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Let $V = \mathbb{P^4}$ denote the space of quartic polynomials, with the $L^2$ inner product $$\langle p,q \rangle = \int^1_{-1} p(x)q(x)dx.$$ Let $W = \mathbb{P^2}$ be the subspace of quadratic polynomials. Find a basis for and the dimension of $W^{\perp}$.

The answer is $$t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35};\,\, \dim (W^{\perp}) =2$$

How did they get that?

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"Space of quartic polynomials"? Do you mean the space of (real, complex, rational...) polynomials of degree $\,\leq 4\,$ and zero? –  DonAntonio Nov 6 '12 at 2:57
    
@DonAntonio yes i do.. –  diimension Nov 6 '12 at 3:01
    
Which one? Real? Complex? –  EuYu Nov 6 '12 at 3:01
    
@EuYu it doesnt specify but im sure is real –  diimension Nov 6 '12 at 3:06
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It must be real or there would be a conjugate inside the inner product... –  copper.hat Nov 6 '12 at 5:06
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4 Answers

up vote 1 down vote accepted

Well, for a finite dimensional vector space we have $\dim W + \dim W^\perp = \dim V$ so that covers the dimension part. For the basis of the orthogonal complement, we have $$\int_{-1}^1 ax^4 + bx^3 + cx^2 + dx + e\ dx = 0$$ $$\int_{-1}^1 x(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$ $$\int_{-1}^1x^2(ax^4 + bx^3 + cx^2 + dx + e)\ dx = 0$$ Because the standard basis of $\mathbb{P}^2$ must satisfy the orthogonality conditions. Therefore we get $$\frac{a}{5} + \frac{c}{3} + e = 0$$ $$\frac{b}{5} + \frac{d}{3} = 0$$ $$\frac{a}{7} + \frac{c}{5} + \frac{e}{3} = 0$$ Solving this system yields $$a = \frac{35}{3}e,\ \ b=-\frac{5}{3}d,\ \ c=-10e$$ with two parameters to vary. Your solutions follows by taking $(a=0,\ b=1)$ and $(a=1,\ b=0)$ respectively.

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Thank you, but I am a bit confused I see how this is an orthogonal complement basis but how does it relate to $t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35}$?? –  diimension Nov 6 '12 at 3:22
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Take $a=0$ and $b=1$ and solve for $c,\ d,\ e$. See what you get. Remember $a,\ b,\ c,\ d,\ e$ are the coefficients of the standard basis. –  EuYu Nov 6 '12 at 3:25
    
Well, for example choose, as written clearly in the above answer, $$a=1\,,\,b=0\Longrightarrow e=\frac{3}{35}\,,\,d=0\,,\,c=-10e=-\frac{6}{7}\Longrightarrow\, \text{one of the pol's is }\, x^4-\frac{6}{7}x^2+\frac{3}{35}$$ and etc. –  DonAntonio Nov 6 '12 at 3:26
    
Yup, I understand now with you guys help! One last question, how did you know that a = 1, b = 0 , and vice versa? –  diimension Nov 6 '12 at 3:31
    
Well, the coefficients kinda have to match. Both your polynomials are monic so those are the natural choices. –  EuYu Nov 6 '12 at 3:34
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Let $v_k(x) = x^k$, $k=0,...,4$. Then $v_k$ is a basis for $\mathbb{P}^4$. (To see this note that any quartic can be written in terms of $v_k$, and if $\sum \alpha_k v_k = 0$, then by differentiating and evaluating at $x=0$ we can see that $\alpha_k = 0$, hence they are linearly independent.)

By the same token, $v_k$, $k=0,1,2$ is a basis for $\mathbb{P}^2$. It follows that $\dim \mathbb{P}^2 = 3$, and since $\mathbb{P}^4 = \mathbb{P}^2 \oplus (\mathbb{P}^2)^\bot$, we see that $\dim (\mathbb{P}^2)^\bot = 2$.

We can find a basis for $(\mathbb{P}^2)^\bot$ by projecting the $v_k$ onto $(\mathbb{P}^2)^\bot$. Clearly $v_k$, $k=0,1,2$ will project to zero. So the only remaining basis elements that need to be projected are $v_3,v_4$.

Note in passing that $\langle v_j, v_k \rangle = \frac{1}{j+k+1}(1-(-1)^{j+k+1})$.

To compute the projection of $x$ onto $(\mathbb{P}^2)^\bot$, we need to determine $\alpha \in \mathbb{R}^3$ such that $\langle x-\sum_{k=0}^2 \alpha_k v_k, v_j \rangle = 0$ for $j=0,1,2$. This is just the linear system $\langle x, v_j \rangle = \langle \sum_{k=0}^2 \alpha_k v_k, v_j \rangle$, or $$ \begin{bmatrix} \langle v_0, v_0 \rangle & \langle v_1, v_0 \rangle & \langle v_2, v_0 \rangle \\ \langle v_0, v_1 \rangle & \langle v_1, v_1 \rangle & \langle v_2, v_1 \rangle \\ \langle v_0, v_2 \rangle & \langle v_1, v_2 \rangle & \langle v_2, v_2 \rangle \end{bmatrix} \alpha = \begin{bmatrix} \langle x, v_0 \rangle \\ \langle x, v_1 \rangle \\ \langle x, v_2 \rangle \end{bmatrix} $$ Grinding through the computations gives $\alpha = \frac{1}{5} (0,3,0)^T$ when $x=v_3$ and $\alpha = \frac{1}{35} (-3, 0, 30)^T$ when $x=v_4$.

Hence a basis for $(\mathbb{P}^2)^\bot$ is $x \mapsto x^3-\frac{3}{5}x$, $x \mapsto x^4+\frac{3}{35}-\frac{6}{7}x^2$.

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Wow, thank you very much! So there are many ways to do it, instead of just taking the $L^2$ inner product, i could have instead used gram matrix. Thank you for pointing that out , Copper.hat –  diimension Nov 6 '12 at 4:00
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You are welcome. Note that you are still taking the inner products, they just happen to be easy to compute for my particular choice of basis, $v_k$. –  copper.hat Nov 6 '12 at 5:08
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Let

$$p(x):=ax^4+bx^3+cx^2+dx+e\in W^\perp$$

Since $W:=\operatorname{Span}\{1,x,x^2\}\,$ , we get:

$$(1)\;\;\;\;\;\;0=\langle\,p\,,\,1\,\rangle=\int_{-1}^1p(x)dx=\frac{2}{5}a+\frac{2}{3}c+2e$$

$$(2)\;\;\;\;\;\;\;\;\;\;\;0=\langle\,p\,,\,x\,\rangle=\int_{-1}^1xp(x)dx=\frac{2}{5}b+\frac{2}{3}d$$

$$(3)\;\;\;\;\;\;\;\;0=\langle\,p\,,\,x^2\,\rangle=\int_{-1}^1x^2p(x)dx=\frac{2}{7}a+\frac{2}{5}c+\frac{2}{3}e$$

The above relies on the easy results that the integral on a symmetric (above zero) interval of an even function is twice the value of its primitive on either of the two limits, whereas the same integral of an odd function is zero.

Now solve the above linear system.

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Thank you, but I am a bit confused I see how this is an orthogonal complement basis but how does it relate to $t^3 - \frac{3}{5}t, t^4 - \frac{6}{7}t^2 + \frac{3}{35}$?? –  diimension Nov 6 '12 at 3:22
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Read my comment below Euyu's answer: it applies exactly the same here. –  DonAntonio Nov 6 '12 at 3:27
    
I understand now with guys help. Thank you very much , DonAntonio! –  diimension Nov 6 '12 at 3:35
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Well, one way would be to use Gram-Schmidt to produce an orthogonal basis, starting with the basis $1$, $t$, $t^2$, $t^3$, $t^4$. The result will be five polynomials $p_r(x)$ for $r=0,\ldots,4$ where $p_r$ has degree $r$. So $p_0$, $p_1$ and $p_2$ will span the space of quadratic polynomials and $p_3$ and $p_4$ will span a 2-dimensional space orthogonal to the quadratics. Since we know that the dimension of $W^\perp$ is two (because $\mathrm{dim}(W)+\mathrm{dim}(W^\perp)=5$), we see that $p_3$ and $p_4$ are a basis for $W^\perp$.

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Thank you for explaining that. –  diimension Nov 6 '12 at 3:23
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