# Homework Help: Vector Space R^2

1. Feb 28, 2012

### bugatti79

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

Consider the vector space R^2, Decide whether or not the following are norms defined on R^2. If they are, verify all axioms of a norm, if not, demonstrate by counter example some axioms which fails. For x=(x_1,x_2) in R^2

2. Relevant equations

(i) || ||_#: R^2 defined by ||x||_#=|x_1|+2|x_2|

(ii) || ||_3:R^2 defined by ||x||=3|x_1|

3. The attempt at a solution

How do I decide whether or not the above are norms on R^2?

2. Feb 28, 2012

### CompuChip

What is the definition of a norm?

3. Feb 28, 2012

### bugatti79

Definition of a norm is a norm on a vector space V is a function || ||: V m to R satisfying for x,y in V and for all scalar alpha the following

1) ||x||>=0
2)||x||=0 iff x=0
3)||aX||=|a|||x||
4) The triangle in equality

But what do I need to do to decide whether or not they are norms?

4. Feb 28, 2012

### Staff: Mentor

Does each of your two norms satisfy all four requirements in the definition above? If so, it's a norm. If not, it isn't.

5. Feb 28, 2012

### bugatti79

but how does one check this?

I mean we don't know the value of x1 and x2, they could both be easily less than 1 and hence it wouldn't satisfy axiom 1?
The value of x1 and x2 could both be 0 and then axiom 2 is satisfied.

I don't know how to argue axiom 3 and 4.........?

If it is a norm, how does one 'verify' it?

thanks

6. Feb 29, 2012

### bugatti79

Any tips on this guys? Thanks

7. Feb 29, 2012

### Fredrik

Staff Emeritus
What do you mean? You just do. Say something like "let x be an arbitrary member of ℝ2" and then use the definition of the function that may or may not be a norm to see if it satisfies the first condition. If it's satisfied, then you move on to the second condition. If it looks like it's not, you try to find an example that proves that it's not satisfied.

I don't know what you're talking about here. Both of the functions you mentioned in your post very obviously satisfy axiom 1.

Axiom 2 is a "for all" statement, so you can't test it with one specific value of x, unless you're trying to prove that it's not satisfied. Note that all of the axioms are "for all" statements.

8. Feb 29, 2012

### HallsofIvy

Why would being "less than 1" contradict "greater than or equal to 0"?

9. Feb 29, 2012

### bugatti79

Sorry, this is a typo, I meant the value of x could be less than 0 not 1...and then it would contradict axoim 1..?

10. Feb 29, 2012

### Fredrik

Staff Emeritus
Assuming that you made another mistake here, and meant x_1 and x_2 when you said "x", then no, it wouldn't, because of the absolute value signs in the definitions. If you had been working with another definition, one that implies that there's an x such that $\|x\|<0$, then the axiom is violated, but that doesn't make it harder to check if the axioms are satisfied. It makes it easier, because then all you have to do is to find that x, show that $\|x\|<0$, and you're done.

11. Feb 29, 2012

### bugatti79

Ok heres my attempt,

1) since x=(x1,x2) is in R^2 and we know |x1|>=0 and |x2|>=0 implies ||x||>=0 for all x in R^2 hence N1 holds

2) ||x||=0 iff both x1=0 and x2=0, ie x=(0,0)=0 the zero vector implies N2 holds

3) ||a x || = |a| |x|
=|a|(|x1|+2|x2|)
= |a| |x1| + 2 |a| |x2| (all im doing is throwing a on both side of equation so of course it holds, right?)

4) The triangle inequality is ||x+y|| <= ||x||+ ||y||
I don't know how to check this one....

Thanks

12. Feb 29, 2012

### Fredrik

Staff Emeritus
1. You can't say "Since x is in R^2" when you haven't even mentioned x before. You need to start with something like "Let x be an arbitrary member of ℝ2". You also need to make it absolutely clear what definitions you're using. I don't see anything that indicates which of the functions you're trying to prove something about.

2. What is x in the iff statement? Why does that statement hold. Which one(s) of the functions are you talking about?

3. This doesn't make any sense. The first equality is what you're trying to prove, except that the right-hand side should say $|a|\,\|x\|$, not |a| |x|. You're obviously not allowed to use what you're trying to prove. The calculation you did after that adds nothing at all. You need to start with $\|ax\|=\dots$, then use the definition, and end it with $\dots=|a|\,\|x\|$.

You must always make it clear if your variables are part of "for all" or "there exists" statements, or if they are defined to represent some specific member of the set you're dealing with. You must always make it absolutely clear how you're using the definitions of the terms or symbols that you're trying to prove something about.

13. Mar 1, 2012

### bugatti79

I am referring to part (i). I am not sure what the subscript 'hatch' or 'star' means

1) Since we are given that x=(x1,x2) is in R^2 and we have that ||x||_#=|x1|+2|x2|, then ||x||>=0 because we have that |x1| and |x2| are both >=0 implies ||x||>=0 for all x in R

2)||x||_#=0? We have ||x||=|x1|+2|x2|=0 iff |x1|=0 and |x2|=0 for all x in R

3) ||ax||_#=....? I didn't know one was not allowed use what we are trying to prove. Then what else can we do?

4) ?

Thanks

14. Mar 1, 2012

### Fredrik

Staff Emeritus
It's just a notation for that specific function.

This is OK, but don't say that you're given that x is in $R^2$. You're not given an x, or any variable at all. You're just supposed to prove a statement of the form "For all v in $R^2$, we have P(v)", where P(v) is a statement that involves v. In this case, P(v) is the statement "$\|v\|\geq 0$". This is how I'd do it:

Let $x\in\mathbb R^2$ be arbitrary. By definition of $\|\ \|_{\#}$, we have
$$\|x\|_{\#}=\underbrace{|x_1|}_{\geq 0}+2\underbrace{|x_2|}_{\geq 0}\geq 0.$$
This is OK too. It's sufficient to say that since $\|x\|_{\#}=|x_1|+2|x_2|$, it's obvious that $\|x\|_{\#}=0$ if and only if $x=0$.

I answered that in my previous post, and I'm doing it again below.

Same idea as in all of these problems. Start by saying "Let $x\in\mathbb R^2$ be arbitrary". Then write down the left-hand side of the equality or inequality you want to prove. Then use the definition of $\|\ \|_{\#}$ and things you know about $\mathbb R^2$ until you end up with the right-hand side of the result you want to prove.

Last edited: Mar 1, 2012
15. Mar 3, 2012

### bugatti79

3) Let $x \in \mathbb R^2.$ By defintion we have $\|x \|_{\#} = \|x_1\|+2\|x_2\|$

$\|a x \|_{\#}=\|ax_1\| +2 \|ax_2\|=|a| ( |x_1|+2|x_2|)=|a| \| x \|_{\#} \forall a, x \in \mathbb R^2$ N3 holds

4) Let $y \in \mathbb R^2$ where $\|y\|_{\#}=|y_1|+2|y_2|$

Therefore $\|x+y\|= |x_1|+2|x_2| + |y_1||+2|y_2|$

By defintion of the inequality $\|x+y\| \le \|x\|+\|y\|$ we have

$\|x+y\|_{\#} \le |x_1|+2|x_2|+|y_1|+2|y_2|$
$$\le \|x\|_{\#}+\|y\|_{\#} \forall x,y \in \mathbb R^2$$ N4 holds..............?

Tackle part 2 in same way?

Thanks

16. Mar 3, 2012

### Fredrik

Staff Emeritus
The first sentence should say "Let $x\in\mathbb R^2$ be arbitrary". The definition of $\|\ \|_{\#}$ tells us that $\|x\|_{\#}=|x_1|+2|x_2|$.

This is mostly OK. The norm symbols after the first equality sign should of course be absolute value signs. a is not a member of $\mathbb R^2$. This is what you should have said: For all $a\in\mathbb R$ and all $x\in\mathbb R^2$, we have
$$\|ax\|_{\#}=\|(ax_1,ax_2)\|_{\#}=|ax_1|+2|ax_2|=|a|(|x_1|+2|x_2|)=|a|\|x\|_{\#}.$$ Note that the "for all" statement makes it unnecessary to also say "Let ... be arbitrary". The point of "let ... be arbitrary" statements is just that they enable us to break up a long sentence into several short sentences.

This is not what the definition of $\|\ \|_{\#}$ says. Try again. The proof should start like this: For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=\dots$$ Then you just use the definitions of x+y and $\|\ \|_{\#}$, and whatever else you need.

Last edited: Mar 3, 2012
17. Mar 3, 2012

### bugatti79

Ok, thanks for the tips.

4) $\forall x,y \in \mathbb R^2$ we have $x+y = x_1+2x_2 +y_1 +2y_2$ therefore

$\|x+y|\_{\#}= |x_1|+2|x_2| +|y_1|+2|y_2|$ but by defintion of triangle inequality we have

$\|x+y\|_{\#} \le |x_1|+2|x_2|+|y_1|+2|y_2| \le \|x\|_{\#}+\|y\|_{\#}$.....?

18. Mar 3, 2012

### Fredrik

Staff Emeritus
Surely, you must have meant to put something else on the right-hand side. The left-hand side is a member of ℝ2, and the right-hand side a member of ℝ.

This is wrong. The right-hand side is equal to $\|x\|_{\#}+\|y\|_{\#}$, but it's certainly not true that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all x,y in ℝ2. For example, when x=(0,1) and y=(0,-1), we have $\|x+y\|_{\#}=0$ and $\|x\|_{\#}+\|y\|_{\#}=4$.

19. Mar 3, 2012

### bugatti79

4) $\forall x,y \in \mathbb R^2$ and $x_1, y_1 \in \mathbb R$

we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

Ok,so basically this question is not a norm in R^2 because axiom 4 does not hold. You have demonstrated that by counter example. but how do you know that axoim 4 does not hold without having to play around with actual values of x and y to test it?

20. Mar 3, 2012

### micromass

No, Fredrik gave a counterexample to the equality. That doesn't mean that axiom 4 doesn't hold.

21. Mar 3, 2012

### micromass

Are you just guessing now??

22. Mar 3, 2012

### Fredrik

Staff Emeritus
No, we don't. The left-hand side isn't even in the same set as the right-hand side (as I have already explained).

As micromass said, I only proved that it's not the case that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. You still have to determine if $\|x+y\|_{\#}\leq\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. (Since 0≤4, my choice of x and y doesn't contradict this inequality).

23. Mar 3, 2012

### HallsofIvy

If you mean that $x= <x_1, x_2>$ and $y= <y_1, y_2>$ then you should say that. Also, the result would be $x+ y= <x_1+ 2y_1, x_2+ 2y_2>$.

24. Mar 3, 2012

### jgens

The first sentence was fine as it was.

25. Mar 3, 2012

### bugatti79

No, I was just building up the equations bit by bit and then convert it into a norm (putting it inside the vertical brackets as shown on the following line of that thread) and then test it. Obviously Im wrong.

"we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?"
I dont understand why the LHS would not be = the RHS when I specify for the LHS that x,y are in R^2 and for the RHS x1 and y1 are in R.

I have an example where the triangle inequality is tested for a taxi cab norm ie || ||_1 and I understand it but I dont know how to do it for a || ||_#......?

Last edited: Mar 3, 2012