每週問題 April 7, 2014

這是線性泛函的問題。

Define a non-zero linear functional f on \mathbb{C}^3 such that if \mathbf{u}=(1,1,1) and \mathbf{v}=(1,1,-1), then f(\mathbf{u})=f(\mathbf{v})=0.

參考解答:

定義於 \mathbb{C}^3 的任一線性泛函可表示為

\displaystyle  f(x_1,x_2,x_3)=c_1x_1+c_2x_2+c_3x_3

(1,1,1)(1,1,-1) 代入上式,

\displaystyle\begin{aligned}  c_1+c_2+c_3&=0\\  c_1+c_2-c_3&=0.  \end{aligned}

解得 c_1=\alphac_2=-\alphac_3=0,其中 \alpha 是任意數。因此,線性泛函 f(x_1,x_2,x_3)=\alpha x_1-\alpha x_2\alpha\neq 0,即為所求。

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