Inner Product - Functions.

• Jun 17th 2008, 05:39 PM
pearlyc
Inner Product - Functions.
Show that the following does not define an inner product on $\displaystyle R^3$:
$\displaystyle <(x,y,z),(a,b,c)> = xa + 2xb + 2ay + yb + 4zc$

Define an inner product on $\displaystyle P_2$ by

$\displaystyle <p,q> = \int^1_{-1} p(x)q(x) dx$

Let $\displaystyle U \subseteq P_2$ be the subspace spanned by {$\displaystyle x,x^2$}.

(i) Find an orthornormal basis for U.
(ii) Find the polynomial in U that is as close as possible to 1. (for the norm corresponding to the above inner product)

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And a little vector space question,

Determine whether the polynomial

$\displaystyle p(x) = x^2 + x + 2$

belongs to span {$\displaystyle {p_1(x), p_2(x),p_3(x)}$} where

$\displaystyle p_1(x) = 2x^2 + x + 2, p_2(x) = x^2 - 2x, p_3(x) = 5x^2 - 5x + 2$

Thanks for your help guys. (:
• Jun 17th 2008, 06:16 PM
o_O
Quote:

Originally Posted by pearlyc
Determine whether the polynomial

$\displaystyle p(x) = x^2 + x + 2$

belongs to span {$\displaystyle {p_1(x), p_2(x),p_3(x)}$} where

$\displaystyle p_1(x) = 2x^2 + x + 2, p_2(x) = x^2 - 2x, p_3(x) = 5x^2 - 5x + 2$

Your set of polynomials (vectors) span p(x) if p(x) can be written as a linear combination of the vectors $\displaystyle p_1(x), p_2(x), p_3(x)$, i.e.
$\displaystyle p(x) = ap_1(x) + bp_2(x) + cp_3(x)$
$\displaystyle x^{2} + x + 2 = (2a + b + 5c)x^{2} + (a - 2b - 5c)x + 2(a+c)$

So you have to determine whether or not the system of equations (constructed by equating the coefficients of each term) is consistent or not:
$\displaystyle \begin{array}{ccccccc} 2a & + & b & + & 5c & = & 1 \\ a & - & 2b & - & 5c & = & 1 \\ a & & & + & c & = & 1 \end{array}$

Note that you don't actually have to find the solutions. The determinant will determine whether or not your system is consistent (implying your set of vectors span p(x)).
• Jun 17th 2008, 10:27 PM
Reckoner
Quote:

Originally Posted by pearlyc
Show that the following does not define an inner product on $\displaystyle R^3$:
$\displaystyle <(x,y,z),(a,b,c)> = xa + 2xb + 2ay + yb + 4zc$

All of the axioms should hold except this one: $\displaystyle \langle\textbf{v},\;\textbf{v}\rangle\geq 0$. Think about it. It should be fairly easy to find a counterexample.