Let $\prod_3$ be the space of polynomials of degree at most 3 and define the following scalar product on $\prod_3$:

< $a_0$ + $a_1x$ + $a_2x^2$ + $a_3x^3$, $b_0$ + $b_1x$ + $b_2x^2$ + $b_3x^3$ > = $a_0b_0$ + $a_1b_1$ + $a_2b_2$ + $a_3b_3$ .

Define the hyperplane H = { p $\in$ $\prod_3$ : p(1) = 0 }.

  1. Find a basis of H.
  2. Find an orthonormal basis of H.
  3. Find the orthogonal projection on H of the polynomials p and q given by p(x) = x and q(x) = x-1.
  4. Find the distance of p to H.

I am having a hard time understanding how H works. How exactly do I go about finding the basis of H?

  • 1
    $\begingroup$ It might help to realize that your $\Pi_3$ is isomorphic to the vector space $\mathbb{R}^4$ with the usual inner product. Would you be able to find a basis for the hyperplane $\{\vec{x}: x_1 + x_2 + x_3 + x_4 = 0\} \subset \mathbb{R}^4$ ? $\endgroup$ – pjs36 Feb 17 '15 at 20:23
  • $\begingroup$ Yes, I am able to find the basis for that, but I'm not sure what it means for $\Pi_3$ to be isomorphic. How will the hyperplane look then? $\endgroup$ – And Feb 17 '15 at 21:07
  • $\begingroup$ Saying $\Pi_3$ and $\mathbb{R}^4$ are isomorphic basically just means that they behave exactly the same, the elements are only called different things. In this example, the polynomial $a_0 + a_1x + a_2x^2 + a_3x^3$ behaves as a vector in $\Pi_3$, just like the vector $\langle a_0, a_1, a_2, a_3\rangle$ in $\mathbb{R}^4$. $\endgroup$ – pjs36 Feb 17 '15 at 21:19
  • $\begingroup$ Many thanks for your help. Do you know what is meant by p(1) = 0? I am assuming it means plugging 1 into p(x) ($a_0$ + $a_1x$ + $a_2x^2$ + $a_3x^3$) yields 0. $\endgroup$ – And Feb 17 '15 at 21:31
  • $\begingroup$ Yes, exactly, and you're quite welcome. It just means that $p(1) = a_0 + a_1(1) + a_2(1)^2 + a_3(1)^3 = 0$. Hopefully that will explain why I brought up finding a basis for the hyperplane $\{\vec{x}: x_1 + x_2 + x_3 + x_4 = 0\} \in \mathbb{R}^4$. $\endgroup$ – pjs36 Feb 17 '15 at 21:37

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