Linear Algebra: Vector space axioms

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The discussion centers on evaluating whether the set of all polynomials of degree three or higher, including zero, qualifies as a vector space. A key axiom, 1*x = x, is not satisfied due to the absence of a zero vector in the set, which is essential for closure under vector addition. Participants clarify that while zero is included, the set fails to maintain closure since adding certain polynomials results in values outside the defined set. The conclusion is that the polynomial space does not fulfill the requirements to be considered a vector space. Understanding these properties is crucial for proper classification in linear algebra.
preet
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Homework Statement


One of the fundamental axioms that must hold true for a set of elements to be considered a vector space is as follows:
1*x = x

I was given a particular space: The set of all polynomials of degree greater than or equal to three, and zero, and asked to evaluate whether or not it was a vector space or not. The one that doesn't hold true is 1*x=x (according to the solutions), but I don't understand why. I can't find a situation where the above axiom holds true. Could anyone help me out?

Thanks
Preet
 
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I'd check the solutions again. The problem is closure under vector addition.
 
The set of vectors and the vector addition must form an abelian group. That is true for those polynomials. However, your polynomial space lacks the "null/zero vector"; since we know that the scalar zero times any vector from your set should give the zero vector. Since the zero vector is not an element of the space, it follows that the polynomials' space plus vector addition plus scalar multiplication do not yield a vector space.

Daniel.
 
Daniel, the original post says that zero is an element of the set. The problem is indeed vector addition, since x^3 + (-x^3 + 1) = 1, for exampl.
 
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