Orthogonal Complements Examples 1

# Orthogonal Complements Examples 1

Recall from the Orthogonal Complements page that if $V$ is an inner product space and $U$ is a subset of $V$ then the orthogonal complement of $U$, $U^{\perp}$ is defined to be the subspace of $V$ such that:

(1)
\begin{align} \quad U^{\perp} = \{ v \in V : <v, u> = 0 \: \forall u \in U \} \end{align}

We will now look at some examples regarding orthogonal complements.

## Example 1

**Consider the vector space $\wp_3 ( \mathbb{R})$, and define an inner product on this vector space by $<p(x), q(x)> = \int_{-1}^{1} p(x) q(x) \: dx$. Let $U = \mathrm{span} (x^2)$. Find a basis for $U^{\perp}$.**

Recall that $V = U \oplus U^{\perp}$ and so $\mathrm{dim} (V) = \mathrm{dim} (U) + \mathrm{dim} (U^{\perp})$. We have that $\mathrm{dim}(U) = 1$ clearly, and $\mathrm{dim} (\wp_3 (\mathbb{R}) = 4$ which implies that $\mathrm{dim}(U^{\perp}) = 3$.

Let $q(x) \in U$. From the definition of the orthogonal projection operator we have that

(2)
\begin{align} \quad U^{\perp} = \{ p(x) \in \wp_3 ( \mathbb{R}) : <p(x), q(x)> = 0 \: \forall q(x) \in U \} \\ \quad U^{\perp} = \{ a_0 + a_1x + a_2x^2 + a_3x^3 : <a_0 + a_1x + a_2x^2 + a_3x^3, cx^2> = 0 \: a_1, a_2, a_3, c \in \mathbb{R} \} \end{align}

We then have that:

(3)
\begin{align} \quad 0 = <a_0 + a_1x + a_2x^2 + a_3x^3, x^2> \\ \quad 0 = \int_{-1}^{1} (a_0 + a_1x + a_2x^2 + a_3x^3)(x^2) \: dx \\ \quad 0 = \int_{-1}^{1} (a_0x^2 + a_1x^3 + a_2x^4 + a_3x^5) \: dx \\ \quad 0 = \left [ a_0 \frac{x^3}{3} + a_1 \frac{x^4}{4} + a_2 \frac{x^5}{5} + a_3 \frac{x^6}{6} \right ]_{x=-1}^{x=1} \\ \quad 0 = \left ( \frac{a_0}{3} + \frac{a_1}{4} + \frac{a_2}{5} + \frac{a_3}{6} \right ) - \left ( - \frac{a_0}{3} + \frac{a_1}{4} - \frac{a_2}{5} + \frac{a_3}{6} \right ) \\ \quad 0 = \frac{2a_0}{3} + \frac{2a_2}{5} \\ \quad \frac{2a_2}{5} = -\frac{2a_0}{3} \\ \quad \frac{a_2}{5} = - \frac{a_0}{3} \\ \quad a_2 = -\frac{5 a_0}{3} \end{align}

Thus:

(4)
\begin{align} \quad p(x) = a_0 + a_1x + -\frac{5 a_0}{3}x^2 + a_3x^3 \\ \quad p(x) = a_0 \left (1 - \frac{5}{3} \right )x^2 + a_1x + a_3x^3 \end{align}

Therefore a spanning set of $U^{\perp}$ is $\left \{x, 1 - \frac{5}{3}x^2, x^3 \right \}$, and this spanning set has the right length of vectors, so it is a basis of $U^{\perp}$.