Simple Linear Algebra Problem... is the book wrong?

**paupsers** · May 23rd 2009, 04:40 PM

Let V denote the set of ordered pairs of real numbers. If (a, b) and (c, d) are elements of V and x is an element of R, define:

(a, b) + (c, d) = (a+c, bd)
and
x(a, b) = (xa, b).

Is V a vector space over R? Justify.
----------------------
Well, it isn't, but the book claims this is due to the fact (which is referred to by (VS4)), that for each element a in V there exists an element b in V such that
a + b = 0.

However, if c = -a, and d = 0, then (a+c, bd) = (0, 0) always. Am I missing something here?

**Sampras** · May 23rd 2009, 06:52 PM

Originally Posted by paupsers

Let V denote the set of ordered pairs of real numbers. If (a, b) and (c, d) are elements of V and x is an element of R, define:

(a, b) + (c, d) = (a+c, bd)
and
x(a, b) = (xa, b).

Is V a vector space over R? Justify.
----------------------
Well, it isn't, but the book claims this is due to the fact (which is referred to by (VS4)), that for each element a in V there exists an element b in V such that
a + b = 0.

However, if c = -a, and d = 0, then (a+c, bd) = (0, 0) always. Am I missing something here?

An element of $V$ is an ordered pair. So for all $(x_1,x_2) \in V$ there exists a $(y_{1}, y_{2}) \in V$ such that $(x_{1}, x_2) + (y_1, y_2) = \bold{0}$ .

**paupsers** · May 23rd 2009, 09:41 PM

But the book says that there does NOT exist an ordered pair such that the sum equals zero, thereby justifying that the set is not a vector space. The book is wrong, then, right?

**Swlabr** · May 24th 2009, 01:17 AM

Originally Posted by paupsers

But the book says that there does NOT exist an ordered pair such that the sum equals zero, thereby justifying that the set is not a vector space. The book is wrong, then, right?

Firstly, note that a vector space forms an Abelian group under +. The zero is called the identity of the group, and the b such that a+b=0 is called the inverse of a.

You are quite right, every element has another elements such that $x+y=0$ . However, this element is not unique. Taking $(0,0)$ to be the identity for the group, then $(-a, 0)$ is the additive inverse for the set $\{(a, b): b \in \mathbb{R} \}$ .

In a group, and so in a vector space whilst under +, the inverse is always unique. If $x+y=0$ and $x+z=0$ then $y = y+(x+z) = (y+x) + z = z$ . Thus, your set set cannot form an Abelian group under +, and so is not a vector space.

The underlying reason for this not working is the fact that you have no identity. You assumed it was $(0,0)$ , but actually it should be $(0,1)$ (and it it, in fact, neither):

Let us examine the case of $(a,b), b \neq 0$ . Then, $(a,b)+(0,0) = (a+0, b*0) = (a,0) \neq (a,b)$ , so $(0,0)$ is clearly not our identity. It is obvious, however, that $(0,1)$ is the unique identity for this set. Therefore, $(0,1)$ is the unique identity for the set $\{ (a,b), a \in \mathbb{R}, b \in \mathbb{R} \setminus \{0\} \}$ under +, which is a subset of our set.

However, if this is the case then set $\{(a, 0): a \in \mathbb{R} \}$ has no additive inverse, $(a,0)+(c,d) = (a+c,0) \neq (0,1) \forall c, d \in \mathbb{R}$ . Thus, the entire set has no single additive inverse, which contradicts the additive Abelian group axioms.

What you have is a fine example of a semigroup, but I'm afraid it's not a group.

Hope that makes sense!

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