6.1 解答

6.1

(a) 直交系とは互いのベクトルが直交している集合より，

$\displaystyle (1,3) \cdot (6,-2) = 6 - 6 = 0$

よって $(1,3) \perp (6, -2)$
また正規直交系とはそれぞれが単位ベクトルで直交系をなしているものより，正規直交系に直すと

$\displaystyle \{\frac{(1,3)}{\sqrt{10}}, \frac{(6,-2)}{2\sqrt{10}} \} \ \framebox{終}$

(b)

$\displaystyle (1,2,2) \cdot (-2,2,-1) = -2 + 4 - 2 = 0$

$\displaystyle (1,2,2) \cdot (2,1,-2) = 2 + 2 -4 = 0$

$\displaystyle (-2,2,-1) \cdot (2,1,-2) = -4 + 2 + 2 = 0$

より

$\displaystyle (1,2,2) \perp (-2,2,-1), (1,2,2) \perp (2,1,-2), (-2,2-1) \perp (2,1,-2) .$

また正規直交系に直すと

$\displaystyle \{\frac{(1,2,2)}{3},\frac{(-2,2,-1)}{3},\frac{(2,1,-2)}{3} \} \ \framebox{終}$

(c) ${\bf i} - 2 {\bf j} + 3{\bf k} \cdot 2{\bf i} - \frac{1}{2}{\bf j} - \frac{1}{3}{\bf k} = 2 + 1 -1 \neq 0$ よって直交系でない $\framebox{終}$

$\displaystyle 0 \leq \vert\vert(f - \lambda g)\vert\vert^{2}$	$\displaystyle =$	$\displaystyle (f - \lambda g , f - \lambda g )$
	$\displaystyle =$	$\displaystyle \vert\vert f\vert\vert^{2} - 2\lambda (f,g) + \lambda^{2}\vert\vert g\vert\vert^{2}$

これは $\lambda$ についての2次式で0より大きいので，その判別式 $\Delta$ は0以下になる．よって

$\displaystyle \Delta = \vert(f,g)\vert^{2} - \vert\vert f\vert\vert^{2} \ \vert\vert g\vert\vert^{2} \leq 0$

これより

$\displaystyle \vert(f,g)\vert \leq \vert\vert f\vert\vert \ \vert\vert g\vert\vert \ \ \ \framebox{終}$

(a)

$\vert\vert f\vert\vert = \{\int_{a}^{b}[f(x)]^{2}dx \}^{1/2}$ より

$\displaystyle \vert\vert f\vert\vert = \{\int_{0}^{2}x^{2}dx\}^{1/2} = \{\frac{x^3}{3}\vert _{0}^{2}\}^{1/2} = \sqrt{\frac{8}{3}} \ \ \framebox{終}$

(b)

$\vert\vert f\vert\vert = \{\int_{a}^{b}[f(x)]^{2}dx \}^{1/2}$ より

$\displaystyle \vert\vert f\vert\vert$	$\displaystyle =$	$\displaystyle \{\int_{0}^{2}[\sin{\pi x}]^{2}dx\}^{1/2} = \{\int_{0}^{2}\frac{1 - \cos{2\pi x}}{2}dx \}^{1/2}$
	$\displaystyle =$	$\displaystyle \{\frac{x}{2} - \frac{\sin{2\pi x}}{4\pi} \vert _{0}^{2} \}^{1/2} = 1 \ \ \framebox{終}$

$\displaystyle \vert\vert f\vert\vert$	$\displaystyle =$	$\displaystyle \{\int_{0}^{2}[\cos{\pi x}]^{2}dx\}^{1/2} = \{\int_{0}^{2}\frac{1 + \cos{2\pi x}}{2}dx \}^{1/2}$
	$\displaystyle =$	$\displaystyle \{\frac{x}{2} + \frac{\sin{2\pi x}}{4\pi} \vert _{0}^{2} \}^{1/2} = 1 \ \ \framebox{終}$

$\displaystyle (P_{0},P_{1})$	$\displaystyle =$	$\displaystyle \int_{-1}^{1}xdx = \frac{x^2}{2}\mid_{-1}^{1} = 0$
$\displaystyle (P_{0},P_{2})$	$\displaystyle =$	$\displaystyle \int_{-1}^{1}\frac{3x^2 - 1}{2} dx = \frac{x^3 - x}{2}\mid_{-1}^{1} = 0$
$\displaystyle (P_{1},P_{2})$	$\displaystyle =$	$\displaystyle \int_{-1}^{1}x\frac{3x^2 - 1}{2} dx = \frac{3x^4}{8} - \frac{x^2}{4}\mid_{-1}^{1} = 0$

より $P_{0}, P_{1}, P_{2}$ は直交型をなすことがわかる． $\framebox{終}$