$\langle p,q\rangle=\int_{-1}^1p(x)q(x)\,\mathrm{d}x \quad, (p,q\in R_3[x])$\,where $R_3[x]$ is the vector space of all real polynomial of degree less than equal to 3.
If $W=\{\,p \in R_3[x]:p(0)=0\}$ then what is $W$ perpendicular i.e the perpendicular subspace with respect to the above inner product.
I just managed to get that $W$ is a three dimension subspace of $R_3[x]$, and think since $W^\perp$ is the compliment (orthogonal complement) of $W$ it has to be one dimenssional but what would it be i am not able to understand. I tried to solve the integral but it became very complicated .please help?
Note that $W=\{a_3x^3+a_2x^2+a_1x|a_i\in\mathbb{R}\}$. Then, $\dim W=3$. Now, since $x,x^2$ and $x^3$ are l.i. and elements of $W$, we conclude that $\{x,x^2,x^3\}$ is a base for $W$. Moreover, $\dim W^\perp=\dim P_3-\dim W=4-3=1$.
Therefore, we must find only one vector orthogonal to $x,x^2$ and $x^3$. Let $p(x)=ax^3+bx^2+cx+d\in W^\perp$.
Then $\langle p(x),x^k\rangle=0$ for $k=1,2,3$. That is
$$\left\{\begin{array}{ccc}\displaystyle\int_{-1}^1ax^4+bx^3+cx^2+dx\,\mathrm{d}x&=&0\\\displaystyle\int_{-1}^1ax^5+bx^4+cx^3+dx^2\,\mathrm{d}x&=&0\\\displaystyle\int_{-1}^1ax^6+bx^5+cx^4+dx^3\,\mathrm{d}x&=&0\\\end{array}\right.\Leftrightarrow\left\{\begin{array}{ccc}\frac{a}{5}+\frac{c}{3}&=&0\\\frac{b}{5}+\frac{d}{3}&=&0\\\frac{a}{7}+\frac{c}{5}&=&0\\\end{array}\right.,$$
which yield us $a=0=c$, $b=1$ and $d=\frac{-3}{5}$. That is, $W^\perp=\mathrm{span}\{x^2-\frac{3}{5}\}$}
PD: The integrals are not difficult. You must note that $\int_{-1}^1(\mbox{odd functions})dx=0$. Then, $\int_{-1}^1x^k\,\mathrm{d}x=0$ for odd $k$.