Question

Exercise 3-8 refer to \({{\bf{P}}_{\bf{2}}}\) with the inner product given by evaluation at \( - {\bf{1}}\), 0, and 1. (See Example 2).

8. Compute the orthogonal projection of q onto the subspace spanned by p, for p and q in Exercise 4.

Short answer

The orthogonal projection is \( - \frac{3}{2}t + \frac{1}{2}{t^2}\).

Step-by-step solution

Write the results from Exercise 4

\(\begin{align*}p\left( { - 1} \right) &= 3\left( { - 1} \right) - {\left( { - 1} \right)^2}\\ &= - 3 - 1\\ &= - 4\end{align*}\)

\(\begin{align*}p\left( 0 \right) &= 3\left( 0 \right) - {\left( 0 \right)^2}\\ &= 0\end{align*}\)

\(\begin{align*}p\left( 1 \right) &= 3\left( 1 \right) - {\left( 1 \right)^2}\\ &= 3 - 1\\ &= 2\end{align*}\)

And,

\(\begin{align*}q\left( { - 1} \right) &= 3 + 2{\left( { - 1} \right)^2}\\ &= 5\end{align*}\)

\(\begin{align*}q\left( 0 \right) &= 3 + 2{\left( 0 \right)^2}\\ &= 3\end{align*}\)

\(\begin{align*}q\left( 1 \right) &= 3 + 2{\left( 1 \right)^2}\\ &= 5\end{align*}\)

Find the inner product of q and p

The inner product \(\left\langle {q,p} \right\rangle \) can be calculated as follows:

\(\begin{align*}\left\langle {q,p} \right\rangle &= \left\langle {p,q} \right\rangle \\ &= p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 0 \right)q\left( 0 \right) + p\left( 1 \right)q\left( 1 \right)\\ &= \left( { - 4} \right)\left( 5 \right) + \left( 0 \right)\left( 3 \right) + \left( 2 \right)\left( 5 \right)\\ &= - 10\end{align*}\)

Find the inner product of p and p

The inner product \(\left\langle {p,p} \right\rangle \) can be calcaulted as follows:

\(\begin{align*}\left\langle {p,p} \right\rangle &= p\left( { - 1} \right)p\left( { - 1} \right) + p\left( 0 \right)p\left( 0 \right) + p\left( 1 \right)p\left( 1 \right)\\ &= \left( { - 4} \right)\left( { - 4} \right) + \left( 0 \right)\left( 0 \right) + \left( 2 \right)\left( 2 \right)\\ &= 16 + 0 + 4\\ &= 20\end{align*}\)

Find the orthogonal projection of q onto subspace spanned by p

The orthogonal projection can be calculated as follows:

\(\begin{align*}\hat q &= \frac{{\left\langle {q,p} \right\rangle }}{{\left\langle {p,p} \right\rangle }}p\\ &= - \frac{{10}}{{20}}\left( {3t - {t^2}} \right)\\ &= - \frac{3}{2}t + \frac{1}{2}{t^2}\end{align*}\)

Thus, the orthogonal projection is \( - \frac{3}{2}t + \frac{1}{2}{t^2}\).