1
$\begingroup$

Consider $P_2$ together with the inner product $(p(x), q(x)) = p(0)q(0) + p(1)q(1) + p(2)q(2)$. Find the projection of $p(x)=x$ onto the subspace $W=\operatorname{span}\{-x+1,x^{2}+2\}$.

How do you solve this question? I don't get what it means to find the projection of $p(x)=x$ onto $W$. I'm only used to finding the projection of a vector onto $W$.

I did, however, find the orthogonal basis for the subspace being $\{-x+1, x^{2}-2x+4\}$. But I don't know how to apply this to the projection equation.

Can someone please help me?

  • 0
    If you have an **orthonormal** basis $\beta=\{v_1,\ldots,v_k\}$ for a subspace, then given a vector $u$, the projection onto the subspace of $u$ is $\langle u,v_1\rangle v_1 + \cdots + \langle u,v_k\rangle v_k$. If you have an **orthogonal** basis, then you can turn it into an orthonormal basis by dividing each vector by its norm.2012-06-24
  • 0
    P.S. "I'm only used to finding the projection of a vector onto $W$". $p(x)$ **is** a vector of the vector space $P_2$.2012-06-24
  • 0
    I geuss $W$ is a subspace of $P_2$ with orthogonal basis ${−x+1,x^{2}−2x+4}$2012-06-24
  • 0
    @Erik: Before it was stated what $W$ was (the problem was just edited), the alleged orthogonal basis appeared *ex nihilo*; no way to verify if it was correct or not.2012-06-24

1 Answers 1

1

Suppose $\mathbf{V}$ is an inner product vector space, and $\mathbf{W}$ is a subspace. If $\beta=\{\mathbf{w}_1,\ldots,\mathbf{w}_k\}$ is an orthonormal basis for $\mathbf{W}$, then the orthogonal projection onto $\mathbf{W}$ can be computed using $\beta$: given a vector $\mathbf{v}$, the orthogonal projection onto $\mathbf{W}$ is $$\pi_{\mathbf{W}}(\mathbf{v}) = \langle \mathbf{v},\mathbf{w}_1\rangle \mathbf{w}_1+\cdots + \langle \mathbf{v},\mathbf{w}_k\rangle \mathbf{w}_k.$$

If you only have an orthogonal basis, then you need to divide each factor by the square of the norm of the basis vectors. That is, if you have an orthogonal basis $\gamma = \{\mathbf{z}_1,\ldots,\mathbf{z}_k\}$, then the projection is given by: $$\pi_{\mathbf{W}}(\mathbf{v}) = \frac{\langle\mathbf{v},\mathbf{z}_1\rangle}{\langle \mathbf{z}_1,\mathbf{z}_1\rangle}\mathbf{z}_1 + \cdots + \frac{\langle\mathbf{v},\mathbf{z}_k\rangle}{\langle\mathbf{z}_k,\mathbf{z}_k\rangle}\mathbf{z}_k.$$

Here, you have a subspace for which you say you already have an orthogonal basis. And you have your vector: $\mathbf{v} = x$. So all you have to do is use the usual formula with these vectors and this inner product. For example, with $\mathbf{v}=x$ and $\mathbf{z}_1 = -x + 1$, we have: $$\langle x,-x+1\rangle = (0)(-0+1) + (1)(-1+1) + (2)(-2+1) = 0+0-2 = -2.$$

Etc.

  • 0
    Btw, for $z_2$=$x^{2}-2x+4$, would it be $\frac{11}{41}$($x^{2}$-2x+4}?2012-06-24
  • 0
    **What** would "it" be? There are so many things to compute, I don't know which one you mean. The projection of $x^2-2x+4$? The normalization of $x^2-2x+4$ onto $\mathbf{W}$? The inner product of $\mathbf{v}$ with $\mathbf{z}_1$? The projection of $\mathbf{v}$ onto $\mathrm{span}(\mathbf{z}_1)$? Something else? Impossible to tell.2012-06-24
  • 0
    Sorry for being so ambiguous. I meant the projection of $x^{2}-2x+4$ on W.2012-06-24
  • 0
    @Ashely: Since $x^2-2x+4$ is *in* $\mathbf{W}$, it's orthogonal projection on $\mathbf{W}$ is itself. So if you really meant that, then I have no idea what you calculated or where $\frac{11}{41}$ came from.2012-06-24
  • 0
    I guess I mixed it up. The inner product of v with $z_2$=$x^{2}-2x+4$ would be what I was to say.2012-06-24
  • 0
    @Ashley: You are still mixed up: the inner product of two vectors must be a scalar. What you wrote was a vector. So it mostcertainly cannot be "the inner product of $v$ with $\mathbf{z}_2$". (If you don't know what the thing you are writing down is, then it is hardly surprising you are having so much trouble getting these problems right...)2012-06-24
  • 0
    I guess I should review this again. What I should say is: is the projection of v onto the vector $z_2$=$x^{2}-2x+4$ $\frac{11}{41}($x^{2}-2x+4}? Or, is the inner product of v and $z_2 \frac{11}{41}$?2012-06-24
  • 0
    What you've written is the result of computing $$\frac{\langle v,\mathbf{z}_2\rangle}{\langle \mathbf{z}_2,\mathbf{z}_2\rangle}\mathbf{z}_2$$which is the orthogonal projection of $v$ onto $\mathrm{span}(\mathbf{z}_2)$.2012-06-24
  • 0
    Alright, thanks2012-06-24