −I_3 thus −A=(A^2 −I_3 )−I_3 ⇔0_(M_3 (R)) =A^2 −2A+I_3 −I_3 +A ⇔0_(M_3 (R)) =A^2 −A in a word: the polynomial (X−1)X cancels therefore, let μ_A be the minimal polynomial of A μ_A ∈{X, X−1, (X−1)X} μ_A =X ⇒ A=0_(M_3 (R)) but 0^2 +^t 0≠I_3 μ_A =X−1 ⇒ A=I_3 ⇒ tr(A)=3≠0 μ_A =(X−1)X⇒ A is diagonalizable with 1 and 0 as eigenvalues ⇒∃G∈Gl_3 (R) ∣ GAG^(−1) ∈{ ((1,0,0),(0,1,0),(0,0,0) ), ((1,0,0),(0,0,0),(0,0,0) )} ⇒tr(A)∈{2,1} ⇒tr(A)≠0 So the answere is no, there are no such matrices$
	$let A be a matrix such that : A^{2} +^{t} A = I_{3}$ $-^{t} A = A^{2} - I_{3}$ $so - A =^{t} A^{2} - I_{3}$ $thus - A = (A^{2} - I_{3}) - I_{3}$ $\Leftrightarrow 0_{M_{3} (R)} = A^{2} - 2 A + I_{3} - I_{3} + A$ $\Leftrightarrow 0_{M_{3} (R)} = A^{2} - A$ $in a word : the polynomial (X - 1) X cancels$ $therefore, let μ_{A} be the m i n i m a l p o l y n o m i a l of A$ $μ_{A} \in {X, X - 1, (X - 1) X}$ $μ_{A} = X \Rightarrow A = 0_{M_{3} (R)} but 0^{2} +^{t} 0 \neq I_{3}$ $μ_{A} = X - 1 \Rightarrow A = I_{3} \Rightarrow tr (A) = 3 \neq 0$ $μ_{A} = (X - 1) X \Rightarrow A is d i a g o n a l i z a b l e with 1 and 0 as e i g e n v a l u e s$ $\Rightarrow \exists G \in G l_{3} (R) ∣ G A G^{- 1} \in {(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}), (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})}$ $\Rightarrow tr (A) \in {2, 1}$ $\Rightarrow tr (A) \neq 0$ $So the answere is n o, there are no such matrices$
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