Norm Verification for Vectors in R^2

[Mark44]

To elaborate on what Fredrik said, how is addition usually defined for vectors in R²? That's basically what he's asking about for the expression x + y. This has nothing to do with norms.

 

[bugatti79]

To elaborate on what Fredrik said, how is addition usually defined for vectors in R²? That's basically what he's asking about for the expression x + y. This has nothing to do with norms.

Yes, but before you focus on $\|x+y\|_{\#}$, you should focus on x+y. Do you understand what the notation $\mathbb R^2$ means? What sort of objects are members of $\mathbb R^2$. How do you add one of those objects to another? Do you at least understand that the notation x+y is completely meaningless until it has been defined? What is the standard definition of x+y for $x,y\in\mathbb R^2$?

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

If the value of y is 0, isn't it still possible that the inequality holds? ie, the LHS can still be less than the RHS ie

$\|x\| \le \|x_1\|+ \|2x_2 \|$

So I am saying that axiom 4 holds. Is this how one examines it? Thanks

 

[micromass]

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

If the value of y is 0, isn't it still possible that the inequality holds? ie, the LHS can still be less than the RHS ie

$\|x\| \le \|x_1\|+ \|2x_2 \|$

So I am saying that axiom 4 holds. Is this how one examines it? Thanks

No. What you wrote makes no sense.

We asked you several times, how is x+y defined??

Let $x=(x_1,y_1)$ and $y=(y_1,y_2)$. Then how did we define

$$(x_1,x_2)+(y_1,y_2)$$

?

 

[Fredrik]

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

Try doing just one thing at a time. For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=...$$ The first step is to use the definition of + to rewrite x+y as an ordered pair of real numbers. The second step is to use the definition of $\|\ \|_{\#}$. Then there's a third step. Don't try to do them all at once. Do them one at a time.

You also need to think about when to use the notation |something| and when to use the notation $\|\text{something}\|_{\#}$. Also, in each step, use an equality sign if what you have on the left is the same thing as what you have on the right. Use ≤ only if you have reason to think that the thing on the right may be greater than the thing on the left.

 

[Mark44]

In addition, in this problem ||x + y|| is completely irrelevant. The norm you appear to be working with is ||x + y||_#. Use the definition of this norm to first say what this expression equals, and then work on showing that the triangle inequality holds.

 

[bugatti79]

In addition, in this problem ||x + y|| is completely irrelevant. The norm you appear to be working with is ||x + y||_#. Use the definition of this norm to first say what this expression equals, and then work on showing that the triangle inequality holds.

Try doing just one thing at a time. For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=...$$ The first step is to use the definition of + to rewrite x+y as an ordered pair of real numbers. The second step is to use the definition of $\|\ \|_{\#}$. Then there's a third step. Don't try to do them all at once. Do them one at a time.

You also need to think about when to use the notation |something| and when to use the notation $\|\text{something}\|_{\#}$. Also, in each step, use an equality sign if what you have on the left is the same thing as what you have on the right. Use ≤ only if you have reason to think that the thing on the right may be greater than the thing on the left.

$x+y=(x_1+y_1,2x_2+2y_2)$?

 

[Mark44]

$x+y=(x_1+y_1,2x_2+2y_2)$?

NO! Do you not understand how to add two vectors together? You should not be attempting to work problems about norms if you don't understand the basics of vector operations.

 

[Fredrik]

$x+y=(x_1+y_1,2x_2+2y_2)$?

No, this is wrong. Note that we're just doing addition here. Those functions that may or may not be norms have nothing to do with it.

 

[micromass]

Bugatti, I think we could help you better if you would explain us your situation. What course are you taking now? What courses did you already take??
I'm asking this because I feel you miss some preliminary knowledge. We can help you rectify it by suggesting things you should look at.

 

[bugatti79]

NO! Do you not understand how to add two vectors together? You should not be attempting to work problems about norms if you don't understand the basics of vector operations.

If v=(v1,v2) and w=(w1,w2) then v+w=(v1+w1,v2+w2)? Thats all I have done above?

No, this is wrong. Note that we're just doing addition here. Those functions that may or may not be norms have nothing to do with it.

Bugatti, I think we could help you better if you would explain us your situation. What course are you taking now? What courses did you already take??
I'm asking this because I feel you miss some preliminary knowledge. We can help you rectify it by suggesting things you should look at.

I am studying topics in analysis part time so I may be over tired from work. I have done PDE's and Calculus 1,2 last year.

 

[Mark44]

If v=(v1,v2) and w=(w1,w2) then v+w=(v1+w1,v2+w2)? Thats all I have done above?

No, it isn't. You were mixing vector addition with the || ||_# norm, producing a meaningless mush. Here is what I'm referring to.

x+y=x₁ +2x₂ +y₁ +2y₂

 

[bugatti79]

No, it isn't. You were mixing vector addition with the || ||_# norm, producing a meaningless mush. Here is what I'm referring to.

So for x,y in R^2 where x=(x1,x2) and y=(y1,y2)

$x+ y= <x_1+ 2y_1, x_2+ 2y_2>$?

 

[micromass]

So for x,y in R^2 where x=(x1,x2) and y=(y1,y2)

$x+ y= <x_1+ 2y_1, x_2+ 2y_2>$?

No! You wrote it correctly in in your post 45. Why do you keep insisting on the 2?

 

[bugatti79]

No! You wrote it correctly in in your post 45. Why do you keep insisting on the 2?

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

 

[Fredrik]

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

Here you're adding x and y correctly, but why don't you just insert the result you've found for x+y into $\|x+y\|_{\#}$, and then use the definition of $\|\ \|_{\#}$? What you wrote doesn't make any sense. The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

Edit: Why is there a comma in there? Did you mean $\|(|x_1|+|y_1|,2|x_2|+2|y_2|)\|_{\#}$? That doesn't make much sense either. I mean, it it's not a nonsense expression, but I have no idea why you would consider an ordered pair with those components.

 

[Mark44]

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

Did you use the definition?

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

Doesn't look like it to me.

 

[bugatti79]

Did you use the definition?


Doesn't look like it to me.

Here you're adding x and y correctly, but why don't you just insert the result you've found for x+y into $\|x+y\|_{\#}$, and then use the definition of $\|\ \|_{\#}$? What you wrote doesn't make any sense. The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

Edit: Why is there a comma in there? Did you mean $\|(|x_1|+|y_1|,2|x_2|+2|y_2|)\|_{\#}$? That doesn't make much sense either. I mean, it it's not a nonsense expression, but I have no idea why you would consider an ordered pair with those components.

Now I am really confused. I know this is a big ask but is it not possible for one of you to write the answer and then I will query why you did it that way?. I have spent 3.5 hours on this trivial problem.

 

[bugatti79]

The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

OK, should it be

$\| x+y\|_{\#}= \| \|x\|+\|y\| \|_{\#}$

 

[Mark44]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

 

[Mark44]

OK, should it be

$\| x+y\|_{\#}= \| \|x\|+\|y\| \|_{\#}$

Way off.

 

[bugatti79]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

At what stage do I play with the inequalities?

 

[micromass]

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

At what stage do I play with the inequalities?

Line 2 to 3 is wrong. You won't have an [STRIKE]inequality[/STRIKE] equality. But perhaps you can use

$$|a+b|\leq |a|+|b|$$

for $a,b\in \mathbb{R}$.

 

[bugatti79]

That's against forum rules.

Here's an example, though.
Suppose x = <3, 2> and y = <-1, 5>
||x + y||_# = ||<2, 7>||_# = |2| + 2|7| = 2 + 14 = 16

I can justify every step above. Can you provide a reason for each step?

See if you can use this as a template to calculate a more general result: ||x + y||_#, with x = <x₁, x₂> and y = <y₁, y₂>

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

 

[Mark44]

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|
=|x1|+|y1|+2|x2| +2|y2|
=|x1|+2|x2|+|y1|+2|y2|
= ||x||_#+||y||_#

I am not sure how to justify going from line 2 to line 3...?

As already noted by micromass, it's not justifiable.

Your first couple of steps were a good start, though, giving justifications for those steps. If you think about things from this perspective (giving a justification - definition, theorem, etc. - for each step) you will have made an important step toward thinking mathematically.

 

[Mark44]

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

Yes, exactly right. This is precisely what we have been trying to get you to do.

 

[Fredrik]

I have spent 3.5 hours on this trivial problem.

I haven't counted the hours I've spent trying to help you, but I would guess that it's at least that many, just for this thread.

x=(x1,x2), y=(y1,y2) both in R^2

line 1 ||x+y||_#=||(x1+y1),(x2+y2)||_# then using definition of the norm || ||_# we have
= |x1+y1|+2|x2+y2|

Finally! This is what I've been trying to get you to do the whole time. Note that there's nothing fancy going on here. It's just the definition of +, followed by the definition of $\|\ \|_{\#}$.

However, the notation $||(x1+y1),(x2+y2)||_{\#}$ is a bit messed up. It should say $\|(x_1+y_1,x_2+y_2)\|_{\#}$.

At what stage do I play with the inequalities?

When you go from line 2 to line 3. Micromass gave you the details.

 

[bugatti79]

Line 2 to 3 is wrong. You won't have an inequality. But perhaps you can use

$$|a+b|\leq |a|+|b|$$

for $a,b\in \mathbb{R}$.

From ||x + y||_# = ||<2, 7>||_# because we have x+y=(x1+y1, x2+y2)

From ||<2, 7>||_# = |2| + 2|7| because we have the definition of the norm ||x ||_# as |x1 |+ 2 | x2|

As already noted by micromass, it's not justifiable.

Your first couple of steps were a good start, though, giving justifications for those steps. If you think about things from this perspective (giving a justification - definition, theorem, etc. - for each step) you will have made an important step toward thinking mathematically.

Given $x=(x1,x2), y=(y1,y2)$ both in R^2

$\|x+y\|_{\#}=\|(x1+y1,x2+y2)\|_{\#}$ where $x+y=(x1+y1, x2+y2)$ by definition of vector addition in R^2

then using definition of the norm || ||_# given we have
$\|x+y\|_{\#}= |x1+y1|+2|x2+y2|$

but by the triangle inequality we have |a+b| <= |a| +|b| hence

$\|x+y\|_{\#}\le|x1|+|y1|+2|x2| +2|y2|$
$\le|x1|+2|x2|+|y1|+2|y2|$
$\le \|x\|_{\#}+\|y\|_{\#}$ Therefore axiom 4 holds

 

[bugatti79]

Yes, exactly right. This is precisely what we have been trying to get you to do.

Line 2 to 3 is wrong. You won't have an [STRIKE]inequality[/STRIKE] equality. But perhaps you can use

$$|a+b|\leq |a|+|b|$$

for $a,b\in \mathbb{R}$.

I haven't counted the hours I've spent trying to help you, but I would guess that it's at least that many, just for this thread.


Finally! This is what I've been trying to get you to do the whole time. Note that there's nothing fancy going on here. It's just the definition of +, followed by the definition of $\|\ \|_{\#}$.

However, the notation $||(x1+y1),(x2+y2)||_{\#}$ is a bit messed up. It should say $\|(x_1+y_1,x_2+y_2)\|_{\#}$.


When you go from line 2 to line 3. Micromass gave you the details.

Ok guys, thanks for your help. I really appreciate it considering you could be spending your time on less trivial tasks. I will tackle part 2 in thread 1 tomorrow. night!

 

[Fredrik]

Given $x=(x1,x2), y=(y1,y2)$ both in R^2

$\|x+y\|_{\#}=\|(x1+y1,x2+y2)\|_{\#}$ where $x+y=(x1+y1, x2+y2)$ by definition of vector addition in R^2

then using definition of the norm || ||_# given we have
$\|x+y\|_{\#}= |x1+y1|+2|x2+y2|$

but by the triangle inequality we have |a+b| <= |a| +|b| hence

$\|x+y\|_{\#}\le|x1|+|y1|+2|x2| +2|y2|$
$\le|x1|+2|x2|+|y1|+2|y2|$
$\le \|x\|_{\#}+\|y\|_{\#}$ Therefore axiom 4 holds

Correct. Here's how I would do it:

For all $x,y\in\mathbb R^2$, we have
\begin{align}
\|x+y\|_{\#} &=\|(x_1+y_1,x_2+y_2)\|_{\#}=|x_1+y_1|+2|x_2+y_2| \leq|x_1|+|y_1|+2|x_2|+2|y_2|=\|x\|_{\#}+\|y\|_{\#}.
\end{align} As you can see, it's not so hard if you just use the definitions and do one step at a time.

 

[bugatti79]

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

Is || ||_3 a special function like || ||_1, || ||_2 and || ||_infty etc?
Or do I proceed as usual to test the axioms?

 

[Mark44]

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

The last part should be ||x||_3 = 3|x_1|. That line is defining how this norm works

Is || ||_3 a special function like || ||_1, || ||_2 and || ||_infty etc?

No. The || ||₁ and || ||₂ norms are special functions. This norm is what is being defined in the top line here.

Or do I proceed as usual to test the axioms?

 

[bugatti79]

Decide whether or not the following is a norm defined on R^2, hence verify if it is or prove by counter example if it is not.
Q2 ii
(ii) || ||_3: R^2 to R defined by || x ||_3= 3 |x_1| for x=(x_1,x_2) in R^2

a) Let x be an element in R^2.


Since |x_1| >=0 then by definition we have ||x||_3=3|x_1| >=0 for x_1 in R.

b) ||x||_3=3|x_1|=0 iff x=0

c) Let a be an element in R, then

||ax||_3=3|ax_1|
= |a|(3|x_1|)
=|a| ||x||_3

d) let y=(y_1) be an element in R^2, then

||x+y||_3=||(x_1+y_1)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
<=3(|x_1|+|y_1|)
<=3|x_1|+3|y_1|
<=||x||_3+||y||_3

Therefore all 4 axioms hold implies this norm is defined in R^2...?

 

[Fredrik]

a) Correct
b) If what you're saying is true, then (0,1)=(0,0). This would imply that 1=0.
c) Correct. I would break up the first step into two, for clarity: $\|ax\|_3=\|(ax_1,ax_2)\|_3=3|ax_1|=\cdots$.
d) What do you mean by y=(y_1)? An ordered pair isn't equal to its first component. Also, x+y is not equal to x_1+y_1. The latter is a real number, not a vector, so you also can't use it as input for the function $\|\ \|_3$. Also, the last two inequalities are actually equalities, so replace <= with = there.

 

[bugatti79]

Decide whether or not the following is a norm defined on R^2, hence verify if it is or prove by counter example if it is not.
Q2 ii
(ii) || ||_3: R^2 to R defined by || x ||_3= 3 |x_1| for x=(x_1,x_2) in R^2

a) Let x be an element in R^2.


Since |x_1| >=0 then by definition we have ||x||_3=3|x_1| >=0 for x_1 in R.

b) ||x||_3=3|x_1|=0 iff x=0

c) Let a be an element in R, then

||ax||_3=3|ax_1|
= |a|(3|x_1|)
=|a| ||x||_3

d) let y=(y_1) be an element in R^2, then

||x+y||_3=||(x_1+y_1)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
<=3(|x_1|+|y_1|)
<=3|x_1|+3|y_1|
<=||x||_3+||y||_3

Therefore all 4 axioms hold implies this norm is defined in R^2...?

a) Correct
b) If what you're saying is true, then (0,1)=(0,0). This would imply that 1=0.
c) Correct. I would break up the first step into two, for clarity: $\|ax\|_3=\|(ax_1,ax_2)\|_3=3|ax_1|=\cdots$.
d) What do you mean by y=(y_1)? An ordered pair isn't equal to its first component. Also, x+y is not equal to x_1+y_1. The latter is a real number, not a vector, so you also can't use it as input for the function $\|\ \|_3$. Also, the last two inequalities are actually equalities, so replace <= with = there.

b) I am actually in 2 conflicting thoughts here and I don't know which is wrong. My first thought is this:
ok, so this axiom does not hold therefore I need to demonstrate by counter example. So if I use this example instead of yours.
since x=(x1,x2). Let x=(1,0) then ||x||_3=||(x1)||_3=3|1|=3. Implies ||x||_3 not = 0 for all x_1 in R

My other thought is this:
Based on what I originally wrote which is this ||x||_3=3|x_1|=0 iff x=0
Wouldn't ||x||_3 =0 always because when I say x=0 I mean x=(x_1,x_2)=0, ie x_1 and x_2 are both 0 in R

d) Ah, because we are dealing in R^2, y should be y=(y_1,y_2).

||x+y||_3=||(x_1+y_1, x_2+y_2)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
=3(|x_1|+|y_1|)
=3|x_1|+3|y_1|
=||x||_3+||y||_3

I don't understand why you leave out the <?

 

[Fredrik]

since x=(x1,x2). Let x=(1,0) then ||x||_3=||(x1)||_3=3|1|=3. Implies ||x||_3 not = 0 for all x_1 in R

I don't know why you write $\|x\|_3=\|(x_1)\|_3$. x is an ordered pair of real numbers. x₁ is a real number. Also, once you have specified that x=(1,0), then there's no need to use the notation x₁. If x=(1,0), then clearly $\|x\|=\|(1,0)\|$. That should be your starting point if you want to try to use (1,0) as a counterexample.

Also, if you have already specified that x=(1,0), which implies that x₁=1, you can't say "for all x₁ in ℝ".

Based on what I originally wrote which is this ||x||_3=3|x_1|=0 iff x=0
Wouldn't ||x||_3 =0 always because when I say x=0 I mean x=(x_1,x_2)=0, ie x_1 and x_2 are both 0 in R

Clearly x=0 implies $\|x\|_3=\|(0,0)\|_3=3|0|=0$. What you need to answer after making this observation is if it's also true that $\|x\|_3=0$ implies that $x=0$.

d) Ah, because we are dealing in R^2, y should be y=(y_1,y_2).

Yes.

||x+y||_3=||(x_1+y_1, x_2+y_2)||_3=3|x_1+y_1| we have |a+b|<=|a|+|b| therefore
=3(|x_1|+|y_1|)
=3|x_1|+3|y_1|
=||x||_3+||y||_3

I don't understand why you leave out the <?

You don't. I said "the last two inequalities", not "the last three inequalities". What I'm saying is that it looks weird to write a≤b when you know for sure that a=b. For example, 5≤3+2 is correct, but it still looks weird. So I'm suggesting that you use equality signs when you know that the two numbers are equal.