2

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?

Blue
  • 75,673
Roy Huang
  • 91
  • 5
  • Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Aug 15 '22 at 13:57
  • 1
    @John Douma: What I wrote in my first comment, at the end: "The zero vector is the real number 1." What you replied: "@ArturoMagidin In that case, the real number 1 acts as the zero in the vector space." You repeated my words at me, and added nothing else. And you keep misusing the equal sign. "Corresponds to" and "is isomorphic to" is not the same as "actually equal to". In $\mathbf{P}_2$, real polynomials of degree at most two, what you wrote would be $\mathbf{0}=0\cdot 1 + 0\cdot x + 0\cdot x^2 = (0,0,0)$. The last equality is false. – Arturo Magidin Aug 15 '22 at 13:59

4 Answers4

6

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 $\Bbb R^8$.

calc ll
  • 8,427
  • I have a policy of explaining downvotes: This answer is probably not wrong per se. But it's a bit like answering someone who asks "how do I hit a tennis ball over the net?" with "well, just hit a drop shot with extra spin and that will go over the net". –  Aug 15 '22 at 03:26
  • Good answer, +1. Since the OP has misconceptions about vector spaces, maybe it is worth giving a concrete example (I would choose a translation but there could be a better choice). – Taladris Aug 15 '22 at 03:33
  • @Taladris Arturo has already given a specific example. Since any two finite dimensional vector spaces over the same field are isomorphic, it can be considered transport of structure from the usual $\Bbb R$. – calc ll Aug 15 '22 at 03:48
  • @Noobie: this example was a comment. Comments are meant to be deleted eventually – Taladris Aug 15 '22 at 03:49
  • @Taladris Anonymous M's answer is also a special case of transport of structure. The bijection being $\varphi (0)=\text{squirrel}$. – calc ll Aug 15 '22 at 04:17
  • There is no structure being transferred in my example! Some authors may decide to denote squirrel by 0, but it is still squirrel. –  Aug 15 '22 at 04:23
3

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.

1

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".

Seeker
  • 3,594
-5

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 $x$.

Suzu Hirose
  • 11,660
  • I don't think the first sentence follows the definition of vector space given at least in the free version of the mentioned book: Page 8 https://linear.axler.net/LinearAbridged.pdf It's true that the additive identity is always unique in a vector space and that the standard vector space structure on e.g. $\mathbb{R}^n$ has additive identity $(0,\dots,0)^T$ –  Aug 15 '22 at 02:38
  • 5
    Take the vector dpace of pairs $(a,b)$ with $a$ and $b$ both positive real numbers, using operations as in my comment above... You are implicitly assuming that the entries lie in the field, that the vector addition is defined coordinatewise using field addition, and that scalar multiplication is defined coordinatewise using field multiplication. That' a lot of assumptions that are not being made explicit, and that make the assertions far less useful than you seem to believe. – Arturo Magidin Aug 15 '22 at 02:45
  • 1
    This answer may not look like as mathy as others. But I personally do like it. Thanks Suzu. – Roy Huang Aug 15 '22 at 04:10
  • 1
    @RuoyunHuang I'm glad this answer was useful. Unfortunately I think some people are playing oneupmanship rather than trying to answer the question appropriately at the level of the questioner. Argumentativeness is a kind of occupational disease of mathematicians, so don't worry about it. – Suzu Hirose Aug 15 '22 at 04:14
  • 2
    @SuzuHirose I generally approve of your contributions to this site, and have upvoted many of them. But here we are really not playing any games. The question was specifically about the form of the neutral element, when details like this come to the fore. The neutral elements of algebraic structure really come in very diverse guises. When you advance in your studies you will encounter things like the collection of contractible paths and the family of matrix algebras $M_n(K)$, $n>0$ assuming the role of the neutral element. Take care, and continue to enjoy the site. – Jyrki Lahtonen Aug 16 '22 at 14:12