Proving the Associative Property for Polynomials in Linear Algebra

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SUMMARY

The discussion focuses on proving the associative property for polynomials in linear algebra, specifically the axiom \(\vec{A} + (\vec{B} + \vec{C}) = (\vec{A} + \vec{B}) + \vec{C}\) for polynomials of the form \(a_0 + a_1 x + a_2 x^2\). Participants clarify that polynomials can be treated as vectors since they form a vector space under standard operations. The proof involves demonstrating that the addition of these polynomials satisfies the properties of vector spaces, confirming their status as vectors.

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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone interested in understanding the relationship between polynomials and vector spaces.

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Homework Statement



Show that the axiom \vec{A} + (\vec{B} + \vec{C}) = (\vec{A} + \vec{B}) + \vec{C} holds for polynomials of the form a_0 + a_1 x + a_2 x^2

The Attempt at a Solution


I'm pretty new to writing proofs for linear algebra so my first question is should I be treating the polynomials as the vectors? That is, should I write something like,

a^A_0 + a^A_1 x + a^A_2 x^2 + (a^B_0 + a^B_1 x + a^B_2 x^2 + a^C_0 + a^C_1 x + a^C_2 x^2) = (a^A_0 + a^A_1 x + a^A_2 x^2 + a^B_0 + a^B_1 x + a^B_2 x^2) + a^C_0 + a^C_1 x + a^C_2 x^2

?
I don't think this is correct since the polynomials aren't really vectors (right?). But I'm not sure how else to place these polynomials into the axioms.
 
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Yes, the polynomials under their usual operations are vectors ( since they form a vector space, and this is what you are trying to show ). It's simple: any element of a space that is a vector space, is a vector.
The key point is that you are proving this with respect to a specific operation, and a specific set of objects ( you already know how to add and subtract these elements, you just have to show that they *also* satisfy these other vector space properties )
 
effectively you can treat a_0 + a_1 x + a_2 x^2 as (a_0,a_1,a_2)^T as the polynomials 1,x,x^2 are linearly independent
 

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