Finding a 3rd polynomial to create a basis.

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SUMMARY

The discussion focuses on finding a third polynomial, \( t_3 \), in the vector space \( V = P_3(\mathbb{R}) \) such that the set \( \{t_1, t_2, t_3\} \) forms a basis for the subspace \( T \), defined by the condition \( t(1) = 0 \). The polynomials provided are \( t_1 = 3x^3 - x - 2 \) and \( t_2 = x^3 - 3x + 2 \). The challenge arises from the fact that while \( V \) has a dimension of 4, the subspace \( T \) only requires 3 polynomials due to the imposed condition.

PREREQUISITES
  • Understanding of polynomial vector spaces, specifically \( P_3(\mathbb{R}) \)
  • Knowledge of basis and dimension concepts in linear algebra
  • Familiarity with polynomial evaluation and conditions such as \( t(1) = 0 \)
  • Ability to manipulate and derive polynomials from given conditions
NEXT STEPS
  • Explore methods for finding polynomial roots and their implications on basis formation
  • Study linear combinations of polynomials in vector spaces
  • Learn about the properties of polynomial spaces and their dimensions
  • Investigate the implications of constraints on polynomial bases in linear algebra
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra and polynomial functions, will benefit from this discussion. It is also relevant for educators looking to enhance their understanding of polynomial vector spaces.

Harambe1
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Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.
 
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Harambe said:
Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.

Hi Harambe! Welcome to MHB! ;)

Indeed, there are 4 elements, so a basis of $V$ will have 4 polynomials.
However, $T$ has the restriction $t(1)=0$, meaning we will be left with a basis of 3 polynomials.

Do you have any ideas how we might find that 3rd polynomial?
 

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