## 每週問題 February 20, 2017

Let $A$ be a $3\times 3$ real orthogonal matrix and $\det A=1$. Prove that

$(\hbox{trace}A)^2-\hbox{trace}(A^2)=2\,\hbox{trace}A$.

\begin{aligned} \hbox{trace}A&=e^{i\theta}+e^{-i\theta}+1=2\cos\theta+1\\ \hbox{trace}(A^2)&=e^{2i\theta}+e^{-2i\theta}+1=2\cos(2\theta)+1=4\cos^2\theta-1, \end{aligned}

\begin{aligned} (\hbox{trace}A)^2-\hbox{trace}(A^2)&=(2\cos\theta+1)^2-(4\cos^2\theta-1)\\ &=4\cos\theta+2=2\,\hbox{trace}A. \end{aligned}

\displaystyle \begin{aligned} \frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}&=\frac{(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)}{2}\\ &=\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3.\end{aligned}

$\displaystyle \lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3=\begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}+\begin{vmatrix} a_{11}&a_{13}\\ a_{31}&a_{33} \end{vmatrix}+\begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}=\hbox{trace}(\hbox{adj}A)$

$\displaystyle p(A)=-A^3+(\hbox{trace}A)A^2-\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}A+(\det A)I_3=0$

$\displaystyle -A^2+(\hbox{trace}A)A-\frac{(\hbox{trace}A)^2-\hbox{trace}(A^2)}{2}I_3+A^T=0$

$\displaystyle -\hbox{trace}(A^2)+(\hbox{trace}A)^2-\frac{3}{2}\left((\hbox{trace}A)^2-\hbox{trace}(A^2)\right)+\hbox{trace}A=0$

### One Response to 每週問題 February 20, 2017

1. chanjian 說道：

老师，你好，我来自长江流域。对你细致的讲解非常兴奋。