每週問題 February 20, 2017

證明三階旋轉矩陣的一個跡數恆等式。

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.

 
參考解答:

證明 1. 實正交矩陣滿足 A^T=A^{-1} 並知道 \det A=1,可知 A 是一個 3\times 3 階旋轉矩陣,特徵值為 e^{i\theta}, e^{-i\theta}, 1,其中 \theta\in[0,2\pi]i=\sqrt{-1}。因此,

\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}

也就得到

\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}

證明 2. 令 \lambda_1,\lambda_2,\lambda_3A=[a_{ij}] 的特徵值。寫出

\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}

再者,

\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)

因為 A(\hbox{adj}A)=(\det A)I=I,可知 \hbox{adj}A=A^{-1}=A^T,合併以上結果即得證。

證明 3. 令 p(t)=\det(A-tI)A 的特徵多項式。Cayley-Hamilton 定理給出

\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

上式通乘 A^{-1}=A^T,並代入 \det A=1,可得

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

計算跡數,

\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

化簡即得所求。

This entry was posted in pow 特徵分析, 每週問題 and tagged , , , . Bookmark the permalink.

1 Response to 每週問題 February 20, 2017

  1. chanjian says:

    老师,你好,我来自长江流域。对你细致的讲解非常兴奋。

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