Question

If V = P3 with the inner product < f, g >=
R 1
11
f(x)g(x)dx, apply the Gram-Schmidt algorithm
to obtain an orthogonal basis from B = {1, x, x2
, x3}

Answer 1

Solution:

We need to find the Orthogonal Basis by applying Gram-Schmidt algorithm.

Given $\beta =${${1, x, x^2, x^3}$ } and < f, g > = $\int_{-1}^ {1} f(x) \space g(x) dx$

From this

$$\beta_1 = 1\\ \beta_2 = x \\ \beta_3 = x^2 \\ \beta_4 = x^3$$

Next,

the Orthogonal vectors are

$$\alpha_1 = 1$$

$$||\alpha_1||^2 = < \alpha_1, \alpha_1 > = <1, 1> = \int _{-1}^ {1} 1 \times 1 dx = [ x ]_{-1}^1 = 2$$

$$\alpha_2 = \beta_2 - \frac {<\beta_2, \alpha_1>} {||\alpha_1||^2 } \space {\alpha_1}$$

$$\alpha_2 = x - \frac {<x, 1>} {||1||^2 } \space \times {1}$$

$$\alpha_2 = x - \frac {\int _{-1}^{1} x \space dx} {\int_{-1} ^{1}1 \space dx } \space \times {1} = x - \frac {0} {2} = x$$

$$(since \int _{-1}^1 x \space dx = 0 )$$

Now,

$$\alpha_3 = \beta_3 - \frac {<\beta_3, \alpha_1>} {||\alpha_1||^2 } \times \alpha_1 - \frac {<\beta_3, \alpha_2>} {||\alpha_2||^2} \times \alpha_2$$

$$\alpha_3 = \beta_3 - \frac {<x^2, 1>} {||1||^2 } \times 1 - \frac {<x^3, x>} {||x||^2} \times x$$

$$\alpha_3 = \beta_3 - \frac {\int_{-1}^1 x^2 \space dx} {\int _{-1}^1 1 \space dx} \times 1 - \frac {\int_{-1}^1 x^2 \space \times x \space dx} {\int_{-1}^1 x \times x \space dx } \times x$$

$$\alpha_3 = \beta_3 - 2 \times\frac {2} {3} \times 1 - 0 = x^2 - \frac {1} {3}$$ $$\alpha_4 = \beta_4 - \frac {<\beta_4, \alpha_1> } {||\alpha_1||^2 } \alpha_1 - \frac {<\beta_4, \alpha_2> } {||\alpha_2||^2 } \alpha_2 - \frac {<\beta_4, \alpha_3> } {||\alpha_3||^2 } \alpha_3$$

$$\alpha_4 = x^3 - \frac {< x^3, 1> } {||1||^2 } \times1 - \frac {<x^3, x> } {||x||^2 } x^2 - \frac {<x^3, x^2 - \frac {1}{3}> } {||x^2 - \frac {1}{3}||^2 } \times (x^2 - \frac {1}{3})$$

$$= x^3 - \frac {3}{5} x$$

Answer: Orthogonal Basis = {$\alpha_1, \alpha_2, \alpha_3, \alpha_4$ } = {$1, x, x^2 - \frac{1}{3}, x^3 - \frac {3}{5} x$ }