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

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.
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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?

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

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?

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 $\displaystyle x+y=0$. However, this element is not unique. Taking $\displaystyle (0,0)$ to be the identity for the group, then $\displaystyle (-a, 0)$ is the additive inverse for the set $\displaystyle \{(a, b): b \in \mathbb{R} \} $.

In a group, and so in a vector space whilst under +, the inverse is always unique. If $\displaystyle x+y=0$ and $\displaystyle x+z=0$ then $\displaystyle y = y+(x+z) = (y+x) + z = z$. 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 $\displaystyle (0,0)$, but actually it should be $\displaystyle (0,1)$ (and it it, in fact, neither):

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

However, if this is the case then set $\displaystyle \{(a, 0): a \in \mathbb{R} \} $ has no additive inverse, $\displaystyle (a,0)+(c,d) = (a+c,0) \neq (0,1) \forall c, d \in \mathbb{R}$. 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!