# Showing Uniqueness of Elements of a Vector Space

1. Jan 18, 2012

### TranscendArcu

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

http://img267.imageshack.us/img267/8924/screenshot20120118at121.png [Broken]

3. The attempt at a solutionWe have that X = A + B. To show that X is unique, let two such sums be denoted by X1 X2 such that X1X2. We write,

X1 = A + B
X2 = A + B

The equations imply,

X1 - A - B = 0
X2 - A - B = 0

Which imply,

X1 - A - B = X2 - A - B. If we add vectors to both sides,

X1 - A - B + A + B = X2 - A - B + A + B

X1 + 0 + 0 = X2 + 0 + 0
X1 = X2, which contradicts our assertion that X1X2. This shows that such an X is unique.

Last edited by a moderator: May 5, 2017
2. Jan 18, 2012

### lanedance

don't you mean A+X=B?

3. Jan 18, 2012

### lanedance

and though on the right track 3 points,
- i think you need to justify the minus operation by invoking the existence of an additive inverse
- eqauting to zero is an unecessary step, i would start from B=B in this case
- you need to show both existence and uniqueness

4. Jan 18, 2012

### TranscendArcu

Yeah, you're totally right. I just typed X =A + B by accident. As for proving the existence of such an X, how do I begin that? I think proving uniqueness should be, with some adjustment, what I have in #1.

5. Jan 18, 2012

### lanedance

just find a vector that exists that satisfies the requirement
A+X=B
A+X+(-A)=B+(-A)
X+A+(-A)=B+(-A)
X+0=B+(-A)
X=B+(-A)

So somewhat obviously B+(-A) does the job

6. Jan 18, 2012

### lanedance

its a little mechanical, but each one of the above operations is justified by a vector space axiom

7. Jan 18, 2012

### TranscendArcu

Okay, that's pretty clear. Thanks!

Another question I have is why the set of polynomials of degree = 2 is not a vector space. It isn't obvious to me which axiom is not being satisfied here. Could you recommend an axiom to examine more closely?

8. Jan 18, 2012

### lanedance

does it contain the zero vector?

9. Jan 18, 2012

### TranscendArcu

So I understood that degree = 2 indicated polynomials of the form $a_0 + a_1 x +a_2 x^2$ if $a_0$,$a_1$,$a_2 = 0$, then certainly the entire polynomial goes to zero.

I doubt, somehow, that this is the conclusion I am supposed to arrive at. Does degree = 2 indicate rather polynomials of the form $a_2 x^2$? But even if this is so, I still think I could let the coefficient be zero to get a zero vector.

What does "degree = 2" actually mean?

10. Jan 19, 2012

### lanedance

I would interpret degree=2 as $a_2\neq0$..

And degree<=2 as putting no constraints on a_0,a_1,a_2... but thats just my interpretation

Its also probably biased by that i'm pretty sure all polynomials with degree<=2 is a vector space

11. Jan 19, 2012

### HallsofIvy

Staff Emeritus
Also the sum of the two second degree polynomials, $1+ x- x^2$ and $2- 2x+ x^2$, is $3- x$ which is not a second degree polynomial.

(The set of all polynomials of degree less than or equal to is a vector space.)