每週問題 July 1, 2013

這是推導 2\times 2 階矩陣的2-範數與 Frobenius 範數的關係式問題。

Let A be a 2\times 2 real matrix. Show that

\displaystyle  \Vert A\Vert_2^2=\frac{\Vert A\Vert_F^2+\sqrt{\Vert A\Vert_F^4-4(\det A)^2}}{2},

where \Vert A\Vert_2 is the 2-norm of A and \Vert A\Vert_F is the Frobenius norm of A.

 
參考解答:

\lambda_1\ge\lambda_2\ge 0 為半正定矩陣 A^TA 的特徵值。下列性質成立:\lambda_1+\lambda_2=\hbox{tr}(A^TA)\lambda_1\lambda_2=\det(A^TA)=(\det A)^2。Frobenius 範數 \Vert A\Vert_F 等於 \sqrt{\hbox{tr}(A^TA)},且 2-範數 \Vert A\Vert_2 即為 A 的最大奇異值,也就是 \sqrt{\lambda_1}。接下來只要解出 \lambda_1 即可。使用以上結果,寫出 A^TA 的特徵多項式

\displaystyle  p(t)=t^2-(\lambda_1+\lambda_2)t+\lambda_1\lambda_2=t^2-\Vert A\Vert_F^2t+(\det A)^2

解得兩根

\displaystyle\begin{aligned}  \lambda_1&=\frac{\Vert A\Vert_F^2+\sqrt{\Vert A\Vert_F^4-4(\det A)^2}}{2}\\  \lambda_2&=\frac{\Vert A\Vert_F^2-\sqrt{\Vert A\Vert_F^4-4(\det A)^2}}{2}  ,\end{aligned}

\lambda_1=\Vert A\Vert^2_2,因此得證。

This entry was posted in pow 二次型, 每週問題 and tagged , . Bookmark the permalink.

2 Responses to 每週問題 July 1, 2013

  1. jianback says:

    特征多项式漏写一个t。:)

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