Find the basis for the orthogonal complement $U^{\perp}$

Question

Question:

Let $P_{3}(\mathbb{R})$ have the standard inner product and $U$ be the subset spanned by the two vectors (which are polynomials) $u_{1}=1+2x-3x^2$ and $u_{2}=x-x^2+2x^3$. Find the basis for the orthogonal complement $U^{⊥}$.

I honestly have no idea how to approach this question. I know what orthogonal complement and a basis are but I don't understand where to begin or even solve this question. Any help would be much appreciated. Thanks in advance.

Answer 1

HINT

Every finite dimensional inner product vector space $V$ admits the decomposition $V = U\oplus U^{T}$, where $U$ is a linear subspace of $V$.

At your case, $V = P_{3}(\textbf{R})$ and $U = \text{span}\{1 + 2x - 3x^{2}, x - x^{2} + 2x^{3}\}$.

In order to find the basis for the orthogonal complement $U^{T}$, consider the inner product defined on $V$ according to \begin{align*} \langle f(x),g(x)\rangle = \int_{0}^{1}f(x)g(x)\mathrm{d}x \end{align*}

Since $\dim W_{1} + \dim W_{2} = \dim(W_{1}+W_{2}) + \dim(W_{1}\cap W_{2})$, we conclude that \begin{align*} \dim V = \dim(U\oplus U^{T}) & = \dim U + \dim U^{T} - \dim(U\cap U^{T}) = \dim U + \dim U^{T} \end{align*} Given that $\dim V = 4$ and $\dim U = 2$, the basis of $U^{T}$ consists of two vectors. With the purpose of finding $u_{3}$ and $u_{4}$ that spans $U^{T}$, it suffices to solve the system of equations obtained from \begin{align*} \langle u_{1},u\rangle = \langle u_{2},u\rangle = 0 \end{align*} where $u\in P_{3}(\textbf{R})$. Can you take it from here?

Answer 2

Here is a laborious way:

You know $\dim P_3 = 4$ and one basis is $e_k(x) = x^k$, $k=0,...,3$.

Now apply Gram Schmidt to the collection $u_1,u_2,e_1,e_2,e_3,e_4$ (discarding any elements if the Gram Schmidt process reduces it to zero). Suppose the result is $v_1,v_2,v_3,v_4$, then $v_3,v_4$ will span $U^\bot$.