Linear independence of polynomial set.

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SUMMARY

The discussion centers on determining the values of c for which the set {1+cx, 1+cx², x-x²} forms a basis for the polynomial space P₂. It is concluded that there are no real values of c that allow this set to be a basis, as the first polynomial can be expressed as a linear combination of the other two. The augmented matrix method confirms this linear dependence, leading to the conclusion that the set does not span P₂.

PREREQUISITES
  • Understanding of polynomial spaces, specifically P₂.
  • Knowledge of linear independence and basis concepts in linear algebra.
  • Familiarity with matrix operations, including row reduction.
  • Experience with polynomial expressions and their linear combinations.
NEXT STEPS
  • Study the concept of basis in vector spaces, focusing on polynomial spaces.
  • Learn about linear independence and dependence through examples in linear algebra.
  • Explore the method of row reduction for solving systems of linear equations.
  • Investigate the implications of spanning sets in vector spaces.
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Students and educators in mathematics, particularly those focusing on linear algebra and polynomial theory, as well as anyone interested in understanding the properties of vector spaces and bases.

bamuelsanks
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Hi guys,

I've been working on a question which is as follows:

For which real values of c will the set $\{1+cx, 1+cx^2, x-x^2\}$ be a basis for $P_2$?

I'm coming up with the answer as no values of c, but am I really wrong?
I've only checked linear independence, because it would imply that it spans $P_2$ (right?)

I figure one would just create the augmented matrix:
$\left( \begin{array}{ccc} 0 & c & 1 \\ 1 & 0 & 1 \\-1 & 1 & 0\end{array} \right)$

And reduce:
$\left( \begin{array}{ccc} 1 & 0 & \frac{1}{c} \\ 0 & 1 & \frac{1}{c} \\ 0 & 0 & 0\end{array} \right)$

Thanks in advance,
SB
 
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for any c real number
1+cx = 1+cx^2 + c(x-x^2 )

that means the first one can be written from the other two elements (linear compination from other elements)thats means they can't be a basis.
other way to prove that they are basis or not try to generate the standard basis {1 , x , x^2 } from the given basis if that can be done then they are basis
 
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