Question

Consider P2 with the inner product given by evaluation at -1, 0, and 1. That is < p, q >= p(−1)q(−1) + p(0)q(0) + p(1)q(1).
Let p(t) = 3t − t2 and q(t) = 3 + 2t2 .
A) Compute < p, q >, ||p||, and ||q||.
B) Compute the orthogonal projection of q onto the subspace spanned by p.

Answer 1

(A)

Evaluate $p$ and $q$ at the given values of $t$, then plug them into the inner product:

$p(t)=3t-t^2\implies\begin{cases}p(-1)=-4\\p(0)=0\\p(1)=2\end{cases}$

$q(t)=3+2t^2\implies\begin{cases}q(-1)=5\\q(0)=3\\q(1)=5\end{cases}$

Now,

$\langle p,q\rangle=4\cdot5+0\cdot3+2\cdot5=30$

$\|p\|=\sqrt{\langle p,p\rangle}=(-4)^2+0^2+2^2=20$

$\|q\|=\sqrt{\langle q,q\rangle}=5^2+3^2+5^2=59$

(B)

Let $V$ denote the subspace spanned by $p$. We need an orthonormal basis for $V$. Since

$p(t)=3t-t^2=3\cdot t+(-1)\cdot t^2$

we have the basis vectors $\{t,t^2\}$; normalize these vectors to get the orthonormal basis,

$\left\{\dfrac t{|t|},\dfrac{t^2}{|t^2|}\right\}=\left\{\dfrac t{|t|},1\right\}$

Then the projection of $q$ onto $V$ is

$\mathrm{proj}_Vq(t)=\left(q(t)\cdot\dfrac t{|t|}\right)\dfrac t{|t|}+\left(q(t)\cdot1\right)1$

$\mahtrm{proj}_Vq(t)=\dfrac{3t^2+2t^4}{t^2}+(3+2t^2)=\boxed{6+4t^2}$

Answer 2

A).

$< p, q >=-10\\||p||=9\sqrt{2} \\||q||=3\sqrt{2}$

B).The orthogonal projection of q onto the subspace spanned by p is$proj_vq(t)=6+4t^2$

We have $p(t) = 3t - t^2 ,q(t) = 3 + 2t^2$

And < p, q >= p(−1)q(−1) + p(0)q(0) + p(1)q(1).

Now,

A).

$p(-1) = 3.(-1) - (-1)^2 \\=-4\\p(1)=3.1-1^2\\=2q(-1) = 3 + 2.(-1)^2\\=5\\q(1) = 3 + 2.(1)^2\\=5p(0) = 3.0 - 0^2 \\=0\\q(0) = 3 + 2.0^2\\=3\\< p, q >= p(-1)q(-1) + p(0)q(0) + p(1)q(1)\\< p, q >=-4.5+0.3+2.5\\=-10\\||p||=<p,p>\\=\sqrt{(-4)^2+0+2^2} \\=\sqrt{20} \\=2\sqrt{5} \\||q||=<q,q>\\=\sqrt{5^2+3^2+5^2} \\=\sqrt{59}$

B)Let V be the subspace spanned by p.

Now, an orthonormal basis for p is$(t,t^2)$

 Then the projection of q onto the subspace spanned by p is

$proj_vq(t)=(q(t).\frac{t}{|t|} ).\frac{t}{|t|} +(q(t).1)1\\proj_vq(t)=\frac{3t^2+2t^4}{t^2} +(3+2t^2)\\proj_vq(t)=6+4t^2$

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