# Linear Algebra-Fields and axioms

1. Sep 16, 2008

### mirandasatterley

1. The problem statement, all variables and given/known data

Let F be any field, and fix a є F. Equip the set V = F2 with two operations as
(x, y)‡(x', y') := (x + x', y + y' − a), for all x, x', y, y' є F,
and define the scalar multiplication by scalars by
c * (x, y) := (cx, cy − ac + a), for all x, y, c є F.
(i) Prove that F2, with these two operations satisfies the two existence axioms for a vector space over F.

3. The attempt at a solution
i) The existance of a zero vector: there exists a zero vector such that 0 + v = v

So I let (x', y') be the zero vector = (0, 0), and got
(x,y) ‡ (0, 0) := (x + 0, y + 0 - a)
= (x, y-a)

At this point, I'm not sure how to deal with the a, or maybe my whole process is wrong! ANy help is appreciated. I'm having the same problem for the other axiom - existance of a negative.

2. Sep 16, 2008

### Defennder

You can't simply designate (0,0) to be the zero vector. The zero vector has to satisfy the rule (x,y) + (zero vector) = (x,y) where + denotes vector addition defined on this vector space.

Instead just denote the zero vector by (w,z) for example, and we have (x,y) + (w,z) = (x,y). Using the defined addition operation, you should be able to deduce what w,z should be to satisfy that. That is then the zero vector.

The same approach works for determinining the scalar multiplicative identity in this field.

3. Sep 16, 2008

### mirandasatterley

So, if (w,z) is the zero vector and (x,y) is any vector, then

(x,y) + (w,z) = (x,y), and addition is defined as
(x,y) ‡ (w,z) := (x+w, y+z-a)

Can I set these equations equal to each other? to get
(x,y) = (x+w, y+z-a) therefore,

x= x+w, and subtracting x from both sides gives w=0, and we also have
y= y+z-a, and solving this gives z=a, so how can I prove that z must be zero?

4. Sep 16, 2008

### Almanzo

The zero vector is a vector which, if added to another vector, gives this other vector as a result.
(x,y)++(0,a) = (x+0,y+a-a) = (x,y).
So (0,a) seems a reasonable zero vector.

If I multiply (x,y) by 0, I get 0(x,y) = (0x,0y-0a+a) = (0,a),
so that's okay, too.

If I multiply (x,y) by 1, I get 1(x,y) = (1x,1y-1a+a) = (x,y), so 1 is a unit element.
Is it also the unique unit element?
Suppose p(x,y) = (x,y). Then (px, whatever) = (x,y), so px = x, so x =1.

5. Sep 16, 2008

### Defennder

z isn't 0. If it were 0, you wouldn't have the zero vector.

6. Sep 16, 2008

### mirandasatterley

For the existance of a negative: there exists a -v such that v + (-v) = 0, does the following proof make sense? Let me know if I am missing something. Thanks.

(x,y) is a vector v
(w,z) is the vector -z, so I need to prove that (w,z) = -(x,y), so I have the equation,

(x,y) + (w,z) = (0,0) from the axiom, and
(x,y) ++ (w,z) := (x+w, y+z-a) from the definition of addition.

therefore (0,0) = (x+w, y+z-a)

And from this, 0=x+w, which solves to w= -x, and
0= y+z-a, which solves to z= a-y

so in the equation for the definition of addition:

(x,y) ++ (-x, a-y) = (x +(-x), y+(a-y) -a) = (0, (y-y) + (a-a)) = (0,0)

7. Sep 16, 2008

### mirandasatterley

I think I understand, so when I solve z=a, and plug this into the addition definition, I am proving that (0,a) is the zero vector since (x,y) ++ (0,a) := (x,y)?

8. Sep 16, 2008

### HallsofIvy

Staff Emeritus
No, no, no. As you were just told, (0,0) is NOT the additive identity for this set!

9. Sep 16, 2008

### Defennder

Yes, this time you got it.