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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

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{statement}\:{true}\:\mathrm{or}\:\mathrm{not}? \\ $$$$\exists\:\mathrm{A}\in\mathscr{M}_{\mathrm{3}} \left(\mathbb{R}\right)\:\mid\:\mathrm{tr}\left(\mathrm{A}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{A}^{\mathrm{2}} +^{{t}} \mathrm{A}=\mathrm{I}_{\mathrm{3}} \\ $$

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

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

$$\mathrm{let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{a}\:\mathrm{matrix}\:\mathrm{such}\:\mathrm{that}:\:\mathrm{A}^{\mathrm{2}} +^{{t}} \mathrm{A}=\mathrm{I}_{\mathrm{3}} \\ $$$$−\:^{{t}} \mathrm{A}=\mathrm{A}^{\mathrm{2}} −\mathrm{I}_{\mathrm{3}} \\ $$$$\mathrm{so}\:−\mathrm{A}=^{{t}} \mathrm{A}^{\mathrm{2}} −\mathrm{I}_{\mathrm{3}} \\ $$$$\mathrm{thus}\:−\mathrm{A}=\left(\mathrm{A}^{\mathrm{2}} −\mathrm{I}_{\mathrm{3}} \right)−\mathrm{I}_{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{0}_{\mathscr{M}_{\mathrm{3}} \left(\mathbb{R}\right)} =\mathrm{A}^{\mathrm{2}} −\mathrm{2A}+\mathrm{I}_{\mathrm{3}} −\mathrm{I}_{\mathrm{3}} +\mathrm{A} \\ $$$$\Leftrightarrow\mathrm{0}_{\mathscr{M}_{\mathrm{3}} \left(\mathbb{R}\right)} =\mathrm{A}^{\mathrm{2}} −\mathrm{A} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{word}:\:\mathrm{the}\:\mathrm{polynomial}\:\left({X}−\mathrm{1}\right){X}\:\:\mathrm{cancels} \\ $$$$\mathrm{therefore},\:\mathrm{let}\:\mu_{{A}} \mathrm{be}\:\mathrm{the}\:{minimal}\:{polynomial}\:\mathrm{of}\:\mathrm{A}\: \\ $$$$\mu_{{A}} \in\left\{{X},\:{X}−\mathrm{1},\:\left({X}−\mathrm{1}\right){X}\right\} \\ $$$$\mu_{{A}} ={X}\:\Rightarrow\:\mathrm{A}=\mathrm{0}_{\mathscr{M}_{\mathrm{3}} \left(\mathbb{R}\right)} \:\mathrm{but}\:\mathrm{0}^{\mathrm{2}} +^{{t}} \mathrm{0}\neq\mathrm{I}_{\mathrm{3}} \\ $$$$\mu_{{A}} ={X}−\mathrm{1}\:\Rightarrow\:\mathrm{A}=\mathrm{I}_{\mathrm{3}} \:\Rightarrow\:\mathrm{tr}\left(\mathrm{A}\right)=\mathrm{3}\neq\mathrm{0} \\ $$$$\mu_{{A}} =\left({X}−\mathrm{1}\right){X}\Rightarrow\:\mathrm{A}\:\mathrm{is}\:{diagonalizable}\:\mathrm{with}\:\mathrm{1}\:\mathrm{and}\:\mathrm{0}\:\mathrm{as}\:{eigenvalues} \\ $$$$\Rightarrow\exists{G}\in\mathscr{G}{l}_{\mathrm{3}} \left(\mathbb{R}\right)\:\mid\:{G}\mathrm{A}{G}^{−\mathrm{1}} \in\left\{\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{pmatrix},\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{pmatrix}\right\} \\ $$$$\Rightarrow\mathrm{tr}\left(\mathrm{A}\right)\in\left\{\mathrm{2},\mathrm{1}\right\} \\ $$$$\Rightarrow\mathrm{tr}\left(\mathrm{A}\right)\neq\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{answere}\:\mathrm{is}\:{no},\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{such}\:\mathrm{matrices} \\ $$

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