每週問題 July 22, 2013

這是有關可對角化矩陣的問題。

Let A be a 2\times 2 real matrix. If A^2=-I, show that A is diagonalizable.

 
參考解答:

考慮 A 的 Jordan 形式 J=M^{-1}AM,其中 M 是可逆矩陣。因為 A^2=-I,可得

J^2=M^{-1}A^2M=-M^{-1}M=-I

假設 A 不可對角化,寫出 J=\begin{bmatrix}  \lambda&1\\  0&\lambda  \end{bmatrix}=\lambda I+N,其中 N=\begin{bmatrix}  0&1\\  0&0  \end{bmatrix}。據此,N^2=0,則有

J^2=(\lambda I+N)^2=\lambda^2I^2+2\lambda IN+N^2=\lambda^2I+2\lambda N

比較上面兩式,可知 N=0\lambda^2=-1,故 J 的形式如下:

J=\begin{bmatrix}  \pm i&0\\  0&\pm i  \end{bmatrix}

其中 i=\sqrt{-1}。換句話說,A 是可對角化矩陣。

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