How to find the orthogonal complement

Question

The question is, considering

• The vector space $\mathcal{P}_3(\mathbb{R})$ of the polynomials with real coefficients of degree $\leq$ 3
• The inner product defined by $\left<p,q\right>=\int_{-1}^{1}pq$

how to find a basis for the orthogonal complement of the space spanned by $\{x-1, x^2+3\}$?

I tried making $$\int_{-1}^{1}(x-1)p(x)=0$$ and $$\int_{-1}^{1}(x^2+3)p(x)=0,$$ which gave me the spanning sets $\{x^3+\frac{1}{5}, x^2-\frac{1}{3}, x+\frac{1}{3}\}$ and $\{x^3, x^2-\frac{9}{25}, x\}$ respectively. Then I think we may determine a basis for the intersection os these subspaces, but didn't see how.

Answer 1

Write a linear combination of one of the spanning spaces, such as $$ ax^3 + b\left(x^2-\frac{9}{25}\right) + cx $$ and take the inner product with the other polynomial (in this case $x-1$). Set it equal to zero and solve for $c$, then plug it back into the above expression and collect terms proportional to $a$ and $b$ to get a spanning set orthogonal to both.

You could also use the Gram-Schmidt process on the basis $\{x - 1, x^2 + 3, 1, x^3\}$ (the last two vectors are arbitrary elements of the complement of the span of $\{x-1,x^2+3\}$). At the end of the process, the last two vectors will be an orthogonal basis for the the orthogonal complement of the span of the first two.

Answer 2

Let $\textsf{W}=\operatorname{span}(\{x-1,x^2+3\})$ be the subspace of $\textsf{P}_3(\mathbb R)$. Then $p(x)\in \textsf{W}^\perp$ if and only if $$\langle p(x),x-1\rangle=\int_{-1}^1 p(x)(x-1)dx=0$$ $$\langle p(x),x^2+3\rangle=\int_{-1}^1 p(x)(x^2+3)dx=0$$ at the same time, since $\{x-1,x^2+3\}$ is a basis for $\textsf{W}$. Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$ for some scalars $a_0,a_1,a_2,a_3$. Then, the two above equations are equivalent to $$\left\{\begin{align} -2a_0+\frac{2}{3}a_1-\frac{2}{3}a_2+\frac{2}{5}a_3=0 \\ \frac{20}{3}a_0+\frac{12}{5}a_2=0 \end{align}\right.$$ (after perform every integral) and solving this last system gives us $$a_2=-\frac{25}{9}a_0, \qquad a_3=\frac{10}{27}a_0-\frac{5}{3}a_1$$ Making $a_0=t$ and $a_1=s$, we can guarantee that $$\begin{align} \textsf{W}^\perp&=\left\{ \left(\frac{10}{27}t-\frac{5}{3}s\right)x^3-\frac{25}{9}tx^2+sx+t:\, s,t\in \mathbb R\right\} \\ &=\operatorname{span}\left(\left\{ \frac{10}{27}x^3-\frac{25}{9}x^2+1,-\frac{5}{3}x^3+x \right\}\right) \end{align}$$

Answer 3

You are looking for two linearly independent polynomials such that $$\int_{-1}^{1}(x-1)p(x)=0$$ and $$\int_{-1}^{1}(x^2+3)p(x)=0$$

Assuming $$p(x)= a+bx+cx^2+dx^3$$ we get $$\int_{-1}^{1}(x-1)p(x)=-2a+(2/3)b-(2/3)c+(2/5)d=0$$ and $$\int_{-1}^{1}(x^2+3)p(x)dx= (20/3)a+(12/5)c=0$$

The polynomials $$(3/2)x-(5/2)x^3$$ and $$(-1/2)+(3/2)x^2$$ satisfy the orthogonality conditions.

Thus you may choose the $$\{ 3x-5x^3, -1+3x^2\} $$ as your basis.