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$V$ is the vector space $\mathbb{R}_{2}\left [ X \right ]$ of polynomials of degree at most 2 in $X$ with real coefficients.
$\left \langle f(X),g(X) \right \rangle$ is the inner product $\sum_{i=1}^{3}f(i)g(i)$
Let $W$ be the subspace of $V$ spanned by polynomials $1$ and $X$.
Find polynomials $p(X)\in W$ and $q(X)\in W^\perp$ such that $X^2=p(X)+q(X)$

So if $W$ is spanned by $1$ and $X$, $W^\perp$ must be spanned by $X^2$ since $V=W\oplus W^\perp$
This seems too easy, but can I just have $p(X)=0$ and $q(X)=X^2$ as then it fulfills all the conditions?

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Are you sure that $X^2$ is orthogonal to $X$ for the inner product you chose? Isn't there anything in your classnotes which allows you to make $X^2$ orthogonal to $1$ and $X$ ? – Louis La Brocante Jan 5 '13 at 23:52
Oh I see, but I don't know how to find something orthogonal to 1 and X? – user51327 Jan 5 '13 at 23:59
@user51897 Do you know the Gram-Schmidt process? You start off with any basis, and remove portions that are parallel to previous basis elements, to obtain an orthogonal basis. – Calvin Lin Jan 6 '13 at 0:09
@CalvinLin I've heard of it but we haven't been taught it yet. I'll go look it up, thanks! :) – user51327 Jan 6 '13 at 9:40
up vote 1 down vote accepted

Now, any vector perpendicular to $\,W:=\operatorname{Span}\{1,x\}\,$ must be perpendicular to both of $\,1,x\,$ , so we look for $\,q(x):=ax^2+bx+c\in\Bbb R_2[x]\,$ s.t.:

$$I\;\;\;\;0=\langle 1,q(x)\rangle=q(1)+q(2)+q(3)\Longrightarrow 14a+6b+3c=0$$

$$II\;\;\;\;0=\langle x,q(x)\rangle=q(1)+2q(2)+3q(3)\Longrightarrow36a+14b+6c=0$$

Multiplying $\,I\,$ by $\,-2\,$ and adding the result to $\,II\,$ we get:

$$-2I+II=8a+2b=0\Longrightarrow b=-4a\,\,,\,\,\text{subst. in}\,\,I\;\;\;3c=24a-14a\Longrightarrow c=\frac{10}{3}a$$

and we certainly get a $\,1-$dimensional subspace, as expected:

$$W^\perp=\{3ax^2-12ax+10a\;\;;\;\;a\in\Bbb R\}=\operatorname{Span}\{3x^2-12x+10\}\,$$

I'll leave it to you now to find $\,rx+s\in W\;\;,\;3ax^2-12ax+10a\in W^\perp\,$ s.t. their sum equals what you want, namely $\,x^2\,$ ...

BTW, check that indeed $\,x^2\notin W^\perp\,$, as Rolando noted, but of course it must appear in the expression for $\,W^\perp\,$

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