How do you find the coordinates of a polynomial in terms of an orthogonal basis?

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SUMMARY

The discussion focuses on finding the coordinates of the polynomial ##x^2 + x + 1## with respect to the orthogonal basis derived from the set ##S = \{1, x, x^2\}## using the Gram-Schmidt process. The orthogonal set obtained is ##S' = \{1, x - \frac{1}{2}, x^2 - x + \frac{1}{6}\}##. The change of basis matrix is established as $$ \left( \begin{array}{ccc} 1 & \frac{1}{2} & \frac{1}{3} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right)$$. To express the polynomial in terms of the orthogonal basis, one must compute the coefficients directly from the new basis rather than the original set.

PREREQUISITES
  • Understanding of polynomial spaces and inner products, specifically ## = \int_{0}^{1} fg \textrm{ } dx##.
  • Familiarity with the Gram-Schmidt process for orthogonalization.
  • Knowledge of change of basis techniques in linear algebra.
  • Ability to manipulate polynomial expressions and coefficients.
NEXT STEPS
  • Study the Gram-Schmidt process in detail to understand orthogonalization of polynomial sets.
  • Learn about change of basis matrices and their applications in linear algebra.
  • Explore polynomial inner products and their implications in function spaces.
  • Practice expressing various polynomials in terms of different bases to solidify understanding.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on linear algebra, polynomial functions, and orthogonal bases. This discussion is beneficial for anyone looking to deepen their understanding of polynomial representation in vector spaces.

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Homework Statement


Given ##S = \{1, x, x^2\}##, find the coordinates of ##x^2 + x + 1## with respect to the orthogonal set of S.

Homework Equations


Inner product on polynomial space:
##<f,g> = \int_{0}^{1} fg \textrm{ } dx##

The Attempt at a Solution


I used Gram-Schmidt to make ##S## orthogonal and got ##S' = \{1, x - \frac{1}{2}, x^2 - x + \frac{1}{6}\}##.

So the change of basis matrix I got was $$ \left( \begin{array}{ccc}
1 & \frac{1}{2} & \frac{1}{3} \\
0 & 1 & 1 \\
0 & 0 & 1 \end{array} \right)$$

But ##x^2 + x + 1## looks exactly like ##S##, so it would seem like it's the identity matrix so then it wouldn't change anything, which is where I'm stuck.
 
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The problem says "with respect to the orthogonal set of S",and because S is not an orthogonal set,so you should find the components in S' basis.So forget about S and calculate the components w.r.t. S' directly!
Also,x^2+x+1 doesn't look exactly like S,only its coordinates w.r.t. S is (1,1,1).
 
Think about what the change of basis matrix does for you. Consider, for example, the polynomial ##x##. Expressed in terms of the original basis ##S##, we can write ##x = 0(1) + 1(x) + 0(x^2)##, so its coefficient vector in terms of the original basis is ##(\begin{array}{ccc}0& 1& 0\end{array})^T##. To express it in terms of the new basis, we simply multiply the matrix by this vector:
$$\text{new coefficient vector} = \left( \begin{array}{ccc}
1 & \frac{1}{2} & \frac{1}{3} \\
0 & 1 & 1 \\
0 & 0 & 1 \end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) = \left( \begin{array}{c} \frac{1}{2}\\ 1 \\0\end{array} \right)$$
This means that our polynomial ##x## can be expressed in terms of the orthogonal basis using the new coefficients:
$$x = \frac{1}{2}(1) + 1\left(x-\frac{1}{2}\right) + 0\left(x^2 - x + \frac{1}{6}\right)$$
and we can easily see that the left hand side does indeed equal the right hand side.

You can use exactly the same technique for any other polynomial. Start by finding the coefficients of ##1 + x + x^2## in terms of the original basis ##S##.
 

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