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Question:

Let $E = \Bbb{R}_2[X]$ an euclidian space with a dot product $\left\langle P,Q\right\rangle \ = \int^1_0 {P(t)Q(t)dt}$.

Calculate the distance from $X^2$ to $\operatorname{span}(1,X)$.


Answer:

Let $(a,b) \in \Bbb{R}^2$. $p$ is the orthogonal projector. We have:

\begin{equation} a + bX = p^\perp_{\operatorname{span}(1,X)}(X^2) \iff \left\{\begin{array}{@{}l@{}} \langle X^2 - (a+bX), 1\rangle = 0\\ \langle X^2 - (a+bX), X\rangle = 0\\ \end{array}\right.\, \iff \left\{\begin{array}{@{}l@{}} a = -1/6\\ b = 1\\ \end{array}\right.\,. \end{equation}

So the wanted distance is $\left\|X^2-(X-1/6)\right\| = \frac{1}{6\sqrt 5}$.


I am a little bit lost since I don't understand how $X^2$ can be projected on $\operatorname{span}(1, X)$.

Since $X^2$ is not a linear combination of $\operatorname{span}(1, X)$.

  • Why $a + bX$ is not equal to zero?

  • How the dot products $\langle X^2 - (a+bX), 1\rangle \ = 0$ and $\langle X^2 - (a+bX), X\rangle \ = 0$ helps to get the projector? It means that we need to gets $a$ and $b$ where $\left[X^2 - (a+bX)\right] \perp 1$, why?

I would like to have great explanations (like if I didn't have the answer).

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2 Answers 2

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Think about $\mathbb R^3$ for a moment. Suppose you have a plane $P$ spanned by two vectors $u, v$ and you have a third vector $w$ which is not in the plane but is also not orthogonal to the plane. Then it makes sense to project $w$ onto the plane, right? But $w$ is not a linear combination of $u$ and $v$ because $w$ is not contained in the plane. It's the same idea here: as long as $X^2$ is not orthogonal to $\mathrm{span}\{1, X\}$ then $X^2$ has a non-zero projection onto that plane.

Now what does "orthogonal projection" mean? It means precisely the point $s = p^\perp_P(w)$ contained in $P = \mathrm{span}\{u,v\}$ such that the line from $s$ to $w$ is orthogonal to $P$. "The line from $s$ to $w$" is parallel to the vector $w - s$, so that means $$ \langle w-s, u\rangle = 0\quad \langle w-s, v\rangle = 0 \tag{$*$} $$ because $u,v \in P$.

Now apply this when $u = 1$, $v = X$, and $w = X^2$. What the answer does is write $s = a + bX$ for some unknown scalars $a, b$; we can do this because $s \in P = \mathrm{span}\{1, X\}$. Then the equations ($*$) above give us two linear equations for the unknowns $a, b$ which we can then solve. This gives us the orthogonal projection $s = p^\perp_P(X^2)$, and finally the distance between $X^2$ and $P$ is defined to be the distance between $X^2$ and $s$, which is $||X^2 - s||$.

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    $\begingroup$ Good job (+1). I was about to start typing something along these lines. Thanks for saving me the trouble :) $\endgroup$ Commented Feb 23 at 16:43
  • $\begingroup$ Thank you very much for the explanations and the drawing by @Stéphane Jaouen, I understand better, I will study your answer and try to understand perfectly. $\endgroup$
    – Leau
    Commented Feb 23 at 16:58
  • $\begingroup$ I can show you a calculation or two if you want, it's up to you to finish the exercise. $\endgroup$ Commented Feb 23 at 17:04
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    $\begingroup$ I would appreciate @StéphaneJaouen :) $\endgroup$
    – Leau
    Commented Feb 23 at 17:09
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Not much to add to @Nicolas Todoroff. Maybe a drawing that will save OP from possible difficulties with his imagination... :) enter image description here


As the answers have already been given, the following calculations are to help OP complete his exercise:

As @Nicolas explained, we need to have $X^2-(bX+a)$ orthogonal to $\color{grey}{span(}\color{red}1,\color{green}X\color{grey})$, ie $$\langle X^2-bX-a,\color{red}1\rangle=0 $$ for example. This is equal to $$\int_{0}^{1}(t^2-bt-a)\color{red}1dt=\frac{1}{3}-\frac b2-a=0$$ Hence $$3b+6a=2 $$

Then $$\langle X^2-bX-a,\color{green}X\rangle=0 \iff 6a+4b=3$$ Thus, $$\begin{cases} 6a+3b=2\\6a+4b=3 \end{cases} \iff \begin{cases} a=\frac{-1}6\\b=1 \end{cases}$$

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