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).

7. Compute the orthogonal projection of q onto the subspace spanned by p, for p and q in Exercise 3.

Short answer

The orthogonal projection is \(\frac{{56}}{{25}} + \frac{{14}}{{25}}t\).

Step-by-step solution

Write the results from Exercise 3

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

\(\begin{align*}p\left( 0 \right) &= 4 + 0\\ &= 4\end{align*}\)

\(\begin{align*}p\left( 1 \right) &= 4 + 1\\ &= 5\end{align*}\)

And,

\(\begin{align*}q\left( { - 1} \right) &= 5 - 4{\left( { - 1} \right)^2}\\ &= 5 - 4\\ &= 1\end{align*}\)

\(\begin{align*}q\left( 0 \right) &= 5 - 4{\left( 0 \right)^2}\\ &= 5\end{align*}\)

\(\begin{align*}q\left( 1 \right) &= 5 - 4{\left( 1 \right)^2}\\ &= 5 - 4\\ &= 1\end{align*}\)

Find the inner product of q and p

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

\(\begin{align*}\left\langle {q,p} \right\rangle & = \left\langle {p,q} \right\rangle \\ &= p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 0 \right)q\left( 0 \right) + p\left( 1 \right)q\left( 1 \right)\\ &= \left( 3 \right)\left( 1 \right) + \left( 4 \right)\left( 5 \right) + \left( 5 \right)\left( 1 \right)\\ &= 28\end{align*}\)

Find the inner product of p and p 

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

\(\begin{align*}\left\langle {p,p} \right\rangle &= p\left( { - 1} \right)p\left( { - 1} \right) + p\left( 0 \right)p\left( 0 \right) + p\left( 1 \right)p\left( 1 \right)\\ &= \left( 3 \right)\left( 3 \right) + \left( 4 \right)\left( 4 \right) + \left( 5 \right)\left( 5 \right)\\ &= 9 + 16 + 28\\ &= 50\end{align*}\)

Find the orthogonal projection of q onto subspace spanned by p

The orthogonal projection can be calculated as follows:

\[\begin{align*}\hat q &= \frac{{\left\langle {q,p} \right\rangle }}{{\left\langle {p,p} \right\rangle }}p\\ &= \frac{{28}}{{50}}\left( {4 + t} \right)\\&= \frac{{56}}{{25}} + \frac{{14}}{{25}}t\end{align*}\]

Thus, the orthogonal projection is \(\frac{{56}}{{25}} + \frac{{14}}{{25}}t\).