Finding an orthogonal basis w/ inner products

Question

Problem

I am having trouble solving this problem. I can't find a solution and am doubting if I am right. Im thinking I let q = p2 - proj[p0, p1]p2, since that would be orthogonal to both p1 and p0. So I got q = t^2-5 and thus [p0,p1, q] should be the orthogonal basis spanned by [p0, p1, p2]. Is this correct?

Answer 1

Heere $\mathbb{P}_3$ denotes the vector space of all the real polynomials of degree not exceding $3$ over the field $\mathbb{R}$ of real numbers.

Let $W$ be the subspace of $\mathbb{P}_3$ spanned by the polynomials $p_0(t) = 1$, $p_1(t) = t$, and $p_2(t) = t^2$. Then $W$ is the subspace of all the real polynomials of degree at most $2$.

For any polynomials $p, q \in \mathbb{P}_3$, we define $$ \langle p, q \rangle \colon= p(-3) q(-3) + p(-1)q(-1) + p(1)q(1) + p(3) q(3). $$ And it can be easily shown that this formula indeed defines an inner product on $\mathbb{P}_3$.

We apply the Gram-Schmidt orthogonalization process on the ordered set $\left( p_0, p_1, p_2 \right)$ to obtain the ordered set $\left( e_0, e_1, e_2 \right)$ such that the set $\left( e_0, e_1, e_2 \right)$ is orthonormal and we also obtain $$ \mbox{ span } \left\{ p_0 \right\} = \mbox{ span } \left\{ e_0 \right\}, $$ $$ \mbox{ span } \left\{ p_0, p_1 \right\} = \mbox{ span } \left\{ e_0, e_1 \right\}, $$ and $$ \mbox{ span } \left\{ p_0, p_1, p_2 \right\} = \mbox{ span } \left\{ e_0, e_1, p_2 \right\}. $$

Now we note that $$ \left\langle p_0, p_0 \right\rangle = \left(p_0(-3) \right)^2 + \left(p_0(-1) \right)^2 + \left(p_0(1) \right)^2 + \left(p_0(3) \right)^2 = 1 + 1 + 1 + 1 = 4, $$ and so $$ \left\lVert p_0 \right\rVert = \sqrt{ \left\langle p_0, p_0 \right\rangle } = 2. $$ Let us put $$ e_0 \colon= \frac{1}{ \left\lVert p_0 \right\rVert } p_0. $$ Then $e_0$ is a "unit vector" in $\mathbb{P}_3$, and the formula for $e_0$ is given by $$ e_0 (t) = \frac{1}{2}. \tag{Definition 0} $$

Then $$ \begin{align} \left\langle p_1, e_0 \right\rangle &= p_1(-3) e_0(-3) + p_1(-1) e_0(-1) + p_1(1) e_0(1) + p_1(3) e_0(3) \\ &= (-3) \frac12 + (-1) \frac12 + (1) \frac12 + (3) \frac12 \\ &= 0, \end{align} $$ and so $$ p_1^\prime \colon= p_1 - \left\langle p_1, e_0 \right\rangle e_0 = p_1. $$ Now we find that $$ \begin{align} \left\langle p_1^\prime, p_1^\prime \right\rangle &= \left\langle p_1, p_1 \right\rangle \\ &= \left( p_1(-3) \right)^2 + \left( p_1(-1) \right)^2 + \left( p_1(1) \right)^2 + \left( p_1(3) \right)^2 = (-3)^2 + (-1)^2 + (1)^2 + (3)^2 \\ &= 20, \end{align} $$ and thus $$ \left\lVert p_1^\prime \right\rVert = \sqrt{ \left\langle p_1^\prime, p_1^\prime \right\rangle } = \sqrt{ 20 } = 2 \sqrt{5}. $$ Let us now put $$ e_1 \colon= \frac{1}{ \left\lVert p_1^\prime \right\rVert } p_1^\prime = \frac{ 1 }{ 2 \sqrt{5} } p_1 $$ Then $e_1$ is also a unit vector in $\mathbb{P}_3$ and such that $$ \left\langle e_0, e_1 \right\rangle = 0; $$ moreover $e_1$ is given by the formula $$ e_1(t) = \frac{1}{2 \sqrt{5} } t. \tag{Definition 1} $$

Next we find $$ \begin{align} \left\langle p_2, e_0 \right\rangle &= p_1(-3) e_0(-3) + p_1(-1) e_0(-1) + p_1(1) e_0(1) + p_1(3) e_0(3) \\ &= (-3)^2 \frac12 + (-1)^2 \frac12 + (1)^2 \frac12 + (3)^2 \frac12 \\ &= 10, \end{align} $$ and $$ \begin{align} \left\langle p_2, e_1 \right\rangle &= p_2(-3) e_1(-3) + p_2(-1) e_1(-1) + p_2(1) e_1(1) + p_2(3) e_1(3) \\ &= (-3)^2 \frac1{2\sqrt{5}}(-3) + (-1)^2 \frac1{2\sqrt{5}} (-1) + (1)^2 \frac1{2\sqrt{5}}(1) + (3)^2 \frac1{2\sqrt{5}}(3) \\ &= 0, \end{align} $$ and then $$ p_2^\prime \colon= p_2 - \left\langle p_2, e_0 \right\rangle e_0 - \left\langle p_2, e_1 \right\rangle e_1 = p_2 - 10 e_0 - 0 e_1 = p_2 - 10 e_0, $$ and as $$ \begin{align} \left\langle p_2^\prime, p_2^\prime \right\rangle &= \left( p_2^\prime(-3) \right)^2 + \left( p_2^\prime(-1) \right)^2 + \left( p_2^\prime(1) \right)^2 + \left( p_2^\prime(3) \right)^2 \\ &= \left( (-3)^2 - 10 \times \frac{1}{2} \right)^2 + \left( (-1)^2 - 10 \times \frac{1}{2} \right)^2 + \left( (1)^2 - 10 \times \frac{1}{2} \right)^2 + \left( (3)^2 - 10 \times \frac{1}{2} \right)^2 \\ &= 64, \end{align} $$ and thus $$ \left\lVert p_2^\prime \right\rVert = \sqrt{ \left\langle p_2^\prime, p_2^\prime \right\rangle } = 8. $$ Let us now put $$ e_2 \colon= \frac1{ \left\lVert p_2^\prime \right\rVert } p_2^\prime = \frac18 p_2^\prime = \frac18 \left( p_2 - 10 e_0 \right). $$ This $e_2$ is a unit vector in $\mathbb{P}_3$ and is orthogonal to both $e_0$ and $e_1$; moreover $e_2$ is given by the formula $$ e_2(t) = \frac18 \left( t^2 - 5 \right) = \frac{1}{8} t^2 - \frac58. \tag{Definition 2} $$

The vectors $e_0$, $e_1$, and $e_2$ given by the formulas in (Definition 0), (Definition 1), and (Definition 2), respectively, are the vectors in our desired orthonormal basis for the subspace of $\mathbb{P}_3$ spanned by the vectors $p_0$, $p_1$, and $p_2$.

Hope this helps.

Answer 2

You can apply the Gram-Schmidt process to get your basis. Fortunately, $p_0$ and $p_1$ are orthogonal, and so are $p_1$ and $p_2$; this simplifies the computations. If you apply this process to $\{p_0,p_1,p_2\}$, you will get an orthogonal basis (which is even orthonormal) is$$\left\{\frac12,\frac t{2\sqrt5},\frac18(t^2-5)\right\}.$$A simpler answer (orthogonal but not orthonormal) is $\{1,t,t^2-5\}$.