# Find the basis for the orthogonal complement $U^{⊥}$

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.

• What is the inner product? I am not sure I am aware of a standard one for polynomials of a fixed $\max$ degree. – copper.hat Mar 26 at 3:51
• It is a good question. I have assumed that $$\langle f(x),g(x)\rangle = \int_{0}^{1}f(x)g(x)\mathrm{d}x$$ – user1337 Mar 26 at 3:59

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 $$U^\bot$$.

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?