3
$\begingroup$

I have a question regarding vector space, to be more accurate the additive identity axiom. I am used to thinking that additive identity simply means add (0,0,0,...) to a vector and get back the vector. I was however told to not approach a problem that way. How would you guys prove this problem fails to satisfy the additive identity. I think something like $(x_1, y_1)$ + (0,0) = $(x_1, y_1) \neq (x_1 + 0, 0)$ would suffice.

In $\mathbb{R}^2$, consider the following operations:

$(x_1, y_1) \oplus (x_2, y_2) = (x_1 + x_2, 0)$

$\alpha \odot (x,y) = (\alpha * x, y) $

is $\mathbb{R}^2$ with these operations a vector space? list all the vector spaces axioms that fail to be satisfied.

$\endgroup$
  • 2
    $\begingroup$ Are you really dealing with vector fields? It seems you’re more looking at vector spaces. $\endgroup$ – mathcounterexamples.net Sep 8 '19 at 16:20
  • $\begingroup$ You are correct thank you. $\endgroup$ – Josue Sep 8 '19 at 16:22
  • 2
    $\begingroup$ Please update your question accordingly. $\endgroup$ – mathcounterexamples.net Sep 8 '19 at 16:23
6
$\begingroup$

The point about what you've been told is that there may be an identity that is not of the form $(0,0,0,\dotsc)$. E.g. this situation. An even simpler example is $\mathbb{R}_{> 0}$ (positive real numbers) with addition operation $$ a \oplus b = ab $$ and multiplication $$ \lambda \otimes a = a^{\lambda} , $$ which you can verify is a vector space with zero vector $1$.

(Yes, there is a simple isomorphism to an "ordinary" vector space, but that's not the point!)

(This is also an excellent example of a situation where you have to be very careful with the notation, since positive real numbers appear in both the scalar field and the vector space.)

$\endgroup$
  • $\begingroup$ thank you, I liked the example very much. $\endgroup$ – Josue Sep 8 '19 at 16:32
3
$\begingroup$

Additive identity means there is something that you can add to any vector, and get the same vector back. That something is then called the additive identity.

For the usual component-wise addition, it turns out that the vector $(0,\ldots,0)$ has that property. When using a different definition of addition, there may well be a different additive identity.

So to prove that there is no additive identity for a certain operation, it doesn't suffice to prove that $(0,\ldots,0)$ is no additive identity, you have to prove that no vector is.

A simple example with additive identity different from $(0,\ldots,0)$ would be the definition $$(v_1,v_2,\ldots,v_n)\oplus(w_1,w_2,\ldots,w_n) = (v_1+w_1-1, v_2+w_2-2,\ldots,v_n+w_n-n).$$ With that definition, $(0,\ldots,0)$ would not be an additive identity, but there still is one, namely $(1,2,\ldots,n)$.

$\endgroup$
2
$\begingroup$

Take $(x,y)\in\mathbb{R}^2$ s.t. $y\neq 0$. Then $(x,y)\oplus (0,0)=(x+0,0)=(x,0)$.

$\endgroup$
  • $\begingroup$ Basically same thing I did correct? $\endgroup$ – Josue Sep 8 '19 at 16:20
  • 1
    $\begingroup$ Yep (I didn't comment that because I'm a new member). $\endgroup$ – H.Bowers Sep 8 '19 at 16:21
  • $\begingroup$ Same here, thank you. $\endgroup$ – Josue Sep 8 '19 at 16:22
2
$\begingroup$

Suppose we denoted our additive identity to be $(e_1, e_2)$. Then, under the given rule of vector addition we have: \begin{align} (x_1, y_1) \oplus (e_1, e_2) = (x_1, y_1) \\ \implies (x_1 + y_1, 0) = (x_1, y_1)\end{align}

Now on comparison, we find that the only possible vectors that satisfy this must have the second coordinate, that is, $y_1 = 0$.

Because not all vectors in $\mathbf{R} ^2$ satisfy this constraint, we see that these operations don't make $\mathbf{R} ^2$ a vector space.

$\endgroup$
2
$\begingroup$

The additive identity property of a vector space can be broken into two parts. First, the addition operation must allow for an identity to exist. Second, that identity must actually be in the space. The operation dictates what the identity has to be, and the set determines whether that identity is included.

There are a few common errors that students encounter when trying to prove that a potential vector space has this property. The most common error is to assume that the identity of the operation in question is the same as that of some previously-seen operation, which is what you have been warned against. The "typical" addition operation of real space suggests that the vector whose components are all identically zero is the identity. However, if you change what addition means, then you may change which vector is the identity, or even deny the existence of one at all!

In the example that you cite, there is no additive identity for the operation.

  1. Let $\left( x_1 , x_2 \right)$ be an arbitrarily-chosen element of $\mathbb{R}^2$.
  2. Assume that $\exists \left( e_1 , e_2 \right) \in \mathbb{R}^2$ which behaves as an identity of $\bigoplus$.
  3. $\left( x_1,x_2 \right) \bigoplus \left( e_1 , e_2 \right) = \left( x_1,x_2 \right)$ because $\left( e_1 , e_2 \right)$ is the identity of $\bigoplus$.
  4. $\left( x_1,x_2 \right) \bigoplus \left( e_1 , e_2 \right) = \left( x_1 + e_1,0 \right)$ by the definition of $\bigoplus$.
  5. Therefore $\left( x_1,x_2 \right) = \left( x_1 + e_1,0 \right)$, implying that $e_1$ is $0$ and $x_2$ is $0$.
  6. $x_2$ was chosen arbitrarily, and so is not necessarily $0$. This is a contradiction.
  7. The assumption that an identity exists is false.

My students sometimes take this to mean that no operation other than the "typical" one provides an identity, but this is not true.

Consider $\mathbb{R}^2$ with the addition operation: $$\left( x_1 , y_1 \right) \bigoplus \left( x_2 , y_2 \right) := \left( x_1 y_2 + x_2 y_1 , y_1 y_2 \right)$$

This operation does have an identity, and there is even an inverse for most elements of $\mathbb{R}^2$. The identity is $\left( 0,1 \right)$, and the inverse of $\left( x,y\right)$ is $\left( -\frac{x}{y^2} , \frac{1}{y} \right)$. You should make sure that you understand why.

$\endgroup$
  • $\begingroup$ I really liked this, thank you very much. $\endgroup$ – Josue Sep 8 '19 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.