每週問題 February 2, 2015

本週問題關於 Cayley 變換。

Let A be a skew Hermitian matrix. Show that Cayley transformation

\displaystyle  U=(I-A)(I+A)^{-1}=(I+A)^{-1}(I-A)

is a unitary matrix.

 
參考解答:

A^\ast=-A。我們先證明 I+A 是可逆矩陣。設 \mathbf{x}\in N(I+A),即 (I+A)\mathbf{x}=\mathbf{0},或寫成 \mathbf{x}=-A\mathbf{x}。等號兩邊左乘 \mathbf{x}^\ast,即有 \mathbf{x}^\ast\mathbf{x}=-\mathbf{x}^\ast A\mathbf{x},再取共軛轉置,可得 \mathbf{x}^\ast\mathbf{x}=-\mathbf{x}^\ast A^\ast\mathbf{x}=\mathbf{x}^\ast A\mathbf{x}。合併上面兩式,推得 \mathbf{x}^\ast\mathbf{x}=0,故 \mathbf{x}=\mathbf{0},即知 I+A 可逆。寫出恆等式 (I+A)(I-A)=(I-A)(I+A),等號兩邊左右同乘 (I+A)^{-1},即得 U=(I-A)(I+A)^{-1}=(I+A)^{-1}(I-A)。使用以上結果,

\displaystyle\begin{aligned}  U^\ast U&=\left((I-A)(I+A)^{-1}\right)^\ast (I-A)(I+A)^{-1}\\  &=((I+A)^\ast)^{-1}(I-A)^\ast (I+A)^{-1}(I-A)\\  &=(I-A)^{-1}(I+A)(I+A)^{-1}(I-A)\\  &=I.  \end{aligned}

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