Does zero vector only exist in dim 0?

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The zero vector exists in all R^n for n > 0 and is defined as the additive identity, represented as <0, 0, ..., 0>. While it is accurate to say that every vector space has a zero vector, it is misleading to refer to "the" zero vector as a singular entity across different spaces. Each vector space has its own zero vector, which may take different forms depending on the context, such as (0, 0, 0) in R^3 or the function f(x) = 0 in the space of polynomials. However, all vector spaces of the same dimension are isomorphic, meaning their zero vectors correspond under this isomorphism. Thus, while the concept of a zero vector is universal, its representation can vary significantly across different vector spaces.
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Hi,

just curious, is it false to say the vector 0 exists in R^n, n>0 and a natural number?
i.e. does x = [0, 0, 0, 0] exist in R^4, or simply the zero space?
My guess is that the first statement is not false given that the zero space is a subspace of R^n, n>0.

Am I right or wrong?

Thanks
 
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The zero vector is the additive identity; it exists in all Rn, and is defined as &lt;0, 0, ... , 0&gt;.

EDIT: My statement is misleading. "The zero vector" does not exist in every Rn; rather, for every n, there exists an element of Rn which is the additive identity and is called the zero vector.
 
But in what sense is the vector "which is the additive identity and is called the zero vector" NOT the "zero vector"?

But, having written that, I do see your point. It is a mistake to think that there exists a single vector that is in all vector spaces. In, for example, R3, the "zero vector" is (0, 0, 0). But in the vector space of "all polynomials of degree 2 or less" the "zero vector" is the function f(x) such that f(x)= 0.

Every vector space contains a zero vector but it is misleading to talk about "the" zero vector as if all zero vectors were the same thing.

(But all vector spaces of dimension n are isomorphic and that isomorphism maps the zero vector in one to the zero vector in the other. A "natural isomorphism" from R3 to the space of polynomials of degree 2 or less takes (a, b, c) to ax^2+ bx+ c and so maps (0, 0, 0) to 0x^2+ 0x+ 0= 0 for all x.)
 
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