Define an inner product on $\mathbb{P_4}[x]$ over $\mathbb{R}$ as follows:
$$
\langle f,g \rangle = \int_{0}^{1} f(x)g(x) \,\mathrm{d}x
$$
Let $W$ be the subspace of $\mathbb{P_4}[x]$ consisting of $0$ and all polynomials with degree $0$: That is, $W= \mathbb{R}$. Find a basis for $W^{\perp}$.
My attempts:
$\mathbb{P_4}[x]= a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$
Now $p_1 =1$, $p_2 = x$, $p_3 = x^2$, $p_4 = x^3$, $p_5 = x^4$.
Now $p_1= w_1$ so that $\| w_1 \|^2 = (w_1,w_1) = (1,1)$.
Now $\| w_1 \|^2 = \int_{0}^{1} 1 \cdot 1 \,\mathrm{d}x = 1$.
Similarly $w_2 = p_2 – \frac{\langle p_2, w_1 \rangle w_1}{\|w_1||^2} = (x – \frac{1}{2})$.
Is this partial answer correct or not?
Please help me; any hints/solution will be appreciated.
Thanks in advance.
Best Answer
Let $f(x)=k\in W$ and $g\in W^{\perp}$ then
$$\langle f,g\rangle = k\int_{0}^{1} g(x) dx=0$$
that is
$$a_0+\frac12a_1+\frac13a_2+\frac14a_3+\frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^{\perp}$ is given by
$$\{x-1/2, x^2-1/3, x^3-1/4, x^4-1/5\}$$