Is there a vector space whose additive identity is not in the form of $(0,\ldots,0)$? Reading Linear Algebra Done Right.
One question:

Does there exist a vector space $V$, such that its additive identity $0$, is actually not in the form of $(0,\ldots,0)$?

Intuition tells me probably no. But I couldn't figure this out from definitions. Any hints?
 A: With transport of structure you can change the additive identity to a different element of the vector space.
Just choose a bijection $\varphi$ of $V$ with $\varphi (\vec0)\ne\vec0$.
Define $a\oplus b=\varphi^{-1}(\varphi (a)+\varphi (b))$ and $\lambda \cdot a=\varphi^{-1}(\lambda \varphi (a))$.
The new additive identity is $\varphi (\vec 0)$.
The two vector spaces are isomorphic (by $\varphi $).

For a nice example of this idea at work in a different context,  see Milnor's $28$ exotic spheres, different differentiable structures on the seven sphere $S^7$ in $\Bbb R^8$.
A: This is a fairly basic question that's worth thinking about. I dislike answering homework questions, but have concern about a few comments and answers. They give you examples of "mathy" vector spaces whose additive identity is not $(0,0,\dots,0)$. But I think these miss the forest for the trees somewhat on the pedagogical front.
A vector space is a non-empty set $V$ together with a binary operation $+:V\times V\to V$, a field of coefficients $F$, and a function $\cdot:F\times V\to V$, which follow a list of axioms which you can find in your book's definition. One requirement is that there must be at least one additive identity element $v\in V$.
Let $V = \{ \text{squirrel} \}$ be the set consisting of the single element $\text{squirrel}$. There is only one binary operation $+$ which is possible to define on $V$. Let $F$ be the field $\mathbb{R}$. There is only one function $\cdot:F\times V\to V$. The additive identity element is $\text{squirrel}\in \{\text{squirrel}\}$. You can check that $(V,F,+,\cdot,\text{squirrel})$ is a vector space.
If you have already seen the concept of isomorphism of vector spaces, this vector space may look familiar. It is certainly isomorphic to a standard vector space with additive identity usually denoted $0$. Then again, every vector space is isomorphic to one with additive identity denoted $0$. But since the question does not ask about isomorphism, the above vector space is a perfectly fine example answering in the negative.
A: Pretty early on in the book Axler gives an example of the vector space $R^{[0,1]}$. This is the vector space of all functions $f : [0,1]\to R$. The additive identity of this vector space is the function $0 : [0,1]\to R$ defined by $0(x)=0$ for all $x\in [0,1]$.
This is a simple counterexample to the question. See that the elements of the vector space $R^{[0,1]}$ are functions and not tuples.
As another example(also pointed out in the comments), have a look at the vector space of all polynomials with degree at most $m$. He talks about this vector space in the chapter called "Finite Dimensional Vector Spaces".
A: If the vector space is finite-dimensional, and if the elements of the vector space look like $(a,b,c,...)$, then you can prove easily that $\vec{0}=(0,0,0,...)$ by supposing that $\vec{0}=(a,0,0,...)$ and then using the definitions to find a contradiction if $a$ is not zero.
However, there are some vector spaces of infinite dimensions, like the space of differentiable functions, and the elements of these vector spaces don't even take the form $(a,b,c,...)$, so the zero element cannot be of the form $(0,0,0,...)$ in these cases. For example, in the above-mentioned case of the space of differentiable functions, the zero vector is the function $f(x)=0$ for all $x$.
