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Question Number 143774 by TheHoneyCat last updated on 18/Jun/21
Is this statement true or not?  ∃ A∈M_3 (R) ∣ tr(A)=0 and A^2 +^t A=I_3
Isthisstatementtrueornot?AM3(R)tr(A)=0andA2+tA=I3
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
yes
yes
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
letAbeamatrixsuchthat:A2+tA=I3tA=A2I3soA=tA2I3thusA=(A2I3)I30M3(R)=A22A+I3I3+A0M3(R)=A2Ainaword:thepolynomial(X1)Xcancelstherefore,letμAbetheminimalpolynomialofAμA{X,X1,(X1)X}μA=XA=0M3(R)but02+t0I3μA=X1A=I3tr(A)=30μA=(X1)XAisdiagonalizablewith1and0aseigenvaluesGGl3(R)GAG1{(100010000),(100000000)}tr(A){2,1}tr(A)0Sotheanswereisno,therearenosuchmatrices

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