直交tensorの固有値

直交tensorの固有値

$ \top

$ \iff\det(\bm A-(\pm1)\bm I)=\det(\bm A\cdot\bm I-(\pm1)\bm A\cdot\bm A^\top)

$ = \det\bm A\det(\bm I-(\pm1)\bm A^\top)

$ = (-(\pm1))^{{\rm tr}\bm I}\det\bm A\det(\bm A^\top-(\pm1)\bm I)

$ = (-(\pm1))^{{\rm tr}\bm I}\det\bm A\det(\bm A-(\pm1)\bm I)

$ \iff\det(\bm A-(\pm1)\bm I)=0\lor1=(-(\pm1))^{{\rm tr}\bm I}\det\bm A

$ \iff\det(\bm A-(\pm1)\bm I)=0\lor\det\bm A=(-(\pm1))^{{\rm tr}\bm I}

$ \because\det\bm A=\pm1

$ \iff(\det(\bm A-1\cdot\bm I)=0\lor\det\bm A=(-1)^{{\rm tr}\bm I})\land(\det(\bm A-(-1)\bm I)=0\lor\det\bm A=1^{{\rm tr}\bm I}=1)

$ \iff(\lnot\det\bm A\neq(-1)^{{\rm tr}\bm I}\lor\det(\bm A-1\cdot\bm I)=0)\land(\lnot\det\bm A\neq1\lor\det(\bm A-(-1)\bm I)=0)

$ \iff(\lnot\det\bm A=-(-1)^{{\rm tr}\bm I}\lor\det(\bm A-1\cdot\bm I)=0)\land(\lnot\det\bm A=-1\lor\det(\bm A-(-1)\bm I)=0)

$ \because\det\bm A=\pm1

$ \iff\begin{dcases}\det(\bm A-1\cdot\bm I)=0&\text{if }\det\bm A=-(-1)^{{\rm tr}\bm I}\\\det(\bm A-(-1)\bm I)=0&\text{if }\det\bm A=-1\end{dcases}

$ \underline{\iff\begin{dcases}\det(\bm A-(\pm1)\bm I)=0&\text{if }\mathrm{tr}\bm I\equiv0\pmod2\land\det\bm A=-1\\\det(\bm A-1\cdot\bm I)=0&\text{if }\mathrm{tr}\bm I\equiv1\pmod2\land\det\bm A=1\\\det(\bm A-(-1)\bm I)=0&\text{if }\mathrm{tr}\bm I\equiv1\pmod2\land\det\bm A=-1\end{dcases}\quad}_\blacksquare

$ \lambda=\frac12(\mathrm{tr}\bm A\pm\sqrt{(\mathrm{tr}\bm A)^2-4})

$ =\frac12\left(I_1^{\bm{A}}\pm\sqrt{\left(I_1^{\bm{A}}\right)^2-4}\right)

$ =\frac12\left(I_1^{\bm{A}}\pm\sqrt{4-4J_2^{\bm{A}}-4}\right)

$ =\frac12I_1^{\bm{A}}\pm\sqrt{-J_2^{\bm{A}}}

$ \arg\lambda=\tan^{-1}\frac{\sqrt{1^2-\left(\frac12I_1^{\bm{A}}\right)^2}}{\frac12I_1^{\bm{A}}}

$ = \cos^{-1}\frac12I_1^{\bm{A}}

$ \det(\pmb A-\lambda\pmb I)=0

$ \iff (\lambda-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2)(\lambda-1)=0

$ \iff(\lambda-(\mathrm{tr}\pmb A-1)\lambda+\det\pmb A)(\lambda-1)=0

$ \iff \lambda=1,\frac12\left(\mathrm{tr}\pmb A-1\pm\sqrt{(\mathrm{tr}\pmb A-1)^2-4\det\pmb A}\right)

$ =1,\frac12\left(\mathrm{tr}\pmb A-1\pm\sqrt{\pmb A:\pmb A-2(\det\pmb A+\mathrm{tr}\pmb A)}\right)

$ \because(\mathrm{tr}\pmb A-1)^2=\lambda_1^2+2\lambda_1\lambda_2+\lambda_2^2-2\mathrm{tr}\pmb A+1=\pmb A:\pmb A+2(\det\pmb A-\mathrm{tr}\pmb A)

$ =1,\frac12\left(\mathrm{tr}\pmb A-1\pm\sqrt{(\mathrm{tr}\pmb A-1)^2\mp4}\right)

$ \because\det\pmb A=\pm1

$ \arg\lambda_1=\tan^{-1}\frac{\sqrt{1^2-\left(\frac12({\rm tr}\bm A-1)\right)^2}}{\frac12({\rm tr}\bm A-1)}

$ =\cos^{-1}\frac12({\rm tr}\bm A-1)

x,y,z軸回転tensorの場合に正しいことはすぐ確認できる

$ {\rm tr}\bm R(\bm n,\theta)={\rm tr}\bm I\cos\theta+{\rm tr}(1-\cos\theta)\hat{\bm n}\hat{\bm n}-{\rm tr}(\sin\theta)\hat{\bm n}\cdot{\Large\bm\epsilon}

$ = 3\cos\theta+(1-\cos\theta)|\hat{\bm n}|^2-0

$ =2\cos\theta+1

$ \therefore\cos^{-1}\frac12({\rm tr}\bm A-1)=\cos^{-1}\cos\theta=\theta

確認完了

これを使って

$ \iff \lambda=1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{(I_1^{\bm A}-1)^2-4\det\bm A}\right)

$ =1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{(I_1^{\bm A}-1)^2-4}\right)

$ =1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{\left(I_1^{\bm A}\right)^2-2I_1^{\bm A}+1-4}\right)

$ =1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{\left(I_1^{\bm A}\right)^2-2I_1^{\bm A}-3}\right)

$ =1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{3(I_1^{\bm A}-J_2^{\bm A})-2I_1^{\bm A}-3}\right)

$ =1,\frac12\left(I_1^{\bm A}-1\pm\sqrt{I_1^{\bm A}-3(J_2^{\bm A}+1)}\right)

22:21:37 大幅に間違えていたので修正

任意次元で固有値が1になる条件を特定できない